# Probability Thread



## CubesOfTheWorld (Apr 8, 2010)

This is just a thread for people to post interesting possibilities of a case. Like, it is a 1/72 chance for a PLL skip. Enjoy.


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## chinesed00d (Apr 8, 2010)

What are the chances of having a cross already done during a scramble?


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## CubesOfTheWorld (Apr 8, 2010)

chinesed00d said:


> What are the chances of having a cross already done during a scramble?



heheh. that is why I made this thread. I don't know this kind of math. That was one I was thinking about.


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## 4Chan (Apr 8, 2010)

At US Nationals 2008, during 3x3 qualification, there was a 1 move cross on white.

True fact.


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## cmhardw (Apr 8, 2010)

I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

1 / [6! * 6!/2] = 1 / 259200

To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552

I thought it was pretty neat.

Chris


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## Lucas Garron (Apr 8, 2010)

CubesOfTheWorld said:


> Before someone says it, it is a 1/1 chance of me being stupid, or a 1/1 chance of this thread being a failure.


Saying that doesn't make your thread more legitimate.

I changed the title, and let's see if we can salvage it:

If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.


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## Tim Major (Apr 8, 2010)

Lucas Garron said:


> If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.



Wow, I didn't know that, and am interested in how you worked that out. How?


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## DT546 (Apr 8, 2010)

not cubing related, but interesting

if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday


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## riffz (Apr 8, 2010)

Lucas Garron said:


> CubesOfTheWorld said:
> 
> 
> > Before someone says it, it is a 1/1 chance of me being stupid, or a 1/1 chance of this thread being a failure.
> ...



I just have to say this:

WOLFRAM ALPHA IS AMAZING

That is all.



DT546 said:


> not cubing related, but interesting
> 
> if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday



That's actually quite funny. But how exactly were the odds of dying calculated?


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## Rinfiyks (Apr 9, 2010)

riffz said:


> That's actually quite funny. But how exactly were the odds of dying calculated?



The odds of winning the lottery jackpot are almost 1 in 14 million.
The population of the usa is 300 million.

300/14=20~, so if more than 20 people die from non-natural causes every day in the usa, DT546's statement is correct.


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## DaijoCube (Apr 9, 2010)

My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.

This might be the luckiest solve you could have on a 4x4x4!


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## TheMachanga (Apr 9, 2010)

DaijoCube said:


> My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.
> 
> This might be the luckiest solve you could have on a 4x4x4!



He scrambled it him self...most likely. It was most likely a crappy scramble.


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## TheMachanga (Apr 9, 2010)

Oh and I had OLL parity and when I finished that it was an LL skip. (qqtimer)


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## DaijoCube (Apr 9, 2010)

TheMachanga said:


> DaijoCube said:
> 
> 
> > My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.
> ...



After solving centers then edges, you most likely can't keep track of the scramble. Scramble was legit.


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## That70sShowDude (Apr 9, 2010)

What are the chances of having a CCT 4x4 scramble that requires 4 or less moves to solve the centers.

I solved the centers in 4 moves on one of em ...


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## eastamazonantidote (Apr 20, 2010)

I want to know the chances of COLL cases and ZBLL cases. Cride5 has tried several times (maybe only once?) to explain this to me with complicated language. I might know what is going on now and I will try to explain.

So for OLL, the chance of Pi, Sune, Anti-Sune, T, U, and L is 1/54 while H is 1/108 and a skip occurs every 1/216 (these statistics are pretty well known). Just taking these cases for EO (setting 1/54 to be the standard):

6 * chance + 1 * .5 chance + 1 * .25 chance = 1

4/27 is the standard chance of occurance.

So the chance of OCLL occurrences after EO:


Spoiler



T: 4/27
U: 4/27
L: 4/27
H: 2/27
Pi: 4/27
Sune: 4/27
Anti-Sune: 4/27
Skip: 1/27



Now for COLLs. PLL probabilities are well known, so I can use those for my data.

Corners correctly placed: 1/18 + 1/18 + 1/36 + 1/72 (U,U,Z,H) = 11/72
Diagonal corner swap: 1/72 + 1/72 + 1/36 + 1/18 + 1/18 (N,N,E,V,Y) = 1/6

Now the others should be divided evenly amongst themselves, if my reasoning is correct (no reason you should trust this). Which means every other case should have a nice 49/432 chance of occurrence. What? Hmmm. Weird number. I guess I'll post it with PLL chances:



Spoiler



T
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

U
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

L
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

H
Correct: 11/972
Diagonal: 1/81
Left: 49/5832
Right: 49/5832
Top: 49/5832
Bottom: 49/5832

Pi
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Sune
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Anti-Sune
Correct: 11/486
Diagonal: 2/81
Left: 49/2916
Right: 49/2916
Top: 49/2916
Bottom: 49/2916

Skip
Aa: 1/486
Ab: 1/486
E: 1/972
F: 1/486
Ga: 1/486
Gb: 1/486
Gc: 1/486
Gd: 1/486
H: 1/1944
Ja: 1/486
Jb: 1/486
Na: 1/1944
Nb: 1/1944
Ra: 1/486
Rb: 1/486
T: 1/486
Ua: 1/486
Ub: 1/486
V: 1/486
Y: 1/486
Z: 1/972



I postulate that ZBLL has the same chance for every edge permutation. There are 12 possible edge permutations and COLL + EPLL generates each edge permutation equally often. Therefore the chance of each ZBLL case is that of the chance of the respective COLL case divided by 12. I don't want to have to post all of those probabilities.

One last thing: am I right?


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## 4Chan (Apr 20, 2010)

Mister eastamazonantidote, I have wondered the same thing. o:
I made a thread about COLL probability, and I'm quite curious myself about ZBLL.


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## ben1996123 (Apr 20, 2010)

Lucas Garron said:


> If you keep doing random alignments and twists on a Square-1, you will hit the square-square shape 2/1839 of the time.



Is that including the middle layer flip or not?


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## Owen (Apr 20, 2010)

What about a edges skip? (on 3x3)


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## kinch2002 (Apr 20, 2010)

Owen said:


> What about a edges skip? (on 3x3)



This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation

Final answer: 1/(12!*2^11) = 9.81*10^11

EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...


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## miniGOINGS (Apr 20, 2010)

eastamazonantidote said:


> One last thing: am I right?



Wow. I actually didn't understand your way of doing the calculations that well, and that's not how I would have done them, so I can't really tell at the moment.


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## mr. giggums (Apr 20, 2010)

kinch2002 said:


> Owen said:
> 
> 
> > What about a edges skip? (on 3x3)
> ...



I think you forgot that two edges can't be switched so you would have to divide that by 2 so it would be 4.91*10^11 I think.


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## kinch2002 (Apr 20, 2010)

mr. giggums said:


> kinch2002 said:
> 
> 
> > Owen said:
> ...



No I didn't, because they can be switched as long as the corners are free - just think of a J perm or something - only 2 edges switched. Thanks anyway


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## mr. giggums (Apr 20, 2010)

kinch2002 said:


> mr. giggums said:
> 
> 
> > kinch2002 said:
> ...



Oh I was assuming (I know I spelled it wrong) that the corners were already solved.


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## miniGOINGS (Apr 20, 2010)

Kinch, I did my calculations the same way as yours (probably everyone else too...) 12*11*10*9*8*7*6*5*4*3*2 for EP, and 2^11 for EO.

479,001,600 * 2048 = 980,995,276,800


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## kinch2002 (Apr 20, 2010)

Yesterday I had a clock scramble where one side of edges (including centre piece) were all matched (but not all at 12 o'clock). The scramble program (as Maarten told me) generates a random position for the clock to be in and then figures out the generator for it. So, to have one particular side solved it would be (1/12^4) - just think about matching all 4 edges to the centre one at a time. So the probability of having at least one side like this is 1-(1-(1/12^4))^2=1/10368 approx. I've only done 1000 solves so I feel nice and lucky to get one


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## Sir E Brum (Apr 20, 2010)

Probability of LL skip after F2L with edges oriented?

Probability of OLL skip after F2L with edges oriented?


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## miniGOINGS (Apr 20, 2010)

EO'd LL Skip: 7 * 1/72 = 1/1944
EO'd OLL Skip: 1/27


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## kinch2002 (Apr 20, 2010)

Sir E Brum said:


> Probability of LL skip after F2L with edges oriented?
> 
> Probability of OLL skip after F2L with edges oriented?



1. I assume AUF is allowed. Easy way: LL skip is 1/15552, so if EO is skipped already (normally 1/8) it's 1/1944.
Slightly harder way: You need to skip EP (doesn't matter where first edge is, so 1/3*1/2 = 1/6), CO (1/3*1/3*1/3 = 1/27) and CP (last 2 corners have no choice because I gave choice on edges so 1/4*1/3 = 1/12) = 1/6*1/27*1/12 = 1/1944
2. OLL skip is 1/216, and EO is normally 1/8 so this gives 1/27 - it's just CO skip really.


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## Sir E Brum (Apr 21, 2010)

kinch2002 said:


> Sir E Brum said:
> 
> 
> > Probability of LL skip after F2L with edges oriented?
> ...



Mmk. Tyvm.


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## Aditya (Apr 28, 2010)

What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.


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## kinch2002 (Apr 28, 2010)

Short and very approximate answer: 1/43 quintillion for each turn

Long answer:


Spoiler



Well I could take this question to mean 'what's the probability that a random state is the solved one?'. This is a fairly well known probability. There are approx. 43 quintillion (4.3*10^19) different states that the cube can be in (not counting positions that can only be achieved by taking pieces out).

This is found most easily by (8!)*(3^7)*(12!)*(2^11)/2. I'm sure there are other places on the forum that explain this probability as well, but do ask if you don't understand how I got those numbers!

In reality your question is slightly different to the one I answered because when the cube is scrambled you are usually at least 15 or so moves from being solved so the next 14 turns have 0 probability of solving it.


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## Lucas Garron (Apr 28, 2010)

Aditya said:


> What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.


I'm not sure. What is the probability of attending a competition?

Good approximation:
http://www.wolframalpha.com/input/?i=expected+value+geometric+distribution+p%3D1/(4.3e19)


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## Aditya (Apr 29, 2010)

Lucas Garron said:


> Aditya said:
> 
> 
> > What is the probability of randomly turning the sides of the cube and solving it? lol. I wonder how you would calculate it.
> ...



Haha lol. How did you come up with that number? Wouldn't the probability of going to a competition be affected if you know that the person is a cuber?


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## Lucas Garron (Apr 29, 2010)

Aditya said:


> Lucas Garron said:
> 
> 
> > Aditya said:
> ...


Yes, it would be. I'm just trying to get you to notice that your question was vague, and badly phrased, so that it almost sounds like it's not really answerable.

http://en.wikipedia.org/wiki/Geometric_distribution
It's a good approximation because the cube is almost completely random after a few dozen moves.


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## mrCage (Apr 29, 2010)

cmhardw said:


> I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.
> 
> The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:
> 
> ...


Maybe you're cubing too much? 
**** happens - and chance happens...

Per


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## Aditya (Apr 30, 2010)

Aditya said:


> Lucas Garron said:
> 
> 
> > Aditya said:
> ...



Yes, my question was phrased badly, and I am sorry if it confused anybody. The geometric distribution looks right, but coming from a Stanford guy, it HAS to be right lol jk.


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## Bayamo Penguin (May 7, 2010)

*"Easy Solve"*

What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
Cross, corner, middle layer, and top edges all solved..
then when orienting the top corners (URul/UruL), it all falls in to place, no mess, no fuss, no hassle


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## riffz (May 7, 2010)

Bayamo Penguin said:


> What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
> Cross, corner, middle layer, and top edges all solved..
> then when orienting the top corners (URul/UruL), it all falls in to place, no mess, no fuss, no hassle



This is way to vague for someone to answer.


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## Bayamo Penguin (May 7, 2010)

riffz said:


> Bayamo Penguin said:
> 
> 
> > What is the probablity of an "easy solve" during a straight LBL (method included in the packaging of the Rubik's)
> ...



Vague in explanation? or Positionwise?


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## Bayamo Penguin (May 8, 2010)

Maybe this is clearer?: This is the position/ situation: F2L complete. Top edges correcly positioned. You still need to position and orient the top corners, and you are using URul, UruL to position them. This is one of those simple LBL solves many people begin with. Then as you complete the algorithm they do fall in to the correct slot, and fall in to correct orientation. I rememebr it would happen when i first started learning, and in fact, some tutorials mention that this may happen. I just was wondering how often that happened.


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## Johannes91 (May 8, 2010)

Skipping orienting corners has probability 1/27, or about 3.7%.


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## eastamazonantidote (May 12, 2010)

I want to know the chances for each X-VHF2L case. On Jason Baum's site, this would be cases: 1, 2, 3, 4, 5, 6, 11, 12, and 16. Assuming complete randomness, what are the chances for each case? If it's easier to just do ZBF2L in general I can work out the VHF2L chances from that. Is this even possible to compute accurately?


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## qqwref (May 12, 2010)

I'd help you out but I don't really understand X-VHF2L. How are you reducing to these cases? Is it an automated process or something like "oh I just intuitively pair" (in which case the probability of each case cannot be computed)?


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## eastamazonantidote (May 23, 2010)

qqwref said:


> I'd help you out but I don't really understand X-VHF2L. How are you reducing to these cases? Is it an automated process or something like "oh I just intuitively pair" (in which case the probability of each case cannot be computed)?



Unfortunately I do my pairing intuitively. I was worried intuition would hurt the odds. You could pair up algorithmically, but I doubt there would be any advantage there. Alright, how about ZBF2L? I can generate the probabilities for VHF2L from the ZBF2L cases.


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## Gold_A (May 27, 2010)

What is the probability of a parity on one layer of the square 1, after having it in cube form and every piece in it's respective layer?


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## Lucas Garron (May 27, 2010)

Gold_A said:


> What is the probability of a parity on one layer of the square 1, after having it in cube form and every piece in it's respective layer?


1/2, under reasonable assumptions.


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## JeffDelucia (Jun 6, 2010)

If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?


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## CuBeOrDiE (Jun 6, 2010)

JeffDelucia said:


> If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?



1/22 I assume, because OLL would be taken care of, and there are 22 PLL cases, one of which is the solved cube. but there is something wrong with your question; if you have all corners oriented and did ELL, you would still have to solve them because ELL orients edges AND corners. correct me if I'm wrong


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## JeffDelucia (Jun 6, 2010)

CuBeOrDiE said:


> JeffDelucia said:
> 
> 
> > If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?
> ...


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## CuBeOrDiE (Jun 6, 2010)

why not?


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## eastamazonantidote (Jun 6, 2010)

CuBeOrDiE said:


> why not?



Because the chances of PLLs occurring is not equally spread. Just look at badmephisto's PLL page for the probabilities. The chance of a PLL skip is 1/72.


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## CuBeOrDiE (Jun 6, 2010)

hmmm... could easily be true, but I wonder why?

edit: how did he get that probability, by many trials or by some kind of mathematics?


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## Johan444 (Jun 6, 2010)

CuBeOrDiE said:


> hmmm... could easily be true, but I wonder why?
> 
> edit: how did he get that probability, by many trials or by some kind of mathematics?



Depends of how many different cases of each PLL you can get. Take i.e. T-perm, the corner swap can be on right, left, back and front side of the up layer. That makes it four different T-perms. A PLL skip only has one case, it's 4 times rarer.


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## Meep (Jun 6, 2010)

Does anyone know the probability of getting a last 4 edges skip in 5x5 reduction?


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## CuBeOrDiE (Jun 6, 2010)

Johan444 said:


> CuBeOrDiE said:
> 
> 
> > hmmm... could easily be true, but I wonder why?
> ...



oh, i didn't think of it that way. thanks!


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## CubesOfTheWorld (Jun 6, 2010)

Johan444 said:


> CuBeOrDiE said:
> 
> 
> > hmmm... could easily be true, but I wonder why?
> ...



H-perm also has one case. J and R perms are like the T-perm. You can use the same algorithm in four different cases.

Whenever I get an H-perm, I think, "Man... Why couldn't that be a PLL skip? It has the exact same chance..."

EDIT: J and R's are not like the T-perm. One _kind_ of R or J is like the T-perm.


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## LewisJ (Jun 6, 2010)

Meep said:


> Does anyone know the probability of getting a last 4 edges skip in 5x5 reduction?



My thoughts say 1/8! (1/40320)

Pretend the last 4 midges don't matter; then, the 8 wings must all be in the correct spot. The 8 wings can be permuted in any possible way (8!) but only one has all the edges solved.


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## JeffDelucia (Jun 6, 2010)

Whats the probability of having LL edges oriented? LL corners?


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## miniGOINGS (Jun 6, 2010)

JeffDelucia said:


> Whats the probability of having LL edges oriented? LL corners?



Edges: 1/8
Corners: 1/27


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## mr. giggums (Jun 6, 2010)

Well the probability of getting a PLL skip after OLL is 1/72 (hence a 1LLL) but you will only use OLL when both corners and edges are misoriented which is 16/19 of the time. So 1/72*16/19=2/171. The chance of getting a OLL skip is 1/216 so a 1LLL. The chance of using COLL 1/8. Then after COLL the chance of edges permutated is 1/12. 1/8*1/12=1/96. Finally the chance of corners oriented 1/27 and after ELL the chance of corners permutated is 1/12. 1/27*1/12=1/324. Now all we have to do is add them all together so 2/171+1/216+1/96+1/324=.029828...
I'm sure I messed up somewhere but thats my answer.


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## JeffDelucia (Jun 6, 2010)

mr. giggums said:


> Well the probability of getting a PLL skip after OLL is 1/72 (hence a 1LLL) but you will only use OLL when both corners and edges are misoriented which is 16/19 of the time. So 1/72*16/19=2/171. The chance of getting a OLL skip is 1/216 so a 1LLL. The chance of using COLL 1/8. Then after COLL the chance of edges permutated is 1/12. 1/8*1/12=1/96. Finally the chance of corners oriented 1/27 and after ELL the chance of corners permutated is 1/12. 1/27*1/12=1/324. Now all we have to do is add them all together so 2/171+1/216+1/96+1/324=.029828...
> I'm sure I messed up somewhere but thats my answer.



I think thats right actually.. so about 3%


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## LewisJ (Jun 6, 2010)

Just my thoughts on that idea - all the algs you need to learn for that add up to a significant portion of ZBLL! COLL + ELL + OLL + PLL = 40 + 30 + 57 + 21 = 148 algs. For a 3% chance...


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## Feryll (Jun 12, 2010)

Bump. But I have a couple of questions:

1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?
2. What is the chance of getting 1 free paired pair while you were about to do the edge pairing (To start off with getting one, not multipairing accidentally)?
3. Free X-cross?
4. 2x1x1 block of anything on the 3x3?

These might have been on some website, but I can't find it.


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## JeffDelucia (Jun 12, 2010)

LewisJ said:


> Just my thoughts on that idea - all the algs you need to learn for that add up to a significant portion of ZBLL! COLL + ELL + OLL + PLL = 40 + 30 + 57 + 21 = 148 algs. For a 3% chance...



Yeah but then use that with EOline Petrus or just fridrich with edge controlling and your chances are much higher.


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## Johannes91 (Jun 12, 2010)

Feryll said:


> 1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?


For a fixed center: (3/23)*(2/22)*(1/21) = *1/1771*
For any center: I'm not sure, will maybe edit soon... [Edit: *1/296*, see new post below]



Feryll said:


> 2. What is the chance of getting 1 free paired pair while you were about to do the edge pairing (To start off with getting one, not multipairing accidentally)?


Chance of getting _at least_ one free pair: *40.6204%*.

Distribution: [[0, 0.593796], [1, 0.3092], [2, 0.0807901], [3, 0.0141328], [4, 0.00186375], [5, 0.000197873], [6, 1.76482e-05], [7, 1.36243e-06], [8, 9.39178e-08], [9, 5.5655e-09], [10, 4.17412e-10], [12, 3.16221e-12]]



Feryll said:


> 3. Free X-cross?


Fixed cross and slot: *1/72,990,720*.
Others: dunno



Feryll said:


> 4. 2x1x1 block of anything on the 3x3?


Do you mean that either a corner-edge pair or an edge-center pair counts?


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## Feryll (Jun 12, 2010)

Johannes91 said:


> Feryll said:
> 
> 
> > 1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?
> ...


wohw, ur 2 smarte 4 me


I meant corner edge pair, but another question would be what are the chances of having no center-edge pair. These things have been in my mind forever


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## Johannes91 (Jun 12, 2010)

Feryll said:


> 1. What is the probability of getting a completed center face on a 4x4, so you skip the very first step?


*0.3379%*, or approximately 1/296

This is very close to a simple estimate (\( 1-(1770/1771)^6 \approx 0.3383\% \)), so I believe it's correct.

Distribution, the first number is the number of faces with 4 centers of the same color and the second is the probability:


```
0 0.996621
1 0.00337051
2 8.6646e-06
3 2.51571e-08
4 1.131e-10
6 2.21766e-13
```


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## That70sShowDude (Jun 15, 2010)

Yesterday I got 2 AUF-less PLL skips in a row. 
What are the chances of this?


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## Kirjava (Jun 15, 2010)

JeffDelucia said:


> If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?




<+Kirjava> the probability doesn't change
<+Kirjava> unless you *know* that using a specific LL technique will give a skip for a specific senario
<+Kirjava> which no-one does/has/ever will

by the way, CLL isn't COLL.


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## cmhardw (Jun 15, 2010)

Kirjava said:


> JeffDelucia said:
> 
> 
> > If you learned full OLL CLL PLL and ELL and when ever you got an all edges oriented OLL you used cll and whenever you got an all corners oriented oll you used Ell what would your probability of a 1lll be?
> ...



Based on Jeff's scenario the probably of a LL skip would change, and in fact it would be more likely. I think we can read COLL instead of CLL, which is what he probably meant. If you have an all edges oriented OLL and do a COLL alg, solving the permutation of the corners with respect to each other would increase the overall chance of a 1LLL.

Also, using ELL would affect the 1LLL chance when corners were both oriented and solved with respect to each other.

I'm about to head in to work, so I don't have time to work out the exact affect on the probability of a 1LLL, but it is certainly going to be more likely because of also knowing COLL and ELL.

Chris


----------



## Kirjava (Jun 15, 2010)

Of course COLL would change the probability. But I decided to interpret the question as if it was CLL - as I think that question is more interesting.

Let me explain my reasoning behind believing that the probability doesn't change. 

I don't like the idea of assuming OLL/PLL is the default method, unless you get a CLL skip in which case you use a subset of LL algs to finish in one look. Knowing OLL/PLL/CLL/ELL myself, I know that I use whichever group of algs will be faster for the first step. Therefore, for the last layer, no specific method is relied on and a specific one is chosen. When the choice of method is made, that choice dictates the probabilities of skipping.

Now, if we take the case of a CLL skip - the method chosen is CLL/ELL. The chances of this one look last layer is still 1/49.

For an OLL skip, the method you will use is OLL/PLL. Chances of this happening were 1/216.

I put forward that the chances of a LL step skip in any given solve is the same, since you have used a specific method to complete that LL and you inherit the probabilities from that method. 

Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.

On a side note, probability for skipping the second step *does not* change, as the probability is dictated by whatever method you use for the first step.


----------



## cmhardw (Jun 16, 2010)

Kirjava said:


> Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.



Yes, but now think of it this way instead. I could actually do out the math of your chance to get a 1LLL, but it is not necessary to show that you have a higher probability of a 1LLL than someone using only OLL/PLL.

Consider the case of a CLL skip. You would surely notice this in your solving, perform the necessary ELL alg to solve the cube, and be done with the LL in 1 look. However, a OLL/PLL solver would see on the "X" OLL case and perform that, with 71/72 probability of still needing to execute a PLL case.

This shows that your method has a higher chance of a 1LLL than OLL/PLL used only. This is equivalent to saying that if you did 1000 solves using your LL method, and a OLL/PLL solver did 1000 solves, then you are likely to have more 1LLL solves than the other person. No matter how you phrase it, you have an improved chance to get a 1LLL over a OLL/PLL only solver.



Kirjava said:


> On a side note, probability for skipping the second step *does not* change, as the probability is dictated by whatever method you use for the first step.



Yes, your probability for skipping the second step does change, *because* you use two methods.

Consider CLL/ELL as your "base" method momentarily. You will "skip" ELL if you are solving, and after F2L you have an already oriented LL (but not permuted). In this case you have not yet done either CLL or ELL, but you would know to execute a PLL alg and thus skip the ELL step. Thus you have a higher chance to skip ELL than a CLL/ELL only solver.

Alternatively phrased you could imagine OLL and PLL as your "base." In this case you skip PLL in the case that all corners are not only oriented, but solved with respect to each other. In this case, rather than executing the "X" OLL, you would simply do the necessary ELL case to solve the cube. So your chance to skip PLL is higher than someone using only OLL/PLL alone.

Keep in mind that this result is very good for you!  No matter how you consider it, your personal chance to skip the second step, or have a 1LLL in any form, is higher than that of anyone using either CLL/ELL alone or OLL/PLL alone.

Chris


----------



## Kirjava (Jun 16, 2010)

cmhardw said:


> Kirjava said:
> 
> 
> > Of course, I understand what you're saying. I'm simply having difficulty consolidating it with what I'm saying ^_^.
> ...




My problem lies in only considering the two methods as a choice, instead of a combined system. I spoke to Johannes about this and he summed it up for me quite well;



> 22:15:03 < jlaire> but yeah, what I was saying, the probability of getting a 1LLL with a given system doesn't change if you learn additional systems, which I think is what you were getting at
> 22:15:28 < jlaire> but the overall probability goes up because a 1LLL with any one system gives you a 1LLL






> This shows that your method has a higher chance of a 1LLL than OLL/PLL used only. This is equivalent to saying that if you did 1000 solves using your LL method, and a OLL/PLL solver did 1000 solves, then you are likely to have more 1LLL solves than the other person. No matter how you phrase it, you have an improved chance to get a 1LLL over a OLL/PLL only solver.




Of course, you are correct. I imagine working out the probability compared to OLL/PLL is actually quite difficult.

Here are some of my other thoughts about this, since I'm too lazy to type them out again;



> 21:57:05 < Kirjava> once you decide, you are bound by the probabilities of that system
> 21:57:36 < Kirjava> if you decide to use CLL/ELL because that would give you a skip
> 21:57:54 < Kirjava> the chances of that skip happening are the same as if you didn't know OLL/PLL at all
> 21:59:04 < Kirjava> I fully accept that I am probably completely wrong
> ...




And I came to some sort of conclusion;



> 22:09:06 < Kirjava> to me it seems like
> 22:09:20 < Kirjava> you're increasing the chances of getting a better LL
> 22:09:51 < Kirjava> yet the probibility of a skip is the same O_O
> 22:10:12 < Kirjava> but but
> ...




Thanks for giving me the clarity needed to understand why my reasoning was incorrect. Despite this, I still think there is some merit to it 



cmhardw said:


> Kirjava said:
> 
> 
> > On a side note, probability for skipping the second step *does not* change, as the probability is dictated by whatever method you use for the first step.
> ...




I'd say that for the former you would be using the OLL/PLL system and CLL/ELL system for the latter. This is because I don't consider CLL/ELL as an extension to OLL/PLL or vice versa, I consider them seperately.

You cannot consider CLL/ELL to be the method used when you know both systems and get an OLL skip, since you automatically use the OLL/PLL system as it's the best one to use in that situation, and the probability of a PLL skip is still 1/72.

However, I think this is less to do with probability and more to do with perspective.



cmhardw said:


> Keep in mind that this result is very good for you!  No matter how you consider it, your personal chance to skip the second step, or have a 1LLL in any form, is higher than that of anyone using either CLL/ELL alone or OLL/PLL alone.




This doesn't mean much when my execution is terrible


----------



## ariasamie (Jun 21, 2010)

someone please answer this guy:


TeddyKGB said:


> *Avg 29.52*
> 
> 30.13
> (25.58)
> ...



this was originally posted on  "NEW" Race to Sub-30!


----------



## JackJ (Jun 21, 2010)

At Minnesota Open 2010, Chester Lian got 4 pyraminx LL skips in a row. What are the odds of that?


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## Rpotts (Jun 21, 2010)

ariasamie said:


> someone please answer this guy:
> 
> 
> TeddyKGB said:
> ...



Probably depends on the solvers SD

EDIT: I enjoyed reading every last bit of the chris/kirjava debate. 
How does CLL/ELL and OLL/PLL overlap affect probabilities, if the discussion is still open.


----------



## eastamazonantidote (Jun 22, 2010)

Assuming a completely random last slot + last layer, what are the chances for each ZBF2L case? I have been thinking about this and I can figure out how many cases intuitively but I can't come up with probabilities.


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## cmhardw (Jun 22, 2010)

Rpotts said:


> EDIT: I enjoyed reading every last bit of the chris/kirjava debate.
> How does CLL/ELL and OLL/PLL overlap affect probabilities, if the discussion is still open.



The reason it affects probabilities is that the two methods have some overlapping situations.

Considering your solving method to be OLL/PLL *and nothing else*, you have the following possibilities for the LL.

1) You must perform an OLL case, followed by the fact that you must perform a PLL case.
2) You skip the OLL step, but must still perform the PLL case.
3) You must perform the OLL case, but you skip the PLL step.
4) You skip both the OLL and PLL steps.

Now for CLL/ELL only and *nothing else* you have the following possibilities:
1) You must perform the CLL case, followed by the fact that you must perform the ELL case.
2) You skip the CLL step, but you must perform the ELL case.
3) You must perform the CLL case, but you skip the ELL step.
4) You skip both the CLL and ELL steps.

So now let's consider the method: OLL/PLL or CLL/ELL. The probabilities are affected because you have some overlap of the cases *between methods* at the OLL step. 

1) You have either a CLL or OLL case, but neither case was skipped. If you execute CLL, then followup with ELL. If you execute OLL, then follow up with PLL.

2) You have either a CLL or OLL case, but neither case was skipped. You perform a CLL case, then the ELL step is skipped.

3) You have either a CLL or OLL case, but neither case was skipped. You perform a OLL case, then the PLL step is skipped. 

Ok the next steps are overlap steps.
4) You skip CLL, but do *not* skip the full OLL. You perform the correct ELL case to solve the LL.

5) You skip OLL, but do *not* skip the full CLL step. You perform the correct PLL to solve the LL.

6) You skip the LL (with the possibility of a necessary AUF to solve the cube).

--------

Ok so now to convince you that the method of combining both methods together has a higher probability of 1LLL I am going to actually calculate the probability of skipping at least one step in each method.

Now for OLL/PLL methods the probability of skipping either OLL, or PLL, or possibly both, is:
1 - (71/72)*(215/216) = 287 / 15552

Now for CLL/ELL methods.
The probability of skipping either CLL, or ELL, or possibly both, is:
1 - (161/162) * (95/96) = 257 / 15552

Ok now for OLL/PLL and CLL/ELL combined method.
The chance for either a 1LLL or a full LL skip is the chance of getting any of either cases: 2, 3, 4, 5 or 6 from my list of the combined method above.

------
Lemma to help for cases 2 and 3.

We need to know the probability of getting a LL that has neither a OLL nor CLL skip. This is:
1 - (1 / 216 + 1 / 162 - 48 / 62208)
1 - (288 / 62208 + 384 / 62208 - 48 / 62208)
1 - (624 / 62208)
1283 / 1296
------

------
Lemma for case #4


> 4) You skip CLL, but do *not* skip the full OLL.



This probability is:
(1/162) * (7/8) = 7 / 1296
------

------
Lemma for case #5


> 5) You skip OLL, but do *not* skip the full CLL step.



This probability is:
(1/216) * (5/6) = 5 / 1296
------

So the probability of a 1LLL (or LL skip) with the CLL/ELL or OLL/PLL combined method is:

P(1LLL) = P(case #2) + P(case #3) + P(case #4) + P(case #5) + P(case #6)

P(1LLL) = (1283/1296) * (1/96) + (1283/1296) * (1/72) + (7/1296) * (95/96) + (5/1296) * (71/72) + (1/15552)

P(1LLL) = 115 / 3456

(115 / 3456) > (287 / 15552) > (257 / 15552)

OR

in prettier form:

3.3% (CLL/ELL or OLL/PLL) > 1.8% OLL/PLL > 1.7% CLL/ELL

So the chance is nearly *double* that you get a 1LLL (or 0LLL) compared to either OLL/PLL method alone, or CLL/ELL method alone.

--edit--
Actually, I just realized that this depends on the frequency (call this q) that you choose between cases 2 and 3 as I list them for CLL/ELL or OLL/PLL methods. I will take a look at this more in depth when I get home tonight. But 3.3% is *NOT* the exact answer using both methods, because of the choice you have between cases 2 and 3. I will fix this tonight.
--edit--

Chris


----------



## cmhardw (Jun 22, 2010)

cmhardw said:


> --edit--
> Actually, I just realized that this depends on the frequency (call this q) that you choose between cases 2 and 3 as I list them for CLL/ELL or OLL/PLL methods. I will take a look at this more in depth when I get home tonight. But 3.3% is *NOT* the exact answer using both methods, because of the choice you have between cases 2 and 3. I will fix this tonight.
> --edit--



Ok I fixed it, and I have a more in depth analysis.

So let's look at the probability of a 1LLL given that you use CLL/ELL and OLL/PLL combined method.

Let q = probability that, given the choice being executing an OLL or a CLL case at the last layer, that you choose to continue the solve by solving the CLL case.

P(1LLL) = q * P(case #2) + (1-q) * P(case #3) + P(case #4) + P(case #5) + P(case #6)

P(1LLL) = q * (1283/1296) * (1/96) + (1-q) * (1283/1296) * (1/72) + (7/1296) * (95/96) + (5/1296) * (71/72) + (1/15552)

P(1LLL) = (8571-1283q) / 373248

And this probability is maximized when q=0. So now that q=0 the probability of a 1LLL using this combined method is:

P(1LLL) = 8571/373248 = 2857/124416 = 2.3% chance

------------

Final Analysis of CLL/ELL and OLL/PLL combined method

This method is best utilized, with a 2.3% chance of a one look last layer or LL skip combined, as long as you always obey the following golden rule:

1) Whenever presented with a last layer that gives you a choice between an unsolved OLL case or an unsolved CLL case, *always* choose to continue by executing the OLL case.

As long as you obey this rule 100% of the time, you will have a 2.3% chance at a 1LLL or a LL skip combined. This is greater than the 1.8% chance using OLL/PLL alone, and both are greater than the 1.7% chance using CLL/ELL alone.

I hope I have cleared this up.

Chris


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## joey (Jun 22, 2010)

cmhardw said:


> This method is best utilized, with a 2.3% chance of a one look last layer or LL skip combined, as long as you always obey the following golden rule:
> 
> 1) Whenever presented with a last layer that gives you a choice between an OLL or a CLL case, *always* choose the OLL case.



What.. I'm confused.


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## Kirjava (Jun 22, 2010)

cmhardw said:


> 1) Whenever presented with a last layer that gives you a choice between an OLL or a CLL case, *always* choose the OLL case.




Sidenote; While the aim is to increase the probability of a skip, this isn't the best technique for having a fast last layer.


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## cmhardw (Jun 22, 2010)

@Joey: The part you quoted, and this is also in line with what Kirjava said, is that strictly from a math point of view, you can increase the chances of a skip in certain situations. The specific situation is that after having solved F2L, you are presented with a LL situation that can be interpretted as *either* an unsolved CLL case, or an unsolved OLL case. It cannot be considered as either a solved CLL or a solved OLL. In these specific cases you should *always* choose to do the OLL case if you want to increase your chances of a skip, with no exception. I changed the wording of the quote to make it a little more clear on this point.

Kirjava brings up a good point that this may not be the fastest case to execute, but it will be the method that increases your chances of a skip *the most*.

--edit--
Ok everyone, final corrections. I got too excited about the result and realized I had written down a number wrong in my reducing of a fraction. Basically, the chance of a skip *is* 2.3% as long as you follow the rule I mentioned in a previous post.

However, I found it interesting to look at what happens if you choose CLL or OLL with a perfect 50-50 split. In this case, the chance of a skip is 2.1% which is still higher than either method alone. So I would say, even if you are choosing CLL and OLL with equal chance, you still noticeably increase your chances for some form of skip during the LL.
--edit--

Chris


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## Tim Reynolds (Jun 22, 2010)

ariasamie said:


> someone please answer this guy:
> 
> 
> TeddyKGB said:
> ...



Let \( f(x) \) be this specific solver's probability of getting a solve of time \( x \). This is a discrete probability distribution since stackmats have limited precision. Then the probability is \( \sum_{x=0.00}^{\infty} f(x)^2 \), where the sum is taken over all numbers \( x>0 \) such that \( 100x \) is an integer. Maybe later tonight I'll try to write out a reasonable probability density and calculate this probability. But I'm kind of too lazy to do that.

The probability that any two consecutive times are the same out of an average of 12 is approximately 11 times this. Of course this overcounts a lot, but \( f(x)^2 \) is probably pretty small, so it's okay.

This also assumes each solve's probability function is independent of all the other times, which is in general a reasonably good assumption I would think.


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## Robert-Y (Jun 22, 2010)

What are the chances of skipping the last two centres on a 5x5x5? (I'm not talking about a supercube)


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## CuBeOrDiE (Jun 22, 2010)

Robert-Y said:


> What are the chances of skipping the last two centres on a 5x5x5?



I think it is : 1/ ((2^4 * 4!) * 2) 
= 1/ (16 * 24 * 2) 
= 1/ (32 * 24) 
= 1/ ((32 * 20) + (32 * 4))
= 1/ (640 + 128) 
= 1 / 768
But I am probably wrong ?

edit: whoops, should have been 1/((2^4*4!)^2)


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## kinch2002 (Jun 22, 2010)

Robert-Y said:


> What are the chances of skipping the last two centres on a 5x5x5?


I would say that you have a 4/8*3/7*2/6*1/5 chance of skipping all 4 xcentres. Same for +centres. So that gives (1/70)^2=1/4900 for skipping the whole last 2 centres


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## Robert-Y (Jun 22, 2010)

kinch2002 said:


> Robert-Y said:
> 
> 
> > What are the chances of skipping the last two centres on a 5x5x5?
> ...


Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)


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## mr. giggums (Jun 22, 2010)

@the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.



mr. giggums said:


> Well the probability of getting a PLL skip after OLL is 1/72 (hence a 1LLL) but you will only use OLL when both corners and edges are misoriented which is 16/19 of the time. So 1/72*16/19=2/171. The chance of getting a OLL skip is 1/216 so a 1LLL. The chance of using COLL 1/8. Then after COLL the chance of edges permutated is 1/12. 1/8*1/12=1/96. Finally the chance of corners oriented 1/27 and after ELL the chance of corners permutated is 1/12. 1/27*1/12=1/324. Now all we have to do is add them all together so 2/171+1/216+1/96+1/324=.029828...
> I'm sure I messed up somewhere but thats my answer.


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## amostay2004 (Jun 22, 2010)

Robert-Y said:


> kinch2002 said:
> 
> 
> > Robert-Y said:
> ...



It isn't at all  Happens quite often for me..though sometimes the last 2 centres need to be swapped


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## kinch2002 (Jun 22, 2010)

amostay2004 said:


> Robert-Y said:
> 
> 
> > kinch2002 said:
> ...


Yeah so 1/35 that you'll get last 2 4x4 centres complete, but half of those times you'll need to swap them.

Btw it's 1/495 that you'll skip the 3rd-to-last centre, giving 1/(70*495)=1/34650 chance of last 3 centres skip.

Even more interestingly, because of the 4 choices of position for your 3rd centre, you have a 1/455 chance of skipping that, which is greater than the chance of skipping the 4th centre! This is all assuming that you have fixed colour for 3rd centre. Obviously if you don't mind which colour you go with then the chance of a skip is even greater (approximately 4*1/455)


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## cmhardw (Jun 22, 2010)

mr. giggums said:


> @the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.
> 
> 
> 
> ...



Can you explain the part in bold more precisely? What exactly do you mean by "when both corners and edges are misoriented"? Do you mean that you don't have both corners and edges oriented at the same time? Because that would be 215/216 chance. I'm afraid I don't see how you got 16/19.

Chris


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## blade740 (Jun 22, 2010)

Robert-Y said:


> Thanks. Hmm... so skipping the last two centres on the 4x4x4 isn't that rare... (but I swear it hasn't happened yet for me!)



You probably assumed you skipped the 3rd and 4th (or 4th and 5th) centers instead. I'd assume that if the 5th and 6th were solved when you solved the 4th, at least one of them was solved before that.


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## Rpotts (Jun 22, 2010)

cmhardw said:


> mr. giggums said:
> 
> 
> > @the 1 LLL debate: I came up with this answer at the top of ther page I don't know if it's right but I couldn't find anything wrong with it.
> ...



I think he's saying that you use OLL unless it's an OLL case that has *either* corners or edges oriented. Which he is claiming is 3/19 of time, when you would use COLL/ELL


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## cmhardw (Jun 22, 2010)

Rpotts said:


> I think he's saying that you use OLL unless it's an OLL case that has *either* corners or edges oriented. Which he is claiming is 3/19 of time, when you would use COLL/ELL



Oh ok, I gotcha. Mr. Giggums can you confirm this?
If this is the case, though, the chance to have either corners or edges (but not both) oriented is:
(1/27) * (7/8) + (26/27) * (1/8) = (7+26)/216 = 11 / 72

I did not allow for both corners and edges to be oriented, because this would constitute an OLL skip, which is handled in another case.

11/72 is approximately = 0.153
3/19 is approximately = 0.158

Hopefully that helps?
Chris


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## mande (Jun 22, 2010)

What is the probability of not getting a single CE pair on a scramble?


----------



## cmhardw (Jun 22, 2010)

Alright, one more post from me about the 1LLL debate, and then I'll let it rest. My previous calculation did not include another case type, and that should be the following:

Another way to get a 1LLL using the combined OLL/PLL and CLL/ELL method is to skip both the CLL and OLL steps, but not the full LL. This would essentially be getting an EPLL after F2L.

Also, I have to give myself one of these --> :fp for not doing the calculation a much easier way. It does not change the result by much, but this result should definitely be correct.

----------------------------

let q = the probability that, when having both an unsolved OLL and an unsolved CLL case after F2L, that you choose to execute a CLL alg.

Again, we are talking about the combined CLL/ELL and OLL/PLL method here. Case #1 refers to case #1 in my previous post.

P(1LLL) = 1 - P(case #1)
P(1LLL) = 1 - [(1283/1296) * q * (95/96) + (1283/1296) * (1-q) * (71/72)]
P(1LLL) = 1 - (q + 364372) / 373248
P(1LLL) = (8876 - q) / 373248

This is maximized when q=0 giving:

P(1LLL) = 8876 / 373248 = 2219 / 93312 = 2.4% approximately.

Pretty close to the same result, but if anyone was following along with the math I left out a case before, *and* I figured out the probability using a very long method when this shortcut existed the whole time.

Chris


----------



## Zane_C (Jun 22, 2010)

This went in the accomplishement thread but it probably belongs here:
16.05, 15.26, (10.13)*[PLL skip]*, 17.92, 16.16*[OLL skip]*, (19.19), 16.91, 17.80, 18.33, 14.93*[PLL skip]*, 11.70*[LL skip]*, 14.69*[PLL skip]*

scrambles were:


Spoiler



1. 16.05 D' L' U R2 B2 R F D' F2 R U2 D F' L2 U' R B2 D R' F' R' F D2 U' L2
2. 15.26 U2 L2 R2 F R U2 F2 B' D2 U2 B' U B U F' L' U B' R2 D2 F' R' B' F L'
3. (10.13)[PLL skip] F2 U2 D L2 R2 B R B R B U' R B' F' R U' L' D B F' D2 U R2 F2 U'
4. 17.92 F L F2 B U' B D2 L F' R' L2 D' R2 D' B' F L D F' U2 R2 L U R2 D'
5. 16.16[OLL skip] R F B' D' R D2 L' B' F L U' L2 D' R' L2 F' R D' L D U' B' L B D
6. (19.19) B L B2 R F' B2 U' L U R2 F' L2 F R2 D' B U' D2 B L2 F2 D2 B' L2 R
7. 16.91 F B D' L2 B' F' U2 R2 D F2 U2 D' R2 F R2 L B U' F2 R2 D2 R F' L' U
8. 17.80 R L F' U B2 L D L2 F' D2 U' L' U' F' U' L2 D B R' D F' L' U2 D' R'
9. 18.33 L2 F' U2 L U' R U2 B2 U' B' L' F U' R2 B' R' U' B' R' L B F R D' R2
10. 14.93[PLL skip] B' L2 D' U' R L' B L U F R2 F2 L' F' U' F R2 U' D2 L' R' B2 R F' B'
11. 11.70[LL skip] U2 L2 U2 D2 L R2 B2 L' D2 U R2 U' D' L D2 U2 L B2 U L R F2 U2 F' B'
12. 14.69[PLL skip] U' F' D2 B2 F' R' L2 D2 F R B2 L F' B2 U' D L B' U2 L F B D U' L2


----------



## Owen (Jun 22, 2010)

Zane_C said:


> This went in the accomplishement thread but it probably belongs here:
> 16.05, 15.26, (10.13)*[PLL skip]*, 17.92, 16.16*[OLL skip]*, (19.19), 16.91, 17.80, 18.33, 14.93*[PLL skip]*, 11.70*[LL skip]*, 14.69*[PLL skip]*
> 
> scrambles were:
> ...



Can someone please calculate this? I tried, but I'm probably wrong.


----------



## Johan444 (Jun 22, 2010)

Owen said:


> Zane_C said:
> 
> 
> > This went in the accomplishement thread but it probably belongs here:
> ...



Calculate what? The probability of getting full step, full step, PLL skip, full step, OLL skip, full step, full step, full step, full step, PLL skip, LL skip, PLL skip, in that order?


----------



## Rpotts (Jun 22, 2010)

Johan444 said:


> Owen said:
> 
> 
> > Zane_C said:
> ...



so full step probability is (215/216)x(71/72) = 15265/15552 ≈ 98.1545%
PLL skip probability is (215/216)x(1/72) = 215/15552 ≈ 1.3824%
OLL skip probability is (1/216)x(71/72) = 71/15552 ≈ .4565%
LL skip probability is (1/216)x(1/72) = 1/15552 ≈ .00643%

btw it all adds up to 15552/15552 so i think my calcs are correct

(15265/15552)x(15265/15552)x(215/15552)x(15265/15552)x(71/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(15265/15552)x(215/15552)x(1/15552)x(215/15552)

6.808388 x 10^-13

.0000000000006808388


----------



## cmhardw (Jun 22, 2010)

Rpotts said:


> so full step probability is (215/216)x(71/72) = 15265/15552 ≈ 98.1545%
> PLL skip probability is (215/216)x(1/72) = 215/15552 ≈ 1.3824%
> OLL skip probability is (1/216)x(71/72) = 71/15552 ≈ .4565%
> LL skip probability is (1/216)x(1/72) = 1/15552 ≈ .00643%
> ...



This is the probability if he must get the exact same kind of skips, in the exact same order that he did during the solve. If you allow the exact same kind of skips, but in any order, you would have to multiply your probability by:
12! / (7! * 3!) = 15840

(6.808388 x 10^-13) * 15840 = approximately 1 x 10^-8
or
0.00000001 probability allowing the exact same scheme of skips to happen in any order.

Chris


----------



## Rpotts (Jun 22, 2010)

Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.


----------



## cmhardw (Jun 22, 2010)

Rpotts said:


> Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.



Yes, your math is good for the situation your were calculating.

Chris


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## DanHarris (Jun 24, 2010)

Chris, you are my hero.


----------



## TeddyKGB (Jun 29, 2010)

Me and a friend were racing yesterday, he's quite a bit slower than me so he solves one cube and I'll solve two, during one particular race, both my cubes had not only the same OLL case but also the same PLL case, just wondering what are the odds of that happening?


----------



## joey (Jun 29, 2010)

Could be 1/15552^2 = 1/241864704.


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## cmhardw (Jun 29, 2010)

TeddyKGB said:


> ... both my cubes had not only the same OLL case but also the same PLL case, just wondering what are the odds of that happening?



Should be \( \sum_{j}\sum_{i} [P(ith OLL)*P(jth PLL)]^2\ \)

I honestly don't have the programming know-ho to do this quickly :-s But given a list of the OLL and PLL probabilities it can definitely be done.



joey said:


> Could be 1/15552^2 = 1/241864704.



Joey I think you are on the right track with this. I think you could argue that a really good estimate to the probability would be 1/15552. You could say that the probability that you get a LL case on the first solve is 1. Then the probability that you get the matching case to that on the second solve is 1/15552. I don't know if the above sum equates to this, and if it does that would be really cool!

Chris


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## Lucas Garron (Jun 29, 2010)

cmhardw said:


> Should be \( \sum_{j}\sum_{i} [P(ith OLL)*P(jth PLL)]^2\ \)




```
o=(Table[RotateRight[#,i]&/@#,{i,4}]//Tally//Length)&/@OLLs;
p=Length/@GatherBy[CanonicalizeLL[StringJoin[#1]]&/@Tuples[{PLLGeneratorCPs,PLLGeneratorEPs}],CubeLL[PerformNewPLLInverse[#]]&];
((o/216)^2//Total)*((p/72)^2//Total)
```

28073/30233088 ≈ 0.000928552




cmhardw said:


> I don't know if the above sum equates to this, and if it does that would be really cool!


Of course it doesn't. What kind of math teacher are you? 

Think of it this way for OLL only: Instead of \( (1/58)^2*58 \), most of the probabilities are just a slight bit larger, and a few of them are quite a bit smaller than 1/58. So, the actual probability for a repeated OLL/PLL/both is always lower than such an estimate. But of course not by much.


----------



## DavidWoner (Jun 29, 2010)

Rpotts said:


> Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.



David Cramer would be proud.


----------



## Rpotts (Jun 29, 2010)

DavidWoner said:


> Rpotts said:
> 
> 
> > Yeah i was calculating it based on johans question, those skips in the same order. Good to know it was correct, if the question is in the same order.
> ...



yay

but wait, doesn't Cramer kinda dislike speedcubing?


----------



## Carson (Jul 11, 2010)

I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?


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## oprah62 (Jul 11, 2010)

Carson said:


> I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?



Jessica's site or badmephistos(I think).

Btw what are the chances of getting 2 solves with the exact same LL and same time? I don't think the time probability can be calculated though. Probably (disregarding time), the same as ^.

edit:here
http://www.ws.binghamton.edu/fridrich/Mike/orient.html


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## Carson (Jul 11, 2010)

oprah62 said:


> Carson said:
> 
> 
> > I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?
> ...



Badmephisto's site has PLL probabilities, but not OLL. I did find it on Macky's site though.


----------



## CubesOfTheWorld (Jul 11, 2010)

oprah62 said:


> Carson said:
> 
> 
> > I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?
> ...



The probability of getting the same time cannot be calculated.


----------



## ben1996123 (Jul 11, 2010)

What is the probability of a first centre skip on 7x7?


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## eastamazonantidote (Jul 12, 2010)

What is the chance of each ZBF2L case occurring? I asked this before but the Hardwick/Kirjava debates took over.


----------



## trying-to-speedcube... (Jul 12, 2010)

The probability for the both-pieces-stuck-in-slot cases is 1/25 total, and I'm pretty sure this is equally distributed. There are 6 such cases, so the probability for any of those 6 is 1/150. 

Now for the edge-stuck-in-slot cases. The probability of an edge being stuck in the slot but not the corner is 1/5*4/5=4/25. There are 6 cases, so the probability for each of those is 4/150.

Same goes for corner-stuck-in-slot cases. 1/5*4/5=4/25. There are also 6 cases, so also 4/150 for all of those.

Then there's the biggest group, where both corner and edge are in the U-layer. The chance of that happening is 4/5*4/5=16/25. There are 24 cases total for this. For ease, let's say the corner is always in UFR. Then there are 4 places where the edge can be, all with 2 orientations. These are all equally distributed. 16/25/24=4/150.

Summary: All the cases have a 4/150 chance of occuring, except for the both-pieces-stuck-in-slot cases, which have a 1/150 chance of occuring.

As for the edge orientation, the cases are always equally distributed when there is at least 1 of the F2L pieces in the U-layer. If not, there is an AUF possible. If the F2L edge is wrongly oriented, the cases have equal probabilities. If the F2L edge is correctly oriented though, the 2 adjacently flipped edges have a probability of twice as high as the 2 oppositely flipped edges.

Going by this list:
In the first group every single case has a probability of 4/150/8=1/300.
In the second group, the first case has 2 cases with a probability of 1/300, the second, third and fourth case have 4 cases with a probability of respectively 1/1200, 1/300, 1/600 and 1/1200 for the skip again. The cases in the fifth and sixth cases all have probabilities of 1/300. The rest of the cases also all have a probability of 1/300 of occuring.


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## irontwig (Jul 12, 2010)

Carson said:


> I am looking for a list of probabilities for each of the OLL's. Does anyone know where I can find this?



Looks the same after y: 1/216
Looks the same after y2: 1/108
The rest: 1/54


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## Rpotts (Jul 12, 2010)

Well put


----------



## whauk (Jul 12, 2010)

ben1996123 said:


> What is the probability of a first centre skip on 7x7?



there are 6 types of center pieces.
there are 24 pieces of each type.
so that makes 24!^6 possibilities.
if one center is solved there are 4!^6 possibilities for one certain center and 4!^6*6 if you want just some random center to be solved.
so thats (4!*6/24!)^6 in the end.
and thats 1.56*10^-130
not very likely


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## That70sShowDude (Jul 13, 2010)

What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?


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## TheTurtleman9 (Jul 13, 2010)

On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?


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## cmhardw (Jul 13, 2010)

That70sShowDude said:


> What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?



\( 
\frac{(4!)^4}{(8!)^2} = \frac{1}{4900} \)

--edit--
Wow.. Interestingly, the probability of every single F2L piece ending up in the LL after finishing the cross is also \( \frac{1}{4900} \)


----------



## Matt S (Jul 13, 2010)

cmhardw said:


> That70sShowDude said:
> 
> 
> > What are the chances of having none of the 8 f2l cubies in the last layer after solving the cross?
> ...



That makes sense, because the number of corners and edges in the LL is equal to the number of each in the F2L once you subtract out the cross. So, having every F2L piece in the LL is equivalent to having every LL piece in the F2L, which is equivalent by symmetry to having every F2L piece in the F2L.


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## cmhardw (Jul 13, 2010)

TheTurtleman9 said:


> On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?



There are 8 corner pieces remaining after the two inner layers have been solved, and all 8! permutations of those pieces are possible. There is no concept of corner orientation, so you can ignore this. Lastly, count how many "solved" states there are, allowing for the possibility of an AUF on either the U or D layers alone, or on both layers. Set this up as a fraction to calculate the total probability.

In case you haven't noticed, I'm purposefully not actually answering your question. The reason for this is that knowing how to calculate probabilities like this one is a good stepping stone for how to calculate the more complex ones. If my hints don't help, reply in this thread or PM me, but try to start from there and see if that helps.

Chris


----------



## jaap (Jul 13, 2010)

whauk said:


> ben1996123 said:
> 
> 
> > What is the probability of a first centre skip on 7x7?
> ...



This isn't correct.
For a particular type of centre piece, there are 24-choose-4 ways for them to be arranged so that the top face is correct, i.e.
24C4 = 24!/(20!4!) = 10626 ways
For all 6 types of top face centre pieces to be correct, that gives
1/(24C4)^6 = 1/ 1439523865867191634829376 = 6.95e-25
This is the probability that the top face centre is solved, regardless of the state of the other face centres.

Assuming for the moment that the six faces are independent and mutually exclusive, the probability of any of the six faces being solved is 6 times that, or about 4.17e-24. To calculate the real probability value you'd probably need to use the inclusion-exclusion principle, but the result won't be far off this approximation.


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## TheTurtleman9 (Jul 14, 2010)

cmhardw said:


> TheTurtleman9 said:
> 
> 
> > On a 2x2x4, if solving center layers then outer, what is the probability of an outer layer skip (both of them)?
> ...



Would it be 1/8! because 8! is all of the permutations and orientations of the corners, and there's only 1 solved state?

Didn't put AUF in though. Would that be 1/(8!*16)?


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## Rpotts (Jul 14, 2010)

This thread should be renamed, Ask Chris Hardwick a question about probability!

LOL


----------



## cmhardw (Jul 14, 2010)

TheTurtleman9 said:


> Would it be 1/8! because 8! is all of the permutations and orientations of the corners, and there's only 1 solved state?
> 
> Didn't put AUF in though. Would that be 1/(8!*16)?



Almost - and *very* close! The only thing to remember is that the final fraction that makes the probability for this type of problem is:

\( \frac{\mbox{number of "solved" states allowing for AUF and/or ADF}}{\mbox{Total number of possible states}} \)

Setup the fraction this way and you'll have your answer.



Rpotts said:


> This thread should be renamed, Ask Chris Hardwick a question about probability!
> 
> LOL



Haha, I'm flattered, but at the same time there are a number of people on this forum whose knowledge of probability dwarf my own. I'm just interested in this type of finite probability, and I'm also a math teacher so I can't get away from math, it's too much fun!  

Chris


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## TheTurtleman9 (Jul 14, 2010)

Oh,so , 16/8! ? 1/2520?


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## cmhardw (Jul 14, 2010)

TheTurtleman9 said:


> Oh,so , 16/8! ? 1/2520?



Yep!


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## eastamazonantidote (Jul 14, 2010)

trying-to-speedcube... said:


> Stuff on ZBF2L



Awesome thanks!


----------



## TheTurtleman9 (Jul 14, 2010)

cmhardw said:


> TheTurtleman9 said:
> 
> 
> > Oh,so , 16/8! ? 1/2520?
> ...



Oh yay.


----------



## Weston (Jul 17, 2010)

Whats the average move count for a CN 2x2x2 block ignoring the corner orientation?


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## hawkmp4 (Jul 19, 2010)

Weston said:


> Whats the average move count for a CN 2x2x2 block ignoring the corner orientation?



That's a pretty complicated problem that's going to require a computer search, not just some probability calculations.


----------



## Zane_C (Jul 27, 2010)

Just wondering what the probability of a complete middle layer edge skip is (LBL). Without any influence such as trying to preserve pairs when inserting corners.


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## hawkmp4 (Jul 27, 2010)

Zane_C said:


> Just wondering what the probability of a complete middle layer edge skip is (LBL). Without any influence such as trying to preserve pairs when inserting corners.



1/8 * 1/7 * 1/6 * 1/5 for edge permutation...1/2^4 for edge orientation.
I believe. I could be wrong. It's really late.
so
1/26880


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## Zane_C (Jul 27, 2010)

Thanks appreciated, just took a look and realised that's not too hard to figure out myself.


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## hawkmp4 (Jul 27, 2010)

You might multiply that by four for four orientations of the cube...but that's the general idea. If you read some posts by Chris Hardwick, you'll pick it up pretty quick. He's the man- at the math, and teaching it.


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## jaap (Jul 27, 2010)

hawkmp4 said:


> You might multiply that by four for four orientations of the cube


No. That is only done with LL calculations because an asymmetrical case can occur in four orientations and therefore represents 4 distinct positions. Rotating the cube as a whole does not change the probabilities in this case because the middle-layer-solved position is fully symmetrical about the y axis.


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## That70sShowDude (Jul 27, 2010)

Probability of a 2x2 scramble containing no "2 bars"?


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## hawkmp4 (Jul 28, 2010)

jaap said:


> hawkmp4 said:
> 
> 
> > You might multiply that by four for four orientations of the cube
> ...


Right! Of course. If I multiplied that by four, that would be calculating the probability of all the middle layer edges being in the middle layer, and having correct permutation relative to each other, yes?


----------



## Christopher Mowla (Jul 30, 2010)

*Probability of Having Inner-Layer Odd Parity on Big Cubes*

I am not very good at probability, but I think I might have just calculated that the probability percentage of having inner-layer odd parity for a cube of degree _n_ equal to:

\( \left( 1-\frac{1}{2^{\left\lfloor \frac{n-2}{2} \right\rfloor }} \right)\times 100 \)


Here is a link to the formula in wolfram.

According to this formula, any cube size beyond 15 has a 99._ percent chance!

The cube sizes under that have the following:

5X5X5 50.0%
7X7X7 75%
9X9X9 87.5%
11X11X11 93.75%
13X13X13 96.875%
15X15X15 98.4375%


Is this right?


----------



## Rinfiyks (Jul 30, 2010)

cmowla said:


> I might have just calculated that the probability percentage of having inner-layer odd parity


Having what? :confused:
Are you referring to parity when you end up with 2 edges left to complete during reduction?


----------



## Lucas Garron (Jul 30, 2010)

cmowla said:


> I was just hoping someone could verify the formula.


What's there to verify? Of course it's correct.

Probability of no parity: 2^-(#orbits)


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## ariasamie (Aug 31, 2010)

After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?


----------



## ~Adam~ (Sep 5, 2010)

cmhardw said:


> I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.
> 
> The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:
> 
> ...



I had a similar but more improbable situation arise. 1st and 2nd F2L pair inserted leaving all remaining pieces in their correct position only leaving 3 corners to be oriented.

Can anyone please tell me the odds of this occurring or better (Solved or 2 edges need flipping) after inserting 2 F2Ls?
Thanks


----------



## hawkmp4 (Sep 5, 2010)

ariasamie said:


> After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?



Orientation is independent of permutation when you use OLL. 1/72.


----------



## ~Adam~ (Sep 5, 2010)

ariasamie said:


> After which OLL is there more chance of getting a PLL skip, or the 1/72 for PLL skip is the same for all of the OLLs?



I would assume the probability is the same for all OLLs.

edit - I got Ninja'd =(


----------



## XXGeneration (Sep 5, 2010)

What are the chances that when you solve your normal cross, an x-cross is also done? Like when you do your normal cross, you manage to do an extended cross inadvertently?


----------



## Stefan (Sep 5, 2010)

XXGeneration said:


> What are the chances that when you solve your normal cross, an x-cross is also done?



About 1%, if I remember correctly.


----------



## Johannes91 (Sep 5, 2010)

StefanPochmann said:


> XXGeneration said:
> 
> 
> > What are the chances that when you solve your normal cross, an x-cross is also done?
> ...



Correct, 1.036368...%.

pairs cases
0 3619921801
1 37715396
2 192606
3 596
4 1
+ 3657830400

Source: http://pastebin.com/yvMyxHGM


----------



## ben1996123 (Sep 5, 2010)

Johannes91 said:


> StefanPochmann said:
> 
> 
> > XXGeneration said:
> ...



So you are 1000 times more likely to get a corners skip than an F2L skip after cross? lol.

EDIT: Some more probabilities that I would like to know (sorry if some have already been posted):

1. 2x2 first layer skip
2. separation skip on 3x3xn where n is 2 or 4+
3. Centers skip on 4x4
4. Megaminx star skip
5. First center skip on 11x11


----------



## StachuK1992 (Sep 6, 2010)

I'm aware that this is the probability thread, so this isn't entirely relevant, but I was wondering if anyone could clear this up from a while ago.



Stachuk1992 said:


> irontwig said:
> 
> 
> > Stachuk1992 said:
> ...



Thanks.


----------



## Lucas (Sep 6, 2010)

ben1996123 said:


> So you are 1000 times more likely to get a corners skip than an F2L skip after cross? lol.
> 
> EDIT: Some more probabilities that I would like to know (sorry if some have already been posted):
> 
> ...



Mmm... let's see... correct me if I am wrong:

1. For one color: 24/(8! / 4! * 3^4) = 24/136080
For any color: 6 times powered? : 1 - (1 - that above)^6 = 0.00106
3. 4!^6/24! = 3.08 * 10^(-16)... a bit difficult 
5. 1 / (24! / 4! * 20!)^20 = 1/3.37*10^80 = 2.97 * 10^(-81)


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## irontwig (Sep 13, 2010)

What's the probability to have these blocks already built:
a: 2x1x1
b: 2x2x1
c: 2x2x2

I assume you can then just half the probabilities if you use non-colour-neutral block F2L.


----------



## Erzz (Sep 14, 2010)

Would it be possible to create a 3 dimensional graph with one variable as corners solved, one as edges solved, and another as probability that the rest of the cube would be solved? Or something similar.

What I mean is, if someone covered the whole cube except the centers and one edge, and you solved that one edge, the probability of the cube being solved after being uncovered. For an LL skip for example, you would solve 8 edges and 4 corners (F2L), the point on the plane where those two meet would be the probability of a LL skip. Or am I too tired to be thinking about math?

If this is possible, what would the equation be?


----------



## eastamazonantidote (Sep 15, 2010)

Erzz said:


> Would it be possible to create a 3 dimensional graph with one variable as corners solved, one as edges solved, and another as probability that the rest of the cube would be solved? Or something similar.
> 
> What I mean is, if someone covered the whole cube except the centers and one edge, and you solved that one edge, the probability of the cube being solved after being uncovered. For an LL skip for example, you would solve 8 edges and 4 corners (F2L), the point on the plane where those two meet would be the probability of a LL skip. Or am I too tired to be thinking about math?
> 
> If this is possible, what would the equation be?



Sounds awesome. Really awesome. I'll get on it at some point, though I'm sure people will beat me to it. It won't be an equation but a series of points (I don't feel like working equations out right now). Get back to me in a week and I should have something worked out.


----------



## Erzz (Sep 15, 2010)

On a related note, what is the probability of a full skip after solving the cross?
Also, the probability of a LL skip without AUF.


----------



## qqwref (Sep 15, 2010)

If x corners and y edges are unknown, and the rest is solved relative to the centers, then there are (x! * y!)/2 * 3^(x-1) * 2^(y-1) possibilities. So the probability it is solved is one over that, assuming all positions are equally likely.

So:
Probability of LL skip (no AUF): x=4, y=4, 1/62208.
Probability of a full cube skip after solving the cross: x=8, y=8, 1/227546313523200.

You can multiply by 4 (for 4 positions considered solved) if you want to also allow anything off by U, U2, U'.


----------



## irontwig (Sep 29, 2010)

What's the probability of the remaining pieces being solvable with two three cycles:
5 corners?
6 corners?
5 edges?
6 edges?
LL?


----------



## bluedasher (Sep 29, 2010)

What is the probability of a PLL skip when using 2 look OLL? I would assume it would be different.
what are the odds of after solving the F2L all edges are oriented on the Upper face?


----------



## That70sShowDude (Sep 29, 2010)

Probability of 2 auf-less PLL skips in a row?


----------



## kinch2002 (Sep 29, 2010)

bluedasher said:


> What is the probability of a PLL skip when using 2 look OLL? I would assume it would be different.
> what are the odds of after solving the F2L all edges are oriented on the Upper face?


1. If you're blinding applying 2 look OLL then it would still be the usual 1/72. Of course you have more options to force a PLL skip as you're doing 2 algs, but this will clearly be difficult unless you learn specific cases, and then it's effectively 1 look OLL with 2 algs anyway.

2. 1/8. This is found by 1/2 chance of each edge being oriented with the last one having no choice. 1/2 x 1/2 x 1/2 = 1/8 


irontwig said:


> What's the probability of the remaining pieces being solvable with two three cycles:
> 5 corners?
> 6 corners?
> 5 edges?
> ...


This one is a tricky one but I'll give it a go.
5 corners: I think this would have to be one 5-cycle. If there was a twisted corner in place then I think it wouldn't be solvable in 2 3-cycles. So the first piece has to go anywhere but to itself (i.e. not twisted), so that's 4/5 (maybe!). Or on second thoughts it might be 6/7 (12/14) if you think of it in terms of which sticker a specific sticker on it needs to go to. So after that, the second piece needs to go to another piece other than the first one, giving you 3/4, or the same 3/4 (9/12) if you think of stickers. The third piece now has to go to another piece other than the first one (it can't go to the second one clearly), giving 2/3 chance, or 2/3 (6/9) in terms of stickers. Now the fourth piece needs to go to the fifth piece, giving 1/2 chance, or 3/5 in terms of stickers (as it can't go back to the same sticker you started with on piece 1 otherwise the fifth piece would be solved rather than twisted in place.
So that gives me 4/5*3/4*2/3*1/2 = 1/5 the first way and 6/7*3/4*2/3*3/5 = 9/35 the second way. I have no idea whether either way is right so maybe someone else could help me out here! I'm not going to start on the others because this twisted pieces thing is doing my head in!



That70sShowDude said:


> Probability of 2 auf-less PLL skips in a row?


 Nice easy one  PLL skip is 1/72, so AUF-less is 1/288. This gives 1/288 x 1/288 = 1/82944 for 2 AUF-less PLL skips in a row.


----------



## That70sShowDude (Sep 29, 2010)

kinch2002 said:


> Nice easy one  PLL skip is 1/72, so AUF-less is 1/288. This gives 1/288 x 1/288 = 1/82944 for 2 AUF-less PLL skips in a row.



O_O I got one of those.


----------



## ben1996123 (Sep 29, 2010)

ben1996123 said:


> 2. separation skip on 3x3xn where n is 2 or 4+
> 4. Megaminx star skip


 
Anyone know any of these?


----------



## qqwref (Sep 30, 2010)

Star skip is easy, assuming you're not color neutral. There are 30 edges with two orientations each, so you just need the probability that those five are solved, which should be (1/60)(1/58)(1/56)(1/54)(1/52) = 1/547223040, assuming a completely random scramble.


I'm not 100% sure what you mean by separation skip; I'll assume you mean you want all the pieces to be in the correct layers. Then, for a 3x3x2, you just need all four top pieces of each type in the four top slots, so the corners and edges each have a probability of 1/(8 choose 4), for a total separation skip probability of 1/4900.

For a 3x3x4 the problem is a little more complicated because in the inner layers you can't tell what layer the edges go to. The probability is the same for the corners, but for the edges we have to choose exactly one edge from each of the four pairs of identical edges. There are (8 choose 4) possible choices, of which 16 choices are acceptable (although I haven't factored in parity). So the inner layers have a separation skip probability of 4/1225. This probability will remain the same for inner layers of any higher 3x3xeven cuboid. The total separation skip probability for the 3x3x4 is 1/1500625.


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## ben1996123 (Oct 1, 2010)

qqwref said:


> I'm not 100% sure what you mean by separation skip; I'll assume you mean you want all the pieces to be in the correct layers. Then, for a 3x3x2, you just need all four top pieces of each type in the four top slots, so the corners and edges each have a probability of 1/(8 choose 4), for a total separation skip probability of 1/4900.


 
Yes that's what I meant. I have had 3 of those on 3x3x2 sim o.o


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## bluedasher (Oct 1, 2010)

for the 3x3, what is the probability of, when using F2LL, you have a cross on top already made? It's more of a tri-cross, but O' well.


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## hawkmp4 (Oct 2, 2010)

bluedasher said:


> for the 3x3, what is the probability of, when using F2LL, you have a cross on top already made? It's more of a tri-cross, but O' well.


 
Do you mean F2L? If so, (1/2)^3 = 1/8.


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## Tim Major (Oct 2, 2010)

hawkmp4 said:


> Do you mean F2L? If so, (1/2)^3 = 1/8.


 
F2LL = Winter Variation. I'm not good with probabilities, and even though this is simple, I don't want to be wrong.


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## hawkmp4 (Oct 7, 2010)

ZB_FTW!!! said:


> F2LL = Winter Variation. I'm not good with probabilities, and even though this is simple, I don't want to be wrong.


 
Okay. Then, from what I know of WV, the idea is that you orient corners while inserting the last pair. And edges are already oriented using another method (says the wiki). In that case, of course, it's probability 1. However, if you don't do anything with edges before you do WV, the chances you'll have all 4 edges oriented is as above, (1/2)^3=1/8.


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## JeffDelucia (Oct 12, 2010)

cmhardw said:


> Joey I think you are on the right track with this. I think you could argue that a really good estimate to the probability would be 1/15552. You could say that the probability that you get a LL case on the first solve is 1. Then the probability that you get the matching case to that on the second solve is 1/15552. I don't know if the above sum equates to this, and if it does that would be really cool!
> 
> Chris


 
Is this including the AUF? He didn't say that both solves had the same AUF so I'm not sure if its correct.


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## ariasamie (Oct 14, 2010)

on a 4x4x4 what is the probability of... duh... I can't describe it in words!
what is the probability of getting AT LEAST on of these:


Spoiler














so i'm only talking about white and yellow which are two opposite centers and they can show up at the same time.
and also if i didn't want to attach pictures, how could I ask it using sentences?


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## Johannes91 (Oct 14, 2010)

Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
Answer: 69%

I can post more statistics if anyone's interested.


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## mrCage (Oct 14, 2010)

Johannes91 said:


> Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
> Answer: 69%
> 
> I can post more statistics if anyone's interested.


How about a V4 with black stickers instead of white??

Per


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## ariasamie (Oct 15, 2010)

Johannes91 said:


> Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
> Answer: 69%
> 
> I can post more statistics if anyone's interested.


 
thank you.
can you tell me how you found that out please?
I may want to explain it to some people.


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## Johannes91 (Oct 15, 2010)

ariasamie said:


> thank you.
> can you tell me how you found that out please?
> I may want to explain it to some people.


I wrote a program that goes through all positions and counts the number of pairs in each one.

It can also be calculated by hand, but at least my way is a bit tedious.



Spoiler



First, the number P of different positions for the 8 center pieces is C(24,4) * C(20,4) = 51482970.

We'll count all positions with no pairs, case by case.

A) W,W,W,W,-,- : 3840 cases
AA) WYY,WYY,W,W,-,- : 6 cases
AB) WYY,WY,WY,W,-,- : 108 cases
AC) WYY,WY,W,W,Y,- : 288 cases
AD) WYY,W,W,W,YY,- : 16 cases
AE) WYY,W,W,W,Y,Y : 64 cases
AF) WY,WY,WY,WY,-,- : 81 cases
AG) WY,WY,WY,W,Y,- : 864 cases
AH) WY,WY,W,W,YY,- : 216 cases
AI) WY,WY,W,W,Y,Y : 864 cases
AJ) WY,W,W,W,YY,Y : 192 cases
AK) W,W,W,W,YY,YY : 4 cases

B) WW,W,W,-,-,- : 1920 cases
BA) WWYY,WYY,W,-,-,- : 2 cases
BB) WWYY,WY,WY,-,-,- : 9 cases
BC) WWYY,WY,W,Y,-,- : 72 cases
BD) WWYY,W,W,YY,-,- : 6 cases
BE) WWYY,W,W,Y,Y,- : 48 cases
BF) WWY,WYY,WY,-,-,- : 12 cases
BG) WWY,WYY,W,Y,-,- : 48 cases
BH) WWY,WY,WY,Y,-,- : 216 cases
BI) WWY,WY,W,YY,-,- : 72 cases
BJ) WWY,WY,W,Y,Y,- : 576 cases
BK) WWY,W,W,YY,Y,- : 96 cases
BL) WWY,W,W,Y,Y,Y : 128 cases
BM) WW,WYY,WYY,-,-,- : 1 case
BN) WW,WYY,WY,Y,-,- : 72 cases
BO) WW,WYY,W,YY,-,- : 12 cases
BP) WW,WYY,W,Y,Y,- : 96 cases
BQ) WW,WY,WY,YY,-,- : 54 cases
BR) WW,WY,WY,Y,Y,- : 432 cases
BS) WW,WY,W,YY,Y,- : 288 cases
BT) WW,WY,W,Y,Y,Y : 384 cases
BU) WW,W,W,YY,YY,- : 12 cases
BV) WW,W,W,YY,Y,Y : 96 cases

C) WW,WW,-,-,-,- : 60 cases
CA) WWYY,WWYY,-,-,-,- : 1 case
CB) WWYY,WWY,Y,-,-,- : 64 cases
CC) WWYY,WW,YY,-,-,- : 16 cases
CD) WWYY,WW,Y,Y,-,- : 192 cases
CE) WWY,WWY,YY,-,-,- : 32 cases
CF) WWY,WWY,Y,Y,-,- : 384 cases
CG) WWY,WW,YY,Y,-,- : 384 cases
CH) WWY,WW,Y,Y,Y,- : 1024 cases
CI) WW,WW,YY,YY,-,- : 24 cases
CJ) WW,WW,YY,Y,Y,- : 384 cases
CK) WW,WW,Y,Y,Y,Y : 256 cases

The comma-separated lists tell how many white/yellow center each face contains.

Total cases for A: 3840 * (6 + 108 + 288 + ...) = 10379520
Total cases for B: 1920 * (2 + 9 + 72 + ...) = 5245440
Total cases for C: 60 * (1 + 64 + 16 + ...) = 165660

Total cases with no 1x2 blocks: N = 10379520 + 5245440 + 165660 = 15790620

Probability that there is at least one 1x2 block: (P - N) / P ≈ *69%*


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## Alcuber (Oct 16, 2010)

mrCage said:


> How about a V4 with black stickers instead of white??
> 
> Per


Lanlan 4x4


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## bluedasher (Oct 18, 2010)

how many moves can a "star" on the megaminx always be solved in? Just curious because I ordered mine today and I want to know some statistics prior to solving it.


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## bluecloe45 (Nov 5, 2010)

Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')


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## bluedasher (Nov 5, 2010)

bluecloe45 said:


> Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')



I'm not really good with probabilities, but here it goes. According to Jessica Fridrich's page...
http://www.ws.binghamton.edu/fridrich/Mike/orient.html
the probability of that algorithm occurring is 1/54. then *3 = 1/162


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## That70sShowDude (Nov 5, 2010)

1/157464 for that oll 3 times in a row


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## mr. giggums (Nov 5, 2010)

bluedasher said:


> I'm not really good with probabilities, but here it goes. According to Jessica Fridrich's page...
> http://www.ws.binghamton.edu/fridrich/Mike/orient.html
> the probability of that algorithm occurring is 1/54. then *3 = 1/162


 
Actually you would do 1/54 ^3 not *3 which would = 1/157464

Edit: Ninja'd twice in less than 5 minutes but mine is more detailed


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## bluedasher (Nov 5, 2010)

That70sShowDude said:


> 1/157464 for that oll 3 times in a row



That sounds more like it.


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## bluecloe45 (Nov 5, 2010)

That70sShowDude said:


> 1/157464 for that oll 3 times in a row


 That just happened LOLOLOOL


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## Rinfiyks (Nov 5, 2010)

bluecloe45 said:


> Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')


 
Well there are 57 OLLs, 58 including the solved case.
20 and 58 have 4 symmetries.
1, 21, 55, 56, 57 have 2 symmetries.
The remaining 51 have 1 symmetry.
2/(4x) + 5/(2x) + 51/x = 1
x = 54
Your OLL has 1 symmetry, so its probability of occurrence is 1/54 (2 symmetries would be 1/108, 4 would be 1/216). Oh bother, I could've just divided 216 by 4 to get 54 instead of adding them all up :fp

Anyway... probability of getting your OLL thrice in a row is 1/54^3 = 1/157464
But of course, it's not less likely that you will get other OLLs 3 times in a row.
So,

1/29 chance you get one of 2 OLLs with 1/216
5/58 chance you get one of 5 OLLs with 1/108
51/58 chance you get one of 51 OLLs with 1/54
Edit: This is all wrong /\
This is right \/
2/216 chance you get one of 2 OLLs with 1/216
5/108 chance you get one of 5 OLLs with 1/108
51/54 chance you get one of 51 OLLs with 1/54

Square it, not cube it, because you don't care which OLL is repeated 3 times, so it is a 1 in 1 chance that you get the first one first try.
2/216 * 1/216^2 + 5/108 * 1/108^2 + 51/54 * 1/54^2
= 551/1679616
= 1 in ~3048
Thanks, Lucas Garron, for pointing that out.



bluecloe45 said:


> That just happened LOLOLOOL


Yeah but you'd be just as pleased to get any other OLL 3 times in a row, so it's not that exciting.
It's like saying
OMG I got D2 B2 D' U' F D2 B2 U F R D2 L' R U' R B2 R B D' scramble! Probability 1 in 43 quintillion!


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## Lucas Garron (Nov 5, 2010)

Rinfiyks said:


> Square it, not cube it, because you don't care which OLL is repeated 3 times, so it is a 1 in 1 chance that you get the first one first try


Sure?

It sort of appears right, but my calculations get 551/1679616:


```
c = Select[Tuples[Table[{0, 1, 2}, {4}]], Mod[Total[#], 3] == 0 &];
e = Select[Tuples[Table[{0, 1}, {4}]], Mod[Total[#], 2] == 0 &];
p = (First[Sort[Table[RotateRight[#, i] & /@ #, {i, 0, 3}]]] & /@ Tuples[{c, e}] // Tally)[[All, -1]]/216;
Total[p^3]
```

EDIT: Oh, duh. Your weights in the sum are off:


Rinfiyks said:


> 1/29 * 1/216^2 + 5/58 * 1/108^2 + 51/58 * 1/54^2!


1/29, eh? 
In practice, since most of the OLLs are the 51, the numbers work out approximately the same, but it's not a good way to do the math.

Correct would be (2 * 1/216) * 1/216^2 + (5 * 1/108) * 1/108^2 + (51 * 1/54) * 1/54^2.
I did 2 * 1/216^3 + 5 * 1/108^3 + 51 * 1/54^3, which is the same, but I think clearer: What's the chance of getting three in a row, for each OLL?


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## Rinfiyks (Nov 6, 2010)

Lucas Garron said:


> EDIT: Oh, duh. Your weights in the sum are off:



Ohhhhhhh yeaaaaaaaah! Stupid mistake. I'll think twice before I post any maths in future.


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## dada222 (Nov 7, 2010)

Suppose you solve the cross on the bottom and top layer, how many algorithms would you need to make up to solve both faces simultaneously? 

Forget it, just realized the number is huge.


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## ben1996123 (Feb 18, 2011)

Bump.

Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.

Would this just be 1/72^6, or would it be 4/72^6 since there are 2 different pll's (H/skip)?

EDIT: next solve forced skip 0.o


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## Johan444 (Feb 18, 2011)

ben1996123 said:


> Bump.
> 
> Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.
> 
> ...


 
In that fixed order it'd be 1/72^5 * P(Forced skip)


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## cuBerBruce (Feb 18, 2011)

ben1996123 said:


> Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.
> 
> Would this just be 1/72^6, or would it be 4/72^6 since there are 2 different pll's (H/skip)?


The standard PLL probabilities assume the permutation of the LL pieces are ignored prior to PLL. If you're sometimes forcing a PLL skip, then clearly you're not always ignoring the permutation of the LL pieces until the PLL step. So without more information on how the PLL is being influenced, accurate probabilities can not be computed.

It's also not clear what exactly you probability you're trying to compute. Do you mean exactly that sequence skip, H, H, skip, skip, skip for a given set of six consecutive solves? (and not skip, H, skip, H, skip, skip; nor skip, skip, H, skip, H, H; nor any other combination of H's and skips)


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## Rpotts (Feb 18, 2011)

What's the probability of getting No AUF LL skip followed by a solve where the last slot + LL is just J perm + AUF? I basically had two LL skips in a row, I inserted the last pair with j perm, giving me a U2 LL.


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## cuBerBruce (Feb 20, 2011)

Rpotts said:


> What's the probability of getting No AUF LL skip followed by a solve where the last slot + LL is just J perm + AUF? I basically had two LL skips in a row, I inserted the last pair with j perm, giving me a U2 LL.


 
Assuming method used was not trying to influence the last layer case...

Probability of no AUF PLL skip: 1/((1/2)*(4!)^2)*(2^3)*(3^3)) = 1/62208

Probability of (Last slot J-Perm with no AUF before) + AUF (after J-Perm, or no AUF):
8/((1/2)*((5!)^2)*(2^4)*(3^4)) = 1/1166400

If possibly an AUF before the J-Perm, then multiply by 4 (i.e., 4 times more likely).

Probability of precisely these two cases in that order on two consecutive solves: 1/(62208*1166400) = 1/72,559,411,200


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## Andrew Ricci (Feb 20, 2011)

How many unique cross cases are there?


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## nccube (Feb 20, 2011)

What is the probability of getting a cubeshape square-1. One scramble from Spanish Championships had it.


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## Mike Hughey (Feb 21, 2011)

nccube said:


> What is the probability of getting a cubeshape square-1. One scramble from Spanish Championships had it.


 
4 in 3678 (1 in 919.5). I have the odds of all of the shapes for square-1 in my blindfold solving page that's listed in my signature.


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## cuBerBruce (Feb 22, 2011)

theanonymouscuber said:


> How many unique cross cases are there?


 
The 4 cross pieces can be arranged 12*11*10*9 ways with 2^4 orientation possibilities for a total of 190080 arrangements.

If you consider rotational symmetry, I calculate this reduces the number of cases to 47652. If you also consider mirror symmetry, I calculate 24098 cases.

If you consider allowing AUF & ADF, I calculate 5582 cases, not counting mirrors as the same; and 2909 cases if you count mirrors as the same.

Subtract 1 from the above numbers if you don't want to count the solved case.


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## Cyrus C. (Feb 22, 2011)

How many total F2L cases are there? Including misplaced edges, corners, etc. Mirrors, rotations, inverses, AUF included.


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## qqwref (Feb 22, 2011)

cuBerBruce said:


> If you consider allowing AUF & ADF, I calculate 5582 cases, not counting mirrors as the same; and 2909 cases if you count mirrors as the same.


Hang on. I'd want to allow AUF *after* the cross (as well as y rotations and mirrors), but I wouldn't want to allow any adjusting moves *before* the cross, just y rotations, since the full case must always be looked at during inspection. How many cases would we have then?


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## Antcuber (Feb 22, 2011)

tough one:

Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?


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## Cyrus C. (Feb 22, 2011)

Antcuber said:


> tough one:
> 
> Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?


 
1 in 43 quintillion, right? There's only one state that the scramble will solve. I assume you mean scrambling a cube, then applying a different scramble to the cube.


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## cuBerBruce (Feb 22, 2011)

qqwref said:


> Hang on. I'd want to allow AUF *after* the cross (as well as y rotations and mirrors), but I wouldn't want to allow any adjusting moves *before* the cross, just y rotations, since the full case must always be looked at during inspection. How many cases would we have then?


 
OK, I assume you want the number of cases for relative cross, without allowing AUF/ADF before cross.

I get 6138 cases (including solved). And by the way, only one of these cases requires 8 moves to make the relative cross.

If you count mirrors as separate cases, then it's 11916 cases (including solved).


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## ben1996123 (Feb 22, 2011)

cuBerBruce said:


> OK, I assume you want the number of cases for relative cross, without allowing AUF/ADF before cross.
> 
> I get 6138 cases (including solved). And by the way, *only one of these cases requires 8 moves* to make the relative cross.





U R L F' B' D2 R L (8f*)


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## cuBerBruce (Feb 22, 2011)

cuBerBruce said:


> OK, I assume you want the number of cases for relative cross, without allowing AUF/ADF before cross.
> 
> I get 6138 cases (including solved). And by the way, *only one of these cases requires 8 moves* to make the relative cross.
> 
> ...


 
Wrong! That's obviously wrong because the U move does not affect the relative arrangement of the cross pieces!

(You need to swap two adjacent cross edges.)


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## hic0057 (Feb 22, 2011)

Pyraminx probability question...

What the probability of having a corner-edge pair on a pyraminx?

What the probabiltiy of having a corner-edge-corner block like one on the first step of petrus made?
http://www.speedsolving.com/wiki/index.php/Pyraminx_Speedsolving_Methods

What is the probability of having a one move corner-edge-corner block?


Thank you so much if you can answer these.


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## cuBerBruce (Feb 22, 2011)

hic0057 said:


> Pyraminx probability question...
> 
> What the probability of having a corner-edge pair on a pyraminx?


[post=395400]See here.[/post]



hic0057 said:


> What the probabiltiy of having a corner-edge-corner block like one on the first step of petrus made?


There are 48407 positions with at least one corner-edge-corner block.

So the probability is 48407/933120 or approximately 5.19%



hic0057 said:


> What is the probability of having a one move corner-edge-corner block?


There are 174136 positions with no corner-edge-corner block that are one move from a position that does have such a block.

So the probability is 174136/933120 or approximately 18.66%.

The probability of needing zero or one move to get such a block is about 23.85%.

There are 445708 positions that require 2 moves to make a corner-edge corner block.


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## Antcuber (Feb 23, 2011)

Cyrus C. said:


> 1 in 43 quintillion, right? There's only one state that the scramble will solve. I assume you mean scrambling a cube, then applying a different scramble to the cube.


 
no i mean for example having a 25 move random sequence that solves itself.


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## irontwig (Feb 23, 2011)

5.19% seems way low, Bruce. It would be 1/24 for a specific corner-edge pair, which is over 4% by itself.


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## cuBerBruce (Feb 23, 2011)

irontwig said:


> 5.19% seems way low, Bruce. It would be 1/24 for a specific corner-edge pair, which is over 4% by itself.


 
I think you're thinking of a 3x3x3 cube, not a Pyraminx.


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## irontwig (Feb 23, 2011)

Oops.


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## clincr (Feb 23, 2011)

Probability of a 2x2 LL skip?


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## Rpotts (Feb 23, 2011)

CP skip - 1/6 x CO skip 1/27 = 1/162

Is that correct? This probability counts a U2/U/U' LL the same as an AUF skip.


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## ilikecubing (Feb 23, 2011)

Probability of a PLL skip?


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## Andrew Ricci (Feb 24, 2011)

ilikecubing said:


> Probability of a PLL skip?


 
1/72


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## toastman (Feb 25, 2011)

Antcuber said:


> tough one:
> 
> Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?


 
I like this. Difficult, but a rough maximum to start the thinking: Assume a 20 move "naive" scramble. Probability that moves 11-20 are the inverse of 1-10 = 1 in 18^10 = 1 in 3.57 Trillion

The actual probability is likely way higher than this, as there's so many possibilities, e.g. where the scramble has junk like R L R' L' in it. I'm guessing by a factor of a hundred.

So, it *could* happen, given millions of cubers doing hundreds of thousands of solves in a lifetime. And I want to see the dude's face when it does. He'd be all like "Holy crap!"


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## Rune (Feb 25, 2011)

Obviously, you are not Bruce?


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## Cyrus C. (Feb 25, 2011)

Couldn't you also think of it as:

That scramble will only solve one state, probability that state is a solved cube: 1 in 43 quintillion.


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## ilikecubing (Feb 25, 2011)

Probability of a PBL skip on 2x2?

probability that corners are solved after finishing F2L on 3x3?


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## irontwig (Feb 25, 2011)

1/6^2
1/27*6


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## Julian (Feb 25, 2011)

toastman said:


> I like this. Difficult, but a rough maximum to start the thinking: Assume a 20 move "naive" scramble. Probability that moves 11-20 are the inverse of 1-10 = 1 in 18^10 = 1 in 3.57 Trillion
> 
> The actual probability is likely way higher than this, as there's so many possibilities, e.g. where the scramble has junk like R L R' L' in it. I'm guessing by a factor of a hundred.
> 
> So, it *could* happen, given millions of cubers doing hundreds of thousands of solves in a lifetime. And I want to see the dude's face when it does. He'd be all like "Holy crap!"


This is wrong because scambles aren't just random strings of letters. First, a random combination is selected, then a scramble is generated that leads to that combination.
So the probability is one in ~43 quintillion.


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## Julian (Feb 25, 2011)

Sorry for the double post, but I'm writing this from my iPod at school, and when I tried to edit my last post, It wouldn't let me scroll though it to get to the end.
anyway...
What the probability of a relative cross solved on any colour?
How many combinatons are there when scambling using <R,U>?
After solving a 2x2x3 block in Petrus, how many combinations are there left?


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## Kynit (Feb 25, 2011)

Julian said:


> After solving a 2x2x3 block in Petrus, how many combinations are there left?


 
6 corners to permute (6!) * 3 orientations for all but one corner (3^5) * 7 edges to permute (7!) * 2 orientations for all but one edge (2^6) / Inability to get PLL parity (2)

=~ 28 billion - does this sound right?


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## Johannes91 (Feb 25, 2011)

Antcuber said:


> Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?


Yes.

To get a probability, you need to define "random scramble". For random state the answer is obvious. But for a random move sequence of 20 or more moves the exact answer seems practically impossible to determine (feel free to prove me wrong!).



Julian said:


> What the probability of a relative cross solved on any colour?


I'm not 100% sure what you mean by relative, but see http://www.cubezone.be/crossstudy.html.



Julian said:


> How many combinatons are there when scambling using <R,U>?


6*5*4 * 3^5 * 7! / 2 = *73,483,200*



Julian said:


> After solving a 2x2x3 block in Petrus, how many combinations are there left?


6! * 3^5 * 7! * 2^6 / 2 = *28,217,548,800*


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## Schmidt (Feb 25, 2011)

After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??


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## Kynit (Feb 25, 2011)

1/4: there are 4 ways to AUF, and 1 is right.


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## cuBerBruce (Feb 26, 2011)

Schmidt said:


> After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??


 


Kynit said:


> 1/4: there are 4 ways to AUF, and 1 is right.



I think the question perhaps needs a little clarification.

Generally a PLL has 4 angle cases and 4 AUF cases. For example, for the A-Perms, the algs I use require same color corner stickers facing me. If not, I have to rotate the cube, or at least the U layer so that same color corner stickers are facing me. I now have the correct angle case. After executing the A-Perm alg, I still might not have the U layer aligned properly with the bottom two layers. This can be handled by an AUF after the algorithm or by rotating the bottom two layers before the alg. If the Schmidt only cares about if the angle case is correct, the answer would be a little larger than 1/4. This is because the N-Perms and H-Perm only have 1 angle case, and Z -Perm and E-Perm have 2 angle cases. I view PLL skip as having 1 angle case and 4 AUF cases.

If we assume there is no attempt to influence the PLL case, and we only care about having the U layer correct for applying the alg, and don't care about if there will need to be an AUF or not, then the probabilty I get is 11/36 (or 85/288 if you want to consider AUF adjustments on PLL skips to matter).

If the question is for having to do no adjustment for either the angle case or for the AUF case, the probability becomes (11/36)*(1/4) = 11/144.

Of course, there are a number of assumptions here, such as you're not trying to influence the PLL case and you only use one particular alg for each PLL case (always applied from the same angle) regardless of which angle or top/bottom alignment situation you might get.


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## ben1996123 (Mar 15, 2011)

Probability of a 2 move scramble QTM in 2x2?


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## Johan444 (Mar 15, 2011)

ben1996123 said:


> Probability of a 2 move scramble QTM in 2x2?


 
Probably only Ryan Heise knows. E-mail him here: webmaster at hi-games.net


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## Rinfiyks (Mar 16, 2011)

Johan444 said:


> Probably only Ryan Heise knows. E-mail him here: webmaster at hi-games.net


Why? Unless I've misunderstood the question, just divide the number of 2-move QTM scrambles (I think it's 27) by the number of combinations?


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## riffz (May 15, 2011)

What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)


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## Marcell (May 15, 2011)

I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?


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## irontwig (May 15, 2011)

ben1996123 said:


> Probability of a 2 move scramble QTM in 2x2?


 
1/408 240


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## riffz (May 15, 2011)

Marcell said:


> I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?



I mean only corners. And yes, completely ignoring twisted corners as if they are solved. 

x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10


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## Marcell (May 15, 2011)

Right, thanks. Now let me think about it


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## cuBerBruce (May 15, 2011)

riffz said:


> What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)


 


riffz said:


> I mean only corners. And yes, completely ignoring twisted corners as if they are solved.
> 
> x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10


 
What about four 2-cycles?
And does it matter if the buffer is not in a cycle?

Anyway, I used GAP to get the distribution of cycle structures for permutations of 8 objects.

```
Even Permutations
{ 7 }		5760
{ 6, 2 }	3360
{ 5 }		1344
{ 5, 3 }	2688
{ 4, 4 }	1260
{ 4, 2 }	2520
{ 3, 3 }	1120
{ 3, 2, 2 }	1680
{ 3 }		 112
{ 2, 2, 2, 2 }	 105
{ 2, 2 }	 210
{  }		   1

Odd Permutations
{ 8 }		5040
{ 6 }		3360
{ 5, 2 }	4032
{ 4, 3 }	3360
{ 4, 2, 2 }	1260
{ 4 }		 420
{ 3, 3, 2 }	1120
{ 3, 2 }	1120
{ 2, 2, 2 }	 420
{ 2 }		  28
```


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## toastman (May 16, 2011)

Awesome stats! Thanks dude.

So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle. The rest do. I hate when that happens. I have to go back and count all the corners by putting my fingers on them.

if my maths are correct, according to his signature, Amostay2008 "pwns" on 3360/40320 or 8.3% of corner solves.


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## cuBerBruce (May 16, 2011)

toastman said:


> So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle.



To get 16065, I see that you added up all the cases where there is at most one cycle. I note that if you use a fixed buffer, that buffer may not always be part of a cycle. For example, take the 7-cycle case. There are 5760 such permutations. 7/8 of them or 5040 will have the buffer in the cycle, and in the remaining 720 cases, the buffer piece is in its correct place (not part of the cycle). If you consider it to be "breaking into a new cycle" in this case when the buffer is not in a cycle to start with, but you still have a cycle to solve, then you have overcounted the number of cases of "not breaking into a new cycle." 

Under these assumptions, the number of cases I calculate reduces to 13700 cases, or just a little under 34%.


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## amostay2004 (May 16, 2011)

toastman said:


> if my maths are correct, according to his signature, Amostay2008 "pwns" on 3360/40320 or 8.3% of corner solves.


 
Who the hell is that


----------



## riffz (May 18, 2011)

cuBerBruce said:


> What about four 2-cycles?
> And does it matter if the buffer is not in a cycle?
> 
> 
> ...


----------



## cuBerBruce (May 18, 2011)

riffz said:


> cuBerBruce said:
> 
> 
> > What about four 2-cycles?
> ...


 
OK, I admit I should have read your post more carefully before I asked those questions. But basically, I also wanted to know how to calculate the number of pieces you have to shoot in both of the cases when the buffer piece is initially in its correct position and when it is not.

It looks to me that the answer to that would be this. If N is the total number of corners incorrectly placed, and C is the number of cycles, then I believe the number of pieces you have to shoot is N + C - 2 if the buffer piece is not solved (i.e. not in its correct position, ignoring orientation), and N + C if the buffer piece is solved.

I would also need to know about how you are handling parity. For now, I will go with the assumption you are considering all even and odd parity cases for the corners, and (for the purposes of this question), you want to solve all corners ignoring the fact that it may be necessary to permute some edges (well, two anyway) in the process.

Then the distribution of cases would be (according to my calculations):

```
number
  of
shoots   cases
   0        1
   1        7
   2       42
   3      231
   4     1015
   5     3430
   6     8379
   7    13083
   8    10408
   9     3409
  10      315
```
This results in an average of about 6.97 pieces to shoot.


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## ilikecubing (May 18, 2011)

Probability of *not* getting a last two edges skip on 4x4?


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## riffz (May 19, 2011)

cuBerBruce said:


> Then the distribution of cases would be (according to my calculations):
> 
> ```
> number
> ...


 
Your assumption about parity is correct. This is exactly what I was looking for. Thank you!


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## toastman (May 19, 2011)

amostay2004 said:


> Who the hell is that


 
Your little brother?


----------



## toastman (May 19, 2011)

riffz said:


> cuBerBruce said:
> 
> 
> > What about four 2-cycles?
> ...


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## riffz (May 20, 2011)

toastman said:


> If I had Four 2-cycles I'd DNF it  *Especially if the damn buffer was solved.*


 
Good thing that's not possible.


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## qqwref (May 21, 2011)

ilikecubing said:


> Probability of *not* getting a last two edges skip on 4x4?


Easy: 23/24 normally, and 7/8 if they are already paired.


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## timeless (May 21, 2011)

riffz said:


> cuBerBruce said:
> 
> 
> > What about four 2-cycles?
> ...


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## EnterPseudonym (May 22, 2011)

4 Z-perms in a row?


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## miniGOINGS (May 22, 2011)

1/36^4 or 1/1,679,616


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## EnterPseudonym (May 22, 2011)

miniGOINGS said:


> 1/36^4 or 1/1,679,616


 
Nice, I thought that was how it was calculated but I wasn't sure.


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## riffz (May 24, 2011)

timeless said:


> this has 12 shoots?


 
No? I don't understand if you are asking a question or trying to correct me.


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## ilikecubing (Jul 6, 2011)

probability of atleast 3 white corners being in the U layer after i make a white cross on D layer


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## Julian (Jul 6, 2011)

ilikecubing said:


> probability of atleast 3 white corners being in the U layer after i make a white cross on D layer


(4/8)*(3/7)*(2/6) = 0.0714285714 ≈ 7%

...I think.


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## qqwref (Jul 6, 2011)

I get (1 + 4*4)/(8!/(4!^2)) ~= 24.3%.

The 1 and 4*4 are the number of positions of the white and yellow corners that leave four and three (respectively) white corners on top; the 8!/(4!^2) is the total number of positions of the white and yellow corners.


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## Julian (Jul 6, 2011)

You're probably right, but could you explain the flaw in my reasoning?

My thought was that there are 8 corners that can occupy UFL, and 4 of them are white. 4/8
Assuming UFL is a white corner, there are 7 corners that can occupy UFR, and 3 of them are white. 3/7
Assuming blah, there are 6, 2 are white. 2/6
And the last doesn't matter.


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## qqwref (Jul 6, 2011)

I think you just calculated the probability that three specific top corners were white - basically you're ignoring the fact that there are four different ways to place three white corners on the top. For every one of your fractions, it's possible that that particular corner was yellow, and that the rest of the corners were white, but you didn't consider that possibility.


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## Stefan (Jul 6, 2011)

qqwref said:


> I get (1 + 4*4)/(8!/(4!^2)) ~= 24.3%.



In other words, (1 + 4*4) / 8C4. How did you get 4*4? Just curious, as I see two different ways.


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## qqwref (Jul 7, 2011)

My reasoning for 4*4 was that the top layer and bottom layer each have 4 possibilities. So I would write the "full" top layer as 1 + 4*4 + 6*6 + 4*4 + 1 (which fortunately is equal to 8 choose 4).


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## Stefan (Jul 7, 2011)

Ok, that's one of the two. Because you had only mentioned the top layer and because you had explicitly mentioned 8! (hinting at distinguishing the different white corners), I thought you might've meant 4 possibilities to pick 3 white corners to be in the top layer, then 4 possibilities to place them there (disregarding the placement in the bottom layer). Which btw also leads to 1 + 4*4 + 6*6 + 4*4 + 1.


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## Jorghi (Jul 13, 2011)

O_O Whats the probability of finding a completed F2L PAIR during the 15 second inspection? Two pairs?? Three? If you only solve on the white cross, and if you are color neutral??

And I mean just finding the pair inside the cube, not joined with a cross or anything.


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## Jorghi (Jul 13, 2011)

For white, there are 4 total correct F2L pairs for white. And a total of (12 Left edges in the scramble).

So only 1/12 of those edges would match with ONE f2l pair corner. And there are 4 possible pairs, so (1/12) * 4 -> So .333 is the probability that AN edge is correct with a corner. 
(SOMEONE correct me if I'm wrong.. Remember no cross is made so its 12 possible edges(including yellow))

_____________________________________________________________________________________________________________________________________________________________________

Now if you are color neutral, there are 6 possible sides. So there are going to be that many more possible f2l pairs, 6*4-> 24 F2L pairs Possible NOW.
But there is another variable to be careful for, instead of 12 possible edges to match with each f2l pair, its 6*12 -> 72

24/72 = 33% again!
WRONG!! I don't think that is correct xD  because each of the probabilities for a single color(white) are independent of each other.
We want to find the full probability TOGETHER!

.333^6 is the probability that EACH cross color will have 33% chance of having an f2l pair = .001, or highly improbable. 

But what is the probability that the cross colors combined together will have of .A. free f2l pair given that they have a 33% chance for 1?

(AGain plz correct me if I'm wrong I'm just saying stuff)


_____________________________________________________________________________________________________________________________________________________________________

I used Wikipedia I think I found out how to find out O_O

(6 choose 1) * .333^1 * .666^5

WHAT?? Only 26% chance for 1 f2l pair?? ok I did something wrong plz fix it someone

WAIT! I think I forgot to take in to account the probability that if you have 1 CE pair found, that the probability for another CE pair will INCREASE!


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## Jorghi (Jul 13, 2011)

O_O I think I'm right O_O

Color Neutral:
Zero pairs found: 0.0872660613 -> 8.7%
One pair found: 0.261798184 -> 26%
Two pairs found: 0.32724773 -> 33%
Three pairs found: 0.218165153 -> 22%
Four pairs found: 0.0818119325 -> 8%
Five pairs found: 0.0163623865 -> 1.6%
Six pairs found: 0.00136353221 -> .1%

Not taking into account this "WAIT! I think I forgot to take in to account the probability that if you have 1 CE pair found, that the probability for another CE pair will INCREASE!"
because I don't really know how to and whether it is already happening because the probability looks like a curve thing.


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## qqwref (Jul 13, 2011)

Where do those numbers come from?


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## Jorghi (Jul 13, 2011)

(6 choose X) * .333^X * .666^(6-X)


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## Jorghi (Jul 13, 2011)

The probabilities should be A LOT higher for pairs found, but I didn't make it so finding an CE pair will lower the # of edges possible for the next pair to 11, 10, 9 ect.


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## MTGjumper (Jul 13, 2011)

I hope you don't use that writing style when answering questions in maths exams.


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## qqwref (Jul 13, 2011)

Jorghi said:


> (6 choose X) * .333^X * .666^(6-X)


 
So the probability of finding 7 pairs is 0 * (1/3)^7 * (2/3)^-1? Something tells me your equation is completely wrong.


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## Jorghi (Jul 13, 2011)

Wow I forgot orientation.

White cross solver:
8 possible matches for a corner, only 1 is correct.
So 1/8 chance of matching the corner, but there are 4 white-corners, so 
50% chance(ignoring orientation) of having 1 f2l CE pair for white.

There is a 33% chance of the Corner being oriented the same way as the edge.
So a 1/6 chance of having a true F2l Pair for Color Neutral.

___________________________________________________________________________________________________



White cross solver:
8 possible matches for a corner, only 1 is correct.
So 1/8 chance of matching the corner, but there are 4 white-corners, so 
50% chance(ignoring orientation) of having 1 f2l CE pair for white.

There is a 33% chance of the Corner being oriented the same way as the edge.
So a 1/6 chance of having a true F2l Pair for Color Neutral.

One F2L Pair:
1/8(Es left) * 4(Cs left) -> 50% -> 17% (With orientation)

Two F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
0.214285714 -> 7% (With Orientation)

Three F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
1/6(Es left) * 2(Cs left)
0.0714285714 -> 2% (With Orientation) 

Four F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
1/6(Es left) * 2(Cs left)
1/5(Es left) * 1(Cs left)
0.0142857143 -> .0047 (With Orientation)


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## Ltsurge (Jul 25, 2011)

just get off the screen and actually cube...


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## cubersmith (Aug 1, 2011)

Chances of a LL Skip


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## RNewms27 (Aug 1, 2011)

cmhardw said:


> I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.
> 
> The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:
> 
> ...


 


cubersmith said:


> Chances of a LL Skip


 
Mission accomplished.


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## Sa967St (Aug 17, 2011)

What's the likelihood of having at least one 2x1x1 block on a 3x3x3 scramble?


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## Owen (Aug 17, 2011)

This thread makes me very sad that I know nothing about mathematics.


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## Johannes91 (Aug 17, 2011)

Sa967St said:


> What's the likelihood of having at least one 2x1x1 block on a 3x3x3 scramble?


About 15%. *[Edit: No it's not. Read the following posts.]*

(Based on some code I wrote over 2 years ago.)


----------



## cubersmith (Aug 17, 2011)

Owen said:


> This thread makes me very sad that I know nothing about mathematics.


 
Hehe. Me too.


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## HelpCube (Aug 17, 2011)

This was posted on here a while ago but ignored. What is the max amount of moves a megaminx cross can be solved in?


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## Cool Frog (Aug 18, 2011)

Johannes91 said:


> About 15%.
> 
> (Based on some code I wrote over 2 years ago.)


 
I inspect for 1x2x1 block every scramble, and this just seems wrong to me... 

(however I couldn't proove you wrong, or proove I am right for that matter. Just my many inspections)


----------



## mr. giggums (Aug 18, 2011)

Cool Frog said:


> I inspect for 1x2x1 block every scramble, and this just seems wrong to me...
> 
> (however I couldn't proove you wrong, or proove I am right for that matter. Just my many inspections)


 
I inspect for it too and it feels like two third. I'll do a test by doing random scrambles and get the average

1 = there is a 1x1x2, 0 = there isn't

1101101111100101001111011010111010011001101111111110111100111011011010111111110110000111001110000110

There was 100 scrambles 65 of them had at least one 1x1x2 pair. So that is 65% which is really close to my estimated 2/3. This isn't entirely accurate but it is close.


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## Johannes91 (Aug 18, 2011)

It's all joey's fault, he said that 15% seems reasonable. I suspected it might be wrong, I probably misremember what the code does.

Edit: Yepp, it only looks for ce-pairs that are connected _and_ solved with respect to centers.


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## Robert-Y (Aug 18, 2011)

I tried 100 scrambles myself using http://www.ryanheise.com/cube/speed.html

and I got 68/100 with at least one 1x1x2 pair. Which of course, is very close to Mr Giggums' estimation of 2/3 too.


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## cubersmith (Aug 28, 2011)

Chance of totally disassembling the cube and re-assembling it in a solvable position?


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## cubernya (Aug 28, 2011)

That's well known. 1/12

1/2 for EO
1/3 for CO
1/2 for Permutation


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## cubersmith (Aug 28, 2011)

theZcuber said:


> That's well known. 1/12
> 
> 1/2 for EO
> 1/3 for CO
> 1/2 for Permutation


 
Cool, can we use this idea to answer any question like that? 
So if I said, if 5 corners and 4 edges are taken out, whats the chance of it being solvable?


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## cubernya (Aug 28, 2011)

Assuming you still assemble it randomly (like blindfolded ) 1/12...as long as you take out 2 corners and 1 edge and assemble randomly it's 1/12


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## cubersmith (Aug 28, 2011)

theZcuber said:


> Assuming you still assemble it randomly (like blindfolded ) 1/12...as long as you take out 2 corners and 1 edge and assemble randomly it's 1/12


 
Haha thanks. I'm learning about probability. Could you tell me what ^ and * mean so I can understand this thread better?


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## cubernya (Aug 28, 2011)

* is multiplication
^ is exponents

Example 5*2=10
5^2=25


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## cubersmith (Aug 28, 2011)

theZcuber said:


> * is multiplication
> ^ is exponents
> 
> Example 5*2=10
> 5^2=25


 
Explain the second one?


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## cubernya (Aug 28, 2011)

You didn't learn that?

5^2 = 5*5
5^3 = 5*5*5
5^4 = 5*5*5*5


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## cubersmith (Aug 28, 2011)

Oh, okay thanks. Yeah I did we just dont call it that in Scotland


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## Chrisalead (Aug 30, 2011)

Probability of a OLL + PBL skip on the 2x2 ? (It happened to me solving with Ortega last week).


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## kinch2002 (Aug 30, 2011)

Chrisalead said:


> Probability of a OLL + PBL skip on the 2x2 ? (It happened to me solving with Ortega last week).


If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162


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## Godmil (Aug 30, 2011)

cubersmith said:


> Oh, okay thanks. Yeah I did we just dont call it that in Scotland


 
*Phew, I thought it was just me who didn't recognise that name (despite knowing what it was).


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## cubernya (Aug 30, 2011)

You probably were thinking of power?


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## Chrisalead (Aug 30, 2011)

kinch2002 said:


> If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162


 
Doh ! I should have noticed that ! Thanks ^^. Fun that the probability for a 2x2 LL skip is so high compared to a 3x3 LL skip ! 96 times more chances in fact !


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## michaelfivez (Aug 31, 2011)

kinch2002 said:


> If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162



With Ortega you don't know the permutation of the first layer? Or am I missing something?
A OLL+PBL skip is 1/162 * 1/6 then?


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## MostEd (Aug 31, 2011)

i keep getting PLL skips on 7x7... while none on smaller size


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## kinch2002 (Aug 31, 2011)

michaelfivez said:


> With Ortega you don't know the permutation of the first layer? Or am I missing something?
> A OLL+PBL skip is 1/162 * 1/6 then?


Surely you can plan the whole first face, and therefore would be able to 'predict' what the permutation of it will be?


----------



## cubersmith (Aug 31, 2011)

kinch2002 said:


> If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162



But I think what he's asking is what are the chances of both layers being fully oriented and permuted after you make a face.



MostEd said:


> i keep getting PLL skips on 7x7... while none on smaller size


 
I might be wrong but I think this is just chance.


----------



## Sebastien (Aug 31, 2011)

So, if you really just build one side and totally ignore the permutation of the bottom layer, then of course the chance of a skip after that layer is 1/6*1/162 = 1/972 which is actually not that rare.


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## kinch2002 (Dec 8, 2011)

If I force edges to be solved at the end of F2L so that I have a maximum of 4 corners left, what the probability that:
a) At least one of them will be solved
b) It's a 3-cycle left (3 twisted is not allowed)

I came out with the answers 31/108 and 31/162 respectively (~0.287 and ~0.191) but I'm really not sure whether I made an error somewhere so I'd appreciate it if someone else had a go.


----------



## cmhardw (Dec 8, 2011)

kinch2002 said:


> If I force edges to be solved at the end of F2L so that I have a maximum of 4 corners left, what the probability that:
> a) At least one of them will be solved
> b) It's a 3-cycle left (3 twisted is not allowed)
> 
> I came out with the answers 31/108 and 31/162 respectively (~0.287 and ~0.191) but I'm really not sure whether I made an error somewhere so I'd appreciate it if someone else had a go.


 
I got the same thing for part a)

For part b) I got 2/9



Spoiler



4*(2*3^2) = 72

1) 4 choices for the solved (permuted and oriented) corner.
2) 2 possible derangement permutations of 3 remaining corners
3) 3^2 possible orientations of the 3 unsolved corners

4*3*3^3 = 324 total possible positions for F2L and LL cross solved.

72/324 = 2/9


----------



## kinch2002 (Dec 9, 2011)

Thanks Chris - I can see a massive hole in my working for b) now 
Now I know how lucky I should expect to get when I'm doing this sort of thing in FMC


----------



## antoineccantin (Dec 29, 2011)

What is the probability of getting a Last Layer skip on a pyraminx?


----------



## Sa967St (Dec 29, 2011)

antoineccantin said:


> What is the probability of getting a Last Layer skip on a pyraminx?


1/12. 
1/3 for having them all permuted correctly and 1/4 for having them all oriented correctly.


----------



## LeighzerCuber (Dec 29, 2011)

What is the probability of getting a random scramble of a 7x7 with one center fully solved? 4x4?


----------



## Christopher Mowla (Dec 29, 2011)

LeighzerCuber said:


> What is the probability of getting a random scramble of a 7x7 with one center fully solved? 4x4?


I think all you need to do to calculate this for the nxnxn is just take out the positions that one composite center contributes to the total number of positions on the nxnxn. To see how I got the following piece (in the denominator of the fraction below) of the formula for the number of positions on the regular nxnxn, see this post.

\( \frac{20!^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }\left( \frac{1}{4!^{5}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }}{24!^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }\left( \frac{1}{4!^{6}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \)

There are \( \left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor \) orbits (sets) of 24 non-fixed center pieces. 4 are in each face. Therefore we change 24! to 20! (24-4=20). For the other factor, we change 4!^6 to 4!^5 because we are pretending to have 5 composite centers instead of 6. 

\( =\left( \frac{\frac{20!}{4!^{5}}}{\frac{24!}{4!^{6}}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\left( \frac{20!}{4!^{5}}\frac{4!^{6}}{24!} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor } \)\( =\left( \frac{20!}{24\left( 23 \right)\left( 22 \right)\left( 21 \right)20!}\frac{4!^{6}}{4!^{5}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\left( \frac{1}{\left( 23 \right)\left( 22 \right)\left( 21 \right)} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\frac{1}{10626^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \)

All you have to do is substitute a value for _n_, corresponding to the cube size you want. See this link for the probability of a random composite center for cube sizes 2 through 11 (you can change the value of _n_ for larger cube sizes).


----------



## cubersmith (Jan 5, 2012)

True story,

Today I got 3 E-Perms in a row. I thought "what are the chances of that?" Later, I got two PLL skips in a row. Can anyone tell me what were the chances of getting both of these things happening on the same day. Considering there are 365 days in a year.


----------



## Julian (Jan 5, 2012)

cubersmith said:


> True story,
> 
> Today I got 3 E-Perms in a row. I thought "what are the chances of that?" Later, I got two PLL skips in a row. Can anyone tell me what were the chances of getting both of these things happening on the same day. Considering there are 365 days in a year.


How many days there are in a year doesn't matter. However, How many solves you did today does.


----------



## cubersmith (Jan 5, 2012)

Julian said:


> How many days there are in a year doesn't matter. However, How many solves you did today does.


 
Around 100


----------



## blakedacuber (Jan 5, 2012)

(chances of Eperm)^3 x (chances of pll skip)^2 x 100


i think its 1/24186470400

not entirely though


----------



## Sa967St (Jan 5, 2012)

blakedacuber said:


> (chances of Eperm)^3 x (chances of pll skip)^2 x 100
> i think its 1/24186470400
> not entirely though


You have to consider that the 3 consecutive E perms and 2 consecutive PLL skips can start and end anywhere in the 100 solves. Also it's possible to have more than 3 or 2 in a row (implied from the question, unless he meant exactly 3 and exactly 2).



cubersmith said:


> True story,
> 
> Today I got 3 E-Perms in a row. I thought "what are the chances of that?" Later, I got two PLL skips in a row. Can anyone tell me what were the chances of getting both of these things happening on the same day. Considering there are 365 days in a year.


For 100 solves in a day:

P (>2 E perms in a row in 100 solves)= 
\( 1- (35/36)^{100} \)\( - 100(99/\binom {100}{2})(1/36)(35/36)^{99} \)\( - \binom {100}{2} (98/\binom {100}{3}) (1/36)^2(35/36)^{98} \)
P (>1 PLL skip in a row in 100 solves)= 
\( 1- (71/72)^{100} - 100(71/\binom {100}{2})(1/72)(71/72)^{99} \)

The probability of getting at least 3 E perms in a row and 2 PLL skips in a row on the same day will be close to the sum of the two above probabilities. Since you can't have an E perm and a PLL skip in the same solve, there will be a bit of error (anyone know how to get it without error?).


----------



## tasguitar7 (Jan 7, 2012)

Am I correct that the probability of getting a cll skip if using cll ell for last layer is 1/162?
162 = 4!*(3^3)/4
4! permutations (can be odd permutation for corners because edges are unsolved)
3^3 orientations (3 orientations per corner except for the 4th whose orientation is dependent on the other 3)
/4 for AUF


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## IanTheCuber (Jan 8, 2012)

I did an AO5 recently, and I got 3 PLL skips. What are the chances of that?


----------



## cubersmith (Jan 8, 2012)

IanTheCuber said:


> I did an AO5 recently, and I got 3 PLL skips. What are the chances of that?


 
Depends how many averages of 5 you did.


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## Czery (Jan 8, 2012)

If you did just one average then...

Chances of PLL skip ==> 1/72
chances getting X 3/5 times ==> 5 C 3 = 10

so 3 PLL skips in of one avg 5 neglecting order = (1/72) / 10 = 1/720


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## Sa967St (Jan 8, 2012)

IanTheCuber said:


> I did an AO5 recently, and I got 3 PLL skips. What are the chances of that?



P (exactly 3 PLL skips in 5 solves) 
= \( \binom {5}{3} (1/72)^3(71/72)^2 \)
= \( 25205/967458816 \)

About 0.00002605%, assuming all the PLL skips were 1/72 (no CxLL, OLLCP or other ways of forcing easier PLL cases used during OLL).

edit: If you meant >2 PLL skips in 5 solves it would be
= \( \binom {5}{3} (1/72)^3(71/72)^2 + \) \( \binom {5}{4} (1/72)^4(71/72)^1 + (1/72)^5 \)
= \( 25205/967458816 + 355/1934917632 + 1/1934917632 \)
= \( 8461/322486272 \)

Which is about 0.00002624%


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## cubernya (Jan 8, 2012)

Czery said:


> If you did just one average then...
> 
> Chances of PLL skip ==> 1/72
> chances getting X 3/5 times ==> 5 C 3 = 10
> ...


 
I'm sorry, but that's not even close.

Odds of getting a skip : 1/72
Odds of not getting a skip : 71/72
Solves with a skip : 3/5

1/72^3 * 71/72^2 * 5/3 = 1/226,845

I would think that's how you would do it, correct?


----------



## Sa967St (Jan 8, 2012)

theZcuber said:


> I'm sorry, but that's not even close.
> 
> Odds of getting a skip : 1/72
> Odds of not getting a skip : 71/72
> ...


 
You got ninja'd, but I got a different final answer. Did you mean 5C3 instead of 5/3? I have no idea where you got the 226 845 from. 

Also odds aren't the same thing as probability.
The odds of a PLL skip are 1:71 whereas the probability/likelihood/chance of a PLL skip is 1/72.


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## cubernya (Jan 8, 2012)

Sa967St said:


> You got ninja'd, but I got a different final answer. Did you mean 5C3 instead of 5/3? I have no idea where you got the 226 845 from.
> 
> Also odds aren't the same thing as probability.
> The odds of a PLL skip are 1:71 whereas the probability/likelihood/chance of a PLL skip is 1/72.


 
Yeah I was thinking 5C3 but for whatever reason put down 5/3. Not sure why


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## avgdi (Mar 4, 2012)

This isn't cubing related but I wanted to know if someone could figure out the probability of something for me.

A couple of years ago I randomly picked up a deck of cards. After shuffling it multiple times I guessed what color the top card would be before flipping it over. I got it right, so next I guessed which suit the second card would be. After getting that right I guessed which number the third card was going to be. I got that one right too and was rather surprised, so then I guessed what the exact card would be. Some how I got that one right too.

It was pretty crazy and me and my friends spent the next hour trying to do it again, but no one ever got it.

I was wondering what the probability of that happening is.


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## qqwref (Mar 4, 2012)

First card's color: 1/2
Second card's suit: 1/4
Third card's number: 1/13
Fourth card exactly: 1/52
Since they are all independent events, you can get the probability by just multiplying everything together: 1/2 * 1/4 * 1/13 * 1/52 = 1/5408.

(This is approximate, though, because each card you already saw does influence the information of the next cards. For instance, if the first card was red, the second card would be more likely to be a spade or club than a heart or diamond. And the fourth card would actually be chosen out of 49 cards since you already know three cards it can't be. However, since I don't know what your actual guesses were, these approximate numbers give a pretty good estimate.)


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## StachuK1992 (Mar 14, 2012)

Not really probability, but how many distinct cases are there for the last 2 slots? (last 2 slot multislot.)


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## jonlin (Mar 17, 2012)

Could you do the math for me?
There are 80 competitors at a competition.
What are the chances that all OLLs appear during that competition during all three rounds.
Assume that 24 get into the 2nd round, 12 get into the finals.
Also, for the 2x2 round, assuming there are 40 competitors, and 2 rounds, assuming 12 get into the finals, what is the chance that you get all CLL/EG cases on the last layer?
They don't have to be using CLL or EG.


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## Sa967St (Mar 19, 2012)

Sahid Velji said:


> Could someone please explain to me how 1/72 and 1/216 (?) are the probabilities of getting a PLL/OLL skip?


 
OLL skip:
P(EO skip) = 1/(2^3) = 1/8 
P(CO skip) = 1/(3^3) = 1/27
P(EO+CO skip) = 1/(8*27) = 1/216

PLL skip:
P(CP skip, regarding AUF) = 1/6
P(EP skip | corners are solved) = 1/12 
P(CP+EP skip) = 1/(12*6) =72


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## Stefan (Mar 19, 2012)

Sa967St said:


> P(CP skip) = 1/(4!/2) = 1/12
> P(EP skip) = 1/3! = 1/6



CP and EP both permute four pieces.
P(CP skip) and P(EP skip) should be the same.


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## Sa967St (Mar 19, 2012)

Stefan said:


> CP and EP both permute four pieces.
> P(CP skip) and P(EP skip) should be the same.


For CP I ignored EP, and for EP I took CP in consideration. I know independently that they're both 1 in 12, but together you have to half it because of P parity. I updated my post.


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## Stefan (Mar 19, 2012)

Sa967St said:


> For CP I ignored EP


 
But for CP alone (ignoring EP) it's either 1/24 (if you don't allow AUF) or 1/6 (if you do allow AUF). I don't see how it's 1/12.


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## Sa967St (Mar 19, 2012)

Stefan said:


> But for CP alone (ignoring EP) it's either 1/24 (if you don't allow AUF) or 1/6 (if you do allow AUF). I don't see how it's 1/12.


I don't know, I thought it was 1/12 not allowing AUF, I guess I'm wrong.
How is a PLL skip 1/72 then?


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## Stefan (Mar 19, 2012)

There are 4!*4!/2=288 permutations (EP*CP/parity), of which 4 are skips (because of AUF). And 4/288 = 1/72.


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## Christopher Mowla (Mar 19, 2012)

@Sarah, just use the number of permutations formula for the LL for these. For any of these skips, just start with the fraction:
\( \frac{\frac{1}{2}\left( \overset{\text{Edges}}{\mathop{\left( 4! \right)\left[ 2^{3} \right]}}\, \right)\left( \overset{\text{Corners}}{\mathop{\left( 4! \right)\left[ 3^{3} \right]}}\, \right)}{\frac{1}{2}\left( \overset{\text{Edges}}{\mathop{\left( 4! \right)\left[ 2^{3} \right]}}\, \right)\left( \overset{\text{Corners}}{\mathop{\left( 4! \right)\left[ 3^{3} \right]}}\, \right)} \)

Now, whatever you wish to find the probability of, remove it from the numerator (never touch the denominator). The only time you remove the 1/2 from the numerator is if both EP and CP are removed (because the odd permutation exists only if EP, CP, or both still need to be solved, not if both do not need to be solved). And, if AUF is to be considered, just multiply your answer by 4 (because 1/4 of the permutations can yield the same result. Because that's true and because the number of permutations are in the denominator, you put the 4 in the numerator, i.e., multiply by 4).


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## Czery (Mar 22, 2012)

While solving my Square 1, I got an edge permutation skip! 
Assuming I use full Vandenburgh, what is the probabilty that I will get this skip again?


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## LouisCormier (Mar 22, 2012)

What are the odds of PLL skip on megaminx? OLL skip?


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## TMOY (Mar 22, 2012)

PLL skip is 1/720, OLL skip is 1/1296.


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## CubeRoots (Mar 22, 2012)

I don't know about OLL but PLL is definitely not p=1/720. Maybe you mean 1/72? there are 22 PLL cases (one of which is PLL skip), for almost every case there are 4 ways it could occur (just the way it is oriented), except the symetrical cases. The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?


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## Rpotts (Mar 22, 2012)

CubeRoots said:


> Can anyone verify or fault my logic?


 


LouisCormier said:


> What are the odds of PLL skip on *megaminx*? OLL skip?



There it is.


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## Sa967St (Mar 22, 2012)

Czery said:


> While solving my Square 1, I got an edge permutation skip!
> Assuming I use full Vandenburgh, what is the probabilty that I will get this skip again?


I believe it's 1/576. 1/24 for the U edges and 1/24 for the D edges.



CubeRoots said:


> The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?


Not quite.
Two PLLs are 1/36 (Z and E), three are 1/72 (H, Na, Nb), and the other 16 are 1/18. That leaves 1/72 for a PLL skip.


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## TMOY (Mar 23, 2012)

CubeRoots said:


> I don't know about OLL but PLL is definitely not p=1/720. Maybe you mean 1/72? there are 22 PLL cases (one of which is PLL skip), for almost every case there are 4 ways it could occur (just the way it is oriented), except the symetrical cases. The one that swaps opposite edges is an example, these can only occur in two ways because of the symmetry. I think there are 6 of these, So in total there are 15*4 + 6*2 = 72 cases + PLL skip so I guess it's roughly 1/72, 1/73 to be pedantic. Can anyone verify or fault my logic?


 
Well, it's pretty easy. Assume there are actually 73 different PLL cases, counting different orientations separately. Since 73 is prime, this is only possible if there are 73 pieces of the same kind on the LL; this is obviously not the case, hence your logic must be wrong 

Actually the correct result for the cube ls 1/72. But as Rpotts pointed out, we were talking about the megaminx, for which the correct result is 1/720. (There are 60 possible permutations of corners (5!/2, since only even permutations are possible), same for edges, hence 3600 possibilities disregarding AUF, hence the probability of a PLL skip is 1 out of 3600/5=720.)


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## jonlin (Mar 25, 2012)

TMOY said:


> PLL skip is 1/720, OLL skip is 1/1296.


 
I'm guessing that an LL skip is 1/933120 chance of happening. Now we have to wait for Simon to report back to us to see when he gets an LL skip on megaminx.


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## Riley (Mar 26, 2012)

Are the chances of getting an LL skip with winter variation (as in you have winter variation in the current solve) 1/22? Because there are 21 PLL's you might get and 1 just solved last layer? (don't count AUF's) Thanks.


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## ben1996123 (Mar 26, 2012)

Riley said:


> Are the chances of getting an LL skip with winter variation (as in you have winter variation in the current solve) 1/22? Because there are 21 PLL's you might get and 1 just solved last layer? (don't count AUF's) Thanks.



1/72 probly.


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## IanTheCuber (Mar 28, 2012)

If you're just orienting the corners, then it would be 1/63.

Otherwise, no idea. You might also have to include half the F2L's, since mirrors make no extra contribution.


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## Riley (Mar 28, 2012)

IanTheCuber said:


> If you're just orienting the corners, then it would be 1/63.
> 
> Otherwise, no idea. You might also have to include half the F2L's, since mirrors make no extra contribution.


 
I'm trying to say, if you have a R U' R' insertion with 3 edges on the top layer that are already correct (like yellow facing up, if white was your cross color), what is the chance of an LL skip, using WV, so that the corners orient? No mirrors. And if it's still 63, can you please explain how you get that? Thanks!


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## Rpotts (Mar 28, 2012)

Riley said:


> I'm trying to say, if you have a R U' R' insertion with 3 edges on the top layer that are already correct (like yellow facing up, if white was your cross color), what is the chance of an LL skip, using WV, so that the corners orient? No mirrors. And if it's still 63, can you please explain how you get that? Thanks!


 
Same chance as a PLL skip - 1/72


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## IanTheCuber (Mar 29, 2012)

Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.


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## Godmil (Mar 29, 2012)

IanTheCuber said:


> Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.


 
Hmm, you know you could probably draw a probability distribution, like a gaussian or something with the 1/150 defining the width. Should be possible to get a rough estimate. But I don't know how to do it


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## MTGjumper (Mar 29, 2012)

IanTheCuber said:


> Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.



I doubt that someone who averaged 37 would only get a sub-35 one in 150 solves. But a normal distribution should work, providing you chose the variance such that 1/150 solves was sub-35.


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## ben1996123 (Mar 29, 2012)

IanTheCuber said:


> Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.



Assuming the solve time, X is random and normally distributed:

X ~ N(37, σ)
P(X < 35) = 0.0067
P(Z < -2/σ) = 0.0067
P(Z < 2/σ) = 0.9933
2/σ = 2.48
σ = 2/2.48 = 0.806
X ~ N(37, 0.806)
P(X < 30) = P(Z < (30-37)/0.806) = P(Z < -8.685) = 0.0000000000000001894%


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## ben1996123 (Apr 10, 2012)

After solving the centres on a 4x4, how many possible states are there?

Attempt:

Edge permutations(24!) * Corner permutations(8!) * Corner orientations(3^7) = 54711040793092776444454502400000

So the probability of a centres skip on 4x4 is 54711040793092776444454502400000/7401196841564901869874093974498574336000000000 = \( \frac{1}{135277939046250} \)

Or do I need to divide something by 24 somewhere?


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## TMOY (Apr 10, 2012)

Seems correct to me. Once the centers are solved, the orientation of the cube is fixed, so you don't have to divide by 24.


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## CubeRoots (Apr 10, 2012)

TMOY said:


> Well, it's pretty easy. Assume there are actually 73 different PLL cases, counting different orientations separately. Since 73 is prime, this is only possible if there are 73 pieces of the same kind on the LL; this is obviously not the case, hence your logic must be wrong
> 
> Actually the correct result for the cube ls 1/72. But as Rpotts pointed out, we were talking about the megaminx, for which the correct result is 1/720. (There are 60 possible permutations of corners (5!/2, since only even permutations are possible), same for edges, hence 3600 possibilities disregarding AUF, hence the probability of a PLL skip is 1 out of 3600/5=720.)



I know this was 2 weeks ago but i realised mistake, I thought there were 2 ways to get Hperm but in reality there was only one so i ended up with 73 instead of 72 possibilities


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## Akash Rupela (Apr 18, 2012)

probability that corners are correctly oriented after f2l=?
Probability that corners are correctly solved after f2l?
Probability that no corner is correctly oriented after f2l?
Probability that edges are correctly solved after f2l?
Probability that No edge is correctly oriented after f2l?


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## Rpotts (Apr 18, 2012)

1/27
1/162
6/27
1/96
1/8


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## cuBerBruce (Apr 18, 2012)

Akash Rupela said:


> Probability that corners are correctly solved after f2l?


1/162, allowing AUF (1/648 if not allowing AUF)


Akash Rupela said:


> Probability that edges are correctly solved after f2l?


1/48, allowing AUF (1/192 if not allowing AUF)


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## ben1996123 (Apr 19, 2012)

What is the probability that after a random 3x3 scramble, one side will be solved (not necessarily a layer, just a face)? Just curious because I got this scramble about a week ago: R2 B2 R2 D' L2 D F2 U F2 L2 D2 F U2 F D' F2 R' F D L2 F D'

7 greens on F. I think I had one other scramble like this some time last year too, but I've never had one with 8 or 9.

Edit: Might as well have a go at calculating it myself:

edge permutations: 4! = 24
edge orientations: 2^4 = 16
corner permutations: 4! = 24
corner orientations: 3^4 = 81
centres: 6

1/(24*16*24*81*6) = \( \frac{1}{4478976} \)

not sure if its correct.


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## cuBerBruce (Apr 20, 2012)

ben1996123 said:


> What is the probability that after a random 3x3 scramble, one side will be solved (not necessarily a layer, just a face)? Just curious because I got this scramble about a week ago: R2 B2 R2 D' L2 D F2 U F2 L2 D2 F U2 F D' F2 R' F D L2 F D'


 
Centers can be considered fixed. Corner and edge stickers of a given color must lie on the same face as the center of that color.

Consider a particular face such as the U face. Probability that the ULF sticker is on the U face is 4/24. The probability that the UFR sticker is on the U face, given that we know the ULF sticker already is, is 3/21. Similarly for the URB and UBL stickers, the probabilities can be given as 2/18 and 1/15. Similarly for the four U-color edge stickers, we get 4/24, 3/22, 2/20, and 1/18. Thus, the probability is (4/24)*(3/21)*(2/18)*(1/15)*(4/24)*(3/22)*(2/20)*(1/18) = 4!*4!/(24*21*18*15*24*22*20*18) = 1/(21*18*15*22*20*18) = 1/44906400.

The probability of at least one of the six faces being a solid color should be roughly (but not exactly) 6 times that or somewhere in the neighborhood of 1 in 7.5 million.


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## SweetSolver (Apr 21, 2012)

What is the probability of one *FACE* being solved after a random scramble on a 2x2?


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## cubingawsumness (Apr 21, 2012)

Just curious, has anyone ever estimated the percentage of people in the US who can solve a 3x3?


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## ThomasJE (Apr 21, 2012)

SweetSolver said:


> What is the probability of one *FACE* being solved after a random scramble on a 2x2?



Don't know, but I've had a 2x2 scramble that had 2 opposite faces already solved (that left just PBL to do). I'll try to find that scramble.


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## cuBerBruce (Apr 22, 2012)

SweetSolver said:


> What is the probability of one *FACE* being solved after a random scramble on a 2x2?


 
For at least one face being a solid color, I get 22654 chances in 3674160. That's approximately 1 chance in 162.

For exactly n faces being a solid color, I get the following distribution:


```
faces     count
  0       3651506
  1         22020
  2           597
  3            36
  4             0
  5             0
  6             1
```


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## SweetSolver (Apr 22, 2012)

Very interesting. Thanks for that


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## ncube (Apr 30, 2012)

What is the probability of the corners being permuted after Winter Variation?


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## ben1996123 (Apr 30, 2012)

What percentage of 2x2 states can be solved in 6 moves or less?


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## mande (Apr 30, 2012)

Probability of at least 1 CE pair formed in a scramble?


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## RNewms27 (Apr 30, 2012)

ben1996123 said:


> What percentage of 2x2 states can be solved in 6 moves or less?


 
62360/3674160= 0.016972587~ 1.7%


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## Rpotts (Apr 30, 2012)

ncube said:


> What is the probability of the corners being permuted after Winter Variation?


 
1/6


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## CubeRoots (May 1, 2012)

What is the probability of LL skip given last layer edges are oriented and two opposite edges are correctly permuted. (i.e. after zz-b phasing)


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## nickvu2 (May 1, 2012)

CubeRoots said:


> What is the probability of LL skip given last layer edges are oriented and two opposite edges are correctly permuted. (i.e. after zz-b phasing)


Based on this several posts back:


> Probability that corners are correctly solved after f2l?
> 1/162, allowing AUF (1/648 if not allowing AUF)



So 1/648 if edges are solved, 1/1296 if edges are phased? 

BTW, I've done 3,000+ solves with ZZ-b and I think I got 1 LL skip.


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## JonWhite (May 1, 2012)

What is the probability that you answer this question correctly?
A) 25%
B) 50%
C) 25%
D) 33.3%


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## Carrot (May 1, 2012)

JonWhite said:


> What is the probability that you answer this question correctly?
> A) 25%
> B) 50%
> C) 25%
> D) 33.3%


 

0/4 ~ 0%


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## IanTheCuber (May 1, 2012)

Odder said:


> 0/4 ~ 0%


 
Odder's sarcasm is awesome.

What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
LL Skip?
PLL Skip?
OLL Skip?
Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?


----------



## Rune (May 1, 2012)

IanTheCuber said:


> Odder's sarcasm is awesome.
> 
> What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
> LL Skip?
> ...



Here "odds" is more proper than "probability".


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## CubeRoots (May 1, 2012)

You can't answer that question. And thanks nickvu2. is that your main method?


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## Julian (May 1, 2012)

CubeRoots said:


> You can't answer that question. And thanks nickvu2. is that your main method?


I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.


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## Bob (May 1, 2012)

Julian said:


> I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.


 
You'd also have to consider how many people will have the same solution. There are too many variables here.


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## Julian (May 1, 2012)

Bob said:


> You'd also have to consider how many people will have the same solution. There are too many variables here.


Oh, yeah, I didn't think about that.


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## nickvu2 (May 1, 2012)

CubeRoots said:


> You can't answer that question. And thanks nickvu2. is that your main method?


Yes, ZZ-b in my main method =)


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## ThomasJE (May 1, 2012)

IanTheCuber said:


> Odder's sarcasm is awesome.
> 
> What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
> LL Skip?
> ...



1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
Same maths can be applied to OLL and LL skips.


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## CubeRoots (May 1, 2012)

Julian said:


> I think you can. You'd have to know what everybody is registered for, whether or not there are cutoff times (in which case you can't calculate it). You'd also have to assume that everyone will finish their solves, and that they're all using straight CFOP with no influencing techniques.


 
sorry hould have quoted, It was tge question before i was talking about 

you are right, you could answer that, in theory


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## kinch2002 (May 1, 2012)

IanTheCuber said:


> What is the propability that at Nationals, given there are about 200 people, that at least one would get a(n):
> LL Skip?
> PLL Skip?
> OLL Skip?
> Let's just say they all use Fridrich. I highly doubt it would happen, but it's a legitemate question. Right? RIGHT?


Some assumptions:
300 averages of 5 are done (200 + later rounds)
All scramble/solve combos are independent of each other
All combos have a 1/15552, 1/72, 1/216 chance of the skips respectively. e.g. the scramble state doesn't affect your chances, and you don't do anything to force any skips.
i) ~9.2%
ii) ~99.9999999%
iii) ~99.9%


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## Bob (May 2, 2012)

ThomasJE said:


> 1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
> Same maths can be applied to OLL and LL skips.


 
I certainly hope not. There's so much wrong with the math here that I don't know where to start.


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## Noahaha (May 2, 2012)

Bob said:


> I certainly hope not. There's so much wrong with the math here that I don't know where to start.


 
Seriously. That is the most cringeworthy post I've ever seen. Kill it with fire.


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## MTGjumper (May 2, 2012)

ThomasJE said:


> 1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...



At first I thought the ellipsis indicated that you realised something was wrong.

Then I realised you were being serious.


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## SweetSolver (May 2, 2012)

What are the chances of solving a 2x2 in 2 moves after a random scramble? (I know it sounds weird but I was curious )


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## nqwe (May 2, 2012)

SweetSolver said:


> What are the chances of solving a 2x2 in 2 moves after a random scramble? (I know it sounds weird but I was curious )


 
Chance of solving the 2x2 cube in 2 moves:

Within to two moves you can reach 54 states.
There are 3674160 possible configurations.

54 / 3674160 = 1 / 68040 = 0.00147%


Chance of solving the 2x2 cube in 2 moves or less:

There's one state defined as solved.
Within to one moves you can reach 9 states.
Within to two moves you can reach 54 states.

54 + 9 + 1 = 64

64 / 3674160 = 0.00174%

Quod erat demonstrandum.


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## ThomasJE (May 2, 2012)

MTGjumper said:


> At first I thought the ellipsis indicated that you realised something was wrong.
> 
> Then I realised you were being serious.



You can't get a recurring sign on a keyboard.


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## Eazoon (May 2, 2012)

what is the probability of a 15 move solve that directly follows the fridrich method? feliks can do about 20 tps so it would be a sub-1 solve.


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## RNewms27 (May 2, 2012)

Eazoon said:


> what is the probability of a 15 move solve that directly follows the fridrich method? feliks can do about 20 tps so it would be a sub-1 solve.


 
Fridrich will not give a 15-move solve without skipping a large amount of the solve. The scramble will also need a sub-16 move optimal solution, not to mention the solution must be following Fridrich steps. Hardly possible.

Feliks has 7-8-9tps in solves. (R U R' U') over and over would not be a Fridrich solution.


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## Bob (May 2, 2012)

Eazoon said:


> what is the probability of a 15 move solve that directly follows the fridrich method? feliks can do about 20 tps so it would be a sub-1 solve.


 
I seriously doubt that it would be sub1. By the same logic, since an average Fridrich solve is 56 moves and he can turn 20 tps, all of his solves are sub-3.


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## Eazoon (May 2, 2012)

RNewms27 said:


> Fridrich will not give a 15-move solve without skipping a large amount of the solve. The scramble will also need a sub-16 move optimal solution, not to mention the solution must be following Fridrich steps. Hardly possible.
> 
> Feliks has 7-8-9tps in solves. (R U R' U') over and over would not be a Fridrich solution.


 
First off, this is a probability thread, its hardly possible, not impossible. second, if he can do up to 9 tps, then could he get sub-3? the record fewest moves is 22 i think.


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## Rpotts (May 2, 2012)

Eazoon said:


> First off, this is a probability thread, its hardly possible, not impossible. second, if he can do up to 9 tps, then could he get sub-3? the record fewest moves is 22 i think.


 
The fewest moves record was not achieved with Fridrich. Faz's fastest reconstructed solves have 9+ TPS, but he can turn substantially faster in bursts.

A solve that directly follows the Fridrich method without skipping any steps can't really be less than 28 moves HTM.

1 moves cross
4 3 move pairs
F RUR'U' F'
A/U perm.

1 + 12 + 6 + 9 = 28


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## kinch2002 (May 2, 2012)

Rpotts said:


> The fewest moves record was not achieved with Fridrich.


Actually IMO both 22 move WRs used Fridrich. Ofc steps were skipped, but that doesn't mean it's not Fridrich


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## Rpotts (May 2, 2012)

not cross - f2l fridrich.

freesyle F2L - LL skip ≠ Fridrich imo


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## Kirjava (May 3, 2012)

It wasn't freestyle F2L. I've heard that it was actually cross -> F2L.


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## Rpotts (May 3, 2012)

Oh really? I thought it was blockbuilding... Is the solve posted anywhere?


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## kinch2002 (May 3, 2012)

Here are the solutions. It's a little off-topic to be talking about whether this is Fridrich or not, so whatever. I'd say everything is standard Fridrich apart from from Jimmy starting with 2x2 block and that István kind of switched from green to yellow F2L, while cancelling a pair into the OLL. Reconstructions are posted by others from other threads

Jimmy:
Scramble : U L2 U' R2 B2 U L2 D2 R2 F2 R2 F L F' D2 L' U B' D R2 F'
Solution : U L U R2 U2 (bloc 222)
B D' R2 D' B R' (cross + second F2L)
B2 U' B U (3rd F2L)
D B2 D2 B' D2 B' D' (4th F2L + LL skip)

István:
scramble: D'R2DL2F2DLR2B'D2U2L2DF'L2UL2FU 
Xcross on green: FD'FL'BR2
2nd f2l: RBR'B2DBD' (12 move with R2 R cancel)
and from here he should've done: U'BU'LU2L' (18 moves total, beaten the scramble by 1 move!)
but he did something like: U2BLU'LF'L'FU'L' (22 moves total)


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## Rpotts (May 3, 2012)

Those are both closer to Fridrich than I had thought, still far "directly follows the Fridrich method," but yea, my bad Sheppz.


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## aronpm (May 3, 2012)

ThomasJE said:


> 1/72 for a PLL skip normally. So 200 x 1/72 = 2.777...
> Same maths can be applied to OLL and LL skips.


 
2.777... = 278% chance

On average, if there are 100 US Nationals, 278 of the US Nationals will have someone get a LL skip

...

learn probability before you try to answer a question in this thread, please


----------



## cubernya (May 3, 2012)

kinch2002 said:


> Here are the solutions. It's a little off-topic to be talking about whether this is Fridrich or not, so whatever. I'd say everything is standard Fridrich apart from from Jimmy starting with 2x2 block and that István kind of switched from green to yellow F2L, while cancelling a pair into the OLL. Reconstructions are posted by others from other threads
> 
> Jimmy:
> Scramble : U L2 U' R2 B2 U L2 D2 R2 F2 R2 F L F' D2 L' U B' D R2 F'
> ...


 
I do remember seeing that alternate ending on another thread. IIRC the 18 turns is optimal. Of course, the threads are somewhere on the forums, but I don't have them bookmarked and have no reason to search for them ATM


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## kinch2002 (May 3, 2012)

kinch2002 said:


> Reconstructions are posted by others from other thread





theZcuber said:


> I do remember seeing that alternate ending on another thread. IIRC the 18 turns is optimal. Of course, the threads are somewhere on the forums, but I don't have them bookmarked and have no reason to search for them ATM


As you can see, I said that they were not my own work. I found the original posts and copied the text into my post.


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## cubernya (May 3, 2012)

kinch2002 said:


> As you can see, I said that they were not my own work. I found the original posts and copied the text into my post.


 
:fp on my part


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## Eazoon (May 3, 2012)

Rpotts said:


> The fewest moves record was not achieved with Fridrich. Faz's fastest reconstructed solves have 9+ TPS, but he can turn substantially faster in bursts.
> 
> A solve that directly follows the Fridrich method without skipping any steps can't really be less than 28 moves HTM.
> 
> ...


 
If he gets really lucky, he could get a LL skip and/or f2l pair skip.


----------



## Julian (May 3, 2012)

Eazoon said:


> If he gets really lucky, he could get a LL skip and/or f2l pair skip.


 


Rpotts said:


> A solve that directly follows the Fridrich method *without skipping any steps*...


yeah


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## Eazoon (May 3, 2012)

Julian said:


> **


 
oops.....


----------



## cuBerBruce (May 3, 2012)

mande said:


> Probability of at least 1 CE pair formed in a scramble?


 
I think mande has asked essentially the same thing once before. 

It seems to me that the (approximate) answer to this must have been discussed previously, but so far I haven't found it. The exact answer appears to me to be rather complicated to calculate.

Using GAP, I computed the distribution of the number of CE pairs for ten million random positions. I also did this a 2nd time. Here are the results.


```
pairs  1st distr    2nd distr

  0      3684459      3686510
  1      3671274      3671462
  2      1837467      1835746
  3       613647       613283
  4       155370       155277
  5        31599        31464
  6         5318         5398
  7          745          752
  8          109           99
  9           11            9
 10            1            0
>=11           0            0
```

This suggests a probability of at least 1 CE pair of somewhere around 0.631.

Note that the probability of no pairs is close to 1/e. Why?


----------



## ThomasJE (May 4, 2012)

aronpm said:


> 2.777... = 278% chance
> 
> On average, if there are 100 US Nationals, 278 of the US Nationals will have someone get a LL skip



Exactly. So it's virtually guaranteed. Even though I forgot about later rounds and 5 solves per average :fp. If I had remembered that, it would have been around 2000%


----------



## aronpm (May 4, 2012)

ThomasJE said:


> Exactly. So it's virtually guaranteed. Even though I forgot about later rounds and 5 solves per average :fp. If I had remembered that, it would have been around 2000%


 
NO
ARGHSDIBGERWNNASCNWG

PROBABILITY DOESNT WORK HOW YOU THINK IT WORKS


----------



## mr. giggums (May 4, 2012)

ThomasJE said:


> Exactly. So it's virtually guaranteed. Even though I forgot about later rounds and 5 solves per average :fp. If I had remembered that, it would have been around 2000%


 
Flipping a coin has a 50% chance. According to your logic if I flip a coin twice (.5 * 2 = 1 or 100%) I should always get a least one head. Try it this doesn't happen.


----------



## Noahaha (May 4, 2012)

ThomasJE said:


> Exactly. So it's virtually guaranteed. Even though I forgot about later rounds and 5 solves per average :fp. If I had remembered that, it would have been around 2000%


 
The problem is that probabilities are multiplied, not added. You need to say that the odds someone does NOT get a PLL skip are 71/72, and then the odds that in 200 solves everyone simultaneously does NOT get a PLL skip are (71/72)^200, and then the odds that there is a PLL skip will be:
1-(71/72)^200

Of course, 200 should be replaced with the actual number of solves. No probability can be greater than 1 or less than 0. Hope this helps.


----------



## cubernya (May 28, 2012)

Odds of a solved layer on a scrambled 2x2?


----------



## ben1996123 (May 28, 2012)

theZcuber said:


> Odds of a solved layer on a scrambled 2x2?



It's been asked and answered in this thread before.


----------



## Nico1 (May 30, 2012)

The probability of this is 44,497,945,755,648,000. You are more likely to win the lottery every single time it plays for the rest of your life than have a 3x3 skip!


----------



## ben1996123 (May 30, 2012)

Nico1 said:


> The probability of this is 44,497,945,755,648,000. You are more likely to win the lottery every single time it plays for the rest of your life than have a 3x3 skip!



No.

Lets assume I live for another 65 years. Lottery is played every week, so its played about 52*65 times.

Probability of winning lottery once is \( \frac{1}{13983816} \)

Probability of winning 52*65 = 3380 times in a row is \( \frac{1}{13983816^{3380}} \) =


Spoiler



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\( \approx 0 \)


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## CubeRoots (May 31, 2012)

ben1996123 said:


> No.
> 
> Lets assume I live for another 65 years. Lottery is played every week, so its played about 52*65 times.
> 
> ...


 
some would say that 1/44497945755648000 \( \approx 0 \), too


----------



## ben1996123 (May 31, 2012)

CubeRoots said:


> some would say that 1/44497945755648000 \( \approx 0 \), too



\( \frac{1}{44497945755648000} \) is about \( 2.713*10^{24135} \) times larger than \( \frac{1}{13983816^{3380}} \) and \( 44497945755648000<43252003274489856000<13983816^{3380} \)


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## CubeRoots (May 31, 2012)

ben1996123 said:


> \( \frac{1}{44497945755648000} \) is about \( 2.713*10^{24135} \) times larger than \( \frac{1}{13983816^{3380}} \) and \( 44497945755648000<43252003274489856000<13983816^{3380} \)


 
yes but the difference between the two probabilities is 2.32...E(-20).

and the fact that it is 2.713...E24135 times larger is irrelevant when the probability of the event is so close to zero as you pointed out


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## ben1996123 (May 31, 2012)

CubeRoots said:


> yes but the difference between the two probabilities is 2.32...E(-20).
> 
> and the fact that it is 2.713...E24135 times larger is irrelevant when the probability of the event is so close to zero as you pointed out



whatever. it's not as though it's going to happen to anyone ever. probably.


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## CubeRoots (May 31, 2012)

ben1996123 said:


> whatever. it's not as though it's going to happen to anyone ever. probably.


 
lol. too true. nice probability pun there


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## Julian (May 31, 2012)

Probability of a CP skip for SQ1 = 1/36?


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## ben1996123 (May 31, 2012)

Julian said:


> Probability of a CP skip for SQ1 = 1/36?



4!^2 permutations, divide by 16 for AU/DF = 1/36 probability for skip.


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## cuBerBruce (May 31, 2012)

theZcuber said:


> Odds of a solved layer on a scrambled 2x2?



The probability is 3814 out of 3674160, or approximately 1 in 963 (for having at least a whole layer solved).

Number of positions with:

Exactly one layer solved: 6*(4! * 3^3 - 1 - 3 - 4*5) = 3744
Exactly two layers solved (adj. faces): 12*(6-1) = 60
Exactly two layers solved (opp. faces): 3*3 = 9
All six layers solved: 1

Total: 3744 + 60 + 9 + 1 = 3814



ben1996123 said:


> It's been asked and answered in this thread before.


 
If it was answered earlier in this thread, I can't find it. The probability of having a solid color face was given, but that's not the same thing as a whole layer solved.


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## Applejuice (Jun 2, 2012)

Whats the probability of having 2 centers solved on a 5x5 after scramble? Less or more than having 1 side done on a 3x3?


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## Bob (Jun 3, 2012)

Applejuice said:


> Whats the probability of having 2 centers solved on a 5x5 after scramble? Less or more than having 1 side done on a 3x3?


 
Without even doing the math, I'm confident it will be significantly less likely than having 1 side of a 3x3.


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## CubeRoots (Jun 3, 2012)

5/18818646 for 2 centers, 1/7484400 for one side, i think


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## cuBerBruce (Jun 3, 2012)

Applejuice said:


> Whats the probability of having 2 centers solved on a 5x5 after scramble? Less or more than having 1 side done on a 3x3?


Well, here's what I come up with.

For the centers for two particular faces on the 5x5x5:
((4/24)*(3/23)*(2/22)*(1/21)*(4/20)*(3/19)*(2/18)*(1/17))^2 = 1/(51482970^2) = 1/2,650,496,200,020,900

Assuming we don't care which two faces have the solved centers, the probability is approximately 15 times that (around 1 in 177 trillion).

For the 3x3x3, I assume "one side done" means a whole face layer being solved.

The probability of a particular face layer on the 3x3x3 being solved is:
(1/24)*(1/21)*(1/18)*(1/15)*(1/24)*(1/22)*(1/20)*(1/18) = 1/25866086400.

The probability of any face layer solved is approximately 6 times that, or about 1 in 4.3 billion.

If you only interested in a face being a solid color, then the probability for a particular face would be:
(4/24)*(3/21)*(2/18)*(1/15)*(4/24)*(3/22)*(2/20)*(1/18) = 1/44906400. For any of the six faces, the probability would be approximately 6 times larger, or about 1 in 7.5 million.


----------



## CubeRoots (Jun 4, 2012)

cuBerBruce said:


> Well, here's what I come up with.
> 
> For the centers for two particular faces on the 5x5x5:
> ((4/24)*(3/23)*(2/22)*(1/21)*(4/20)*(3/19)*(2/18)*(1/17))^2 = 1/(51482970^2) = 1/2,650,496,200,020,900
> ...



Okay, we agree on a face on a 3 by 3, but: you made a mistake on the 5*5*5 part. 2 centres solved, for first one there are 6 options, leaving 5 options for the remaining center, so it's 6*5=30 times larger, not 15. I don't know about the rest of your working though so who know's who if either of us is right.


----------



## cuBerBruce (Jun 4, 2012)

CubeRoots said:


> Okay, we agree on a face on a 3 by 3, but: you made a mistake on the 5*5*5 part. 2 centres solved, for first one there are 6 options, leaving 5 options for the remaining center, so it's 6*5=30 times larger, not 15.



No, we're choosing 2 faces out of 6. 6 choose 2 is 15. Of two chosen faces, it doesn't matter which one is considered as the "first."


----------



## applemobile (Jun 4, 2012)

Someone please answer me this. I do 10 blindfolded solves. First one wearing white shoes, second wearing black shoes, third wearing white shoes and then repeating swapping shoes. The shoes are the same, feel the same, and make no difference to the solve. The 1st,3rd,5th and all the solves with odd numbers whilst wearing white shoes are DNF. Before you do the 11th solve, you have the choice of what shoes to wear. You MUST wear either of the pairs. Which shoes do you pick to give you statistically the highest chance of completing the solve?


----------



## Noahaha (Jun 4, 2012)

applemobile said:


> Someone please answer me this. I do 10 blindfolded solves. First one wearing white shoes, second wearing black shoes, third wearing white shoes and then repeating swapping shoes. The shoes are the same, feel the same, and make no difference to the solve. The 1st,3rd,5th and all the solves with odd numbers whilst wearing white shoes are DNF. Before you do the 11th solve, you have the choice of what shoes to wear. You MUST wear either of the pairs. Which shoes do you pick to give you statistically the highest chance of completing the solve?


 
Well in the world of probability it doesn't matter, however you may have lost confidence in your ability to complete a solve wearing white shoes, so in the real world black is probably the way to go.


----------



## applemobile (Jun 4, 2012)

Wearing black shoes on an odd numbered solve? what if that completely blows your 50% streak and you never solve again? Risky move.


----------



## CubeRoots (Jun 4, 2012)

cuBerBruce said:


> No, we're choosing 2 faces out of 6. 6 choose 2 is 15. Of two chosen faces, it doesn't matter which one is considered as the "first."


 
Dammit you're right, I was thinking 6 permute 2


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## Noahaha (Jun 4, 2012)

applemobile said:


> Wearing black shoes on an odd numbered solve? what if that completely blows your 50% streak and you never solve again? Risky move.


 
This is no longer a probability question.

But I see what you're doing.


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## Applejuice (Jun 4, 2012)

Thank Bob, CubeRoots & cuBerBruce for the answer. I thought it was less but wasn't sure.

After solving all centers on a 5x5, what's the probability for all edges to be paired correctly?


----------



## ThomasJE (Jun 4, 2012)

applemobile said:


> Someone please answer me this. I do 10 blindfolded solves. First one wearing white shoes, second wearing black shoes, third wearing white shoes and then repeating swapping shoes. The shoes are the same, feel the same, and make no difference to the solve. The 1st,3rd,5th and all the solves with odd numbers whilst wearing white shoes are DNF. Before you do the 11th solve, you have the choice of what shoes to wear. You MUST wear either of the pairs. Which shoes do you pick to give you statistically the highest chance of completing the solve?


 

```
Solve	Colour shoes	Success?	
1	White		No
2	Black		
3	White		No
4	Black		
5	White		No
6	Black		
7	White		No
8	Black		
9	White		No
10	Black		
11	???		???
```
So, with white shoes on an odd no. solve was always DNF. But, we know nothing about the black shoes. So, there is evidence to say that white shoes will cause a DNF, but no evidence that black shoes will cause a DNF. So, logically, you should use black shoes.


Spoiler: BUT...






applemobile said:


> Someone please answer me this. I do 10 blindfolded solves. First one wearing white shoes, second wearing black shoes, third wearing white shoes and then repeating swapping shoes. *The shoes are the same, feel the same, and make no difference to the solve.* The 1st,3rd,5th and all the solves with odd numbers whilst wearing white shoes are DNF. Before you do the 11th solve, you have the choice of what shoes to wear. You MUST wear either of the pairs. Which shoes do you pick to give you statistically the highest chance of completing the solve?


So, if they make no difference, then it doesn't matter what shoes you wear; they will make no difference to the solve. So, the answer is *neither*.


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## applemobile (Jun 4, 2012)

applemobile said:


> Someone please answer me this. I do 10 blindfolded solves. First one wearing white shoes, second wearing black shoes, third wearing white shoes and then repeating swapping shoes. The shoes are the same, feel the same, and make no difference to the solve. The 1st,3rd,5th and all the solves with odd numbers whilst wearing white shoes are DNF. Before you do the 11th solve, you have the choice of what shoes to wear.* You MUST wear either of the pairs. *Which shoes do you pick to give you statistically the highest chance of completing the solve?



See above.


----------



## ThomasJE (Jun 4, 2012)

applemobile said:


> See above.



The question is:


> Which shoes do you pick to give you statistically the highest chance of completing the solve?



Neither will give you a better chance of completing the solve, since the shoes make no difference.


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## Rune (Jun 4, 2012)

I know it against the rules, but do try with a white shoe on your left foot and a black on your right. And vice versa.
You never know!


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## cuBerBruce (Jun 4, 2012)

Applejuice said:


> After solving all centers on a 5x5, what's the probability for all edges to be paired correctly?



On the 5x5x5, the probability that all the edges would be tripled up correctly is:

1/24! = 1/620,448,401,733,239,439,360,000

For the 4x4x4, the probability of all edges being paired up correctly is: 
1/23!! (where !! is the double factorial operator), or 1/(23*21*19*17*15*13*11*9*7*5*3*1) = 1/316,234,143,225.


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## TheAwesomeAlex (Jun 10, 2012)

whats the probability of getting a 2x2 LL skip?
whats the probability of getting a pyraminx LL skip?


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## Julian (Jun 11, 2012)

TheAwesomeAlex said:


> whats the probability of getting a 2x2 LL skip?


1/162


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## Applejuice (Jun 12, 2012)

Thanks for the answers! 
I'm really starting to enjoy this thread. ;D
What's the probability of having sune as OLL + A perm as pll?


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## Bob (Jun 12, 2012)

Applejuice said:


> Thanks for the answers!
> I'm really starting to enjoy this thread. ;D
> What's the probability of having sune as OLL + A perm as pll?



P(Sune) = 1/54
P(A perm) = 2/18 = 1/9

P(Sune -> A) = 1/54 * 2/18 = 1/486.

Do you count Sune and Antisune the same? In that case, it would be twice as likely, 1/243.


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## cubenut99 (Jun 12, 2012)

cuBerBruce said:


> On the 5x5x5, the probability that all the edges would be tripled up correctly is:
> 
> 1/24! = 1/620,448,401,733,239,439,360,000
> 
> ...



holy crap


----------



## TMOY (Jun 15, 2012)

TheAwesomeAlex said:


> whats the probability of getting a pyraminx LL skip?



It's 1/12 (3 possible permutations, '4 possible orientations for each one of them).


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## Smiles (Jun 19, 2012)

what is the probability of getting an x-cross (unintentionally) after doing a regular cross?


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## Stefan (Jun 19, 2012)

Smiles said:


> what is the probability of getting an x-cross (unintentionally) after doing a regular cross?



About 1.0%. Determined by 10000 computer simulations and this:

Chance of specific F2L pair solved: 1/24 * 1/16.
Chance of specific F2L pair *not* solved: 1 - 1/24 * 1/16.
Chance of all four F2L pairs *not* solved: (1 - 1/24 * 1/16) ^ 4.
Chance of at least one F2L pair solved: 1 - (1 - 1/24 * 1/16) ^ 4 = 1.038%.
Note that the last two aren't exact, because the probabilities of the four pairs aren't independent, because where pieces of one pair are affects where pieces of the other pairs can be. But they seem to be fairly independent, as the experiment confirms.

For a description of the experiment and more results, see page 35 and the pages around it here:
http://www.stefan-pochmann.info/hume/hume_diploma_thesis.pdf


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## mDiPalma (Jul 4, 2012)

this may be a hard one.

what are the probabilities of the fewest amount of misoriented edges [R,L,U,D,F2,B2] in all orientations as opposed to in a fixed orientation?

for example F' U D2 L2 R U R' U' B F' L B2 R2 F L' F R2 F2 R' B' L' B F L U' 
gives 6 bad edges in wg,wb,yg,and yb orientations
but gives 4 bad edges in wr,wo,yr, and yo orientations

so, if i compared the amount of misoriented edges in fixed orientation to the *least* amount of misoriented edges out of *all* orientations (y,x,z,&combinations thereof), all orientations should have a distribution that is skewed to the left of the fixed orientation data.

crider's site gives this data for a fixed orientation

0 bad edges - 1/2048
2 - 66/2048
4 - 495/2048
6 - 924/2048
8 - 495/2048
10 - 66/2048
12 - 1/2048

so i guess im wondering if somebody somehow could fill out a chart like ^^that^^ counting the likelihoods of the FEWEST amount of misoriented edges in all orientations in all scrambles.

does that make sense?


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## Escher (Jul 5, 2012)

Does anybody know of a decent book covering probability?

Just looking for a reasonable 'introductory' text, University level is fine. 

Thanks :3


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## cuBerBruce (Jul 6, 2012)

mDiPalma said:


> what are the probabilities of the fewest amount of misoriented edges [R,L,U,D,F2,B2] in all orientations as opposed to in a fixed orientation?



There are a total of 12! * 2^11 = 980,995,276,800 configurations of the edges. For the purposes of calculating what mDiPalma wants, I believe we can consider the four E-layer edges indistinguishable, the four M-layer edges indistinguishable, and the four S-layer edges to be indistinguishable. This reduces the number of configurations we actually need to consider to 12! * 2^11 / 4!^3 = 70,963,200. Also, I believe three (appropriately chosen) cube orientations suffice for counting the misoriented edges.

Using GAP, I've come up with the following distribution:


```
bad edges   count
    0      103741
    2     6500978
    4    35204527
    6    26728948
    8     2411347
   10       13658
   12           1
```

Note that for 12 bad edges, every edge must be in its proper inner layer. If an edge is in the wrong inner layer, there will be some cube orientation that will make it oriented. Hence, the above distribution has only 1 case with 12 bad edges.


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## mDiPalma (Jul 6, 2012)

cuBerBruce said:


> Using GAP, I've come up with the following distribution:



i love you.

so a color neutral zz solver would have this:

0 bad edges - .146%
2 - 9.161%
4- 49.610%
6 - 37.666%
8 - 3.398%
10 - .019%
12 - ~0%

as opposed to this:

0 bad edges - .049%
2 - 3.223%
4- 24.170%
6 - 45.117%
8 - 24.170%
10 - 3.223%
12 - .049%

cool thanx man <3 <3 <3


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## cubernya (Jul 6, 2012)

cuBerBruce said:


> Note that for 12 bad edges, every edge must be in its proper inner layer. If an edge is in the wrong inner layer, there will be some cube orientation that will make it oriented. Hence, the above distribution has only 1 case with 12 bad edges.



So basically the only position where all edges are bad in any orientation is superflip (corners could be scrambled of course)


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## cuBerBruce (Jul 6, 2012)

theZcuber said:


> So basically the only position where all edges are bad in any orientation is superflip (corners could be scrambled of course)


Edges can also be scrambled, but only within their respective layers (E,M,S). So actually 13824 possible edge configurations.


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## antoineccantin (Jul 28, 2012)

Last 2 Center skip on 5x5?


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## kinch2002 (Jul 28, 2012)

antoineccantin said:


> Last 2 Center skip on 5x5?


Fairly simple just to think through logically.
+ and x centres will clearly behave the same in term of probability, so just calculate the probability of skipping a particular set and square the result.
Pick a piece of colour 1 and place it on a centre: 4/8 (1/2) chance of getting it right
Pick 2nd piece of colour 1 and place it in a free space: 3/7
3rd piece: 2/6 (1/3)
4th piece: 1/5
1/2 x 3/7 x 1/3 x 1/5 = 1/70
(1/70)^2=*1/4900*


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## Christopher Mowla (Jul 29, 2012)

antoineccantin said:


> Last 2 Center skip on 5x5?


Just as kinch2002 said, it's 1/4900.

In general, the probability for getting a C center skip on a cube of size _n_ is (with 6-C centers already solved) is:

\( \frac{1}{\left( \frac{\left( 4C \right)!}{\left( 4! \right)^{C}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \)

Now, this doesn't take into account how many of the other pieces are in their solved positions.

Let me explain.


Spoiler



The 5x5x5 has 48 non-fixed center pieces (+ center pieces and X-center pieces), 8 corners, 12 middle edges, 24 wing edges *and 6 fixed center pieces (on the xyz axis)*.

Since the 5x5x5 cube has fixed centers, then we can actually count how many pieces are not in their exact solved positions by comparing their locations with respect to the fixed centers.

If you use pure reduction, meaning that you do not make an effort to solve any corners, middle edges, or wing edges prior to completing all centers, then it is very unlikely that you will have:

4 or more corners in their correct locations (not to mention oriented correctly in their correct locations).
4 or more wing edges in their solved locations.
5 or more middle edges in their solved locations (not to mention oriented correctly in their correct locations).

And...not to mention that to have two or all three of the above on a cube at once.


These high improbabilities are true for the corners, wings, and middle edges on the 5x5x5 from the scramble to the point to where you intentionally just solve the first 4 centers.

Although this is all my opinion, you can give me a "more realistic" set of restrictions if you think that I should be more lenient.


So the probability of getting a last two center skip is indeed 1/4900 if you include the instances when even all corners, all middle edges, and all wing edges are in their correct locations (the most unlikely extreme case).

Just as a realistic guess of mine (you can request a different amount of wings, middle edges and corners to be in their correct locations), but ignoring the orientations of the corners and the middle edges (regardless whether or not they are in their correct locations or not), suppose that we consider the scenarios where:

No more than 3 wing edges are solved, no more than 3 corners are in their correct locations, and no more than 4 middle edges are in their correct locations.

*Goal*: We are going to calculate a percentage of the number of possible configurations of the 5x5x5 cube which meet the conditions in the above sentence (ignoring the centers completely...just the cage portion of the 5x5x5 cube). We will then multiply this percentage (in decimal form) by 1/4900 to get a smaller probability fraction...because it's less likely to get a last two centers skip than 1/4900.


Spoiler



The formula for the number of positions of the _regular 6-color _nxnxn cube is:

\( F\left( n \right)=\frac{8!\times 3^{7}}{24^{\left( n+1 \right)\bmod 2}}\left( 12!\times 2^{10} \right)^{n\bmod 2}24!^{\left\lfloor \frac{n-2}{2} \right\rfloor }\left( \frac{24!}{4!^{6}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor } \)

Since we are ignoring the permutations which the non-fixed centers contribute, we are left with:

\( F\left( n \right)=\frac{8!\times 3^{7}}{24^{\left( n+1 \right)\bmod 2}}\left( 12!\times 2^{10} \right)^{n\bmod 2}24!^{\left\lfloor \frac{n-2}{2} \right\rfloor } \)

Since the 5x5x5 is an odd cube, we have:

\( F\left( n \right)=\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)24!^{\left\lfloor \frac{n-2}{2} \right\rfloor } \)

and substituting 5 for _n_, the denominator of the fraction which we will use to calculate a percentage is:

\( \left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)\left( 24! \right) \)

For the numerator of the fraction, we will use the same thing as the denominator, but we will substitute a (smaller) value for each of the factorials.

Let \( f\left( p,x \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left\lfloor \frac{k\left( k-2 \right)!}{e}+\frac{1}{2}\cos ^{2}\left( \frac{k\pi }{2} \right) \right\rfloor }{k\left( k-2 \right)!\left( p-k \right)!} \right)} \)

(I derived the expression in the summation in this thread. I did remove the "round" operator and replaced it with the fldoor and cosine function, but it's still the same formula for the integers...which is all that counts with permutation puzzles).

This can be simplified to:

\( f\left( p,x \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( \frac{!k}{k-1} \right)}{k\left( k-2 \right)!\left( p-k \right)!} \right)=\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( k-1 \right)\left( \frac{!k}{k-1} \right)}{k\left( k-1 \right)\left( k-2 \right)!\left( p-k \right)!} \right)}} \)

\( =\sum\limits_{k=p-x}^{p}{\left( \frac{p!\left( !k \right)}{k!\left( p-k \right)!} \right)}=\sum\limits_{k=p-x}^{p}{\left( \begin{matrix}
p \\
k \\
\end{matrix} \right)\left( !k \right)} \)

[Link]​
Where:
_p_ = the number of the piece type. For wing edges, _p_ = 24, for middle edges, _p_ = 12, and for corners, _p_ = 8.
(_p_ is just the total number of each piece type).

No more than _x_ of each piece type are solved.
\( 0\le x\le p \)

(_k_ is just the index of the summation).

Note that \( \sum\limits_{k=p-p}^{p}{\left( \begin{matrix}
p \\
k \\
\end{matrix} \right)\left( !k \right)}=\sum\limits_{k=0}^{p}{\left( \begin{matrix}
p \\
k \\
\end{matrix} \right)\left( !k \right)}=p! \)


So, here are the calculations which must be done to get a percentage:


Spoiler



We must do a calculation for the wing edges, middle edges, and corners separately.

*The Wing Edges*
_p_ is always 24 (in cube sizes 4x4x4 and greater) because there 24 wings per orbit.
Since I gave the scenario that we allow no more than 3 wing edges to be solved, _x_ = 3.

So we have:

\( \sum\limits_{k=24-3}^{24}{\left( \begin{matrix}
24 \\
k \\
\end{matrix} \right)\left( !k \right)} \)
= 608667230147569787984693 [Link]


*The Middle Edges*
_p_ is always 12 (in cube sizes 3x3x3 and greater) because there are 12 middle edges on every odd cube greater than the 1x1x1.
Since I gave the scenario that we allow no more than 4 middle edges to be in their correct positions, _x_ = 4.

So we have:

\( \sum\limits_{k=12-4}^{12}{\left( \begin{matrix}
12 \\
k \\
\end{matrix} \right)\left( !k \right)} \)
= 477248562 [Link]


*The Corners*
_p_ is always 8 (in cube sizes 2x2x2 and greater) because there 8 corners in cube sizes 2x2x2 and greater.
Since I gave the scenario that we allow no more than 3 corners to be solved, _x_ = 3.

So we have:

\( \sum\limits_{k=8-3}^{8}{\left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)\left( !k \right)} \)
= 39549 [Link]


And so our fraction which gives us our percentage is:

\( \frac{\left( \text{39549}\times 3^{7} \right)\left( \text{477248562}\times 2^{10} \right)\left( \text{608667230147569787984693} \right)}{\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)\left( 24! \right)}\approx \text{0}\text{.958731} \)

[Link]


And therefore, with the (realistic) assumption that


> No more than 3 wing edges are solved, no more than 3 corners are in their correct locations, and no more than 4 middle edges are in their correct locations.



the probability of a last two centers skip on the 5x5x5 cube (using pure reduction) (ignoring the orientations of the corners and middle edges) is

\( \approx \frac{1}{4900}\left( \text{0}\text{.958731} \right)\approx \frac{1}{5111} \)

In other words, the official probability claim for a last two center skip on the 5x5x5 with reduction claimed that it is 4.13% more likely to occur than what it actually is (again, by the "realistic" assumptions/restrictions I presumed). And, I think I might have been too lenient at that. So it could well be more than 4.13% off in reality.


----------



## ThomasJE (Aug 4, 2012)

Skip on Guimond 1st step (3/4 of a face with opposite colours) and Ortega step 1 (face)?


----------



## cuBerBruce (Aug 10, 2012)

ThomasJE said:


> Skip on Guimond 1st step (3/4 of a face with opposite colours)


Assuming you're interested in the probability for the color neutral solving...
[post]147363[/post]

So probability is 3097152/3674160 or approximately 84.3%.



ThomasJE said:


> and Ortega step 1 (face)?


[post]152726[/post]
[post]737035[/post]

So probability (again, assuming color neutral) is 22654/3674160 (approximately 0.617% or about 1 in 162).


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## Pokerizer (Aug 10, 2012)

cmhardw said:


> To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
> 1 / 15552
> 
> I thought it was pretty neat.
> ...



well that sux, I had a solve the other day with a LL skip without having to align edges.... So much for seeing that again any time soon lmao


----------



## Ickathu (Aug 13, 2012)

What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
Does that make sense?


----------



## Schmidt (Aug 13, 2012)

antoineccantin said:


> Last 2 Center skip on 5x5?


So far: 1/Lifetime
I was just casually solving while my kids played on the playground, so I wasn't paying 100% attention to the solving, so when I had solved the first 4 centers I was turning the cube around to find the next unsolved centers. After a few x-y-z's I couldn't find any


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## Bob (Aug 13, 2012)

Schmidt said:


> So far: 1/Lifetime
> I was just casually solving while my kids played on the playground, so I wasn't paying 100% attention to the solving, so when I had solved the first 4 centers I was turning the cube around to find the next unsolved centers. After a few x-y-z's I couldn't find any



I think the probability that you counted wrong is higher than the probability that you skipped 2 centers.


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## JasonK (Aug 14, 2012)

Ickathu said:


> What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
> Does that make sense?



Skip = 1/12
U-perm = 1/6
2flip = 1/4
Flipcycle = 1/2


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## mDiPalma (Aug 15, 2012)

this question has likely been answered before, but advanced search has wierded out on me:

what is the distribution for the minimum amount of 2-generator [RU] moves it can take to solve a cube that CAN be solved 2-generator, without rotations? like if you perform an antisune on your cube, then the fewest moves it would take to be solved would be 7, and this state would be tallied in the 7 column.

does that make sense?

is this an impossibly hard question to answer? if it is, then don't bother.


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## Stefan (Aug 15, 2012)

Look for "Analysis of the 3x3x3 <U, R> group" here: http://cubezzz.dyndns.org/drupal/text/fullcube.txt


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## mDiPalma (Aug 29, 2012)

Thanks! I have some more questions!

What is the probability of a last layer skip on a 3x3 given that the cube is in a state that can be solved 2 generator <RU> and has the F2L completed?

How does this compare with a standard LL skip after standard F2L?


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## Bob (Aug 29, 2012)

mDiPalma said:


> Thanks! I have some more questions!
> 
> What is the probability of a last layer skip on a 3x3 given that the cube is in a state that can be solved 2 generator <RU> and has the F2L completed?
> 
> How does this compare with a standard LL skip after standard F2L?



<RU> gen means that edge orientation is solved and corner permutation is solved.
P(OLL skip) = 1/27
P(PLL skip) = 1/12
P(LL skip) = 1/324

For a typical 3x3 solve, P(LL skip) = 1/15552, so it is 48 times more likely with <RU> gen.

Where did my numbers come from?
<RU> OLL: You can count them (4 each of S, As, L, Pi, U, T, 2 H, and 1 solved) or you can calculate them (3^4/3 [each corner has 3 orientations, but divide by 3 because only 1/3 of these is possible])

<RU> PLL: You can count them again (4 Ua, 4 Ub, 2 Z, 1H, 1 solved) or you can calculate them (4!/2 [4 positions for first edge, 3 for second, 2 for 3rd, and 1 for last edge, but divide by 2 because only 1/2 of all PLL are possible])

3x3 LL: P(OLL skip) = 1/216 and P(PLL skip) = 1/72. Multiply for your answer.


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## F perm (Aug 30, 2012)

What is the chance of two N's in a row then a Z, then a H, then an E, in that order. It's all the unlikely PLL's. (This just happened to me!)


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## ben1996123 (Aug 30, 2012)

F perm said:


> What is the chance of two N's in a row then a Z, then a H, then an E, in that order. It's all the unlikely PLL's. (This just happened to me!)



1/483729408, but it's not really that special since there are 4084101 different combinations of PLLs you can get in 5 solves, so its only really ~118 times rarer than most other combinations.


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## Bob (Aug 30, 2012)

ben1996123 said:


> 1/483729408, but it's not really that special since there are 4084101 different combinations of PLLs you can get in 5 solves, so its only really ~118 times rarer than most other combinations.



I think you made a mistake.

P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36

P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352

The number of "different combinations" depends on your definition of what a combination is. Is Ua the same as Ub? Since the OP says only "N" instead of Na and Nb, I assume he groups them together. Therefore, there are only 13 "different" permutations (U, H, Z, A, E, J, T, R, F, G, V, N, and Y) so the number of combinations is 13^5 = 371,293. (I'm ignoring PLL skip as a case because I'm lazy.)

The original sequence of permutations is rarer than the expected combinations, but not 1/118. The PLLs that the OP got were N, Z, H, and E, which makes up 7/72 of all PLL cases. To only get one of these PLLs each time and not any of the others will occur on average (7/72)^5, which is about 1/115125. The most common combinations will involve one of the other PLLs, which account for 8/9 of PLL cases. "Most other combinations" will have only PLLs from that group, and the probability of getting 5 PLLs from that group is (8/9)^5 ~ 5/9, which makes the OP's combination about 63,886 times rarer than "most other combinations."

....I think.


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## F perm (Aug 30, 2012)

Bob said:


> I think you made a mistake.
> 
> P(N) = 1/36
> P(Z) = 1/36
> ...


Yeah, this seems right. Super rare!
I think it was two Na's in a row.
The average sucked though


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## nqwe (Aug 30, 2012)

Bob said:


> I think you made a mistake.
> 
> P(N) = 1/36
> P(Z) = 1/36
> ...



This is the chance of gettting these permutations, but not getting them in a specific order


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## Bob (Aug 30, 2012)

nqwe said:


> This is the chance of gettting these permutations, but not getting them in a specific order



What would be the calculation for these permutations in the specified order, then?


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## nqwe (Aug 30, 2012)

Sorry, I was wrong..
1/120932352 * 5! is the chance of getting these permutations in any order, which includes N-N-Z-H-E. I thought 1/120932352 would give you the chance of getting the permutations in any order.


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## Sajwo (Oct 7, 2012)

I just get 6 R perm in a row (all PLLs was on right hand). What's the chance fot something like that?


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## JasonK (Oct 7, 2012)

Sajwo said:


> I just get 6 R perm in a row (all PLLs was on right hand). What's the chance fot something like that?



For it to happen in those specific six solves:
Chance of R(b)-perm = 1/18
(1/18)^6 = *1/34012224*

But for it to happen in any random set of six consecutive solves, it's a lot higher.


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## Bob (Oct 7, 2012)

JasonK said:


> For it to happen in those specific six solves:
> Chance of R(b)-perm = 1/18
> (1/18)^6 = *1/34012224*
> 
> But for it to happen in any random set of six consecutive solves, it's a lot higher.



...and also more likely if it could have been any permutation 6 times.


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## PoHos1 (Oct 24, 2012)

*4x4 lucky..*

hey all .. I have mabye a big lucky because I solve rubiks cube 4x4 and first solve is with PLL skip and lucky second solve is with pll skip and wery lucky 3. solve is agein witch pll skip 
so how its possibly ? 
how is big % to get 3 times.. ?


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## vd (Oct 24, 2012)

Probability of PLL skip on 3x3 is 1/72. On 4x4, there is parity possible, so probability is just 1/144. So getting 3 PLL skips in the row on 4x4 has probability of 1/144*144*144 = 1/2 985 984. Very unusual i guess.


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## PoHos1 (Oct 24, 2012)

what the **** so its wery big lucky 

hele půjdeš na facebook můžeme pokecat


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## AndersB (Oct 24, 2012)

OMG so lucky, that's almost as lucky as solving a 2x2 by only doing random moves!


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## nqwe (Oct 24, 2012)

AndersB said:


> OMG so lucky, that's almost as lucky as solving a 2x2 by only doing random moves!



This is by far easier to "achieve".


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## Bob (Oct 25, 2012)

nqwe said:


> This is by far easier to "achieve".


How so?


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## mitch1234 (Dec 2, 2012)

*Probablilty of Getting God's Number for an Optimal Solution*

I decided that I would do some simple calculations and figure this out for myself. Probablity for was rounded to two significant figures, so its a bit off but who cares about exact numbers. So here is what I have come up with: 

Well there are 43,252,003,274,489,856,00 permutations for the cube. And all of those permutations all of them can be solved in 20 moves or less. No cube can be solved in >20 moves with an optimal solution. 20 moves is less likely an optimal solution than 8 is. So I decided I would calculate just how likely it is to happen, then put it in perspective. 

For solving the cube in 20 moves there is a 3.0 x10^8 cases. These cases can not be solved in anything less than 20. So the odds of picking a random cube (you will need a lot of cubes for this) is ~1 / 144,173,344,248. Big number, well I told you would need a lot of cubes. 

These cases practically don't exist, but we do have one really well know case of it, being the super flip.

Now to put it in perspective. There are about 7.056 billion people that live on earth currently. And if we had the entire population of earth get a cube and randomly scramble it a bit then all at once stop scrambling, we would need 20 more populations of the earth just to have 1 person have 1 cube that is exactly gods number.

Okay I'm done, time to sleep. There are probably some errors in this but just tell me so if you think there are, I'll gladly check my math.


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## kinch2002 (Dec 2, 2012)

mitch1234 said:


> Probablilty of Getting God's Number for an Optimal Solution etc


Thanks, that was interesting. Just wondering where you got the number 3 x 10^8 for the number of 20 move optimal cases from?


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## Ranzha (Dec 2, 2012)

kinch2002 said:


> Thanks, that was interesting. Just wondering where you got the number 3 x 10^8 for the number of 20 move optimal cases from?



3×10^8 is the approximate figure from cube20.org, for one thing =)


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## DrJohnFever (Dec 10, 2012)

What's the probability of taking your cube apart, putting the pieces back in completely randomly, and having the cube be solvable?


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## Noahaha (Dec 10, 2012)

DrJohnFever said:


> What's the probability of taking your cube apart, putting the pieces back in completely randomly, and having the cube be solvable?



1/12

1/3 chance corners are oriented, 1/2 chance of even PLL parity and 1/2 chance of edges flipped the right way.


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## cuBerBruce (Dec 15, 2012)

Is there some reason this thread was moved (to Speedcubing Help/Questions)?


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## ThomasJE (Dec 15, 2012)

cuBerBruce said:


> Is there some reason this thread was moved (to Speedcubing Help/Questions)?



Probably because people are commonly asking many questions here; like the one I am about to ask.

After OLL/OCLL/WV etc., what is the chance of the corners being solved (counting H-perm as cornered solves, and discounting AUF's)?


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## kinch2002 (Dec 15, 2012)

ThomasJE said:


> Probably because people are commonly asking many questions here; like the one I am about to ask.
> 
> After OLL/OCLL/WV etc., what is the chance of the corners being solved (counting H-perm as cornered solves, and discounting AUF's)?


1/6
Think of it like this:
1. Pick any corner. Doesn't matter where it is because you're allowing AUFs (at least that's how I'm interpreting your question)
2. The corner to the right of it needs to be a specific piece out of the 3 corners left. So that's 1/3 chance.
3. The next corner needs to be a specific piece out of the the 2 corners last. So that 1/2 chance.
4. Both steps 2 and 3 need to happen for corners to be solved so it's 1/3*1/2=1/6


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## ThomasJE (Dec 15, 2012)

kinch2002 said:


> 1/6
> Think of it like this:
> 1. Pick any corner. Doesn't matter where it is because you're allowing AUFs (at least that's how I'm interpreting your question)
> 2. The corner to the right of it needs to be a specific piece out of the 3 corners left. So that's 1/3 chance.
> ...



Ah, thanks. Now I think of it, I could have done it myself, since PLL skip is 1/72, and EPLL skip is 1/12, so 72 divided by 12 is 6; therefore the chance of CPLL skip is 1/6.


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## cuBerBruce (Dec 15, 2012)

ThomasJE said:


> Probably because people are commonly asking many questions here; like the one I am about to ask.


And people like me will probably ignore this thread from now on.


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## Smiles (Dec 16, 2012)

What is the probability of all edges being oriented in a ZZ solve before starting...
with a fully fixed colour scheme?
being fully colour neutral?

I got a 3 move cross on white with all the edges oriented (qqtimer scramble). I wasn't doing ZZ, but it was interesting that I never turned F or B during the solve.


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## ThomasJE (Dec 16, 2012)

Smiles said:


> What is the probability of all edges being oriented in a ZZ solve before starting...
> with a fully fixed colour scheme?
> being fully colour neutral?
> 
> I got a 3 move cross on white with all the edges oriented (qqtimer scramble). I wasn't doing ZZ, but it was interesting that I never turned F or B during the solve.



Fixed colour scheme is 1/2048/~0.05% (source), but full colour neutral; I have no idea.


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## JDogg (Mar 9, 2013)

*OLL Case Probabilities*

I've searched around the forum and Google and can't seem to find the right answer here. Are the probabilities typically listed for all the OLL cases wrong? I did the math and something isn't adding up right. There are several places that list the probabilities for each of the 58 cases (57 + the 1 solved case), like the OLL document on this page. Cubefreak shows the same probabilities and I've seen the same numbers listed in a couple other places too.

Of the 58 cases:
2 of them have a 1/216 probability.
4 of them have a 1/108 probability.
52 of them have a 1/54 probability.

With common denominator of 216, that's 2/216 + 8/216 + 208/216, which equals 218/216, or 100.926%. There's an extra 1/108th there. Those probabilities can't possibly be right.

Is there something obvious I'm missing here? Does anyone actually know how those probabilities are calculated? I assume it's a calculation based simply on the combination of the number of edges that are flipped and then the number of corners that are rotated a certain way (which is why the majority of cases end up being equally likely). But I wouldn't know where to begin trying to calculate them myself.


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## whauk (Mar 9, 2013)

JDogg said:


> 4 of them have a 1/108 probability.



i counted 5. and with 51 having 1/54 probability it should turn out correctly.

on calculating these: one OLL with a certain AUF has 3^3 possibilities for CO (every corner has 3 and the last one is fixed) and 2^3 possibilities for EO (same). that turns out to be 216 OLL cases with a certain AUF. however we dont mind AUF and consider one OLL "equal" with another if U,U',U2 makes them exact same cases. most OLLs look different when doing U,U',U2 so they actually have a chance of 4/216=1/54. (because all 4 possibilities of {(none),U,U',U2} were counted as different cases before) then again there are some that look identical after U2 and therefore have a chance of 2/216=1/108. and even some cases look identical after U,U' and U2 so they have a chance of 1/216.


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## Sa967St (Mar 9, 2013)

Without ignoring AUF, there are 3^3=27 possible corner orientations and 2^3=8 edge orientations. The common denominator comes from 27x8=216. 

Within these 216 cases, most of the OLL cases appear four times, once for each y rotation. These are the OLLs that have a 4/216=1/54 probability of occurring. The ones that appear two times have a 2/216=1/108 probability and the ones that appear only once have a 1/216 probability. 

There are five OLLs (1, 21, 55, 56, 57) that have a 1/108 probability and two (20, solved) that have a 1/216 probability.

Essentially, the probabilities depend on their reflections. The ones that are 1/108 look the same when you do a y2 and the ones that are 1/216 look the same when you do y, y2 and y'.


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## JDogg (Mar 9, 2013)

whauk said:


> i counted 5. and with 51 having 1/54 probability it should turn out correctly.
> 
> on calculating these: one OLL with a certain AUF has 3^3 possibilities for CO (every corner has 3 and the last one is fixed) and 2^3 possibilities for EO (same). that turns out to be 216 OLL cases with a certain AUF. however we dont mind AUF and consider one OLL "equal" with another if U,U',U2 makes them exact same cases. most OLLs look different when doing U,U',U2 so they actually have a chance of 4/216=1/54. (because all 4 possibilities of {(none),U,U',U2} were counted as different cases before) then again there are some that look identical after U2 and therefore have a chance of 2/216=1/108. and even some cases look identical after U,U' and U2 so they have a chance of 1/216.



Jeez. I triple-checked my math by counting up the probabilities on 3 different websites and missed the 1/108 for the single dot case on all 3 sites. So to answer my question - yes, there _was_ something obvious I was missing.

Interesting about the calculations though. It works different than I thought, but it's still neatly simple.


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## guythatlikesOH (May 19, 2013)

I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.


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## ~Adam~ (May 19, 2013)

If anyone fancies doing the math I recently had a 4 move last 4 centres while solving a 4x4 using Yau.


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## Bestsimple (May 19, 2013)

guythatlikesOH said:


> I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.



May i ask how you got that?


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## ben1996123 (May 19, 2013)

Bestsimple said:


> May i ask how you got that?



luck, of course


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## mark49152 (May 19, 2013)

guythatlikesOH said:


> I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.


If you get a PLL skip, the probability of the next two solves also being PLL skips is 1/5184


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## ben1996123 (May 19, 2013)

mark49152 said:


> If you get a PLL skip, the probability of the next two solves also being PLL skips is 1/5184



*1/72 for the original pll skip = 1/373248


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## Renslay (May 19, 2013)

ben1996123 said:


> *1/72 for the original pll skip = 1/373248



That is the probability for _that particular three_ solves. But if you have a PLL skip, then the probability of having three in a row PLL skip (i.e., two more), is (1/72)^2. Conditional probability.

Edit:
Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?


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## Ollie (May 19, 2013)

Renslay said:


> Edit:
> Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?



Why? The number of solves he did prior to the PLL skips has no effect on the probability of getting three in a row.


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## Renslay (May 19, 2013)

Let's say I do N solves today. What is the probability of P = I have at least 3 PLL skips in a row?
Let p denote the probability of a single PLL skip.

N = 1 -> P = 0
N = 2 -> P = 0
N = 3 -> P = p^3
N = 4 -> P = (p^3)(1-p) + (1-p)(p^3) + p^4 (first 3 solves with PLL skip then no PLL skip -OR- first solve with no PLL skip then 3 PLL skip -OR- 4 PLL skips). Note that this is higher than p^3
...
N = k -> (okay, I'm too lazy for that)
...
N = inf -> P = 1

Or question number 2: what is the flaw in the reasoning above?


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## Ollie (May 19, 2013)

Renslay said:


> Let's say I do N solves today. What is the probability of P = I have at least 3 PLL skips in a row?
> Let p denote the probability of a single PLL skip.
> 
> N = 1 -> P = 0
> ...



Not sure if you're deliberately trying to sound condescending, or what you're trying to prove, but the probability is the same regardless of how many solves you do in a day. If you flip a coin 15 times and it comes up heads on all of those, it does not make it more likely that you'll get a tails on the 16th time.


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## CubeRoots (May 19, 2013)

Renslay said:


> That is the probability for _that particular three_ solves. But if you have a PLL skip, then the probability of having three in a row PLL skip (i.e., two more), is (1/72)^2. Conditional probability.
> 
> Edit:
> Okay, technically, the "probability of having 3 PLL-skip in a row" has an important question: out of how many solves?



he said, I _just_ got 3 in a row. The probability that he would _just_ get 3 in a row is 1/373248 as you know.

Your question is interesting but it doesn't change anything about the calculations that were made


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## mark49152 (May 19, 2013)

CubeRoots said:


> he said, I _just_ got 3 in a row. The probability that he would _just_ get 3 in a row is 1/373248 as you know.


Wrong. That is the probability for any three particular solves. The probability of having PLL skips on the first three solves of the day is 1/373248. The probability of having 3 in a row at some unspecified point in a session of many solves is obviously higher.


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## Ollie (May 19, 2013)

mark49152 said:


> Wrong. That is the probability for any three particular solves. The probability of having PLL skips on the first three solves of the day is 1/373248. The probability of having 3 in a row at some unspecified point in a session of many solves is obviously higher.



Irrelevant.



guythatlikesOH said:


> I just got three PLL skips in a row (Which also gave me my best single time, and average). Probability: 1/373248.



No mention of the average occurring at any specific point during a session, no mention that it was done in the first solves of the day and no mention of the number of solves done in the session. Therefore his calculation is correct. Stop over-complicating it.


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## guythatlikesOH (May 19, 2013)

> No mention of the average occurring at any specific point during a session, no mention that it was done in the first solves of the day and no mention of the number of solves done in the session. Therefore his calculation is correct. Stop over-complicating it.



Thank you for clarifying. Three successive solves in a row are an independent event, and therefore the probability of three solves in a row being PLL skips is not affected by any prior or following solves.


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## Renslay (May 19, 2013)

There is a difference between "the probability of having the same outcome three times in a row _for the next three cases_" (dices, coins, PLL skip, etc), and "there will be three identical outcomes in a row in the next K cases" The first one is clearly independent as you say, P = (1/p)^3, however, the latter has lim P = 1 as K -> inf, because it is not specifyed _where_ will be that row in the series. Of course no matter how many cubes I already solved, a chance for a PLL skip (or 3 in a row) is still the same. But that is because we're talking about the _next_ case. But if I'm talking about "what is the chance that I will have a PLL skip today after ten thousand solves", that is pretty high, because I did not specify which solve contains the skip.

That's why I asked how many solves was that day, because I was wondering the chance of having 3 PLL skips in a row _anywhere_ during the solving.

Again, one who has cubing for years has much more chance that he _had_ 3 PLL skips in a row than a fresh cuber. But speaking of the chance that someone _will_ have 3 PLL skips in a row is the same.

Aaaand yes, I'm starting to over-complicate it. Sorry for that.


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## Crowned xerxes (May 19, 2013)

What is the probability of having a move move solve for a cross?


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## Renslay (May 19, 2013)

Crowned xerxes said:


> What is the probability of having a move move solve for a cross?



You mean a one move solve? Depends on. For one particular color, or for any cross?
Anyway, this has the answer:
http://www.cubezone.be/crossstudy.html


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## CubeRoots (May 19, 2013)

Right, it's like this:

The event is: I just got 3 PLL skips in a row. In other words: the last 3 solves I did were PLL skips.

The probability of this event is 1/373248


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## ryanj92 (May 19, 2013)

Surely the phrase 'if I do N solves what is the probability of getting 3 PLL skips in a row' has the following solution:

(N-2)*[(1/72)^3]*[(71/72)^N-3]
for N >= 3
The N-2 factor comes from the fact that the three PLL skips can be solves 1, 2, and 3, or 2, 3, and 4, right up to N-2, N-1, and N. This gives N-2 possibilites. Then the individual probability (in the knowledge that each solve is independent of the last solve) is the probability of one PLL skip cubed, multiplied by the probability that all other solves are not PLL skips.

EDIT: I've seen my own flaw already - my model assumes that ALL other solves have to not have a PLL skip, whereas you should include the other cases really (ie in 5 solves, solves 1-3 are skip, 4 is not skip and 5 is skip). in reality this diifference should be very small as probability of PLL skip < probability of non-PLL skip) so the above should give a good estimation?


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## PedroSabioni (May 24, 2013)

*Last layer skip probability?*

What is the probability of having a LL skip after finishing the F2L?


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## Ollie (May 24, 2013)

PedroSabioni said:


> What is the probability of having a LL skip after finishing the F2L?



1/15552


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## Smiles (May 24, 2013)

last year while i was trying my friend's cube, the very first 2 solves i got were both non-AUF PLL skips.
1/(72*4)^2 = 1/82944.

based on these results, his cube doesn't need PLL. i should get his cube.


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## MorrisKid101 (May 31, 2013)

*Acts of God?*

I was practising my algorithm s earlier and remembered an OLL case I remembered that I hadn't used in ages. I practised the algo until I remembered it again, and since that, it has appeared in 2 look OLL for the last 15-ish solves in a row. I used a scramble given by Speed Cube Timer for Android, and scrambled with yellow on top. I wanted to know what 'acts of God' could have happened, like a number of OLL/PLL/ LL skips in a row?

[moved here from being a separate thread by AvGalen]


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## TheNextFeliks (May 31, 2013)

If you build a random face on 2x2, the probability of: Solved 1/7 Adjacent 4/7 Diagnol 2/7


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## Username (May 31, 2013)

TheNextFeliks said:


> If you build a random face on 2x2, the probability of: Solved 1/7 Adjacent 4/7 Diagnol 2/7



How did you find this?


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## CubeRoots (May 31, 2013)

wrong it's 1/6 solved 1/6 diag 2/3 adj


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## TheNextFeliks (May 31, 2013)

CubeRoots said:


> wrong it's 1/6 solved 1/6 diag 2/3 adj



Oh yeah. You're right whoops.


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## PedroSabioni (Jun 20, 2013)

Why is the H perm so rare? What's the most rare PLL?


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## JasonK (Jun 20, 2013)

PedroSabioni said:


> Why is the H perm so rare? What's the most rare PLL?



If you count mirrored PLLs as the same, then H is the rarest (1/72 chance). If you don't, the N perms also each have a 1/72 chance.


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## ben1996123 (Jun 20, 2013)

PedroSabioni said:


> Why is the H perm so rare? What's the most rare PLL?



because its the same case from all angles (ie. hperm = U hperm U' = U2 hperm U2 = U' hperm U)

same reason why g perms are so common, there are 4 of them and each one is different from every angle


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## PedroSabioni (Jun 20, 2013)

JasonK said:


> If you count mirrored PLLs as the same, then H is the rarest (1/72 chance). If you don't, the N perms also each have a 1/72 chance.



Thanks for the answer! But I would like to understand why some PLL's are rarer than others. When I started learning CFOP I thought every PLL have the same probability.


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## PedroSabioni (Jun 20, 2013)

CubeRoots said:


> it's to do with symmetry. The symetric (in terms of their diagrams) PLLs have less chance of occuring. For example U perms can happen in 4 ways due to the 4 AUFs. However Nperms can happen in 2 ways, since they are symetrical about an axis there are only actually 2 AUFs so they have 2 ways to occur instead of 1. hence occur less frequently. Now with Hperm it is symetrical about both axes and so has 1 AUF and 1 way of occuring. That is why hperm will occur less frequently.



Got it! Good explanation, thanks man!


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## ben1996123 (Jun 20, 2013)

CubeRoots said:


> it's to do with symmetry. The symetric (in terms of their diagrams) PLLs have less chance of occuring. For example U perms can happen in 4 ways due to the 4 AUFs. However Nperms can happen in 2 ways, since they are symetrical about an axis there are only actually 2 AUFs so they have 2 ways to occur instead of 1. hence occur less frequently. Now with Hperm it is symetrical about both axes and so has 1 AUF and 1 way of occuring. That is why hperm will occur less frequently.



thats what i said

also nperms only have 1 auf


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## CubeRoots (Jun 20, 2013)

ben1996123 said:


> thats what i said
> 
> also nperms only have 1 auf



yaaa I was just trying to elaborate further. My bad on the Nperms though.


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## weirdesky (Jun 21, 2013)

*Chance of a single correctly placed Edge/Corner*

So I've been thinking about this for a while, but the only answer I can seem to get is 7/8 for corners and 11/12 for edges. Can someone help me out here, and tell me where I'm going wrong?


----------



## Christopher Mowla (Jun 21, 2013)

weirdesky said:


> So I've been thinking about this for a while, but the only answer I can seem to get is 7/8 for corners and 11/12 for edges.



I assume that:
[1] "Correctly placed" means that the corner/edge may be unoriented or oriented in its location.
[2] You're allowing any random configuration of the 3x3x3 cube (including configurations in which only a few pieces are unsolved).



weirdesky said:


> Can someone help me out here, and tell me where I'm going wrong?


You have to take the cycle classes (cycle types) into account as you're doing a calculation like this. Despite all of the possible cycle types, for this type of problem, we just group all cycle types which involve the same number of pieces with each other. So we can use my formula (which I did research on and developed here...you can see that post to see all of the cycle types, of you're interested):

\( \sum\limits_{k=p-x}^{p}{\left( \begin{matrix}
p \\
k \\
\end{matrix} \right)!k} \)

Where:
_p_ = the total number of the piece type. For wing edges, _p_ = 24, for middle edges, _p_ = 12, and for corners, _p_ = 8.
No more than _x_ of each piece type are solved. \( 0\le x\le p \).
and _k_ is just the index of the summation.

*Corners*
As you probably know, the corners have 8! different permutations. The sum of all of the possible cycle types (and all versions of each cycle type) for 8 objects is equal to this amount. That is, if we let _x_ = 8 (meaning that we sum all cycle types involving 0 objects to all 8 objects...all cycle types), we have:

\( \sum\limits_{k=8-8}^{8}{\left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k}=\sum\limits_{k=0}^{8}{\left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k}=8! \)

[Link]

But since we are just trying to count how many permutations involve exactly 7 pieces (because involving 7 pieces is equivalent to saying that one piece is left alone/solved), we just use the expression \( \left( \begin{matrix}
8 \\
k \\
\end{matrix} \right)!k \) and not the summation. That is, we substitute 7 for _k_ in the expression, and we get 14, 832.

So we simply do (14832/8!)*100 = 36.8%.

*Edges*
We use the same reasoning as we used for the corners, and we also get 36.8%.

*Corners and Edges*
If we want to calculate the chance that only one corner and one edge are correctly placed, we simply multiply them together, but we must be careful to not forget to multiply by two because the parity of the edges and corners must match (cube law of permutations).

\( \left( \frac{\left( \begin{matrix}
8 \\
7 \\
\end{matrix} \right)!7}{8!} \right)\left( \frac{\left( \begin{matrix}
12 \\
11 \\
\end{matrix} \right)!11}{\frac{12!}{2}} \right)\times 100= \) 27%.


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## pipkiksass (Jun 24, 2013)

So I was practicing yesterday and hit the same LL twice in a row with random scrambles. Completely different solves up to that point, but then identical LL, including AUF - both dot OLLs (OLL 2 from the wiki), followed by J perms with no AUF.

I was just thinking about this today, because I had the same OLL 2 solves in a row... which subsequently both turned into an N perm (but this time with different AUF).

So I guess my question is: what is the probability of getting the same LL (CFOP) 2 solves in a row. This is assuming no edge control, so all OLLs are possible. I'd be interested to know how much lower the probability of my first experience was (i.e. identical LL: same OLL, same PLL, same AUF) than the likelihood of just the same OLL and PLL would be (my second experience).

Answers on a postcard! TIA

edit: I realise as I'm rereading my own post that the probability would be different for more/less common OLL/PLL cases - just _how_ to calculate would be very much appreciated.


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## TheCasualPerson (Jun 27, 2013)

What is the chance of a rotationless f2l (no F or B turns)?


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## mDiPalma (Jun 27, 2013)

TheCasualPerson said:


> What is the chance of a rotationless f2l (no F or B turns)?



1/256 before cross
1/16 after cross


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## Frubix (Jun 27, 2013)

What is the chance of in a Pyraminx average of 5, having no tips turned in any of the solves, and what is the chance on a single?


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## TheNextFeliks (Jun 27, 2013)

Frubix said:


> What is the chance of in a Pyraminx average of 5, having no tips turned in any of the solves, and what is the chance on a single?



4 tips * 3 positions =12 configurations 
Only 1 has all tips untouched. So 1/12 for single. 
Avg: 1/12^5=1/248,832


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## Username (Jun 27, 2013)

TheNextFeliks said:


> 4 tips * 3 positions =12 configurations
> Only 1 has all tips untouched. So 1/12 for single.
> Avg: 1/12^5=1/248,832



Nope. 3^4=81 possible configurations. 1/81 chance for single


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## ben1996123 (Jun 27, 2013)

TheNextFeliks said:


> 4 tips * 3 positions =12 configurations
> Only 1 has all tips untouched. So 1/12 for single.
> Avg: 1/12^5=1/248,832



no

1/3^20 wich is lick 0.00000003%


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## TDM (Jun 27, 2013)

TheCasualPerson said:


> What is the chance of a rotationless f2l (no F or B turns)?


1/1 with ZZ (you never said CFOP!)

Does anyone know the chance of a 2GLL skip?


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## mDiPalma (Jun 27, 2013)

TDM said:


> 1/1 with ZZ (you never said CFOP!)
> 
> Does anyone know the chance of a 2GLL skip?



1/(12*3^3)=*1/324*

Also PM me if you're interested in ZZ-d/2gll. I average like 18.


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## TeddyKGB (Aug 9, 2013)

I just did a 4x4 solve and got a LL skip with the exception of OLL parity and PLL parity, are the odds of this the same as a regular LL skip?


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## DuffyEdge (Aug 9, 2013)

TeddyKGB said:


> I just did a 4x4 solve and got a LL skip with the exception of OLL parity and PLL parity, are the odds of this the same as a regular LL skip?


Probably not because there are different kinds of pll parities, and you can get a flipped edge on four possible edges. But I don't know


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## Antonie faz fan (Aug 9, 2013)

What is the change of getting ( in comp ) on 4x4 on OLL OLL parity and the rest solved like 1 edge flipped?
And what is the change of getting edge skip on 4x4 REDUCTION


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## Wassili (Aug 9, 2013)

I'm guessing this has already been asked, but what are the chances of a megaminx ll-skip, and an auf-less ll-skip?


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## cmhardw (Aug 9, 2013)

Wassili said:


> I'm guessing this has already been asked, but what are the chances of a megaminx ll-skip, and an auf-less ll-skip?



There are \( \left(\frac{5!}{2}\right)^2*2^4*3^4=4665600 \) last layer positions.

5 of those are the solve state, but possibly rotated. 1 of them is the AUF-less LL skip.


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## yoshinator (Aug 10, 2013)

cmhardw said:


> There are \( \left(\frac{5!}{2}\right)^2*2^4*3^4=4665600 \) last layer positions.
> 
> 5 of those are the solve state, but possibly rotated. 1 of them is the AUF-less LL skip.



Somebody at Vancouver open got a LL skip (either forte or turbo)


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## JustinJ (Aug 10, 2013)

yoshinator said:


> Somebody at Vancouver open got a LL skip (either forte or turbo)



it was forte


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## yoshinator (Aug 10, 2013)

JustinJ said:


> it was forte



Why couldn't Louis have gotten that at worlds -_-


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## cmhardw (Aug 10, 2013)

TeddyKGB said:


> I just did a 4x4 solve and got a LL skip with the exception of OLL parity and PLL parity, are the odds of this the same as a regular LL skip?



If you _must_ have both OLL and PLL parity, and assuming the edge groups are all paired up, then this is:
\( \frac{16 \left( \begin{array}{ccc} 4 \\ 2 \end{array} \right)}{(4!)^2*2^4*3^3}=\frac{1}{2592} \)



Antonie faz fan said:


> What is the change of getting ( in comp ) on 4x4 on OLL OLL parity and the rest solved like 1 edge flipped?



Assuming you mean just 1 edge group flipped, and the rest of the LL is otherwise solved this would be:
\( \frac{16}{(4!)^2*2^4*3^3}=\frac{1}{15552} \)

Coincidentally, this is the exact same probability of a LL skip on 3x3x3, assuming no partial edge control, corner orientation control, etc. (!)



Antonie faz fan said:


> And what is the change of getting edge skip on 4x4 REDUCTION



Assuming you mean that after solving centers (not using Yau, but simply solving the centers), that every edge group is already paired up for you then this would be:

\( \frac{12!*2^{12}}{24!}=\frac{1}{316234143225} \)



yoshinator said:


> Somebody at Vancouver open got a LL skip (either forte or turbo)



-You're very nearly 300 times more likely to get at least one LL skip on 3x3x3 in an average of 5 than to get a LL skip on one solve of megaminx
-You're very nearly 12 times more likely to get a LL skip on one solve of 3x3x3 than you are to get at least one megaminx LL skip in an average of 5
-You're about 26 times more likely to get a megaminx LL skip on one solve than you are to get at least two 3x3x3 LL skips in an average of 5

That's very exciting!


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## cmhardw (Aug 11, 2013)

\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!}\left[ \left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)*\frac{12!}{(12-k)!}*2^k\right] \) \( \left[\left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)*\frac{(12-k)!}{(12-k-j)!}*2^j*(24-2k-2j)!*(-1)^j\right]=\frac{12}{23} \)

When solving the 4x4x4 using reduction (not Yau method), the expected number of solved edge groups at the beginning of the edge pairing step is \( \frac{12}{23} \) edge group, or about 0.52 edge groups. Basically, about half the time no edge groups will already be solved, and about half the time 1 edge group will already be solved (expected value).

Mathematica is AWEsome!


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## Stefan (Aug 12, 2013)

cmhardw said:


> Mathematica is AWEsome!



The complete opposite of tapatalk. Which replaces each of your formulas by a friggin smiley as if that's funny! Argh!


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## cmhardw (Aug 12, 2013)

\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!*12!*2^{11}} \left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)*\frac{12!}{(12-k)!}*2^k* \) \( \left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)*\frac{(12-k)!}{(12-k-j)!}*2^j*(24-2k-2j)!*(12-k-j)!*2^{(12-k)-1-j}*(-1)^j \)

This simplifies to

\( \sum_{k=0}^{12} \sum_{j=0}^{12-k} \frac{k}{24!}\left( \begin{array}{ccc} 12 \\ k \\ \end{array} \right)\left( \begin{array}{ccc} 12-k \\ j \\ \end{array} \right)(-1)^j(24-2k-2j)!=\frac{1}{46} \)

When solving the 5x5x5 using standard reduction, the expected number of solved tredge groups at the beginning of the edge pairing step is \( \frac{1}{46} \) tredge group, or about 0.0217 tredge groups. Basically, about one in every 46 of your solves will have one tredge group already solved at the start of the edges step (expected value).



Stefan said:


> The complete opposite of tapatalk. Which replaces each of your formulas by a friggin smiley as if that's funny! Argh!



That does sound like a pain  I don't have a smart phone yet, so it'll be a little longer before I have to put up with that.


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## cmhardw (Aug 14, 2013)

*Expected number of semi-pairs at the start of 5x5x5 edge pairing (reduction method)*

\( \sum_{k=0}^{24} \sum_{j=0}^{24-k} \frac{k}{24!} \left( \begin{array}{ccc} 24 \\ k \end{array} \right) \left( \begin{array}{ccc} 24-k \\ j \end{array} \right) (24-k-j)! (-1)^j = 1 \)

When solving 5x5x5 using reduction, at the beginning of the edge pairing step you are expected to have 1 semi-paired edge group (a middle edge paired with exactly one wing edge) at the start of the step. This assumes that you do not use any techniques to either influence or preserve semi-pairs while solving centers.

This also assumes that the utility of a solved tredge is the same as that of two semi-pairs in the strict sense of the term "semi-pair" (basically if you have no preference whatsoever between getting two semi-pairs vs. a solved tredge).


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## Stefan (Aug 15, 2013)

cmhardw said:


> \( \sum_{k=0}^{24} \sum_{j=0}^{24-k} \frac{k}{24!} \left( \begin{array}{ccc} 24 \\ k \end{array} \right) \left( \begin{array}{ccc} 24-k \\ j \end{array} \right) (24-k-j)! (-1)^j = 1 \)
> 
> When solving 5x5x5 using reduction, at the beginning of the edge pairing step you are expected to have 1 semi-paired edge group (a middle edge paired with exactly one wing edge) at the start of the step. This assumes that you do not use any techniques to either influence or preserve semi-pairs while solving centers.
> 
> This also assumes that the utility of a solved tredge is the same as that of two semi-pairs in the strict sense of the term "semi-pair" (basically if you have no preference whatsoever between getting two semi-pairs vs. a solved tredge).



I don't quite understand that formula (haven't really tried, though) but here's how I think about it:

The middle edges provide reference, so we can just look at this whole thing as a permutation of 24 objects (the 24 wings) and ask how many we expect to be at the correct place. Now we could go over all 24! permutations and for each one, count how many of the 24 objects are at the correct place, add them all up, and divide by the number of permutations, 24!. But imagine the 24! permutations as a static huge table with 24! rows and 24 columns. What we've done is count "correct places" anywhere in the table, going over the table row by row. Rows have different numbers of correct places, from 0 (much of the time) to 24 (just once). Now instead of row by row, let's count the correct places anywhere in the table, but going over the table column by column. Due to symmetry, column 1 contains object 1 just as often as any other of the 24 objects, so 1/24th of the time, column 1 has a correct place. So unlike rows, all 24 columns have the same number of correct places, namely (1/24)*(24!). Summing them all up (i.e. multiplying with 24) and dividing by 24! of course gives you: 1. No Mathematica needed for this one 

Btw: *{n \choose k}* and *\binom{n}{k}* are much nicer ways to get \( {n \choose k} \).


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## cmhardw (Aug 15, 2013)

Stefan said:


> I don't quite understand that formula (haven't really tried, though) but here's how I think about it:
> 
> The middle edges provide reference, so we can just look at this whole thing as a permutation of 24 objects (the 24 wings) and ask how many we expect to be at the correct place. Now we could go over all 24! permutations and for each one, count how many of the 24 objects are at the correct place, add them all up, and divide by the number of permutations, 24!. But imagine the 24! permutations as a static huge table with 24! rows and 24 columns. What we've done is count "correct places" anywhere in the table, going over the table row by row. Rows have different numbers of correct places, from 0 (much of the time) to 24 (just once). Now instead of row by row, let's count the correct places anywhere in the table, but going over the table column by column. Due to symmetry, column 1 contains object 1 just as often as any other of the 24 objects, so 1/24th of the time, column 1 has a correct place. So unlike rows, all 24 columns have the same number of correct places, namely (1/24)*(24!). Summing them all up (i.e. multiplying with 24) and dividing by 24! of course gives you: 1. No Mathematica needed for this one
> 
> Btw: *{n \choose k}* and *\binom{n}{k}* are much nicer ways to get \( {n \choose k} \).



Stefan, you never cease to amaze me. That is a very clear explanation of how to count the number of times a piece ends up at the correct location. That is a really elegant proof!

My calculation is an application of the inclusion/exclusion technique.

Sum the number of "correct" pieces, k, as k goes from 0 correct to 24 correct. For each k, calculate using index j the number of derangement permutations of the remaining (24-k) "unsolved" pieces. This gives me 24 terms over the k index. Multiply each of the 24 terms by k and divide by 24! to provide it's contribution to the expected value result.

Also, thank you for those ways to produce a binomial term more elegantly than with arrays!


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## advincubing (Aug 15, 2013)

I apologize if this has already been asked (I didn't see it in a quick search): If 4x4 OLL parity probability is 50% and 4x4 PLL parity probability is 50%, is it accurate to say that there is a 25% chance of having no parity (and 25% of having both parities)? Asked differently, are the two parities independent or somehow dependent?


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## cmhardw (Aug 15, 2013)

advincubing said:


> I apologize if this has already been asked (I didn't see it in a quick search): If 4x4 OLL parity probability is 50% and 4x4 PLL parity probability is 50%, is it accurate to say that there is a 25% chance of having no parity (and 25% of having both parities)? Asked differently, are the two parities independent or somehow dependent?



Your thoughts are correct. The parities are independent of each other in reduction solving.


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## TheCasualPerson (Sep 15, 2013)

I was dumbfounded 2 minutes ago when I got a scramble with correct edge orientation *and* a completed, correctly positioned line!?! My question is what are the chances of skipping the EOline stage from a scramble???


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## Renslay (Sep 15, 2013)

TheCasualPerson said:


> I was dumbfounded 2 minutes ago when I got a scramble with correct edge orientation *and* a completed, correctly positioned line!?! My question is what are the chances of skipping the EOline stage from a scramble???



All edges oriented: 1/2^11 = 1 / 2048
DF and DB are in place: 1/12 * 1/11 = 1 / 132
Overall: 1 / 270336


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## TheCasualPerson (Sep 15, 2013)

Renslay said:


> All edges oriented: 1/2^11 = 1 / 2048
> DF and DB are in place: 1/12 * 1/11 = 1 / 132
> Overall: 1 / 270336



Wow that is incredible! So basically I will never get a scramble like that again? ^_^


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## Renslay (Sep 15, 2013)

TheCasualPerson said:


> Wow that is incredible! So basically I will never get a scramble like that again? ^_^



Well, there is always a chance to get it again.


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## Jaycee (Oct 16, 2013)

kinch2002 said:


> If you knew that the first layer permutation would be solved then you can't really call it a PBL skip. In which case, it's just a LL skip which is 1/162



I was trying to figure what this probability was on my own during my study hall today because the person I've been teaching to solve some puzzles (he's got Pyraminx down so I'm teaching him 2x2) asked me. In my head, it came out to be 1/648. I knew that couldn't be right so I told him I'd tell him tomorrow. Can somebody explain to me *why* the probability of a LL skip on 2x2 is 1/162? I know it has to do with the possible orientations and permutations of the pieces, I just clearly went about my calculation the wrong way.

Oh and by the way I'm completely disregarding that Daniel's post was responding to something to do with Ortega and PBL


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## Christopher Mowla (Oct 16, 2013)

Jaycee said:


> I was trying to figure what this probability was on my own during my study hall today because the person I've been teaching to solve some puzzles (he's got Pyraminx down so I'm teaching him 2x2) asked me. In my head, it came out to be 1/648. I knew that couldn't be right so I told him I'd tell him tomorrow. Can somebody explain to me *why* the probability of a LL skip on 2x2 is 1/162? I know it has to do with the possible orientations and permutations of the pieces, I just clearly went about my calculation the wrong way.
> 
> Oh and by the way I'm completely disregarding that Daniel's post was responding to something to do with Ortega and PBL


You have got to divide 648 by 4 to get 162 because of AUF. So you did the work correctly.


----------



## Jaycee (Oct 16, 2013)

Ahh, I see. Thanks.


----------



## ben1996123 (Oct 17, 2013)

Jaycee said:


> I was trying to figure what this probability was on my own during my study hall today because the person I've been teaching to solve some puzzles (he's got Pyraminx down so I'm teaching him 2x2) asked me. In my head, it came out to be 1/648. I knew that couldn't be right so I told him I'd tell him tomorrow. Can somebody explain to me *why* the probability of a LL skip on 2x2 is 1/162? I know it has to do with the possible orientations and permutations of the pieces, I just clearly went about my calculation the wrong way.
> 
> Oh and by the way I'm completely disregarding that Daniel's post was responding to something to do with Ortega and PBL



3^4 orientations, 4! permutations /3 for corner twist parity, /4 for AUF
1/(3^4*4!/(3*4)) = 1/162


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## Antonie faz fan (Oct 17, 2013)

Probeltie for no tips scrambled At pyraminx


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## Username (Oct 17, 2013)

Antonie faz fan said:


> Probeltie for no tips scrambled At pyraminx



4 tips, all of them have 3 orientations: 3x3x3x3 = 3^4 = 81 

so 1/81

I'm not 100% sure though


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## ben1996123 (Oct 17, 2013)

Username said:


> 4 tips, all of them have 3 orientations: 3x3x3x3 = 3^4 = 81
> 
> so 1/81
> 
> *I'm not 100% sure though*



yeb ittuce 1/81


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## advincubing (Dec 5, 2013)

*void cube parity*

I thought I had seen this before, but I can't find it on this thread: *What is the probability of void cube parity?* 1 in 2?

Chris' explanation in this thread of void cube parity was really helpful.


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## kcl (Dec 6, 2013)

advincubing said:


> I thought I had seen this before, but I can't find it on this thread: *What is the probability of void cube parity?* 1 in 2?
> 
> Chris' explanation in this thread of void cube parity was really helpful.



Almost positive it's 50%.


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## Lucas Garron (Dec 6, 2013)

advincubing said:


> I thought I had seen this before, but I can't find it on this thread: *What is the probability of void cube parity?* 1 in 2?
> 
> Chris' explanation in this thread of void cube parity was really helpful.



That depends how you solve. Roux solvers can basically avoid it.

However, if you pick an orientation for the centers at random (e.g. by solving cross and F2L without caring about the other pieces), the chance of parity is 1/2 for the usual reasons.


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## penguinz7 (Dec 12, 2013)

I don't know what the most rare is, but I find Z Perms fairly rare in my solves.


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## pipkiksass (Dec 12, 2013)

penguinz7 said:


> I don't know what the most rare is, but I find Z Perms fairly rare in my solves.



PLLs? Pretty sure it's H perms and N perms at 1/72 and Z, E and maybe others are 1/36. Something about rotational symmetry??


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## ThomasJE (Dec 12, 2013)

pipkiksass said:


> PLLs? Pretty sure it's H perms and N perms at 1/72 and Z, E and maybe others are 1/36. Something about rotational symmetry??



I believe you're right.

EDIT: http://www.kungfoomanchu.com/guides/andy-klise-3x3x3-speedcubing-guide-v4.pdf
Second page, on the right


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## bubbagrub (Jan 12, 2014)

Sorry if this has been asked here already, but what are the probabilities of the PBL cases in Ortega? I seem to get the one with two T-perms about 75% of the time (i.e., the one that is described as UF/DF here: http://www.cubewhiz.com/ortegapbl.php)


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## ThomasJE (Jan 18, 2014)

bubbagrub said:


> Sorry if this has been asked here already, but what are the probabilities of the PBL cases in Ortega? I seem to get the one with two T-perms about 75% of the time (i.e., the one that is described as UF/DF here: http://www.cubewhiz.com/ortegapbl.php)



Here they are:
http://www.kungfoomanchu.com/guides/andy-klise-2x2x2-speedcubing-guide.pdf
Under XLL.


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## JackJ (Jan 24, 2014)

Not cubing related but interesting. What are the odds of 3 baby buncos in a row? Best guess I have is 6^(9)(5/6) = 8398080 but I'm pretty sure I'm wrong as I have next to no experience with probability.


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## Bob (Jan 24, 2014)

JackJ said:


> Not cubing related but interesting. What are the odds of 3 baby buncos in a row? Best guess I have is 6^(9)(5/6) = 8398080 but I'm pretty sure I'm wrong as I have next to no experience with probability.


What is a baby bunco?


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## kcl (Jan 24, 2014)

Bob said:


> What is a baby bunco?



Good, I'm not the only one. Jack, if you can tell me what it is I can probably help.


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## JackJ (Jan 24, 2014)

Oh, I guess it's a Wisconsin thing. It's a dice game where the player player rolls three dice at once. If you're on the 3rd of 6 games, and you roll all 3s, thats a bunco worth 21 points. A baby bunco is essentially the same thing but the number rolled doesn't match the game you're on. So if you're on the 3rd game still and all 3 dice land on 1,2,4,5, or 6, that's a baby bunco worth 5 points. 

Three baby buncos in a row happened to a friend of mine and I'm curious.


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## tseitsei (Jan 24, 2014)

JackJ said:


> Oh, I guess it's a Wisconsin thing. It's a dice game where the player player rolls three dice at once. If you're on the 3rd of 6 games, and you roll all 3s, thats a bunco worth 21 points. A baby bunco is essentially the same thing but the number rolled doesn't match the game you're on. So if you're on the 3rd game still and all 3 dice land on 1,2,4,5, or 6, that's a baby bunco worth 5 points.
> 
> Three baby buncos in a row happened to a friend of mine and I'm curious.



OK.
So you roll three dices and all have the same number. Is that right?

And this happens 3 times in a row...

Let's first calculate the propability of one babu bunco:
First dice can be anything. so P=1
Second must be the same as first so P=1*(1/6)=1/6
Third must be the same as first two so P=(1/6)*(1/6)=1/36
1/36 for one baby bunco
for three in a row it would then be (1/36)³= 1/46656~0.0021%


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## Bob (Jan 24, 2014)

tseitsei said:


> OK.
> So you roll three dices and all have the same number. Is that right?
> 
> And this happens 3 times in a row...
> ...



Instead of 1 it's 5/6 because rolling all 3s is not a baby.


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## tseitsei (Jan 24, 2014)

Bob said:


> Instead of 1 it's 5/6 because rolling all 3s is not a baby.


True. stupid me


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## Rune (Jan 24, 2014)

P(baby bunco)= (5/6)*(1/6)^2


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## Bob (Jan 24, 2014)

Rune said:


> P(baby bunco)= (5/6)^2



Not true.

P(baby bunco) = (5/6) * (1/6) * (1/6) = 5/216
P(3 baby buncos) = (5/216)^3


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## Rune (Jan 24, 2014)

My bad, I just forgot the factor 1/6^2


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## cubemaste r (Jan 24, 2014)

how many cases does l3c on a 4x4 have for yau(AUF is aloud only on one of the centers)


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## Phillip1847 (Jan 26, 2014)

How many solves would you have to do to get a 50% chance of seeing every last layer case? With AUF? W/O AUF? 90%?


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## RageCuber (Jan 29, 2014)

Just thought of something stupid but, What are the chances of solving the cross then getting a f2l skip? (no OLL or PLL skip)


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## Renslay (Jan 29, 2014)

RageCuber said:


> Just thought of something stupid but, What are the chances of solving the cross then getting a f2l skip? (no OLL or PLL skip)



corner permutations: 1/ 8*7*6*5
corner orientations: 1 / 3^4
edge permutations: 1 / 8*7*6*5
edge orientations: 1 / 2^4

All of them at once (multiplication) = 1 / 3657830400. You have a much higher chance for winning a lottery with a single ticket (5 numbers out of 90, chance for winning is 1 / 43949268 ).

Note that these numbers are only valid if you solve the cross "blindly", so you don't look at other edges, corners, or already matched edge-corner pairs. In a similar lucky case you probably would see and try to solve xx-cross or xxx-cross intentionally (and get the last one or two corner-edge pairs by chance, hence getting an "F2L skip"). So, in practice, the chance for an F2L skip is much higher.


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## JackJ (Jan 30, 2014)

Tried calculating the probability of a LL skip on 4x4 while using K4. 

OLL skip: 3^(3)*2^(3)*2^(3)= 1/1728
PLL skip: 4!*4^(2)*3^(2)= 1/3456. That's an AUFless PLL skip so it's really 1/864. 
LL skip: 1728*864= 1/1492992

Nearly 1 in 1.5 million?! Surely that's not right. It's probably something to do with parity but I'm not sure how to implement it. Help anyone?


----------



## Christopher Mowla (Jan 30, 2014)

JackJ said:


> Tried calculating the probability of a LL skip on 4x4 while using K4.
> 
> OLL skip: 3^(3)*2^(3)*2^(3)= 1/1728
> PLL skip: 4!*4^(2)*3^(2)= 1/3456. That's an AUFless PLL skip so it's really 1/864.
> ...


When solving the 4x4x4 K4 style, all steps through F3L will solve all pieces except for:

*4 Corners*, which have (4!)(3)(3)(3) positions.
*8 Wing Edges*, which have 8! positions.

The parity of corners and wing edges are independent of one another (and thus all permutations of each are allowed independently), and so we just multiply these two numbers together. We also divide by 4 for AUF since all unsolved pieces are in the same layer and thus we can apply this rotational symmetry reduction.

1/6,531,840.


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## CyanSandwich (Feb 10, 2014)

What's the probability of getting at least 1 LL skip in 15522 solves?


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## Kit Clement (Feb 10, 2014)

CyanSandwich said:


> What's the probability of getting at least 1 LL skip in 15522 solves?



I'm going to assume you meant 15552, not 15522. Also assuming no edge/OLL control of any sort.

Simpler question - what's the chance of none happening? In one solve, 15551/15552. Take that to the 15552nd power to get the probability of no LL skips in that many solves.

(15551/15552)^15552 = .368 or 36.8%

Thus, the probability of at least one happening is the complement of that.

1 - .368 = .632 or 63.2%


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## CyanSandwich (Feb 10, 2014)

Thanks. I don't know why I couldn't figure it out.


----------



## guysensei1 (Feb 10, 2014)

Just now, I was doing a solve, and after I finished 3 F2L pairs, I ended up with a sort of sideways J perm on the R face.

Whats the probability of that?


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## cmhardw (Feb 10, 2014)

CyanSandwich said:


> What's the probability of getting at least 1 LL skip in 15522 solves?



Getting at least 1 success in n trials with a probability of 1/n can be closely approximated by \( 1-\frac{1}{e} \) for large n.


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## Sajwo (Feb 10, 2014)

What is the propability of getting 7 moves solution on skewb? FL 3 and LL 4?

Or getting pure U/H/Z perm?


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## TDM (Feb 10, 2014)

How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?


----------



## cmhardw (Feb 10, 2014)

TDM said:


> How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?



For \( n \in \mathbb{N} \)

\( 1-\left(\frac{15551}{15552}\right)^n > \frac{1}{2} \)

Since n is a natural number this solution can be written:

\( n \geq 10780 \)

----------edit----------



Phillip1847 said:


> How many solves would you have to do to get a 50% chance of seeing every last layer case? With AUF? W/O AUF? 90%?



I am _really_ interested in this question. I haven't had time to really sit and think about this yet, but I really want to work on this question!


----------



## Logiqx (Feb 11, 2014)

I got my first cross skip at the weekend using Prisma Puzzle Timer... no moves, no alignment required.

I've probably done about 5,000 timed solves and at most 10,000 solves in total.

I'm aware of the cross study but I decided to have a go at the maths myself. Have I got this right?

Probabilty of white cross = 1/190080
1/(24*22*20*18)

Probability of white / yellow cross ~ 1/95042
1/(24*22*20*18) + 1/(24*22*20*18) - 1/(24*22*20*18*16*14*12*10)
p(white cross) + p(yellow cross) - p(white cross + yellow cross)

Probability of any colour cross ~ 1/31704
I'm lazy and just looked at http://www.cubezone.be/crossstudy.html

In comparison to LL skip:
LL skip without needing AUF = 1/62208
LL skip needing AUF = 1/15552 - I've had one of these in an untimed solve

It's moderately interesting that they are in the same ball-park.


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## ThomasJE (Feb 11, 2014)

TDM said:


> How many solves would you have to do before the probability of having had at least one LL skip is more than 50%?



(15551/15552)^n <= 0.5 because that is the probabity of no LL skips.

I would use trial and error; but I can't right now. There's probably a quicker way though.

EDIT: Just worked it out on paper and with my phone calculator. 10780 solves are needed to have a 50% chance of at least 1 LL skip.


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## uberCuber (Feb 11, 2014)

ThomasJE said:


> (15551/15552)^n <= 0.5 because that is the probabity of no LL skips.
> 
> I would use trial and error; but I can't right now. *There's probably a quicker way though.*



Chris Hardwick already posted the answer above, but, Logarithms:

\( \ln { \left( \frac{15551}{15552} \right)^n } < \ln{0.5} \)

\( n \ln {\frac{15551}{15552}} < \ln{0.5} \)

\( n \geq 10780 \)


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## ThomasJE (Feb 11, 2014)

uberCuber said:


> Chris Hardwick already posted the answer above, but, Logarithms:
> 
> \( \ln { \left( \frac{15551}{15552} \right)^n } < \ln{0.5} \)
> 
> ...



Tapatalk sees three smilies. I can't see the fractions at all.

Did I get it right?


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## uberCuber (Feb 11, 2014)

ThomasJE said:


> Tapatalk sees three smilies. I can't see the fractions at all.
> 
> Did I get it right?



Lol, yes, 10780 is the right number


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## Stefan (Feb 11, 2014)

Logiqx said:


> 1/(24*22*20*18) + 1/(24*22*20*18) - 1/(24*22*20*18*16*14*12*10)
> p(white cross) + p(yellow cross) - p(white cross + yellow cross)



This is not quite right.


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## Logiqx (Feb 11, 2014)

Stefan said:


> This is not quite right.



Are you referring to the way I wrote it (somewhat lazy as it should really be 1/24 x 1/22 x 1/20, etc) or the final answer?

My calculation comes to the same result as the cross study, which if I understand correctly is 53759 / 5109350400.

My reasoning is that p(white cross) and p(yellow cross) intersect so I used the subtraction as a way of avoiding double-counting.


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## Stefan (Feb 12, 2014)

Sorry, I made thinking/coding mistakes. It indeed gives the correct result, although I'll have to think about why it does. How would you do "any colour cross"?


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## Methuselah96 (Feb 12, 2014)

Stefan said:


> Sorry, I made thinking/coding mistakes. It indeed gives the correct result, although I'll have to think about why it does. How would you do "any colour cross"?



He's just combining permutation and orientation of cross edges by distributing the 2 from a calculation that normally has them separate like: 2^4 x (12 x 11 x 10 x 9).
So then he would probably do a neutral cross like: 24 x 22 x 20 x 18 x 16 x 14 x 12 x 10 x 8 x 6 x 4 x 1 (not multiplying the last number by two because the orientation is already determined) derived from (2 orientation x 12 permutations) (2 orientation x 11 permutations)...(2 orientation x 2 permutation)(1 orientation x 1 permutation).



Stefan said:


> I made mistakes.


Is this a first?


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## AFatTick (Feb 12, 2014)

This has most of the probabilities of pll and oll cases. http://badmephisto.com/badmephisto-speedcubing-method.pdf


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## Logiqx (Feb 12, 2014)

Stefan said:


> How would you do "any colour cross"?



My approach would be to figure out the probability of all six crosses, then >=3 crosses (includes 6 cross probability), then >=2 crosses (includes 3 cross probabilities), then >= 1 cross (includes 2 cross probabilities).

I just had a quick go at it and my first result was slightly off... 1/31703.96091 instead of 1/31703.94349.

I can get the correct answer with a kludge in my calculation (re: chance of getting >= 2 or 3 crosses). The kludge uses the probability of six crosses but more times than I'd expected. I haven't got my head around why the kludge works yet. I might think about it again at lunch time...

I'd better do some actual work, tempting as it is to solve this little brain teaser!


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## Logiqx (Feb 12, 2014)

Please forgive the strange notation. I've treated orientation and permutation as a single property.

I realise it could be written using factorials and distinguishing between orientation + permutation. 

Six crosses - 1 case (11 edges correct, meaning the twelth is also correct)
1 / (24*22*20*18*16*14*12*10*8*6*4)

Five or Four crosses are not possible.

Three crosses - 20 cases
Two opposite crosses (10 edges, 12 cases) = 12 / (24*22*20*18*16*14*12*10*8*6)
+ Three adjacent crosses (9 edges, 8 cases) = 8 / (24*22*20*18*16*14*12*10*8)

Two crosses - 15 cases
Opposite crosses (8 edges, 3 cases) = 3 / (24*22*20*18*16*14*12*10)
+ Adjacent crosses (7 edges, 12 cases) = 12 / (24*22*20*18*16*14*12)

One cross - six cases
6 / (24*22*20*18)

Combining everything...
p(one cross)-(p(two crosses)-(p(three crosses)-10*p(six crosses))) = 1/31703.94349

This is the correct answer but it includes a kludge. I don't understand what the multiplication by 10 is doing!

I think it accounts for intersections of some of the three cross cases.


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## RageCuber (Feb 12, 2014)

I just got two PLL skips in one avg12, no edge control or anything like that.
whats the probability?


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## Kit Clement (Feb 12, 2014)

RageCuber said:


> I just got two PLL skips in one avg12, no edge control or anything like that.
> whats the probability?



Probability of PLL skip with no forcing skips - 1/72. Let X be the random variable for the number of PLL skips. This follows a Binomial(12, 1/72) distribution.

\( P(X = 2) = \binom{12}{2}(1/72)^2(71/72)^{10} = 0.011 \)

But you're probably more interested in cases of 2 PLL skips or more, as this probability does not include this.

\( P(X \geq 2) = 1 - P(X = 1) - P(X = 0) \)
\( = 1 - \binom{12}{1}(1/72)^1(71/72)^{11} - \binom{12}{0}(1/72)^0(71/72)^{12} = 0.0116 \)


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## Christopher Mowla (Feb 12, 2014)

RageCuber said:


> I just got two PLL skips in one avg12, no edge control or anything like that.
> whats the probability?


I'm guessing that you wanted to know the probability for your specific event. If you just wanted to know the probability of getting 2 PLL skips in 12 solves in general, then Kit Clement's answer, \( \left( \begin{matrix}
12 \\
2 \\
\end{matrix} \right)\left( \frac{1}{72} \right)^{2}\left( \frac{71}{72} \right)^{10} \), is correct.

However, the probability of getting 2 PLL skips one after the other in 12 solves is different than if your first solve and 12th solve are PLL skips, for example.

So in general, for 12 solves, 2 of which are PLL skips,

The probability when there are _x_ non-PLL skips in between the 2 PLL skip solves = \( \left( 12-\left( x+1 \right) \right)\left( \frac{1}{72} \right)^{2}\left( \frac{71}{72} \right)^{10},\text{ 0}\le x\le 10. \) [Link]


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## Kit Clement (Feb 13, 2014)

cmowla said:


> I'm guessing that you wanted to know the probability for your specific event. If you just wanted to know the probability of getting 2 PLL skips in 12 solves in general, then Kit Clement's answer, \( \left( \begin{matrix}
> 12 \\
> 2 \\
> \end{matrix} \right)\left( \frac{1}{72} \right)^{2}\left( \frac{71}{72} \right)^{10} \), is correct.
> ...



Is that not what he wanted though? He never specified that the two PLL skips came in a row.


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## Christopher Mowla (Feb 13, 2014)

Kit Clement said:


> Is that not what he wanted though? He never specified that the two PLL skips came in a row.


It's kind of hard to tell exactly. Certainly what you gave him is what he _asked for_, but if I just got 2 PLL skips in a average of 12 and wanted to know the probability of that happening, I would want to know the exact probability of that particular situation (because otherwise to me it would be inaccurate). Therefore I just added on to what you did to make sure our friend's curiosity has been met.


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## Stefan (Feb 13, 2014)

Got more time now...



Stefan said:


> Sorry, I made thinking/coding mistakes.



... meaning I thought the white and yellow crosses aren't independent, and not just because of the cases where both of them are solved, but also where one being unsolved also influences the other (like white center not being solved because a white edge is in the yellow layer, immediately ruling out the yellow center as well). But I was over-thinking it. Subtracting P(white and yellow) does handle it properly. And my coding mistake is funny, I used Python's Decimal class with many digits precision, but then used it too late in the process, after floats had already caused rounding errors. Should have just used WolframAlpha.



Methuselah96 said:


> So then he would probably do a neutral cross like: 24 x 22 x 20 x 18 x 16 x 14 x 12 x 10 x 8 x 6 x 4 x 1



That's the easy part, the interesting part is counting, discounting and recounting the crosses properly, as seen in his new posts.



Methuselah96 said:


> Is this a first?



Of course!



Logiqx said:


> p(one cross)-(p(two crosses)-(p(three crosses)-10*p(six crosses))) = 1/31703.94349



I confirm, though prefer integers and case counting instead of fractions and probabilities.
http://ideone.com/hetM2d



Logiqx said:


> This is the correct answer but it includes a kludge. I don't understand what the multiplication by 10 is doing!



Well, we want to count each case with any solved cross exactly once. From my code/perspective:

c1 = number of cases with *at least one* solved cross
c2 = number of cases with *at least two* solved crosses
c3 = number of cases with *at least three* solved crosses
c6 = number of cases with *at least six* solved crosses

total = c1 - c2 + c3 - 10*c6


The cases with *exactly one* solved cross are counted by using c1.
The cases with *exactly two* solved crosses are counted twice by c1, so we have to discount them once (i.e. subtract c2).
The cases with *exactly three* solved crosses are counted thrice by c1 and discounted thrice by c2, so we have to count them once again (i.e. add c3).
The cases with *exactly six* solved crosses (*) are counted six times by c1, discounted 6c2=15 times by c2, and recounted 6c3=20 times by c3, so we have counted them 6-15+20=11 times and thus have to discount them 10 times (i.e. subtract 10*c6).
(*) yes, there's just one such case


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## Logiqx (Feb 13, 2014)

Thanks for sparing your time Stefan. It's nice to have a final resolution.



Stefan said:


> I was over-thinking it.



I started along those lines then decided there must be a simpler approach, easily visualised with a two-circle venn diagram. It just got a little harder to visualise with more than three circles!



Stefan said:


> That's the easy part, the interesting part is counting, discounting and recounting the crosses properly, as seen in his new posts.



I hope you enjoyed it. I haven't found it described elsewhere on this forum.



Stefan said:


> Of course!



My greatest cubing achievement, lol.



Stefan said:


> Well, we want to count each case with any solved cross exactly once. From my code/perspective



Excellent. Thanks for that explanation and I can now see my oversight.

I couldn't understand why I could get so close to the correct answer without using x*c2 and y*c3 and z*c6 , just 10*c6.

I'd overlooked how counting and discounting might end up with x=1, y=1 and z=10.


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## Stefan (Feb 13, 2014)

Logiqx said:


> I hope you enjoyed it.



I did, good stuff. Also thanks for your calculations, I might not have tried it without. And the mysterious 10 intrigued me, I didn't understand it at first, either.


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## cubemaste r (Feb 14, 2014)

How many possible permutations does a 3x3x3x3 have?


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## Jokern (Feb 14, 2014)

cubemaste r said:


> How many possible permutations does a 3x3x3x3 have?


I was trying to figure this out bymyself, but after some googling turns out I would be wrong. But anyway:
Short answer: about 10^120 (a few more than a 6x6x6)
long answer: http://www.superliminal.com/cube/permutations.html


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## Saxsoprano (Feb 14, 2014)

Anyone knows the chance of an LL skip using always WV with ZZ method? Only curiosity.


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## uberCuber (Feb 14, 2014)

Saxsoprano said:


> Anyone knows the chance of an LL skip using always WV with ZZ method? Only curiosity.



If you're using ZZ, EO is always solved, and if you're using WV, CO is always solved, so all that is left is PLL. An "LL skip" then would be equivalent to a PLL skip, which has a probability of 1/72


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## Saxsoprano (Feb 14, 2014)

Ok, thanks!!


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## JackJ (Feb 15, 2014)

So I celebrated my 18th birthday yesterday. My cousin also welcomed his first son yesterday. Additionally, another one of my cousins turned 10 yesterday. So are the odds of three specific people being born on the same day 1/(365^2) since the first person has a 100% chance of being born on random day of a 365 day calendar and the other people need two need to follow suit?


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## Stefan (Feb 15, 2014)

JackJ said:


> So I celebrated my 18th birthday yesterday. My cousin also welcomed his first son yesterday. Additionally, another one of my cousins turned 10 yesterday. So are the odds of three specific people being born on the same day 1/(365^2) since the first person has a 100% chance of being born on random day of a 365 day calendar and the other people need two need to follow suit?



Apparently the days around September have a higher probability (at least in USA), so no. Also, don't forget February 29.


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## JackJ (Feb 16, 2014)

Stefan said:


> Apparently the days around September have a higher probability (at least in USA), so no. Also, don't forget February 29.



I noticed this too. I think this happening may be more of a trend than a mathematical probability though. 

Also, leap year shouldn't complicate the problem too much. Since my cousin was born in 2004, a leap year, the new equation should look something like this:

1/(365*366)

But hey, I don't know too much about math of this sort and wikipedia has an entire page dedicated to explaining just how complicated these types of problems can be.


----------



## Stefan (Feb 16, 2014)

JackJ said:


> I think this happening may be more of a trend than a mathematical probability though.



What do you mean with "trend", and why wouldn't it be a mathematical probability?

This is one analysis I came across:
http://thedailyviz.com/2012/05/12/how-common-is-your-birthday/
http://thedailyviz.com/2012/05/18/how-common-is-your-birthday-pt-2/

Also, if you look at the source data at
https://docs.google.com/spreadsheet/ccc?key=0AjlIKRG8DtTqdDJ4S0FFSktma1dOWkhXZnRpV0tocmc#gid=0
then you'll see that births didn't just go up around September overall, but in every single one of those ten years.


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## JackJ (Feb 16, 2014)

Stefan said:


> What do you mean with "trend", and why wouldn't it be a mathematical probability?
> 
> This is one analysis I came across:
> http://thedailyviz.com/2012/05/12/how-common-is-your-birthday/
> ...



A quick Google search yields a good definition of trend: A general direction in which something is developing or changing. 

I didn't say it wasn't a mathematical probability either, it just seemed like more of a trend to me. The graphs you presented me are, for the most part, uniform.

Assuming the date of conception between my parents and the other two couples was random, I think my approximation is pretty solid.


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## Smiley (Feb 16, 2014)

I had a 2x2 scramble yesterday: U R' F R2 F R2 F R' U'
It is impossible to match up 2 colors in one move, how awesome is that???
anyway, it made me wonder what the chances are to start with no 2 colors matched up.


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## Rune (Feb 16, 2014)

"I wouldn’t say that they hold it in over Christmas, but they force them out before New Years. Gotta get the tax deduction on the books".


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## Stefan (Feb 16, 2014)

JackJ said:


> A quick Google search yields a good definition of trend: A general direction in which something is developing or changing.



Yeah but how is it developing/changing, and how does that make it less of a probability?



JackJ said:


> The graphs you presented me are, for the most part, uniform.



Not only are they not (although it's unclear what you mean with "for the most part" and what a uniform graph is), but they don't even appear to be based on a uniform distribution.


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## Jaycee (Feb 23, 2014)

2 OLL skips in a row, both leading to EPLLs.

(1/216 * 1/6)^2 --- is that correct?


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## TDM (Feb 23, 2014)

Jaycee said:


> 2 OLL skips in a row, both leading to EPLLs.
> 
> (1/216 * 1/6)^2 --- is that correct?


Yes.


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## kcl (Feb 24, 2014)

Odds of a complete cross skip?


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## Jaycee (Feb 24, 2014)

kclejeune said:


> Odds of a complete cross skip?



(1/12)*(1/11)*(1/10)*(1/9)*((1/2)^4) ------- that last term is really 1/16 so it can be written as (1/(12*11*10*9*16))

1/190080.

Then again it's 1 AM and I'm tired so I could be wrong.


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## kcl (Feb 24, 2014)

Jaycee said:


> (1/12)*(1/11)*(1/10)*(1/9)*((1/2)^4) ------- that last term is really 1/16 so it can be written as (1/(12*11*10*9*16))
> 
> 1/190080.
> 
> Then again it's 1 AM and I'm tired so I could be wrong.



Does that change if I say odds of a complete cross skip on white? I suck at probability lol.


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## Jaycee (Feb 24, 2014)

kclejeune said:


> Does that change if I say odds of a complete cross skip on white? I suck at probability lol.



I was just about to edit the post saying that I think I did the math assuming one cross color, so I don't think so. :3


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## Tim Major (Feb 24, 2014)

Kit Clement said:


> I'm going to assume you meant 15552, not 15522. Also assuming no edge/OLL control of any sort.
> 
> Simpler question - what's the chance of none happening? In one solve, 15551/15552. Take that to the 15552nd power to get the probability of no LL skips in that many solves.
> 
> ...



.501 = (15551/15552)^x

x = 10780

So if one was to do 10780 solves, it is more likely for them to have had an LL skip, than for them to have not.

And considering most people will atleast use SOME basic edge control, in an average of 10000 it is likely one will have an LL skip.

On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;

how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)


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## TDM (Feb 24, 2014)

Tim Major said:


> On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;
> 
> how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)


I've seen it before.


Spoiler: Answer



Isn't it 23 or something? I might have remembered it wrong, but I'm 99% sure that's it. Well, I'm at least 50% sure anyway  I was explaining this to my brother the other day and he didn't believe me, neither did anyone else who heard, so idk if it is actually 23.


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## CyanSandwich (Feb 25, 2014)

TDM said:


> I've seen it before.
> 
> 
> Spoiler: Answer
> ...





Spoiler



http://en.wikipedia.org/wiki/Birthday_problem


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## Logiqx (Feb 25, 2014)

kclejeune said:


> Odds of a complete cross skip?



http://www.speedsolving.com/forum/s...ability-Thread&p=951837&viewfull=1#post951837

Explanations are in following posts.


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## Logiqx (Feb 25, 2014)

Tim Major said:


> On an unrelated note, this reminded me of the birthday pairs problem, for those who haven't seen it;
> 
> how many people would you need to randomly pick out of a crowd before it is more likely than not (>50% chance) of two of the subjects sharing a birthday (as in, 24th of Feb, not 24th of Feb in a certain year)





TDM said:


> I've seen it before.



Me too... I often consider it during my life as a software developer.

Applied to cubing: If you take 6 random solves there's >50% chance of getting the same PLL twice.

You can illustrate this with six random numbers between 1 and 18:
http://www.random.org/integers/?num=6&min=1&max=18&col=6&base=10&format=html&rnd=new

The odds of at least 2 identical numbers are around 55%. This is an over-simplification because some PLL cases aren't 1/18 chance but it is close enough!


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## ChickenWrap (Feb 25, 2014)

Logiqx said:


> Me too... I often consider it during my life as a software developer.
> 
> Applied to cubing: If you take 6 random solves there's >50% chance of getting the same PLL twice.
> 
> ...



Aren't there 21 plls?


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## AmazingCuber (Feb 25, 2014)

ChickenWrap said:


> Aren't there 21 plls?



But most cases have a probability of 1/18. Check badmephisto's PLL page for more info.


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## Carrot (Feb 25, 2014)

TDM said:


> I've seen it before.
> 
> 
> Spoiler: Answer
> ...



It's 23. 


Spoiler



ceil(sqrt(2*n*log(2))+(3-2*log(2))/6) = birthday problem for years with n days.
put 365 in it and you get 23 ;-) (source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)



for the how many times should I solve a cube before I have more than 50% chance of getting a PLL/OLL/LL skip: ceil(log(50%)/log((#-1)/#)) where #=15552 for LL skip, figure out how to use it


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## TDM (Feb 25, 2014)

CyanSandwich said:


> Spoiler
> 
> 
> 
> http://en.wikipedia.org/wiki/Birthday_problem


Oh cool, didn't know there was an entire Wikipedia page on it. ty.


Carrot said:


> It's 23.
> 
> 
> Spoiler
> ...


I have no idea of what ceil or log means, but if #=15552 for LL skip, I don't think I need to  The only problem is that it doesn't work in Google, which I usually use as a calculator, and I don't think my calculator has ceil. Do you know an online calculator that has it/if it's shown as something else on other calculators?


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## cmhardw (Feb 25, 2014)

TDM said:


> I have no idea of what ceil or log means...



ceil and log are functions. The ceiling function is often used to help with precisely calculating an integer answer when your intermediate results are decimal numbers. The Logarithm function is essential when working with exponents.


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## TDM (Feb 25, 2014)

cmhardw said:


> ceil and log are functions. The ceiling function is often used to help with precisely calculating an integer answer when your intermediate results are decimal numbers. The Logarithm function is essential when working with exponents.


Oh cool, there's not actually that complicated. ty!


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## Logiqx (Feb 25, 2014)

Carrot said:


> It's 23.
> (source: David Brink: "A (probably) exact solution to the Birthday Problem" it's an interesting paper)



There are several approximations for the birthday problem but I like the two below. The second one is pretty easy to remember:



Spoiler



0.5 + sqrt(0.25 - 2 * N * log(0.5))

0.5 + 1.17741 * sqrt(N)



Here's a little Python script with a selection of values for N (re: cubing, computing and birthday):


```
import math
for n in [18, 54, 256, 365, 15552, 65536, 4294967296]:
    approx1 = 0.5 + 1.17741 * math.sqrt(n)
    approx2 = 0.5 + math.sqrt(0.25 - 2 * n * math.log(0.5))
    approx3 = math.sqrt(2 * n * math.log(2)) + (3 - 2 * math.log(2)) / 6
    print n, approx1, approx2, approx3
```

Results:


Spoiler



18 5.49532757142 5.52028868693 5.26427860676
54 9.15216115415 9.166596535 8.92111225942
256 19.33856 19.3451945187 19.1075113001
365 22.9943864654 22.9999431512 22.7633378354
15552 147.332043776 147.332897895 147.100997524
65536 301.91696 301.917380472 301.685916704
4294967296 77163.24176 77163.2432372 77163.0121865


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## cuber952 (Feb 26, 2014)

This might be stupid but, do some OLLs stand a better chance of getting a PLL skip than others? If so, which OLLs give PLL skips most often?

- fedora


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## Jaycee (Feb 26, 2014)

I'd think not, because AFAIK orientation and permutation are independent of each other.


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## Stefan (Feb 26, 2014)

cuber952 said:


> This might be stupid but, do some OLLs stand a better chance of getting a PLL skip than others? If so, which OLLs give PLL skips most often?
> 
> - fedora



Who is fedora?


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## guysensei1 (Feb 26, 2014)

What is the probability of getting 1 ortega face solved on 2x2? What about a LBL type face?

I guess this isn't probability, but what are the god's numbers for ortega faces and LBL faces?


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## uberCuber (Feb 26, 2014)

I was going to write this in a PM so as not to go off-topic in this thread, but your inbox is full. :s



Stefan said:


> Who is fedora?



That's coming from a game started in a private cubing group on facebook. Each person participating is assigned another one as a target. Every single time a participant makes a post, whether it be in that facebook group, elsewhere on facebook, or even on speedsolving, the post has to contain a particular word (for the first round of this game quite awhile ago, the word was 'cube', and now this time it's 'fedora'). If a participant catches his target making a post that doesn't contain that word, he can post BLAMMO!, and the target is out. Last person left wins.

And there are a few other rules to make this actually work, but that's the gist.


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## Stefan (Feb 26, 2014)

Oh. John has been to 40 competitions and is one of the fastest cubers, I couldn't imagine him asking those questions and it looked like he was quoting some noob to mock him or someone else using his account to ask them 

Although, about his second question: Some OLLs in fact do give PLL skips more often than others (someone guess which).


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## Logiqx (Feb 26, 2014)

Carrot said:


> for the how many times should I solve a cube before I have more than 50% chance of getting a PLL/OLL/LL skip: ceil(log(50%)/log((#-1)/#)) where #=15552 for LL skip, figure out how to use it



I've never really looked closely at the "my birthday" problem.

There is an easy approximation for cubers - 0.69314 * n ... rounding the result upwards. Remembering 69% is pretty easy.

I ran a few cubing numbers through Python using the original formula and my approximation:


```
import math
for n in [18, 54, 72, 216, 15552, 31704, 95042, 190080]:
    approx1 = int(math.ceil(math.log(0.5) / math.log((n - 1.0) / n)))
    approx2 = int(math.ceil(0.69314 * n))
    print '1 /', n, ':', approx1, 'solves (approx1) or', approx2, 'solves (approx2)'
```

Results - i.e. solves required for 50% chance of a specific event:



Spoiler



1 / 18 : 13 solves (approx1) or 13 solves (approx2) - J-Perm
1 / 54 : 38 solves (approx1) or 38 solves (approx2) - T-OLL
1 / 72 : 50 solves (approx1) or 50 solves (approx2) - PLL skip
1 / 216 : 150 solves (approx1) or 150 solves (approx2) - OLL skip
1 / 15552 : 10780 solves (approx1) or 10780 solves (approx2) - LL skip
1 / 31704 : 21976 solves (approx1) or 21976 solves (approx2) - CN cross skip
1 / 95042 : 65878 solves (approx1) or 65878 solves (approx2) - W/Y cross skip
1 / 190080 : 131754 solves (approx1) or 131753 solves (approx2) - W cross skip


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## Carrot (Feb 26, 2014)

Logiqx said:


> There are several approximations for the birthday problem but I like the two below. The second one is pretty easy to remember:
> 
> 
> 
> ...



It's right that your approximations are easier to remember etc, but they are approximations with errors  The one I posted is an approximation with no known errors. (or at least no one has any instances and even if they did you could extend it using his paper to get it even more exact.)


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## Tim Major (Feb 26, 2014)

Tipping point (over .5) for a dice is roughly 4 rolls.
(5/6)^x>0.5
Has to be 4ish rolls.

But logically, it should be a tipping point after 3 rolls.

So using my own maths I get 4 rolls, using intuition I get 3. Can someone explain?

The reason I ask is I explained the LL skip tipping point to a friend and we got in an argument and he suggested simplifying it to a die and I can no longer feel comfortable in my answer even though mathematically I'm right.


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## Carrot (Feb 26, 2014)

Tim Major said:


> Tipping point (over .5) for a dice is roughly 4 rolls.
> (5/6)^x>0.5
> Has to be 4ish rolls.
> 
> ...



Roll with some dices and see for yourself...

jk. 

Your friend is using the "expectance" instead.

It's true you are expected 0.5 6's in 3 throws, but the distribution just happens to be:
0: 125 cases
1: 75 cases
2: 15 cases
3: 1 case

So do an weighed average of that and you'll get 0.5 6's

But the question was 50% chance of getting at least one, which is the same as 50% chance of NOT getting one, then you suddenly have:
125/216 which is less than 50%, meaning you didn't hit the tipping point yet


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## Logiqx (Feb 26, 2014)

Carrot said:


> It's right that your approximations are easier to remember etc, but they are approximations with errors  The one I posted is an approximation with no known errors. (or at least no one has any instances and even if they did you could extend it using his paper to get it even more exact.)



That's fair enough... still close to within 1/4 of a solve on the cubing problems. One of my friends can probably do the second approximation to 5 decimal places in his head but he's a certified freak, lol.


----------



## Jaycee (Feb 26, 2014)

Stefan said:


> Oh. John has been to 40 competitions and is one of the fastest cubers, I couldn't imagine him asking those questions and it looked like he was quoting some noob to mock him or someone else using his account to ask them
> 
> Although, about his second question: Some OLLs in fact do give PLL skips more often than others (someone guess which).



-It felt weird to be answering a question of his because frankly I'm still a "noob" and I definitely know who he is >_>

-The only thing I can think of is whatever OLL algorithm has the permutation of more common PLL (for example if your OLL algorithm does a G-perm would you be more likely to get a PLL skip than if it did an H-perm?)


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## whauk (Feb 26, 2014)

Stefan said:


> Some OLLs in fact do give PLL skips more often than others (someone guess which).



I have no idea to explain this. Please enlighten us


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## cmhardw (Feb 26, 2014)

Stefan said:


> Some OLLs in fact do give PLL skips more often than others (someone guess which).





Jaycee said:


> -The only thing I can think of is whatever OLL algorithm has the permutation of more common PLL (for example if your OLL algorithm does a G-perm would you be more likely to get a PLL skip than if it did an H-perm?)



After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).


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## Carrot (Feb 26, 2014)

cmhardw said:


> After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).



yours would require learning multiple OLL's... If you take the 4 flipped edges you would due to symmetry be able to apply same alg from 4 different angles and therefore increase your chance of PLL skip by factor 4 

Also, it seems all PLLs have same probability if you take AUF into account (correct me if I'm wrong)


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## Stefan (Feb 26, 2014)

Spoiler: Why some OLLs give PLL skips more often than others



Because some OLLs occur more often than others.


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## TDM (Feb 26, 2014)

cmhardw said:


> After reading Jaycee's post my guess would be a particular X-OLL alg that does a G perm. If you assume that the solver knows the reflections/inverses of that alg as well, then s/he could skip PLL 2/9 the time that OLL came up using AUF and reflection/inverse of the OLL alg (when necessary).


If you're only using one alg, then you could use this and AUF before the alg to force a skip. If you're not including mirrors/inverses it doesn't have to be a G perm, just any PLL without rotational symmetry (N/H/Z).


Carrot said:


> Also, it seems all PLLs have same probability if you take AUF into account (correct me if I'm wrong)


Yeah, I think your right.


Stefan said:


> Spoiler: Why some OLLs give PLL skips more often than others
> 
> 
> 
> Because some OLLs occur more often than others.


Also a possible was of answering... although with Chris' method the least common OLLs, the ones with rotation symmetry, can be used to force a PLL skip more often, giving them an equal chance of a PLL skip, so you're not necessarily right.


----------



## Stefan (Feb 26, 2014)

TDM said:


> Also a possible was of answering... although with Chris' method the least common *PLLs*, the ones with rotation symmetry, can be used to force a PLL skip more often, giving them an equal chance of a PLL skip, so you're not necessarily right.



(I think you mean OLLs.) Yes, I very much like that idea. But I'd say you're doing more than just OLL then, as you're inspecting and reacting to the permutation. Also, the "solved OLL" case is one of those with rotation symmetry but you can't exploit that symmetry in that way.


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## Rubiks560 (Feb 26, 2014)

What is the probability of getting a PLL skip if you use OLLCP every solve?
And can you explain to me how you got the answer?

Nvm. 1/12 right?


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## kcl (Feb 26, 2014)

Rubiks560 said:


> What is the probability of getting a PLL skip if you use OLLCP every solve?
> And can you explain to me how you got the answer?



I'm tempted to say it's 1/12 because that's what it is with COLL and COLL is a subset..


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## TDM (Feb 26, 2014)

Stefan said:


> (I think you mean OLLs.) Yes, I very much like that idea. But I'd say you're doing more than just OLL then, as you're inspecting and reacting to the permutation. Also, the "solved OLL" case is one of those with rotation symmetry but you can't exploit that symmetry in that way.


Yeah, sorry I meant OLLs. And you can say that a G perm is a way of "solving the solved OLL", because technically it is as the OLL is solved after doing it  I see what you're saying though.


Rubiks560 said:


> What is the probability of getting a PLL skip if you use OLLCP every solve?
> And can you explain to me how you got the answer?
> 
> Nvm. 1/12 right?





kclejeune said:


> I'm tempted to say it's 1/12 because that's what it is with COLL and COLL is a subset..


Yes, the chance of an EPLL skip is 1/12. Chances of all EPLLs:
Ua: 4/12 = 1/3
Ub: 4/12 = 1/3
Z: 2/12 = 1/6
H: 1/12
Solved: 1/12


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## cmhardw (Mar 4, 2014)

*This will seem to be a strange question. It includes edits based on me having miscalculated the original question, and instead calculating the answer to a different scenario. Everything is now corrected, and the answer given is for the question posed.*

Here's a fun question: Fully scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. What is the probability that none of the 6 centers are oriented correctly (as they are in the solved state)?

Answer below:


Spoiler



\( \frac{365}{2048} \)

\( \frac{{6 \choose 0}*4^5*2-{6 \choose 1}*4^4*2+{6 \choose 2}*4^3*2-{6 \choose 3}*4^2*2+{6 \choose 4}*4*2-{6 \choose 5}*2+{6 \choose 6}}{4^5*2} \)



--edit--

And a cool followup question.

Fully scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. What is the expected number of correctly oriented centers?

Hint: There is a short way and a long way to calculate this. Both give the same answer.



Spoiler



Easy way: 6 centers each with probability \( \frac{1}{4} \) to be solved. \( 6*\frac{1}{4}=\frac{3}{2} \) or 1.5 solved centers expected.

Long way: (calculation done before hand)
Number of states with:
0 centers oriented: 365
1 center oriented: 726
2 centers oriented: 615
3 centers oriented: 260
4 centers oriented: 75
5 centers oriented: 6
6 centers oriented: 1

\( 0*\frac{365}{2048}+1*\frac{726}{2048}+2*\frac{615}{2048}+3*\frac{260}{2048}+4*\frac{75}{2048}+5*\frac{6}{2048}+6*\frac{1}{2048}=\frac{3}{2} \)


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## TDM (Mar 4, 2014)

cmhardw said:


> Here's a fun question: Fully scramble a 3x3x3 supercube. What is the probability that none of the 6 centers are oriented correctly (as they are in the solved state)?
> 
> Answer below:
> 
> ...


Sorry, but I don't understand this... surely it would be (3/4)^6? Our answers aren't far from each other, but I just don't see how it can be anything else.


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## cmhardw (Mar 4, 2014)

TDM said:


> Sorry, but I don't understand this... surely it would be (3/4)^6? Our answers aren't far from each other, but I just don't see how it can be anything else.



Wow, you're right. Hmm... Now I am wondering what exactly it is I calculated, especially since the case count still adds up to 2048.

Hmmm.... I'm at work now, but I'll think on that more when I get home. It could just be coincidence and I might have just calculated nonsense.


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## uberCuber (Mar 4, 2014)

Oh good, so it wasn't just me who was looking at that expression and having no idea where it came from :confused:


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## cmhardw (Mar 4, 2014)

TDM said:


> Sorry, but I don't understand this... surely it would be (3/4)^6? Our answers aren't far from each other, but I just don't see how it can be anything else.



I edited the original post. I was calculating a different scenario and didn't realize it. See the original post for the new question (to which the answer I gave is the right answer).


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## cmhardw (Mar 4, 2014)

This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:

*Question 1)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?

*Question 2)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?


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## TDM (Mar 4, 2014)

cmhardw said:


> This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:
> 
> *Question 1)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?
> 
> *Question 2)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?





Spoiler: Answer to Q1



0?


E: Just realised this is completely wrong... :fp


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## qqwref (Mar 4, 2014)

A possibly simpler answer to
"Here's a fun question: Fully scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. What is the probability that none of the 6 centers are oriented correctly (as they are in the solved state)?":



Spoiler



All centers are misoriented either 1, 2, or 3 turns clockwise. But the number of centers that are 2 turns clockwise must be even (i.e. 0, 2, 4, or 6); and there are two ways to choose each of the remaining centers. So out of 4^6/2 possibilities, this can be done in
\( {6 \choose 0}*2^6+{6 \choose 2}*2^4+{6 \choose 4}*2^2+{6 \choose 6}*2^0 = 365 \)
ways. Thus the probability is 365/2048.



And similarly, for
"Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?", we have


Spoiler



Again, all centers are misoriented either 1, 2, or 3 turns clockwise. But now the parity is odd, so the number of centers that are 2 turns clockwise must be odd (i.e. 1, 3, or 5); and there are two ways to choose each of the remaining centers. So out of 4^6/2 possibilities, this can be done in
\( {6 \choose 1}*2^5+{6 \choose 3}*2^3+{6 \choose 5}*2^1 = 364 \)
ways. Thus the probability is 364/2048 = 91/512.


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## cmhardw (Mar 4, 2014)

cmhardw said:


> This gives me an idea. I am double posting because I think this one deserves its own post. Here are two more fun questions:
> 
> *Question 1)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the probability that none of the centers are in their correct orientations?
> 
> *Question 2)* Fully Scramble a 3x3x3 supercube. Solve the corners and edges paying no particular attention to the orientations of the centers. Now perform one random quarter turn on an outer slice of the cube. What is the expected number of correctly oriented centers?



Before looking at TDM's answer I got:


Spoiler



Q1: \( \frac{364}{2048}=\frac{91}{512} \)

Q2: \( \frac{3}{2} \)



Spoiler



Q2:
\( 0*\frac{364}{2048}+1*\frac{732}{2048}+2*\frac{600}{2048}+3*\frac{280}{2048}+4*\frac{60}{2048}+5*\frac{12}{2048}+6*\frac{0}{2048} \)

It seems this can also be calculated by reasoning that the expected number of oriented centers is 1.5 for a randomly scrambled 3x3x3 supercube. If you fix center orientation parity to even then the expected value of oriented centers is 1.5 centers (see my earlier post). Therefore the expected number of oriented centers if you fix odd orientation parity must also be also be 1.5 centers since odd and even parity both have probability 1/2


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## Stefan (Mar 5, 2014)

About _"no centers solved"_: Ignoring edges/corners (i.e. ignoring parity) there are 3^6=729 cases with no centers solved. Fix an order of the six centers and interpret the cases as ternary numbers from 000000 to 222222. Adding 1 turns each even parity case except 222222 into an odd parity case and this is a one-to-one correspondence. So there's one more even parity case than there are odd parity cases, meaning the 729 split into 365 for even parity and 364 for odd parity.


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## DeeDubb (Mar 19, 2014)

*43 Quintilian Possibilities, how many are solved?*

This is kind of a weird question, but one that's been bugging me. I remember having a smiley face cube a long time ago, and I solved it, but, the centers were twisted upside down on 2 of the faces. I could never fix it, so it looked kind of goofy. If that were a standard cube with plain stickers, that would be solved. So I'm wondering how many solved possibilities there are among the 43 Quintilian possible permutations. If it's only one, then what about those other times when the centers might be twisted?

Also, I remember seeing special stickers for a cube that would simulate my smiley issue, where they had horizontal lines across all of the stickers, so not only would you have to solve, but also fix the twisted centers issue (and maybe other issues that I'm unaware of. First of all, what is that cube called, and does it have MORE than 43 quintilian possible combinations?


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## CyanSandwich (Mar 19, 2014)

I think the 43 quintillion. combinations don't include spun centers. I'm basing that on the fact that they don't change anything on a Rubik's cube and the equation doesn't take it into account.


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## DeeDubb (Mar 19, 2014)

CyanSandwich said:


> I think the 43 quintillion. combinations don't include spun centers. I'm basing that on the fact that they don't change anything on a Rubik's cube and the equation doesn't take it into account.



Yeah, that's what I was thinking. So, 43 quintillion and only 1 solved. I guess guess rubik's cubes that aren't solved with spun centers must have much more than 43 quint.


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## DeeDubb (Mar 19, 2014)

cmowla said:


> You guys are absolutely correct. In fact, there are (click me) solved positions for the 2x2x2 through 11x11x11 cubes, for example. The cubes which have stickers on the centers which makes them distinguishable (big cube centers) or helps you tell if they are rotated (fixed centers, like those on the 3x3x3) are called supercubes. So the answer to your question is 2048 (although 2048 is not counted as part of the 43 quintillion).



Thank you. That will help me sleep at night :tu


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## cmhardw (Mar 19, 2014)

DeeDubb said:


> This is kind of a weird question, but one that's been bugging me. I remember having a smiley face cube a long time ago, and I solved it, but, the centers were twisted upside down on 2 of the faces. I could never fix it, so it looked kind of goofy. If that were a standard cube with plain stickers, that would be solved. So I'm wondering how many solved possibilities there are among the 43 Quintilian possible permutations. If it's only one, then what about those other times when the centers might be twisted?



If you've ever wondered the same thing about the 20x20x20 cube, you can check out how many ways there are to rearrange a solved 20x20x20 cube and it still appear solved.


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## Rocky0701 (Mar 19, 2014)

Yesterday i got 3 Z perms in a row. Probability: 1/46656, but i think something was just wrong with my scramble generator or something.


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## Kit Clement (Mar 19, 2014)

Rocky0701 said:


> Yesterday i got 3 Z perms in a row. Probability: 1/46656, but i think something was just wrong with my scramble generator or something.



There's also just lot of improbable events in cubing, and if you do enough solves, something improbable has to happen eventually.


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## Rocky0701 (Mar 19, 2014)

Kit Clement said:


> There's also just lot of improbable events in cubing, and if you do enough solves, something improbable has to happen eventually.



Yeah, that is true, but it was still really weird.


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## Saxsoprano (Apr 2, 2014)

If I always use phasing (with ZZ), and after I use COLLs taken from ZZLLs, wich is the probability of a PLL skip? Thanks!


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## TDM (Apr 2, 2014)

Saxsoprano said:


> If I always use phasing (with ZZ), and after I use COLLs taken from ZZLLs, wich is the probability of a PLL skip? Thanks!


You're preserving phasing, so the relative probabilities of the outcomes are the same as the probabilities of the EPLLs without including U perms.
H perm: 1/12 -> 1
Z perm: 2/12 -> 2
Solved: 1/12 -> 1
Chance of skip = 1/4


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## Saxsoprano (Apr 4, 2014)

Ok, thanks!!


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## Ollie (Apr 7, 2014)

Just a silly thing - this weekend I got 3 PLL skips in three different rounds, all from the same alg x U R' U' L' U R U' r'?. I think I managed to predict the latter 2 skips! Any idea what the probability of this is from the 27 sighted solves I did this weekend? (Ignoring 2x2x2 because I know CLL)


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## guysensei1 (Apr 7, 2014)

What is the probability of a solved face on 2x2? 
What about a solved layer?


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## thejerber44 (Apr 27, 2014)

How many combinations are possible with a 2-gen scramble?


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## TheNextFeliks (Apr 27, 2014)

thejerber44 said:


> How many combinations are possible with a 2-gen scramble?



I think it should be (7!/2)*((3^6)/3)= 612,360 but not sure.


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## cmhardw (Apr 27, 2014)

thejerber44 said:


> How many combinations are possible with a 2-gen scramble?



\( 7!*\frac{6!}{3!*2}*3^5=73483200 \)

You may find this page interesting.


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## thejerber44 (Apr 27, 2014)

cmhardw said:


> \( 7!*\frac{6!}{3!*2}*3^5=73483200 \)
> 
> You may find this page interesting.



Thanks for the quick reply  The website is awesome! A friend told me that he was doing some 2-gen solves (hand scrambles) and when he finished scrambling, the cube was a mere U2 from being solved. He's either really lucky, bad at scrambling, or lying


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## cmhardw (Apr 29, 2014)

thejerber44 said:


> Thanks for the quick reply  The website is awesome! A friend told me that he was doing some 2-gen solves (hand scrambles) and when he finished scrambling, the cube was a mere U2 from being solved. He's either really lucky, bad at scrambling, or lying



Yeah, odds are in favor of a poor scramble or a fib  However, the truth is still possible if your friend got very lucky!


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## StarOfDoom (May 2, 2014)

What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)


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## kcl (May 2, 2014)

StarOfDoom said:


> What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)



Depends if you use any influencing of last slot, WV, COLL, etc.


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## Jaycee (May 2, 2014)

This is just my attempt at finding an answer to the question. I assume one of people more knowledgeable about this will give the correct at some point, but I think I'll try anyway.

Chance of an LL skip on any given single solve: 1/15552 = 0.0000643004115

Chance of not getting an LL skip on any given single solve: 15551/15552 = 0.9999356995885

3 solves had an LL skip out of 17000: (3(0.0000643))/17000 = something google can't calculate so I'll do this next thing instead

16997 solves did not have an LL skip: (16997(0.9999356995885)/17000 = 16995.9071 / 17000 (apparently google doesn't wanna go out that many decimal places)

16995.9071/17000 = chances of having 16997 non-LL-skips in 17000 solves = 99.9759241%

Inverse of that = *0.0240759% chance of having 3 LL skips in 17000 solves.*

This is all assuming, of course that no influencing methods are used, which is probably not true. Oh well.

7 minutes later edit: Maybe not? Both things happened, (3 LL skips and 16997 non-LL-skips), so I think the probabilities get multiplied. In which case

2.40701035% chance of 3 LL skips in 17000 solves? Maybe. Eh. I stopped making sense a while ago.


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## uberCuber (May 2, 2014)

I believe this should just be a binomial probability, right?

(17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.


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## GuRoux (May 2, 2014)

uberCuber said:


> I believe this should just be a binomial probability, right?
> 
> (17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.



That seems correct.


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## StarOfDoom (May 2, 2014)

Jaycee said:


> This is all assuming, of course that no influencing methods are used, which is probably not true. Oh well.



I do not do any influencing. I have enough trouble just finding and getting the f2l pairs inserted in the short amount of time, let alone partial edge control... so I don't think any influencing was involved.


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## Jaycee (May 2, 2014)

Ah. In any case my math was wrong and uberCuber's was correct. Don't know why I was adding probabilities instead of multiplying them, that was pretty derp.


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## LucidCuber (May 7, 2014)

Does anyone know the probability of getting 4 edges skipped after centres on a 4x4x4?


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## cmhardw (May 7, 2014)

LucidCuber said:


> Does anyone know the probability of getting 4 edges skipped after centres on a 4x4x4?



The probability that at least 4 edge groups are paired after completing centers is (assuming you are not using Yau or any other approach to solve edges during the solving of centers):

P(at least 4 edge groups paired)=1-P(no edge groups paired)-P(1 edge group paired)-P(2 edge groups paired)-P(3 edge groups paired)

P(at least 4 edge groups paired) = \( 1-\frac{1}{24!}\sum_{i=0}^{3} \sum_{j=0}^{12-i} \binom{12}{i}*\frac{12!}{(12-i)!}*2^i*\binom{12-i}{j}*\frac{(12-i)!}{(12-i-j)!}*2^j*(-1)^{j}*(24-2i-2j)! \)

*This is an approximately 0.208% chance.*

In fraction form that would be about \( \frac{1}{481} \)

For those interested in the details, here is the distribution that this creates:


Spoiler



Each number represents the total number of all wing edge states that have the described characteristic:

exactly 0 edge groups paired: 368420070161105761075200
exactly 1 edge group paired: 191842564893283713024000
exactly 2 edge groups paired: 50126103570256861593600
exactly 3 edge groups paired: 8768672584653864960000
exactly 4 edge groups paired: 1156363172135829504000
exactly 5 edge groups paired: 122770256129792409600
exactly 6 edge groups paired: 10949790800019456000
exactly 7 edge groups paired: 845319706037452800
exactly 8 edge groups paired: 58271119441920000
exactly 9 edge groups paired: 3453103374336000
exactly 10 edge groups paired: 258982753075200
exactly 11 edge groups paired: 0
exactly 12 edge groups paired: 1961990553600

Notice that the summation correctly predicts that it is not possible to have exactly 11 edge groups paired, since having 11 edge groups paired means that the 12th is also paired.

Also notice that for exactly 12 edge groups paired, which has 1961990553600 positions that fit this characteristic, that this number is:
1961990553600=12!*2^12 which is exactly what we would expect.

Also notice that all of the numbers above sum to 620448401733239439360000 which is equal to 24!, which is the total number of wing edge permutations on the 4x4x4.


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## LucidCuber (May 7, 2014)

Thanks for that. I haven't had a LL skip in a year and a half so I was due some luck.


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## cmhardw (May 7, 2014)

LucidCuber said:


> Thanks for that. I haven't had a LL skip in a year and a half so I was due some luck.



Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong  I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.

The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.

See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.


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## Christopher Mowla (May 7, 2014)

cmhardw said:


> Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong  I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.
> 
> The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.
> 
> See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.


Interesting post Chris!

This reminded me of something. Knowing what you know about the structure of the 24! permutations of wing edges, would it be easy for you to confirm/find a counterexample to condition [2] in my conjecture in this post? The motivation for this, of course, is to be able to claim that at most 6 inner slices are required to pair the dedges (if the permutation is even. I didn't think it through, but it also might be true for odd permutations as well, but the purpose of it being an even permutation is, of course, for fewer moves).

I've been wanting to start a thread in puzzle theory about this conjecture, but I have been too busy with you know what. But if you might be able to confirm/prove condition [2] always can be met, then I definitely want to start a thread so that you can post your proof there!


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## LucidCuber (May 7, 2014)

cmhardw said:


> Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong  I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.
> 
> The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.
> 
> See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.




Would it be possible to get each case (0 edges 1,2,3 edges etc) with a fractional probability? Those numbers are to big to comprehend at that level


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## cmhardw (May 7, 2014)

LucidCuber said:


> Would it be possible to get each case (0 edges 1,2,3 edges etc) with a fractional probability? Those numbers are to big to comprehend at that level



Sure.

*For clarity, all stats below are for the 4x4x4:*

Probability of no edge groups paired after centers: \( \approx \frac{59}{100} \)

Probability of at least 1 edge group paired after centers: \( \approx \frac{41}{100} \)

Probability of at least 2 edge groups paired after centers: \( \approx \frac{1}{10} \)

Probability of at least 3 edge groups paired after centers: \( \approx \frac{1}{62} \)

Probability of at least 4 edge groups paired after centers: \( \approx \frac{1}{481} \)

Probability of at least 5 edge groups paired after centers: \( \approx \frac{1}{4609} \)

Probability of at least 6 edge groups paired after centers: \( \approx \frac{1}{52327} \)

Probability of at least 7 edge groups paired after centers: \( \approx \frac{1}{683837} \)

Probability of at least 8 edge groups paired after centers: \( \approx \frac{1}{10009627} \)

Probability of at least 9 edge groups paired after centers: \( \approx \frac{1}{167054487} \)

Probability of at least 10 edge groups paired after centers: \( \frac{1}{2377700325} \)

Probability of all 12 edge groups paired after centers: \( \frac{1}{316234143225} \)


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## Filipe Teixeira (May 7, 2014)

whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?


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## cmhardw (May 7, 2014)

filipemtx said:


> whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?



On any side (color neutral)? On either of 2 opposite sides (opposite color neutral)? On only 1 side (no color neutrality)?


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## Filipe Teixeira (May 7, 2014)

on one side


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## cmhardw (May 7, 2014)

filipemtx said:


> whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?





filipemtx said:


> on one side



\( \frac{23}{12!*2^{11}} = \frac{23}{980995276800} \)

--edit--
This calculation is incorrect, and was corrected by Daniel Sheppard here.


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## Filipe Teixeira (May 7, 2014)

thanks! one more thing, how much percent is that, please?


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## Rocky0701 (May 8, 2014)

filipemtx said:


> thanks! one more thing, how much percent is that, please?


You couldn't of just typed in: 23 divided by 980995276800 into Google? Anyway, it is 2.34455767E-11% of a chance. AKA a .00000000234455767% chance of happening.


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## Filipe Teixeira (May 8, 2014)

Rocky0701 said:


> You couldn't of just typed in: 23 divided by 980995276800 into Google? Anyway, it is 2.34455767E-11% of a chance. AKA a .0000000000234455767% chance of happening.



thanks I didn't know how to calcute the percentage though


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## 10461394944000 (May 8, 2014)

filipemtx said:


> thanks I didn't know how to calcute the percentage though



...just multiply it by 100


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## Carrot (May 8, 2014)

filipemtx said:


> thanks I didn't know how to calcute the percentage though



jeez... percentage... derives from 'per cent' (kind of ish) which roughly means 'hundredths'. so basically multipy any number by 100 and you have the number represented as percentage. (0.25 = 25%, 0.01 = 1%, 25 = 2500%) It's THAT simple

EDIT: Ben, y u ninja?


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## Filipe Teixeira (May 9, 2014)

> jeez... percentage... derives from 'per cent' (kind of ish) which roughly means 'hundredths'. so basically multipy any number by 100 and you have the number represented as percentage. (0.25 = 25%, 0.01 = 1%, 25 = 2500%) It's THAT simple



sorry, I'm dumb


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## kinch2002 (May 9, 2014)

StarOfDoom said:


> What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)





uberCuber said:


> I believe this should just be a binomial probability, right?
> (17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.



Indeed, but your question is needlessly specific, resulting in an answer that exaggerates your true 'luckiness'. Less misleading would be the probability of getting *at least* 3 LL skips in 17k solves, which is 9.82%


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## Robert-Y (May 10, 2014)

^People need to stop forgetting this 

On a related note... I was wondering how unlucky you have to be to get x solves in row without a LL skip. Here are the probabilities that you will receive at least one LL skip using the standard CFOP method without any special tricks, for certain x number of solves in a row:

100 solves: 1 - (15551/15552)^100 ~= 0.64%
500 solves: 1 - (15551/15552)^500 ~= 3.16%
1000 solves: 1 - (15551/15552)^1000 ~= 6.23%
5000 solves: 1 - (15551/15552)^5000 ~= 27.5%
10000 solves: 1 - (15551/15552)^10000 ~= 47.43%
50000 solves: 1 - (15551/15552)^50000 ~= 95.98%
100000 solves: 1 - (15551/15552)^50000 ~= 99.84%

So you could say that if you've done at least 50000 solves in your lifetime, you have probably received at least one LL skip hopefully 
But if not, it doesn't mean you're *that* unlucky since the chances that you haven't received one in 50000 solves is ~= 4% or 1/25.


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## advincubing (May 29, 2014)

What is the probability of the last F2L corner + edge set being already paired in the top layer, without manipulating the earlier inserts to force pairing? (Relates to question of how often partial edge control -- sledgehammer vs. RU'R' and their mirrors -- can be applied through the last insert.) The calculation is easy enough if someone can point me to a summary of F2L case probabilities....

Thanks.


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## Tempus (May 29, 2014)

advincubing said:


> What is the probability of the last F2L corner + edge set being already paired in the top layer, without manipulating the earlier inserts to force pairing?


Assuming I understand you correctly, there should be five corners and five edges in an indeterminate state. As for the corner piece, there's a 4/5 probability that the relevant corner is in the top layer and a 2/3 probability that the cross color on the relevant corner isn't facing up. As for the edge piece, there's now a 1/5 probability that it's against the correct face of the corner piece and a 1/2 probability that it's oriented correctly.

Multiply them all together and you get p=4/75, or about a 5.333% chance that they are matched up together in the top layer.


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## FailCuber (May 29, 2014)

What's the probablity of getting a PLL skip on a 2x2?


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## ryanj92 (May 29, 2014)

FailCuber said:


> What's the probablity of getting a PLL skip on a 2x2?



1 in 6, as there are only 6 possible permutations of the top layer  (front/back/left/right adjacent swap, diagonal swap and the solved case)


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## tseitsei (May 29, 2014)

FailCuber said:


> What's the probablity of getting a PLL skip on a 2x2?



There are 4! = 24 possible permutations of LL pieces so 

AUFless skip is only 1 permutation of those so 1/24
With AUF 4/24=1/6


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## advincubing (May 29, 2014)

Tempus said:


> Assuming I understand you correctly, there should be five corners and five edges in an indeterminate state. As for the corner piece, there's a 4/5 probability that the relevant corner is in the top layer and a 2/3 probability that the cross color on the relevant corner isn't facing up. As for the edge piece, there's now a 1/5 probability that it's against the correct face of the corner piece and a 1/2 probability that it's oriented correctly.
> 
> Multiply them all together and you get p=4/75, or about a 5.333% chance that they are matched up together in the top layer.



Thank you for the response. That methodology makes perfect sense, but the result seems surprisingly low. I feel like I get a made pair in the top layer more like 25-40% of the time. I wonder if the difference is due to something I didn't think to address/describe in my question: Often the last F2L edge and corner are setup so that they can be paired before being inserted. This would be a case like F2L #5 where you would do (U' R U R') to pair the edge and corner in the U layer, AUF, then have the option to insert with R U' R' or a hedgeslammer. Those cases are in contrast to the ones that are solved and inserted in one fell swoop with, say, R U R'.

That complicates things quite a bit, I know. Not sure the best methodology for calculating with that nuance....


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## goodatthis (May 29, 2014)

Okay, kind of a complicated question here, but is there any difference between the probability of getting a PLL skip when you use OLL (non edges oriented) vs OCLL? Basically what I mean is, if a PLL skip is just a 1LLL case (or ZBLL) that you happened to get as your OLL, and since OLLs permute pieces as well, do non edges oriented OLLs permute in a more common pattern than OCLLs? (Or less common?) 

For example, let's say that a certain OLL permutes in a U perm like fashion. U perms are common, so another OLL that permutes in an H perm (less common) like fashion should be less likely to cause an PLL skip, right? I'm sorry if I'm a noob at this kind of thing, haha.


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## ryanj92 (May 29, 2014)

goodatthis said:


> Okay, kind of a complicated question here, but is there any difference between the probability of getting a PLL skip when you use OLL (non edges oriented) vs OCLL? Basically what I mean is, if a PLL skip is just a 1LLL case (or ZBLL) that you happened to get as your OLL, and since OLLs permute pieces as well, do non edges oriented OLLs permute in a more common pattern than OCLLs? (Or less common?)
> 
> For example, let's say that a certain OLL permutes in a U perm like fashion. U perms are common, so another OLL that permutes in an H perm (less common) like fashion should be less likely to cause an PLL skip, right? I'm sorry if I'm a noob at this kind of thing, haha.



No, because during OLL the permutation of your LL pieces is generally random, so the chance of the OLL and OCLL actually causing a PLL skip is pretty slim 
Maybe the nuances you mention have some minute effect on the probabilities - however, it will definitely be minute and calculating the probabilities won't give us any profound results

Also, not everybody uses the same OLL's or even OCLL's anyway, so the probabilities will vary from person to person


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## qqwref (Jul 8, 2014)

a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?
b) Same as a, but the 6 faces are a random permutation of the 6 colors.
c) How about on a 3x3x3?
d) How about on a 4x4x4?


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## IRNjuggle28 (Jul 9, 2014)

qqwref said:


> a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?
> b) Same as a, but the 6 faces are a random permutation of the 6 colors.
> c) How about on a 3x3x3?
> d) How about on a 4x4x4?



I lack the knowledge to answer these, sadly. My thoughts for A are that the ones where it could be identified with UFRL are the ones where the UFR and UFL corners don't share any colors. If, for example, the blue/yellow/orange corner and the white/green/red corner were the two UF corners, other corners could be determined from only seeing two colors because, for example, if the DFL corner had blue on L and yellow on F, the other color has to be red because there are only two pieces with blue and yellow, and one of them already has all 3 colors visible on UFRL. The less variety of UFR and UFL colors, the harder it'll be to identify. It wouldn't shock me if the answer is 100% and I simply can't see how it would be done, though.

I doubt I said anything you don't already know.  

B is stated a bit unclearly. Can you clarify what you're asking? 

I have no idea about C. I feel confident that D is 0%, since the centers on B and D are completely undetectable, and could be in any configuration. Even if the cube looked completely solved only looking at UFRL, it could still be not solved.


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## kinch2002 (Jul 9, 2014)

filipemtx said:


> whats the probability that the 4 cross edges are oriented but not permuted, ie all on D layers but not correctly permuted?





filipemtx said:


> on one side





cmhardw said:


> \( \frac{23}{12!*2^{11}} = \frac{23}{980995276800} \)


Just been reading through recent posts in this thread and I think for once Chris is wrong 
23 = Number of ways the 4 cross pieces can be arranged to create the case we are interested in
12! * 2 ^ 11 = Number of different edge configurations
I believe for the second number you actually want 24*22*20*18 = 190080 = Number of different configurations for those 4 edges
This gives 23/190080 = 0.0121%


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## cmhardw (Jul 9, 2014)

kinch2002 said:


> Just been reading through recent posts in this thread and I think for once Chris is wrong
> 23 = Number of ways the 4 cross pieces can be arranged to create the case we are interested in
> 12! * 2 ^ 11 = Number of different edge configurations
> I believe for the second number you actually want 24*22*20*18 = 190080 = Number of different configurations for those 4 edges
> This gives 23/190080 = 0.0121%



That's true, I see that now.

Using the approach I used, my first calculation should have been:
\( \frac{23*8!*2^7}{12!*2^{11}}=\frac{23}{12*11*10*9*2^4}=\frac{23}{190080} \)

Good catch!


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## uberCuber (Jul 9, 2014)

IRNjuggle28 said:


> If, for example, the blue/yellow/orange corner and the white/green/red corner were the two UF corners, other corners could be determined from only seeing two colors because, for example, if the DFL corner had blue on L and yellow on F, the other color has to be red because there are only two pieces with blue and yellow, and one of them already has all 3 colors visible on UFRL.



If you can see two of the stickers on a corner piece, knowing your color scheme will automatically tell you the third color; looking at other corners is unnecessary. DBL and DBR are the only potentially ambiguous pieces because you can only see one sticker of each.


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## IRNjuggle28 (Jul 10, 2014)

uberCuber said:


> If you can see two of the stickers on a corner piece, knowing your color scheme will automatically tell you the third color; looking at other corners is unnecessary. DBL and DBR are the only potentially ambiguous pieces because you can only see one sticker of each.



Very true. I'm thinking that with each DB corner, you have a 1/3 chance of the color showing that differentiates the corners, so if the two corners at DBL and DBR were a pair of corners that shared two colors with each other, there would be a 4/9 chance that neither corner would have that color showing. So, it may be that (odds of DBL and DBR corners sharing 2 colors)X(4/9) is the answer. I'm guessing it's more complex, though.

Well, the answer is not 100%. I have found two permutations that cannot be differentiated from each other by UFLR. A pure flip T case versus a T case with the two corners that are not oriented swapped. Algs below.

R U2 R' U' R U' R' L' U2 L U L' U L y' x
cannot be differentiated from
R U R' U' R' F R2 U' R' U' R U R' F' R U2 R' U' R U' R' L' U2 L U L' U L y' x


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## Tempus (Jul 10, 2014)

qqwref said:


> a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?


I think that this is 16/21, or about 76.19%, but I'm not completely certain that I haven't missed some esoteric parity complication. Needs more thought...



qqwref said:


> b) Same as a, but the 6 faces are a random permutation of the 6 colors.


I've been thinking about this one for a while, and it seems that the more I think about it, the more complex the question turns out to be. Every time I think I'm beginning to get a handle on it, I discover some new level of complexity behind the previous one.



qqwref said:


> c) How about on a 3x3x3?


I'm not even going to touch this one.



qqwref said:


> d) How about on a 4x4x4?


I'm pretty sure that IRNjuggle28 is correct about this one being 0%.


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## Phillip1847 (Jul 10, 2014)

IRNjuggle28 said:


> Well, the answer is not 100%. I have found two permutations that cannot be differentiated from each other by UFLR. A pure flip T case versus a T case with the two corners that are not oriented swapped. Algs below.
> R U2 R' U' R U' R' L' U2 L U L' U L y' x
> cannot be differentiated from
> R U R' U' R' F R2 U' R' U' R U R' F' R U2 R' U' R U' R' L' U2 L U L' U L y' x


the red sticker on UB?


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## uberCuber (Jul 10, 2014)

Phillip1847 said:


> the red sticker on UB?



The discussion is about 2x2.


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## Phillip1847 (Jul 10, 2014)

uberCuber said:


> The discussion is about 2x2.



Sorry, I assumed he was addressing the 3x3 question as the algs are more 3x3.


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## Tempus (Jul 10, 2014)

Tempus said:


> I think that this is 16/21, or about 76.19%, but I'm not completely certain that I haven't missed some esoteric parity complication. Needs more thought...


Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.

Given a scrambled 2x2 that is stickered the standard way, one can see all three sides of the UFL and UFR pieces, so we know their permutation and orientation. Similarly, we see two sides each of the UBR, DFR, UBL, and DFL pieces, and since we know that the color scheme is standard, we can determine the permutation and orientation of them as well. This leaves only the DBL and DBR pieces, of which we only see one sticker each. Since we know by process of elimination what pieces occupy those two positions, we need only figure out their permutation (2 possibilities) and orientation (3 possibilities taking parity into account).

There are three main scenarios for these two semi-hidden pieces. The first is that they are corners that share no colors, i.e. opposite corners. Each piece only has one such counterpart out of 7 available, so this scenario has a 1/7th chance of occurring. In this case, knowing the parity state of the rest of the cube, seeing one sticker is enough to determine both the orientation and permutation of both pieces. We can see two stickers, so in this scenario there is zero chance of confusion.

The second scenario is that the two pieces share exactly one color, i.e. they are corners that belong on the same face but are otherwise opposite. Each piece has 3 such counterparts out of 7 available, so this scenario has a 3/7ths chance of occurring. In this case, any sticker that isn't the one color they share will instantly give the game away, so only the orientation where the common color is exposed on both pieces will work. In 2/3rds of these cases, parity will make this arrangement impossible, and in the remaining third, there is a 1/3rd chance that the common color will be the one exposed. (1/3)*(1/3)*(3/7)=1/21.

The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.

This leaves a third and final parity state where either A and A can be in view, or B and B can be in view, or the distinct colors can be in view, with equal probability. A and A will confuse, as will B and B, and so there is a 2/3rds chance of confusion, but only a 1/3rd chance of this parity state. Multiply in the probability of having exactly two colors in common, and you get (2/3)*(1/3)*(3/7)=2/21.

Add them all up and you get 0+1/21+2/21=1/7. A 1 in 7 chance of the cube state being confusable with another, or a 6/7 chance of it being uniquely identifiable merely by viewing the U, F, L, and R sides.

Mind you, this is just for question A. Question B, which I believe allows for non-standard color schemes, is one I am unprepared to answer.


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## elrog (Jul 10, 2014)

Tempus said:


> I'm pretty sure that IRNjuggle28 is correct about this one being 0%.



But you could have a single dedge flip on the DB edge? There are also different ways you can move the centers. 

I think it would be interesting to see these stats for both regular cubes and super cubes.


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## goodatthis (Jul 10, 2014)

What is the probability of a one move XCross on white/yellow cross? Just white? Full color neutral?

I actually had a scramble like that on the third solve I ever did on TTW. Unfortunately I copied the scramble down but lost it.


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## guysensei1 (Jul 10, 2014)

How many L6E cases are there? 100s? 1000s?


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## TDM (Jul 10, 2014)

guysensei1 said:


> How many L6E cases are there? 100s? 1000s?


Corners (AUF): 4
Centres: 4
EO: 2^5 = 32
EP: 6!/2 = 360

4*4*360*32 = 184320


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## guysensei1 (Jul 10, 2014)

TDM said:


> Corners (AUF): 4
> Centres: 4
> EO: 2^5 = 32
> EP: 6!/2 = 360
> ...


Whoa.

Didn't expect that.


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## TDM (Jul 10, 2014)

guysensei1 said:


> Whoa.
> 
> Didn't expect that.


That's why LSE skips don't happen often  If you make sure the centres are either solved or a M2 away from being solved before/during CMLL, like I do, then it goes to under 100,000 (92160 to be precise). LSE skips still aren't going to be happening very often.


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## guysensei1 (Jul 10, 2014)

TDM said:


> That's why LSE skips don't happen often  If you make sure the centres are either solved or a M2 away from being solved before/during CMLL, like I do, then it goes to under 100,000 (92160 to be precise). LSE skips still aren't going to be happening very often.



If we defined an LSE skip to include the U/U'/U2 or M/M'/M2 away from solved cases, what's the probability then?
(This should go in the probability thread but meh)


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## Renslay (Jul 10, 2014)

guysensei1 said:


> How many L6E cases are there? 100s? 1000s?



Assuming you fix the corners and centers AUF (so you have to deal only with the edges):

Permutation: 6!/2 = 360
Orientation: 2^5 = 32

Which means 11520 cases.

Edit: okay, didn't see TDM's reply... Which is the same, but together with AUFs (which I think can be a bit misleading: the number of PLLs is also counted without AUFs).


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## TDM (Jul 10, 2014)

Renslay said:


> Assuming you fix the corners and centers AUF (so you have to deal only with the edges):
> 
> Permutation: 6!/2 = 360
> Orientation: 2^5 = 32
> ...


Yeah, I should probably have thought to not include AUF. So with the M slice solved it's more likely to get a LSE skip than to get a LL skip in CFOP, but if you have it either solved or an M2 out then it's less likely than a LL skip.


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## cmhardw (Jul 10, 2014)

qqwref said:


> a) What's the probability that a random 2x2x2 position can be completely identified by only seeing the U, F, R, and L faces?



My attempt before reading other responses:


Spoiler



My answer: \( \frac{16}{21} \) or approximately 76.2% chance.

We can completely identify all corners of the U layer as well as DFL and DFR. We cannot identify the corners at DBL or DBR if:
1) Their stickers share exactly 2 colors, and _either_ of those shared colors from _both_ corners is showing at *L*DB and *R*DB
2) Their stickers share exactly 1 color, and that color is the one showing at *L*DB and *R*DB.


Let's compute the probability that the DBL and DBR corners are unidentifiable:

Case1:
There are 12 possible choices of a pair of corners that share 2 colors on their sets of stickers (i.e. corners that are adjacent on an edge of the cube). Call the corners A and B. Let's say that colors A_1 and B_1 are the same, A_2 and B_2 are the same, and A_3 and B_3 are opposite colors.

If either of A_3 or B_3 is visible at *L*DB or *R*DB then both corners are identifiable. If neither of A_3 or B_3 is visible then the corners are ambiguous.

There are 4*2=8 ways to permute those corners such that neither A_3 nor B_3 is visible. Since the orientations of the corners must be a legal state, then after permuting the first corner (4 ways) there is only a 2/3 probability that the corner orientation doesn't _force_ the last corner to show the identifiable color. Also, the last sticker doesn't have two choices anymore, only 1 since it is forced.

The probability of such a case is:
\( \frac{12}{\binom{8}{2}}*\frac{4*\frac{2}{3}}{6}=\frac{4}{21} \)

Case 2:
There are 12 pairs of corners that share exactly one color in their sets of stickers. These are corners on a diagonal of a face (for example UBR and UFL).

Call the corners A and B. Stickers A_1 and B_1 are the same color. Stickers A_2 and B_2 are opposite colors, and so are stickers A_3 and B_3.

If any of the stickers A_2, B_2, A_3, B_3 are visible at *L*DB and *R*DB, then both corners are identifiable. If both stickers A_1 and B_1 are visible, then the corners are ambiguous.

There are 2 ways for both stickers to be visible. Since the corner orientations must be in a legal state, we can permute the first sticker in 2 ways, but then there is only a 1/3 probability that the other unidentifiable sticker can be legally permuted to the visible location.

The probability of such a case is:
\( \frac{12}{\binom{8}{2}}*\frac{2*\frac{1}{3}}{6}=\frac{1}{21} \)

Quick note:
If the two corners at DBL and DBR are corners that belong in diagonally opposite corners across the cube, then they are always identifiable at DBL and DBR

Adding the two case probabilities we get:
\( \frac{4}{21}+\frac{1}{21}=\frac{5}{21} \)

Therefore the probability that the 2x2x2 *is* identifiable is \( \frac{16}{21} \) or approximately 76.2% chance.


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## cmhardw (Jul 10, 2014)

Tempus said:


> Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.



Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?



Tempus said:


> The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, *but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.*



I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.

Am I not following your argument? Am I making a mistake in my logic somewhere?


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## Tempus (Jul 11, 2014)

cmhardw said:


> Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?


I'm talking about orientation parity. Perhaps I am using the wrong term, since true parity should only be either odd or even, whereas corner orientation is tristate, but I don't know of a better term to use, so I'm sticking with that one for now. If you turn just one corner clockwise 120 degrees, you reach an illegal cube state. If you then turn any other corner counter-clockwise 120 degrees, the cube state is made legal again. This is what I'm talking about.



cmhardw said:


> I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.
> 
> Am I not following your argument? Am I making a mistake in my logic somewhere?


Okay, let me give some examples. I have here a solved 2x2 with yellow on top, blue in front, and red on the right. Grab your 2x2 and you can follow along at home. For the purposes of this discussion, X will mean the left side sticker of the DBL corner, and Y will mean the right side sticker of the DBR corner.

As I look at the solved cube, X is orange and Y is red, and since I know the DB corners contain only one red sticker and one orange sticker, I know the full state of the cube without looking at the D or B faces.

Now apply this *swap algorithm*: U F U' F D R D2 R F D2 F. It swaps the DB corners while preserving which stickers are exposed at X and Y.
The result is that now X is red and Y is orange. This is still a uniquely identifiable state.

Now apply this *rotation algorithm*: R F' D R' D R D2 F2 R' D'. It rotates the DBR corner clockwise 120 degrees and the DBL corner counterclockwise 120 degrees. (Between these two algorithms, we can bring about any of the six legal states of those two corners without affecting the rest of the cube.)

As we look at the cube, X and Y are both green. If we were to execute the above swap algorithm again, they would still both be green, and so this state is not uniquely identifiable.

Now execute the rotation algorithm again. X and Y are now both white, and the same argument applies. Because we can use the swap algorithm to reach a new state that looks the same, this position is also confusable. So far, we have operated entirely inside one orientation parity state of the DB corner pair.

Now we will execute a new algorithm which I will call the *parity algorithm*: R U2 R' U R2 U' R2 U R' U2. This one changes the parity state of the DB corners by twisting the UFR corner. This change in parity is readily observable without viewing the B or D faces.

If you've followed along properly, you should see that X is now white and Y is now green. White exists on both of the DB corners, and green exists on both of the DB corners. But I assert that this state is still uniquely identifiable without viewing the D or B faces. To demonstrate why, we will cycle through all six possible DB corner pair states for the current orientation parity state:


*State**X**Y**Notes*#1WhiteGreenThis is the starting state. Now perform the *rotation* algorithm.#2RedWhiteNow perform the *rotation* algorithm again.#3GreenOrangeNow perform the *swap* algorithm.#4OrangeGreenNow perform the *rotation* algorithm again.#5WhiteRedNow perform the *rotation* algorithm one last time.#6GreenWhiteThis is the sixth and final state.

Notice that each of these six states has a distinct set of values for X and Y. This means that if you see X is white and Y is green, you know that the DB corners are in state #1, whereas if you see that X is green and Y is white, you know that they are in state #6. There is no ambiguity here because of parity.

In order to reach the other White/Green states, you would have to change the orientation of one of the other six corners, which would be visible without viewing the D or B faces, and so the observer, understanding parity perfectly, would be able to figure out what you had done and would still not be fooled. This is why the answer to question A is 7/6 (~85.714%) and not 16/21 (~76.19%).

Does this sufficiently answer your question?


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## IRNjuggle28 (Jul 11, 2014)

elrog said:


> But you could have a single dedge flip on the DB edge? There are also different ways you can move the centers.
> 
> I think it would be interesting to see these stats for both regular cubes and super cubes.



What you said only supports the conclusion that you can determine a 4x4 permutation from UFRL 0% of the time. Yes. On each solve, there is the possibility of the DB edge either being flipped, or not. Ergo, there will never be a 4x4 permutation that can be identified without looking at the DB edge. Ergo, there are no 4x4 permutations that can be identified from seeing the U, F, L, and R faces.


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## cmhardw (Jul 11, 2014)

Tempus said:


> I'm talking about orientation parity.
> 
> ...
> 
> Does this sufficiently answer your question?



Excellent analysis! Thanks for writing this out, that was very clear!


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## Tempus (Jul 11, 2014)

cmhardw said:


> Excellent analysis! Thanks for writing this out, that was very clear!


You are most welcome. Perhaps the reason that this one was clearer is that I wrote it while fully awake, whereas my previous attempt was written while falling asleep at the keyboard.


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## goodatthis (Jul 26, 2014)

So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes. 

So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
Now let's say you had super cube stickers on an Examinx. How many combinations would this puzzle have? 
I'm not completely sure if minxes can have super cube centers, but I'm just going to assume they can.


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## TDM (Jul 26, 2014)

goodatthis said:


> So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes.
> 
> So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
> Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
> ...


Permutations...
So each type of 'centre piece' (i.e. anything not on the outer layer) has 60 positions it could go in to, so for each one there are 60! possible permutations. There are 30 types, so for centres alone we have *60!^30*, which is a huge number. About 4.03*10^2457. This is assuming this is a supercube, as you said in your post.
There are 30 edges in the middle, so *30!* permutations.
60 permutations for each of the five other types of edge, so *60!^5*.
20 corners, 20! permutations.
And *divide by 2* (is it 2?) for parity.
Orientations...
Centres have no orientation.
Middle edges: 30 edges, so *2^29*.
Other edges have no orientation.
Corners: *3^19*.

In total...
(60!^30*30!*60!^5*20!*2^29*3^19)/2 = *3.23*10^2935*
so almost 3000 digits, unless I made a mistake somewhere


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## cmhardw (Jul 26, 2014)

TDM said:


> And *divide by 2* (is it 2?) for parity.



Think about this more. You're on the right track.


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## TDM (Jul 26, 2014)

cmhardw said:


> Think about this more. You're on the right track.


erm... I barely know anything about just Megaminxes, so I don't really know about this. Is it I don't need to divide by two because it's a supercube (or superdodecahedron), so I would have included parity there?


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## cmhardw (Jul 26, 2014)

TDM said:


> erm... I barely know anything about just Megaminxes, so I don't really know about this. Is it I don't need to divide by two because it's a supercube (or superdodecahedron), so I would have included parity there?



Odd permutation parity, for any piece orbit, is impossible on a megaminx (and it's relatives).


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## TDM (Jul 26, 2014)

cmhardw said:


> Odd permutation parity, for any piece orbit, is impossible on a megaminx (and it's relatives).


So does that mean you don't have to divide by two?


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## cmhardw (Jul 26, 2014)

TDM said:


> So does that mean you don't have to divide by two?



If odd permutation parity is impossible, then you have to divide by 2 _for every piece orbit on the puzzle_. This is the case because you were calculating the cases for a super-minx, where the concept of permutation parity makes sense for all piece types.


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## TDM (Jul 26, 2014)

cmhardw said:


> If odd permutation parity is impossible, then you have to divide by 2 _for every piece orbit on the puzzle_. This is the case because you were calculating the cases for a super-minx, where the concept of permutation parity makes sense for all piece types.


Right, so when you say every orbit, does that mean you divide by 2 six times (i.e. dividing by 64)? I'm really confused...


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## Stefan (Jul 26, 2014)

cmhardw said:


> If odd permutation parity is impossible, then you have to divide by 2 _for every piece orbit on the puzzle_.



Do you mean when they're independent, or is the independence already included in the term "piece orbit"? Because I think the 3x3x3 has *two* piece orbits (corners and edges) and we divide by 2 only *once*.


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## TMOY (Jul 27, 2014)

goodatthis said:


> Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)


11-layered, not 13. We already have:
- megaminx: 3 layers
- gigaminx: 5 layers
- teraminx: 7 layers
-petaminx: 9 layers.
So examinx should have 11.


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## cmhardw (Jul 27, 2014)

TDM said:


> Right, so when you say every orbit, does that mean you divide by 2 six times (i.e. dividing by 64)? I'm really confused...



If an examinx is an 11 layered minx, then you divide by 2^36 for the superexaminx. 
There are 30 center orbits, 1 corner orbit, 1 middle edge orbit, 4 wing edge orbits.



Stefan said:


> Do you mean when they're independent, or is the independence already included in the term "piece orbit"? Because I think the 3x3x3 has *two* piece orbits (corners and edges) and we divide by 2 only *once*.



Independence of the orbits is not necessarily true on the examinx, but that is irrelevant here.

Look at a quarter turn on an examinx. No matter what face type is turned you are either performing a 5 cycle of pieces in an orbit or two 5-cycles of pieces in an orbit (for x-centers). An odd parity permutation is impossible. You must divide the overcount calculation for the examinx by 2^36 (divide by one factor of 2 for each of the 36 orbits on the puzzle).


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## cmhardw (Jul 27, 2014)

For the superexaminx (an 11 layered minx) I get:

(60! / 2)^34 *(20! / 2) (3^19)(30! / 2)(2^29)


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## TDM (Jul 27, 2014)

I get it now... I didn't know what 'orbit' meant. I also looked at the first clear image on google images for 'examinx', which has 13 layers, not 11, so my calculations were wrong anyway. I think I understand what you're saying now, thanks for correcting me.


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## Christopher Mowla (Jul 27, 2014)

qqwref helped me with this in the past, and I recall that the number of positions on the minx of order _n_ (for all _n_>2) is:






​ 

Abbreviating the above product as (a)(b)(c)(d),



Spoiler: (a)



*Numerator of (a)*
The numerator represents the number of permutations and orientations of 20 corners.

*First Factor in the Denominator of (a)*
We divide by (20*3) for even minx's, as we can fix a corner to be solved on any even minx and thus divide the total number of positions by how much of the product any single corner contributes. We may also interpret dividing by 60 for the even minx to be that we instead fix one non-fixed piece or one wing edge, as each of them also contribute a factor of 60 into the total number of positions.

*Second Factor in the Denominator of (a)*
We divide by (3) by the minx law of corner orientations.

*Third Factor in the Denominator of (a)*
We divide by (2) since only even permutation corner permutations are allowed.





Spoiler: (b)



*Numerator of (b)*
The numerator represents the number of permutations and orientations of 30 middle edges.

*First Factor in the Denominator of (b)*
We divide by (2) by the minx law of middle edge orientations.

*Second Factor in the Denominator of (b)*
We divide by (2) since only even permutation middle edge permutations are allowed.





Spoiler: (c)



This factor tells us how many positions of the wing edges there are.

There are 30(2) = 60 wing edges in every wing orbit (that is, there are 2 wing edges of each wing edge orbit in each composite edge), and only half of the permutations are allowed (odd permutations cannot exist on a minx).

The minx^n has the same number of wing edge orbits that the _n_x_n_x_n_ cube has.





Spoiler: (d)



This factor tells us how many positions of the non-fixed centers there are.

There are 5(12) = 60 non-fixed center pieces in every non-fixed center piece orbit (that is, there are 5 non-fixed center pieces in each of the 12 faces per orbit).

Since every face has 5 of the same color centers, there are 5! equivalent positions regarding each of the 12 color sets of non-fixed centers, and this is true for every orbit of non-fixed centers. Thus we divide by 5! raised to the number of non-fixed center orbits.

We do not also divide by 2 because, for example, we can picture 2 discolored centers as either being a 3-cycle (even permutation) or a 2-cycle (odd permutation), since this is not the super minx.

Lastly, the minx^n has the same number of non-fixed center orbits as the _n_x_n_x_n_ cube.



As I did here, the number of positions formula of the minx^n can be written as a product of powers of consecutive prime numbers. This time, a product of all primes from 1 to 59 instead of just 1 to 23.















​ 
For the superminx, I get that we simply multiply the number of positions formula for the minx^n by





(this is the number of "solved positions" of the minx^n). For example minx^11 has this number/ 10^501 solved positions.​
Let the above product be represented by (e)(f).


Spoiler: (e)



This is a factor pertaining to the number of positions the non-fixed centers contribute.

*Numerator of (e)*
Non-fixed centers are distinguishable on the superminx^n. Therefore we do not divide by (5!)^12 raised to the number of non-fixed center orbits anymore. We include this to simply counteract doing this in the number of positions formula for the regular 12 colored minx^n.

*The Denominator of (e)*
We divide by 2 because only even permutations of non-fixed centers are allowed on the superminx^n.





Spoiler: (f)



This is the fixed center of the odd minx^n factor.

There are 12 fixed centers on the odd minx^n, and each can be rotated in 5 ways.

We do not divide by 2 here because if we do an outer face quarter turn to a face on a solved odd layer superminx^n, we can solve back all pieces using commutators (except for the fixed center) without rotating the face itself.


We multiply (a)(b)(c)(d)(e)(f) to get the following formula for the superminx^n (for n>2):


 
Again, we can write this formula as a product of powers of consecutive primes. To make things simple, we can just convert the superminx factor to a product of the following consecutive primes, and we can manually merge the exponents of the common prime numbers both products have to have a formula for the superminx^n as a product of powers of consecutive prime numbers from 1 to 59.



 


goodatthis said:


> So I'm basically asking for trouble here, and I'm not even sure how one would calculate it and/or if someone even would, but here goes.
> 
> So there's something like 10^996 combinations on a Petaminx. (Not sure on the base but I know the exponent)
> Now I've heard rumors and seen photos of an Examinx, a 13 layered minx. (Assuming you consider megaminx as 3 layered)
> ...


10^996 is correct for minx^9.

For minx^11, using the formula above for the regular 12-color minx^n, I get 10^1533. Or, this number of positions.

For the superminx^11, I get 10^2035 or this number of positions.



cmhardw said:


> For the superexaminx (an 11 layered minx) I get:
> 
> (60! / 2)^34 *(20! / 2) (3^19)(30! / 2)(2^29)


Chris, I get:
(60!/2)^24*(20!/2) (3^19) (30!/2) (2^29) (5^12)

(Don't we count fixed center rotations?)


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## cmhardw (Jul 27, 2014)

cmowla said:


> Chris, I get:
> (60!/2)^24*(20!/2) (3^19) (30!/2) (2^29) (5^12)
> 
> (Don't we count fixed center rotations?)



I see what I did wrong. I forgot about centralmost center rotations, but I also thought that the largest outer pentagon of centers had the same number of layers as he minx itself (obviously incorrect). I can see now that the pentagon rings from outside to inside have 8, 6, 4, 2 orbits or 20 center orbits plus the 4 wing orbits for an exponent of 24 and not 34.


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## TDM (Jul 27, 2014)

cmhardw said:


> I forgot about centralmost center rotations


Same. I was thinking about some supercube centres, but not all of them...


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## goodatthis (Jul 28, 2014)

TMOY said:


> 11-layered, not 13. We already have:
> - megaminx: 3 layers
> - gigaminx: *5* layers
> - teraminx: 7 layers
> ...



Well, you may think so, but an Examinx is not the next step up. By the regular orders of magnitude, it goes peta- exa- zetta-, but the one person who actually designed it named it wrong. Also, gigaminxes have 5 layers, not 6. I believe a 6 layered minx is called a super kilominx or star minx or something. 

Examinx:


Spoiler







Based on Cmowla's formula, a super 13 layered minx would have 1.1*10^2934 combinations.


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## 10461394944000 (Jul 28, 2014)

goodatthis said:


> Well, you may think so, but an Examinx is not the next step up. By the regular orders of magnitude, it goes peta- exa- zetta-, but the one person who actually designed it named it wrong.



what?


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## goodatthis (Jul 28, 2014)

10461394944000 said:


> what?



Huh? 

The person who designed the Examinx should have named it the zettaminx, since petaminx is 9 layers, and Examinx is 13.


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## TMOY (Jul 28, 2014)

goodatthis said:


> Well, you may think so, but an Examinx is not the next step up. By the regular orders of magnitude, it goes peta- exa- zetta-, but the one person who actually designed it named it wrong.


If the one person who designed it has no clue about how that naming scheme works, we're not forced to stick to his error forever just because he chose the wrong name in the first place. His puzzle is a zettaminx, not an examinx, period.



goodatthis said:


> Also, gigaminxes have 5 layers, not 6.



Yes, U know, that was a typo, sorry.


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## IRNjuggle28 (Aug 17, 2014)

What are the odds of getting a skip of orientation of the 8 edges other than the cross pieces? Because that's what just happened. I just got a solve that was rotationless, and was <RUL> except for one F move and one D move. This is a CFOP solve, not a ZZ solve. http://tinyurl.com/n5l3hy4

EDIT: actually, that's easy. The answer to getting a L8E orientation skip is just 1/2^7 or something. 
EDIT2: ninja'd

What about the odds of having a scramble that's solvable with only two moves that aren't RUL? I guess another way to ask that would be what are the odds of having an EOline solvable in 2 non RUL moves?


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## uberCuber (Aug 17, 2014)

IRNjuggle28 said:


> What are the odds of getting a skip of orientation of the 8 edges other than the cross pieces? Because that's what just happened. I just got a solve that was rotationless, and was <RUL> except for one F move and one D move. This is a CFOP solve, not a ZZ solve. http://tinyurl.com/n5l3hy4



I believe 1/2^7 = 1/128 = 0.78%. (Each edge has two possible orientations, so 1/2 for each of 7 edges, and the 8th edge's orientation is defined by the other 7).


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## Renslay (Aug 24, 2014)

I was doing some skewb solves, when suddenly I had a last-5-centers skip. I solved 1 center + 4 corners first, then solved the last 4 corners with R' F R F', and then suddenly the skewb was solved. What is the probability for such a skip?


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## Cale S (Aug 24, 2014)

Renslay said:


> I was doing some skewb solves, when suddenly I had a last-5-centers skip. I solved 1 center + 4 corners first, then solved the last 4 corners with R' F R F', and then suddenly the skewb was solved. What is the probability for such a skip?


If you are left with 5 remaining centers, the U center can be in one of 5 places, so the chance of it being solved is 1/5. Given that the U center is solved, the F center can be in one of 4 places, so it has a 1/4 chance of being solved. Given the U and F centers are solved, the R center can be in one of 3 places, so it has a 1/3 chance of being solved. Once the U, F, and R centers are solved, L5C must be solved because there are two left and skewb can't have parity, so the chance of having a last five centers skip is 1/5*1/4*1/3 = 1/60. Another way to think of it is that there are 5 centers, so you could arrange them in 5! = 120 ways, but dividing by 2 (because skewb can't have parity) gives you 60 combinations, 1 of which is solved.

If you know Sarah's intermediate method, this is reduced to 1/12 because the U center is solved during the second step.


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## goodatthis (Sep 11, 2014)

So this is a tricky one.

What are the chances that 3/4 of the 4 f2l slots are simple 3 move inserts+AUF in between? Lets just assume we can optimally solve pair order to force 3 move inserts. But no multislotting algs, simply different pair order.


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## Robert-Y (Sep 11, 2014)

I'm guessing that the request is for at least 3 pairs, and that F2L pair skips are also taken into consideration?


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## Bindedsa (Sep 11, 2014)

Robert-Y said:


> I'm guessing that the request is for at least 3 pairs, and that F2L pair skips are also taken into consideration?



Wouldn't any F2L pair skips after the first pair be considering accidental multislotting?


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## goodatthis (Sep 11, 2014)

Robert-Y said:


> I'm guessing that the request is for at least 3 pairs, and that F2L pair skips are also taken into consideration?


Yep, they are.


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## guysensei1 (Sep 22, 2014)

If you had a set of 3x3 stickers (9 stickers of each sticker color, 6 colors), and you randomly stickered a cube, what are the odds that it will be solvable? ie, anytime there are 9 stickers of the same color on every face, it's 'solved', regardless of the color scheme.

What if we restricted the stickering such that centerpiece stickers must go on center pieces, corner stickers must go on corners, and edge stickers must go on corners (with a fitted sticker set or something in reality)?


What if we restricted 'solved' to only one specific color scheme?


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## Stefan (Sep 22, 2014)

guysensei1 said:


> If you had a set of 3x3 stickers (9 stickers of each sticker color, 6 colors), and you randomly stickered a cube, what are the odds that it will be solvable?



I think about 1 in 3*10[sup]15[/sup]. In other words, don't try it.


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## guysensei1 (Sep 22, 2014)

Stefan said:


> I think about 1 in 3*10[sup]15[/sup]. In other words, don't try it.



Yikes. What about the other restrictions?


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## Stefan (Sep 22, 2014)

guysensei1 said:


> What about the other restrictions?



Come on, my fish came with instructions.


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## goodatthis (Sep 23, 2014)

What is the probability that a 1x2x3 block will form in the LL during the solving of F2L? I realized that when I was at US Nationals, that happened 3 times.


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## Stefan (Sep 23, 2014)

goodatthis said:


> What is the probability that a 1x2x3 block will form in the LL during the solving of F2L? I realized that when I was at US Nationals, that happened 3 times.



You mean just formed during LL and then perhaps broken again, or also still there at the end of F2L?


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## goodatthis (Sep 23, 2014)

Stefan said:


> You mean just formed during LL and then perhaps broken again, or also still there at the end of F2L?



Yeah, just formed at some point during F2L, even if it's in the middle of an F2L alg.


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## Stefan (Sep 23, 2014)

Hmm, I'd say that depends on your personal F2L then. A beginner might take more moves for F2L and thus have more chances to build such a block and thus have a higher probability.


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## Robert-Y (Sep 23, 2014)

Yeah, it depends on how many times the cross has been restored during F2L I think.


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## kinch2002 (Sep 23, 2014)

Definitely too vague to calculate...as Stefan says, it totally depends on your F2L. On another note, I hope you took advantage of it and switched LL color for at least one of them


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## penguinz7 (Sep 24, 2014)

Just got 15 oll paritys in a row. Yes, 15. wut. 1/32768?
0.00003051757% Pretty sure thats a record for number of oll paritys in a row..


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## FJT97 (Oct 9, 2014)

If I do a first face (not a layer), how many positions are possible then?


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## cmhardw (Oct 9, 2014)

FJT97 said:


> If I do a first face (not a layer), how many positions are possible then?



Permute the solid color layer:
(4!)^2

Permute and orient the rest of the cube:
(4!)*3^3 corners
(8!/2)*2^7 edges

For edge permutations you divide by 2 since the edge permutation parity must match the corner permutation parity.

Altogether:
(4!)^2*(4!)*3^3*(8!/2)*2^7 = 963158 999040


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## guysensei1 (Oct 9, 2014)

How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.


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## TMOY (Oct 9, 2014)

guysensei1 said:


> How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.



First we count the total number of cases (considering cases which are identical up to rotation as different); we obtain 3^3 corner orientations, 4! corner permutations, 2^2 edge orientations and 3! edge permutations, dividing by 2 because of parity, we obtain 3^3*4!*2^2*3!/2=7776 positions.

Next we count the number of symmetrical positions. For a position to be symmetrical, the corner on top of the tripod must be in place (not necessarily oriented), and the other three must all have orientations compatible with the orientation of the top corner (only 1 possibility), and their permutation must be a 3-cycle. The edges have only one possible orientation, and their permutation must also be a 3-cycle. Hence 3*3*3=27 symmetrical positions, and 7776-27=7749 asymmetrical positions..

So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.


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## cmhardw (Oct 9, 2014)

guysensei1 said:


> How many cases are there to solve the cube in 1 look if one has built a tripod already? Assume that cases that are simply a cube rotated version of another case can be considered the same.



That's a difficult question. In terms of number of combinations in total there are:
4!*3^3*3!/2*2^2=

I tried using the lemma that is not Burnside's (and I may have used it incorrectly) and got:
1/6 * (7776 + 3*(4(2)) + 1(3(2*2))) = 1302 unique cases, which is more than total LL combinations.

Since the tripod is essentially the last layer with 1 solved edge, I would think it would be less than the 1211 algs required for 1LLL, but maybe since the tripod has fewer symmetries (I count 8 for a layer and 6 for a tripod, but that may be wrong) then there would be more tripod cases?

Can someone with a better math background comment on this?

--edit--
Ninja'd by TMOY


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## guysensei1 (Oct 9, 2014)

Thanks for the calculation! What if we used 2 look tripod? Which set will have less cases: solve corners in 1 look, then edges, or edges first then corners?


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## cmhardw (Oct 9, 2014)

TMOY said:


> So the total number of cases, excluding the solved case which is of course symmetrical, are: 7749/3+26=2609 cases.



My number is off by almost exactly a factor of 2 from yours. Can you help me figure out where I went wrong? I can explain my calculation if you need. It's been so long since I've used the (not) Burnside's lemma and I'd like to make sure I'm not misusing it.


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## G2013 (Oct 9, 2014)

How probable is this to happen?
https://www.youtube.com/watch?v=iaTUkd948JM


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## Renslay (Oct 9, 2014)

G2013 said:


> How probable is this to happen?
> https://www.youtube.com/watch?v=iaTUkd948JM



Well, it happened to me twice a day. No big deal...


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## TMOY (Oct 10, 2014)

cmhardw said:


> My number is off by almost exactly a factor of 2 from yours. Can you help me figure out where I went wrong? I can explain my calculation if you need. It's been so long since I've used the (not) Burnside's lemma and I'd like to make sure I'm not misusing it.



The difference between our two calculations is simply that you're counting two cases which are mirror images of each other as only one while I count them as two.


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## SirWaffle (Oct 10, 2014)

*7x7 Last 4 edge odds?*

I really am unsure the correct search terms to use to find the answer so that's why I am creating a new thread. What are the odds of skipping one of the last 4 edges after realigning the centers on 7x7? This happened to me in a solve so I was curious.


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## cmhardw (Oct 10, 2014)

SirWaffle said:


> I really am unsure the correct search terms to use to find the answer so that's why I am creating a new thread. What are the odds of skipping one of the last 4 edges after realigning the centers on 7x7? This happened to me in a solve so I was curious.



The probability that at least 1 edge group is solved is approximately 0.127%

Clicky

--edit--
I fixed a conceptual math error about 5 minutes after sending this post.


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## Stefan (Oct 10, 2014)

You can also use the four middle edges as reference.

http://www.wolframalpha.com/input/?...)^(i+1)+*+(8-2*i)!^2+/+8!^2+for+i+from+1+to+4


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## cmhardw (Oct 10, 2014)

Stefan said:


> You can also use the four middle edges as reference.
> 
> http://www.wolframalpha.com/input/?i=sum+(4+choose+i)*(-1)^(i+1)+*+(16-4*i)!+/+16!+for+i+from+1+to+4
> 
> Edit: Hmm, just saw your edit and I don't understand your changed answer.



I was treating the 16 wings as 1 orbit of 16 pieces. My correction was to make it two separate orbits, each with 8 pieces.


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## cmhardw (Oct 10, 2014)

Stefan said:


> You can also use the four middle edges as reference.



Wow, I understand what you did there using the edges as reference. I also just realized that in my expression those terms would cancel out, effectively using the edges as reference.

I would not have thought to use inclusion-exclusion like you did where you start by counting the number of states with 1 (really at least 1) edge group solved. That makes sense now that I look at it. For some reason I always want to use it to find derangements, when it's obviously a more powerful method with further uses.

Thanks for the math tips! As always, Stefan, you've got skills!


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## Stefan (Oct 10, 2014)

cmhardw said:


> I was treating the 16 wings as 1 orbit of 16 pieces



Oh crap, I did that, too 
(I developed mine independently from yours, so I made the mistake myself, didn't copy it from you)
(fixed it now)


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## goodatthis (Oct 12, 2014)

What is the probability of a L4E skip on 5x5, assuming freeslice is used.

Asking because it happened to me two days ago. The solve wasn't PB though


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## cmhardw (Oct 12, 2014)

goodatthis said:


> What is the probability of a L4E skip on 5x5, assuming freeslice is used.
> 
> Asking because it happened to me two days ago. The solve wasn't PB though



Using the midges as reference it should be
1/8! = 1/40320

If you used AVG or anything to try to partially build tredges throughout your solve, then the probability increases.


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## MennoniteCuber1 (Oct 13, 2014)

What are the most probable CLL cases for each orientation?


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## kcl (Oct 14, 2014)

MennoniteCuber1 said:


> What are the most probable CLL cases for each orientation?



Anything not pure or diag?


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## rybaby (Oct 14, 2014)

What is the probability of having a solved square in a scramble?


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## Stefan (Oct 14, 2014)

rybaby said:


> What is the probability of having a solved square in a scramble?



1

(assuming you meant 3x3x3 or another puzzle with square stickers)


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## rybaby (Oct 14, 2014)

Stefan said:


> 1
> 
> (assuming you meant 3x3x3 or another puzzle with square stickers)



Why do you even waste your time with answers like these?
Now is there someone who would like to be helpful?


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## cmhardw (Oct 14, 2014)

rybaby said:


> Why do you even waste your time with answers like these?
> Now is there someone who would like to be helpful?



The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)

Clicky
Also clicky


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## rybaby (Oct 14, 2014)

cmhardw said:


> The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 2.2*10^(-12)
> 
> Clicky
> Also clicky



How about a solved 2x2 square permuted properly? (Not necessarily aligned with centers)


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## cmhardw (Oct 14, 2014)

rybaby said:


> How about a solved 2x2 squat permuted properly? (Not necessarily aligned with centers)



Oh, well you should have asked that in the first place 

I will be away from my computer for a bit, but can work on this a little later in the day. This is a tricky question and will likely require very clever math, or a computer program to run an intelligent search search of case types. I think clever math should be able to do it, but there are others on here with a much more solid math background who would know better than I.


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## ChickenWrap (Oct 14, 2014)

Stefan said:


> 1
> 
> (assuming you meant 3x3x3 or another puzzle with square stickers)



I might not have invented my own BLD method, but at least I'm not an a**hole to random people on the internet.


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## cmhardw (Oct 14, 2014)

ChickenWrap said:


> I might not have invented my own BLD method, but at least I'm not an a**hole to random people on the internet.



The question was ambiguous, Stefan made an assumption (and not a bad one) about what the question was referring to.

How is that being an a-hole?

Stefan doesn't like ambiguous questions, and for good reason.


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## rybaby (Oct 14, 2014)

cmhardw said:


> Stefan made an assumption (and not a bad one) about what the question was referring to.



You must really undervalue my intelligence if you think that my original question would be answered by Stefans post.


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## cmhardw (Oct 14, 2014)

rybaby said:


> What is the probability of having a solved square in a scramble?





Stefan said:


> 1
> 
> (assuming you meant 3x3x3 or another puzzle with square stickers)





rybaby said:


> You must really undervalue my intelligence if you think that my original question would be answered by Stefans post.



Your question, as asked, was answered by Stefan's post. Your _intended_ question was not answered by Stefan's post, because you _did not ask your intended question_.

That's the difference. Do you understand now why Stefan's post was not mean? He was, in fact, trying to teach you to ask more specific questions so that your intended questions get answered right away.


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## cmhardw (Oct 14, 2014)

rybaby said:


> How about a solved 2x2 square permuted properly? (Not necessarily aligned with centers)



This is a rough estimation, but will at least give us a basic idea.

A 2x2x1 block is formed when a corner has 2 of it's adjacent edges correctly permuted, as well as the associated center piece.

The number of "positions around a corner" that have at least one 2x2x1 block is:
(3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238

There are 24*22*20*6*4=253440 "positions around a corner"

This means that, in regards to one specific corner, the probability of having no 2x2x1 blocks around it is:
(253440-238)/253440 = 253202/253440

The _approximate_ probability that at least one corner on a 3x3x3 would have at least one 2x2x1 is:
1-(253202/253440)^8 or approximately 0.75% chance.

This is a hack calculation, and is not the true value. This is my estimation to figure out roughly about what is this probability.

--edit--

I find it interesting that this probability is corroborated by:

8*(238/253440) which is also approximately 0.75%

This is the chance that one corner has at least one 2x2x1 block multiplied by the 8 corners.


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## hawkmp4 (Oct 14, 2014)

> The number of "positions around a corner" that have at least one 2x2x1 block is:
> (3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238
> 
> There are 24*22*20*6*4=253440 "positions around a corner"



Could you elaborate on the details of those expressions/is thit correct?
24 possible orientations of the "target' corner?

FWIW I don't think that there's an issue of content, but one of tone. Stefan meant to teach, and OP didn't see that so it came off as abrasive. The dangers of text communication!


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## cmhardw (Oct 14, 2014)

hawkmp4 said:


> Could you elaborate on the details of those expressions/is thit correct?
> 24 possible orientations of the "target' corner?



I'm using inclusion/exclusion in a way that Stefan recently taught me.

In the way I am counting, I am picking an arbitrary corner and fixing it as my reference point. For clarity let's call the reference corner UFR. Around this corner are three adjacent edges, UF, FL, UR. There are also three center pieces that touch the corner on a diagonal (U, F, R) centers.



cmhardw said:


> The number of "positions around a corner" that have at least one 2x2x1 block is:
> (3 choose 1)*10*2*4-((3 choose 2)*1)+((3 choose 3)) = 238



(3 choose 1)*10*2*4 chooses which of the 2x2x1 blocks around the UFR corner to be "solved", then it permutes one of 10 possible edge pieces to the 3rd edge location around the UFR corner (the one that is not part of the 2x2x1 block), orients that recently permuted edge (2) and then there are 4 possible positions the centers can be in. This is because one center is part of the 2x2x1, but the others have 4 possible 90 degree rotations in their slice. This term calculates the number of cases where one 2x2x1 block exists. It is an overcount, and will be handled by the rest of the inclusion/exclusion procedure.

-((3 choose 2)*1) continues the inclusion/exclusion by subtracting the number of ways that there can be two 2x2x1 blocks solved around the UFR corner. There is, in fact, only one way and that is to have all three 2x2x1s around UFR solved forming a 2x2x2 block. However, for the inclusion/exclusion to work I have to count that there are (3 choose 2) ways to pick "two" 2x2x1s to be solved.

+((3 choose 3)) This continues the inclusion/exclusion by adding back the number of ways that three 2x2x1 blocks can be solved. There is only 1 way, and (3 choose 3) = 1.

Since we know that (3 choose 1)*10*2*4 overcounts the number of 2x2x1s when we have the special case of a 2x2x2, let's look at the 2x2x2 case.

(3 choose 1)*10*2*4 counts the 2x2x2 case three times, since there are three 2x2x1s forming the 2x2x2 block, and this term counts every way that a 2x2x1 can be formed.
-((3 choose 2)*1) subtracts the 2x2x2 block case three times (net is that this case has been counted 0 times). It subtracts it three times because there are three ways to look at exactly two sides solved on a 2x2x2.
+((3 choose 3)) adds back all cases where three 2x2x1s are solved. There is only one way, the special case where you have a 2x2x2. Now this special case has been counted a net total of 1 time, which is good because there is only 1 case.

For all cases where exactly one 2x2x1 block is formed (and there is not 2x2x2), these cases are all counted exactly once in the term (3 choose 1)*10*2*4. So this entire expression using inclusion/exclusion has counted every case with at least 1 2x2x1 block uniquely, and is not an overcount.

Once I have the number of cases using one corner as reference, I then view each corner as independent of all others (I view each corner "in its own bubble") and calculate the probability that every corner has no 2x2x1 around it. Take 1 minus that probability for the probability that at least one corner has at least one 2x2x1. This is not the true probability, as you cannot view each corner as completely independent ("in its own bubble"), but it will be a rough estimation of the true probability.


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## Stefan (Oct 15, 2014)

cmhardw said:


> He was, in fact, trying to teach you to ask more specific questions



This, thank you. I just demonstrated that the question as asked doesn't make sense and was hoping he'd clarify. I thought that was obvious.

2x2 square indeed seems rather difficult.



cmhardw said:


> I find it interesting that this probability is corroborated by:
> 
> 8*(238/253440) which is also approximately 0.75%



Yeah, for small probabilities, n*p and 1-(1-p)^n are almost the same:
http://www.wolframalpha.com/input/?i=8*p+vs+1-(1-p)^8+for+p+from+0+to+0.01
(This should show a comparison, a plot with the two functions. Reload if it doesn't (a moment ago that happened to me).)

Edit: Oh neat, this works as well: http://www.wolframalpha.com/input/?i=n*p+vs+1-(1-p)^n,+for+p+from+0+to+0.01,+for+n+from+2+to+10



cmhardw said:


> The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)
> 
> Clicky



I think your "3 adjacent faces" term should end in 2^1, not 2^3.



cmhardw said:


> Clicky



Just a quick question about this one. Why the "/2" in "4!/2*2^3"? The midges can be permuted all 4! ways.


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## hawkmp4 (Oct 15, 2014)

Thanks for the explanation and the link, Chris. Good stuff to know.


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## Stefan (Oct 15, 2014)

I just realized you probably can't use inclusion-exclusion simply like that for the _"3x3 face"_ problem. Inclusion-exclusion works by overcounting at first, then uncounting the cases unintentionally counted more than once, then recounting the cases now counted less than once, etc, until every case has been counted exactly once.

Consider the earlier _"solved 7x7 edge among the last four"_ problem:

Counting all cases with at least 1 solved edge (because we haven't counted those with exactly 1 solved edge yet)...

 Counts all cases with exactly 1 solved edge exactly 1 time (each has now been counted 1 time).
 Counts all cases with exactly 2 solved edges exactly 2 times (each has now been counted 2 times).
 Counts all cases with exactly 3 solved edges exactly 3 times (each has now been counted 3 times).
 Counts all cases with exactly 4 solved edges exactly 4 times (each has now been counted 4 times).
*Un*counting all cases with at least 2 solved edges (because we have double-counted those with exactly 2 solved edges)...

 Counts all cases with exactly 2 solved edges exactly -1 times (each has now been counted 1 time).
 Counts all cases with exactly 3 solved edges exactly -3 times (each has now been counted 0 times).
 Counts all cases with exactly 4 solved edges exactly -6 times (each has now been counted -2 times).
Counting all cases with at least 3 solved edges (because those with exactly 3 solved edges are at 0 again)...

 Counts all cases with exactly 3 solved edges exactly 1 time (each has now been counted 1 time).
 Counts all cases with exactly 4 solved edges exactly 4 times (each has now been counted 2 times). (marked red for later)
*Un*counting all cases with at least 4 solved edges (because after 4-counting them, then (-6)-counting them and then 4-counting them, they're now overall double-counted)...

 Counts all cases with exactly 4 solved edges exactly -1 times (each has now been counted 1 time).

So you can see we counted, uncounted, counted and uncounted. In other words, we added the four numbers multiplied by +1, -1, +1 and -1, respectively. But that only turned out correct because of how "nicely" the countings and uncountings happened to "cancel out".

Now let's get back to the _"3x3 face"_ problem.

For example when handling the _"three adjacent faces"_ subcases of the _"three faces"_ cases, you're *not* unintentionally 4-counting all _"four faces"_ cases (like marked red above in the other problem) but only *2*-counting the _"four faces except two adjacent"_ subcases and even *not* counting the _"four faces except two opposite"_ subcases.

Unless for some "miracle" it's still correct (like the naive +1, -1, +1, -1 turned out correct earlier), I'm afraid your already complicated expression is too simple.


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## guysensei1 (Oct 15, 2014)

What's the probability of a solved 1x2x2 block forming on LL after completing F2L?

What's the probability of a solved 1x1x3 block forming on LL after completing F2L? What's the probability of a flipped 1x1x3 block forming?


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## MennoniteCuber1 (Oct 15, 2014)

Anything. Which permutation case would be most common for Sune, for example. With correctly oriented corner on ULF, would the most common case have adjacent switch on right, back, left, or front? Or if it was diagonal...


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## kinch2002 (Oct 15, 2014)

MennoniteCuber1 said:


> What are the most probable CLL cases for each orientation?





MennoniteCuber1 said:


> Anything. Which permutation case would be most common for Sune, for example. With correctly oriented corner on ULF, would the most common case have adjacent switch on right, back, left, or front? Or if it was diagonal...


There are 43 CLLs (including skip)
39 of them have 4/162 probability
Both double-bar H cases: 2/162
Y perm: 1/162
Skip: 1/162


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## Stefan (Oct 15, 2014)

cmhardw said:


> The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)
> 
> Clicky
> Also clicky



I fixed a few little mistakes and used what I think are the correct multipliers for the inclusion/exclusion. If I didn't make a mistake, there are 5,778,848,497,349 states with a solved face (happens to be a prime number).

Here's my code, I'd appreciate some checking 
http://ideone.com/ESdzhS
https://gist.github.com/pochmann/21408b91d0876c5e5828


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## goodatthis (Oct 15, 2014)

What is the probability of a paired (solved) edge with the use of reduction on...

5x5? 
6x6?
7x7?

And could someone give me the general idea in how to go about finding this? I was thinking something along the lines of this for a 5x5: 

pick a random midge (say the white-red one)
chances that one wing will be attached to it: 1/24 for 24 possible positions for a wing to be in 
chances that another wing will be attached: 1/24
(1/24)*(1/24)= 1/576 for some reason this probability seems rather small, this has happened to me a few times on 5x5 and I've probably done about 600 solves in my lifetime

Is this the way to do it? This is the first time I've really ever dabbled in cube probability, so excuse any mistakes I made.

edit: do any problems I have with this arise from the probability of a connected midge to a wing?


----------



## Cale S (Oct 16, 2014)

goodatthis said:


> What is the probability of a paired (solved) edge with the use of reduction on...
> 
> 5x5?
> 6x6?
> ...


Shouldn't it be (1/24)*(1/23) = 1/552? Once you have one wing paired up there are only 23 positions the other wing can be in.
For 6x6, you could use one wing as a reference, and then it would be (1/24)*(1/23)*(1/23) = 1/12696
7x7 would be (1/24)*(1/23)*(1/24)*(1/23) = 1/304704
This is just the probability of having one specific edge solved, I'm not sure how you would find the chances of having any one edge solved.


----------



## MennoniteCuber1 (Oct 17, 2014)

What is the probability of the same PLL twice in a row?


----------



## cmhardw (Oct 17, 2014)

goodatthis said:


> What is the probability of a paired (solved) edge with the use of reduction on...
> 
> 5x5?
> 6x6?
> ...



I would use inclusion/exclusion as a counting technique for this problem.


----------



## guysensei1 (Oct 17, 2014)

MennoniteCuber1 said:


> What is the probability of the same PLL twice in a row?


There are 3 probabilities for PLLs, 1/18, with 15 PLLs, 1/36, with 2 PLLs and 1/72, with 4 PLLs. 

Thus the probability is (1/18)^2*15+(1/36)^2*2+(1/72)^2*4 which is roughly 4.8%

Am I right?


----------



## cmhardw (Oct 17, 2014)

Here's my calculation for the 5x5x5 for the probability that at least 1 tredge is paired after finishing centers. This assumes you do not try to influence tredge pairing at all during centers.

Clicky

This probability is approximately 2.1%


----------



## Stefan (Oct 17, 2014)

guysensei1 said:


> There are 3 probabilities for PLLs, 1/18, with 15 PLLs, 1/36, with 2 PLLs and 1/72, with 4 PLLs.
> 
> Thus the probability is (1/18)^2*15+(1/36)^2*2+(1/72)^2*4 which is roughly 4.8%
> 
> Am I right?



Better round to 4.9%. But there are 16 PLLs with 1/18 anyway.



cmhardw said:


> Here's my calculation for the 5x5x5 for the probability that at least 1 tredge is paired after finishing centers. This assumes you do not try to influence tredge pairing at all during centers.
> 
> Clicky
> 
> This probability is approximately 2.1%



Got the same, though I like to use "choose":
http://www.wolframalpha.com/input/?...choose+i)+*+(24-2i)!+/+24!+for+i+from+1+to+12


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## guysensei1 (Oct 17, 2014)

Stefan said:


> Better round to 4.9%. But there are 16 PLLs with 1/18 anyway.



Oh shoot let me recalculate
(1/18)^2*16+(1/36)^2*2+(1/72)^2*4 roughly 5.2%


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## Chrizz (Oct 24, 2014)

I doubt anyone is gonna believe this, but I just had a last 2 layers skip, after I finished the cross I only had to do a u' and the cube was solved. What are the chances of this happening?

Edit: some additional information: I was just casually solving, so it was just a hand scramble, it probably wasn't 20 moves, but it must've been more than 12. I don't optimally solve the cross and I don't remember how many moves I took this time.
I don't think the exact probability is calculable because of the unknown scramble length and movecount to solve.


----------



## martinss (Oct 24, 2014)

Chrizz said:


> I doubt anyone is gonna believe this, but I just had a last 2 layers skip, after I finished the cross I only had to do a u' and the cube was solved. What are the chances of this happening?



There are {8! \times 3^7 \times (12!/2) \times 2^{11}} = 43,252,003,274,489,856,000 possibilites for the cube. Divide by 24 for the first edge, by 22 for the second edge, by 20 for the third edge, by 18 for the last edge of the cross
{43,252,003,274,489,856,000 /24 /22 /20 /18} = 2.275,463,1 e+14. So 1/(2.275,463,1e+14). (I'm not sure at all)


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## qqwref (Oct 24, 2014)

Chrizz said:


> What are the chances of this happening?


IF it was a random scramble, four (since you were a D move from solved) out of 8! * 8! * 3^7 * 2^6, or 1 in 227,546,313,523,200 (227 trillion). But as you said, it certainly wasn't a random scramble.


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## cuber8208 (Oct 28, 2014)

I had a go at trying to figure out the probabilities of 'EPLLs' for when DF and DB are swapped.

Can someone have a go at them so I can understand how the calculation works please? 

My idea is (if O and W aren't so frequent) to start cross with 2 opposite edges swapped at the start of the solve for leaving these cases after COLL. Maybe get lucky for an M2 U2 M2 PLL


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## kinch2002 (Oct 28, 2014)

cuber8208 said:


> I had a go at trying to figure out the probabilities of 'EPLLs' for when DF and DB are swapped.
> 
> Can someone have a go at them so I can understand how the calculation works please?
> 
> My idea is (if O and W aren't so frequent) to start cross with 2 opposite edges swapped at the start of the solve for leaving these cases after COLL. Maybe get lucky for an M2 U2 M2 PLL


Opp: 2/12
Adj: 4/12
W: 4/12
O(cw): 1/12
O(ccw): 1/12

Yes, doing oppcross can often be a good idea imo. I do it maybe 1 in every 30 solves or so,.and try a bit harder to force edges/do COLL /OLLCP. I know how to do the above EPLLs and they're certainly no worse than many PLLs


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## cuber8208 (Oct 28, 2014)

kinch2002 said:


> Opp: 2/12
> Adj: 4/12
> W: 4/12
> O(cw): 1/12
> O(ccw): 1/12



Thanks Dan, did you have a source for those stats?
If you did them yourself can you point me in the right direction of how to figure them out by myself?


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## kinch2002 (Oct 28, 2014)

cuber8208 said:


> Thanks Dan, did you have a source for those stats?
> If you did them yourself can you point me in the right direction of how to figure them out by myself?


I did them myself. I do it like this by thinking though how to create these cases by picking a location and working out probabilities of them being in the right places.
For Opp:
Pick a random location.
Option 1 is that it belongs opposite (1/4 chance). If that is the case, look at the location opposite to it, which has to belong back in the original location (1/3 chance).
Option 2 is that it is in the correct place (1/4). If that is the case, look at the location opposite to it, which has to also be solved (1/3).
1/4 * 1/3 + 1/4 * 1/3 = 2/12
Note that because of permutation parity, I don't need to think about where the last 2 pieces are.

That was a super long way of explaining what is a simple logical process to me. :/


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## cuber8208 (Oct 28, 2014)

Thanks, I have a better grasp on it now I think


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## GiraffeCubing (Oct 28, 2014)

I was solving with Roux and after my first two blocks all I needed to do was M2 U2 M2 U2 and the cube was solved. It was a hand scramble and I wasn't timing but it definitely would have been a PB if I was. What is the probability of this case happening? Is is more or less likely that a full last layer skip in CFOP?


----------



## Jakube (Oct 28, 2014)

GiraffeCubing said:


> I was solving with Roux and after my first two blocks all I needed to do was M2 U2 M2 U2 and the cube was solved. It was a hand scramble and I wasn't timing but it definitely would have been a PB if I was. What is the probability of this case happening? Is is more or less likely that a full last layer skip in CFOP?



Well, basically you skipped CMLL and the first two L6E parts:

4! (corner permutation)
/4 (AUF)
*3^3 (corner orientation)
*2^5 (edge orientation)
*6*5 (placing UL and UR edge)
= 1 in every 155520 solves, which is exactly 10 times more unlikely than a LL-skip


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## GiraffeCubing (Oct 29, 2014)

Be right back, gonna go buy a lottery ticket. Also thanks.


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## rybaby (Nov 24, 2014)

What is the probability of a solved cross on a 3x3 scramble? (For Stefan's sake, what is the probability of all edges of at least one layer being correctly oriented and permuted). Also, what is the probability of a cross's pieces being correct relative to each other but a move is needed to correct the centers? I got two solved crosses on scrambles in the past two days, one was completely solved and the other was off by a D move (cross on D).


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## GuRoux (Nov 24, 2014)

rybaby said:


> What is the probability of a solved cross on a 3x3 scramble? (For Stefan's sake, what is the probability of all edges of at least one layer being correctly oriented and permuted). Also, what is the probability of a cross's pieces being correct relative to each other but a move is needed to correct the centers? I got two solved crosses on scrambles in the past two days, one was completely solved and the other was off by a D move (cross on D).



i think its 7920 and 31680, but i'm not too good with probability.


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## obelisk477 (Nov 24, 2014)

GuRoux said:


> i think its 7920 and 31680, but i'm not too good with probability.



inb4 "those aren't probabilities!!!1!11"


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## Stefan (Nov 24, 2014)

GuRoux said:


> i think its 7920 and 31680, but i'm not too good with probability.



Given that probabilities aren't larger than 1, I agree with that last part 

For the solved case, have a look here:
http://www.cubezone.be/crossstudy.html

For "one move is needed", multiply by 3 for a good estimate.


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## GuRoux (Nov 24, 2014)

obelisk477 said:


> inb4 "those aren't probabilities!!!1!11"



one divided by those then.


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## rybaby (Nov 24, 2014)

GuRoux said:


> one divided by those then.



Wow, it's pretty crazy that I got two scrambles like that this weekend. Maybe the cubing gods are punishing me for not using CFOP . I'll prove them wrong.


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## GuRoux (Nov 24, 2014)

rybaby said:


> Wow, it's pretty crazy that I got two scrambles like that this weekend. Maybe the cubing gods are punishing me for not using CFOP . I'll prove them wrong.



i don't think i've ever had a finished cross, or maybe just never noticed it. i've probably done around 100,000 solves though.


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## Stefan (Nov 25, 2014)

GuRoux said:


> i've probably done around 100,000 solves though.



How did you estimate that?


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## GuRoux (Nov 25, 2014)

Stefan said:


> How did you estimate that?



there are two places where i usually do my timed solve: gqtimer and nanotimer. from the recorded times there i estimate i did around 200 timed solves a day for the last 6 months. before that was probably ranged between 50-150. 200*6*30+100*18*30+(un-timed solves)=>90,000=>100,000


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## Berd (Dec 2, 2014)

I just got a cmll skip in roux - probability?


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## GuRoux (Dec 2, 2014)

Berd said:


> I just got a cmll skip in roux - probability?



1 in 162 or something very close to that.


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## Berd (Dec 2, 2014)

GuRoux said:


> 1 in 162 or something very close to that.


Thankyou :*


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## TDM (Dec 2, 2014)

GuRoux said:


> 1 in 162 or something very close to that.


Yes. If you want to calculate the number of corner cases, it's:
((3^3)*4!)/4 = 27*6 = 162.
where 3^3 is the number of corner orientations, 4! is corner permutations, and the dividing by 4 is because of AUF.
(and of course, only one of these is solved)


----------



## obelisk477 (Dec 6, 2014)

Not really sure where to ask this (not having the OAQT is throwing me for a loop):

What are the optimal movecount distributions for 1)EOLine and 2)EOCross from a fixed (or of course y2) orientation.


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## guysensei1 (Dec 12, 2014)

What's the probability that a LL case is able to be AUFed into a 3 corner, 3 edge cycle? Of course G perms are such an example. Another is M U R U R' U' R' F R F' M' U'


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## Username (Dec 12, 2014)

guysensei1 said:


> What's the probability that a LL case is able to be AUFed into a 3 corner, 3 edge cycle? Of course G perms are such an example. Another is M U R U R' U' R' F R F' M' U'



I'm really bad at this stuff, so don't take my word for it

One edge and one corner solved

the three remaining corners can be in 2 different permutations for it to be a 3cycle (the third way they would be permuted correctly)
the three remaining edges can be in 2 different permutations for it to be a 3cycle (the third way they would be permuted correctly)
corners can be in 3^3 orientations, of which only a third is reachable with regular turns (so 3^2=9)
edges can be in 2^3 possible orientations, of which only half are reachable with regular turns (so 2^2=4)

2*2*9*4=144

the probability is 144/15552?


At this point of writing this I realize that I didn't calculate for some cases where you might be able to AUF it in two different ways so I don't know (I don't even know if that matters or not)

E: Should I multiply the 144 by 16? It wouldn't be the same corner and edge every time

2304/15552, or 4/27?


----------



## cuBerBruce (Dec 15, 2014)

Username said:


> I'm really bad at this stuff, so don't take my word for it
> 
> One edge and one corner solved
> 
> ...


No. 15552 is the total cases mod AUF. But you need to consider all 4 possible fixed corners, and all 4 possible fixed edges.






Username said:


> Should I multiply the 144 by 16? It wouldn't be the same corner and edge every time
> 
> 
> 2304/15552, or 4/27?


Yes, but but this is still not the final answer.



Username said:


> At this point of writing this I realize that I didn't calculate for some cases where you might be able to AUF it in two different ways so I don't know (I don't even know if that matters or not)


Yes, it matters. You've overcounted the number of cases because of this.

I've done a brute force enumeration using GAP. The result is that the actual probability is 11/81. There are 1920 cases out of 15552 where there is a single AUF case that satisfies the problem, and 192 cases where there are 2 AUF cases satisfying the problem. (1920 + 192)/15552 = 11/81.

GAP code:



Spoiler





```
U := (1,3,9,7)(2,6,8,4)(10,37,28,19)(11,38,29,20)(12,39,30,21);
D := (16,25,34,43)(17,26,35,44)(18,27,36,45)(46,48,54,52)(47,51,53,49);
L := (1,19,46,45)(4,22,49,42)(7,25,52,39)(10,12,18,16)(11,15,17,13);
R := (3,43,48,21)(6,40,51,24)(9,37,54,27)(28,30,36,34)(29,33,35,31);
F := (7,28,48,18)(8,31,47,15)(9,34,46,12)(19,21,27,25)(20,24,26,22);
B := (1,16,54,30)(2,13,53,33)(3,10,52,36)(37,39,45,43)(38,42,44,40);
 
U2 := U*U;
U3 := U2*U;
D2 := D*D;
D3 := D2*D;
L2 := L*L;
L3 := L2*L;
R2 := R*R;
R3 := R2*R;
F2 := F*F;
F3 := F2*F;
B2 := B*B;
B3 := B2*B;

Aperm := R*B3*R*F2*R3*B*R*F2*R2;
Jperm := B2*L*U*L3*B2*R*D3*R*D*R2;
fururf := F*U*R*U3*R3*F3;
GPLL := Group (U, Aperm, Jperm);
GLL := Group (U, Jperm, fururf);

FaceletCycleLen := function (g, n)
  local i, j;
  i := n;
  j := 1;
  while i^g <> n do
    i := i^g;
    j := j + 1;
  od;
  return j;
end;

CubieCycleLen := function (g, n)
  local i, j, cmap;
  cmap := [
   1, 2, 3, 4, 5, 6, 7, 8, 9,
   1, 4, 7, 13, 14, 15, 52, 49, 46,
   7, 8, 9, 15, 23, 24, 46, 47, 48,
   9, 6, 3, 24, 32, 33, 48, 51, 54,
   3, 2, 1, 33, 41, 13, 54, 53, 52,
   46, 47, 48, 49, 50, 51, 52, 53, 54
  ];
  i := cmap[n];
  j := 1;
  while cmap[i^g] <> cmap[n] do
    i := cmap[i^g];
    j := j + 1;
  od;
  return j;
end;

Check33 := function (g)
  local ok, c1, c3, e2, e4;
  ok := false;
  c1 := CubieCycleLen (g, 1);
  if (c1 = 3) and (FaceletCycleLen (g, 1) = 3) then
    ok := true;
  else
    c3 := CubieCycleLen (g, 3);
    if (c3 = 3) and (FaceletCycleLen (g, 3) = 3) then
      ok := true;
    fi;
  fi;
  if ok then
    ok := false;
    e2 := CubieCycleLen (g, 2);
    if (e2 = 3) and (FaceletCycleLen (g, 2) = 3) then
      ok := true;
    else
      e4 := CubieCycleLen (g, 4);
      if (e4 = 3) and (FaceletCycleLen (g, 4) = 3) then
        ok := true;
      fi;
    fi;
  fi;
  return ok;
end;

Check33All := function (G)
  local g, g2, g3, g4, h, j, c, Lused, Lout, Lc, Le, Lx;
  Lused := [];
  Lout := [];
  for g in G do
    if not (g in Lused) then
      h := g;
      j := 1;
      Lx := [];
      c := 0;
      while j <= 4 do
        if Check33 (h) then
          c := c + 1;
        fi;
        Add (Lused, h);
        h := h*U;
        j := j + 1;
      od;
      Add (Lout, c);
    fi;
  od;
  return Lout;
end;

result := Check33All (GLL);;
nx := Size (result);
n1 := Size (Positions (result, 1));
n2 := Size (Positions (result, 2));
(n1+n2)/nx;
```


----------



## Hssandwich (Dec 16, 2014)

What a are the odds of an under 3 moves last layer for beginner's method on 15 puzzle? I got one earlier


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## TDM (Dec 18, 2014)

Hssandwich said:


> What a are the odds of an under 3 moves last layer for beginner's method on 15 puzzle? I got one earlier


What's the beginner's method?


----------



## Berd (Dec 18, 2014)

TDM said:


> What's the beginner's method?


Bens tutorial maybe? [emoji136]


----------



## TDM (Dec 18, 2014)

Berd said:


> Bens tutorial maybe? [emoji136]


Ok, but 


Hssandwich said:


> What a are the odds of an under 3 moves last layer for beginner's method on 15 puzzle? I got one earlier


What's the 'last layer'?


----------



## Berd (Dec 18, 2014)

TDM said:


> Ok, but
> 
> What's the 'last layer'?


Last 2 layers [emoji136]


----------



## Hssandwich (Dec 18, 2014)

Yeah, sorry, should have been clearer.


----------



## TDM (Dec 18, 2014)

Hssandwich said:


> Yeah, sorry, should have been clearer.


Ok, here's how I would see it, but I could be wrong:
There are 8!/2 = 20160 cases for the last two layers.
Using the beginner's method, your blank can be either were 11 or 12 goes. This means there are four <=3 move ways to solve from there (moving the blank):
D
RD
DR
LDR
4/20160= *1/5040*


----------



## Leo123 (Dec 18, 2014)

TDM said:


> What's the beginner's method?



The Dan Brown tutorial probably is.


----------



## GuRoux (Dec 18, 2014)

Leo123 said:


> The Dan Brown tutorial probably is.



i think he is talking about 15 puzzle beginner's method.


----------



## Hssandwich (Dec 18, 2014)

I'm a 15 puzzle nub, I average sub 20 on an app on my ipad


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## TDM (Dec 18, 2014)

Hssandwich said:


> I'm a 15 puzzle nub, I average sub 20 on an app on my ipad


It's hard to be fast on a touchscreen. Download ben's sim on a computer; it's much easier to use keyboard.
(use two hands; don't do what I did and use OH for half a year, because switching is hard)


----------



## Leo123 (Dec 27, 2014)

GuRoux said:


> i think he is talking about 15 puzzle beginner's method.


Oh!


----------



## PixelWizard (Jan 5, 2015)

How high is the possibility that you can use Winter Variation?


----------



## guysensei1 (Jan 5, 2015)

PixelWizard said:


> How high is the possibility that you can use Winter Variation?



pretty sure that the probability of a solved EO (so you can use WV) is 1/8 (I may be horribly wrong)


----------



## Bindedsa (Jan 5, 2015)

guysensei1 said:


> pretty sure that the probability of a solved EO (so you can use WV) is 1/8 (I may be horribly wrong)



Should be halved, unless you always force a made pair.


----------



## guysensei1 (Jan 5, 2015)

Bindedsa said:


> Should be halved, unless you always force a made pair.



Well it's not too hard to use R U' R' to get a solved pair. I guess. (Maybe we should account for last slot skips too but idk how?)

Also it depends on how you solve your F2L pairs. Maybe all of your F2L solutions don't end with a connected pair.


----------



## Antonie faz fan (Jan 5, 2015)

PixelWizard said:


> How high is the possibility that you can use Winter Variation?



100% ( uf you use edge control, wich is like very easy).


----------



## guysensei1 (Jan 5, 2015)

Antonie faz fan said:


> 100% ( uf you use edge control, wich is like very easy).



if you go through all that effort just to use WV you may as well do OLL/PLL right?


----------



## Antonie faz fan (Jan 5, 2015)

guysensei1 said:


> if you go through all that effort just to use WV you may as well do OLL/PLL right?



maybe, but he didn't ask what was more efficient to do though :3


----------



## PixelWizard (Jan 5, 2015)

guysensei1 said:


> if you go through all that effort just to use WV you may as well do OLL/PLL right?



If I understood right, then you get a oll skip everytime you use WV, so I planned to learn it and use it as much as possible... WV seems to drop the times a lot...

I average about 16-17 seconds. Before learning WV I have to learn OLL completely (around 5 to go).


----------



## marfer (Jan 5, 2015)

*(CFOP) Probabilities for 1LLL*

Hello speedcubers,
I know that there slightly more than 1000 legal configurations for the last layer (don't have the exact number right now but it's there somewhere in the wiki), but has anybody estimated what are the probabilities of getting each of them after Cross + F2L ? I suspect (hope) it's not an uniform distribution, so perhaps the most probable 30 configurations make up for 80% of the probability. That would mean that in 80% of the solves one could do a 1LLL (greatly lowering the turn count), and then the for the rest use classical OLL and PLL. Has this path been investigated too optimize CFOP ?
Thanks.


----------



## waffle=ijm (Jan 5, 2015)

As far as I know, any speedsolvers will tend to sway against 1LLL (though it be nice) for speed due to recognition. Though in terms of pure optimization there's Dr. Morley Davidson who knows full ZBLL and knows some 1LLL for the maths and FMC.

EDIT - He also knows Anti-ZBLL (correct me if I'm wrong on the name of this) which is ZBLL for when edges aren't oriented.


----------



## TDM (Jan 5, 2015)

If the case has rotational symmetry (you can do U2 then the same alg, and it will still solve the same case) then it has either half or a quarter of the chance of a case without symmetry (whether it's a half or a quarter depends on if it's just U2 symmetry, or if it's for all four AUFs i.e. H perm, N perms, skip, and those cases with a 4-edge flip). So for most cases, the probability of getting them is 1/3888. For some it's 1/7776, and for a couple it's 1/15552.


----------



## stoic (Jan 5, 2015)

Ah, Pareto's principle. 
Nice idea


----------



## Bindedsa (Jan 5, 2015)

marfer said:


> I suspect (hope) it's not an uniform distribution, so perhaps the most probable 30 configurations make up for 80% of the probability. That would mean that in 80% of the solves one could do a 1LLL (greatly lowering the turn count)


This can't possible be true because I always pay attention to the permutation of my LL while doing OLL, and I'm pretty sure I would have notice this by now. That being said I do find that some cases seem to come up more commonly, but not enough that I feel it would be worth focusing on them. It's much better to look for cases that are reasonably fast and easy to recognize.



waffle=ijm said:


> As far as I know, any speedsolvers will tend to sway against 1LLL (though it be nice) for speed due to recognition.


Not necessarily, if you get used to recognizing cases while doing your AUF, you don't lose that much time. It just means every now and then you'll end up doing U2 U' U' or something. It means I miss a lot of 1LLLs, but I don't I waste much time trying to recognize a case I don't know


----------



## irontwig (Jan 5, 2015)

marfer said:


> I suspect (hope) it's not an uniform distribution, so perhaps the most probable 30 configurations make up for 80% of the probability.
> Thanks.



No, that's not how it works, most LL cases are equally likely to occur.


----------



## marfer (Jan 5, 2015)

irontwig said:


> No, that's not how it works, most LL cases are equally likely to occur.



Too bad. The initial idea is interesting, but unless the statistical distribution is quite far from the uniform, one will end up needing a few hundred LL patterns to cover the 80% probability mass and that's not practical. I might end up simulating by computer the Cross + F2L stages just too see what exactly is this distribution. I am quite a lazy person though, so don't count on it  Plus I get the feeling that it's not so easy to program, otherwise someone would have already done it.


----------



## guysensei1 (Jan 5, 2015)

waffle=ijm said:


> As far as I know, any speedsolvers will tend to sway against 1LLL (though it be nice) for speed due to recognition. Though in terms of pure optimization there's Dr. Morley Davidson who knows full ZBLL and knows some 1LLL for the maths and FMC.


aaaaaaand then there's Bindedsa 


> EDIT - He also knows Anti-ZBLL (correct me if I'm wrong on the name of this) which is ZBLL for when edges aren't oriented.



Isn't AntiZBLL the set where corners are oriented but edges not (all) oriented?


----------



## Musicalboy2 (Jan 5, 2015)

marfer said:


> Hello speedcubers,
> I know that there slightly more than 1000 legal configurations for the last layer (don't have the exact number right now but it's there somewhere in the wiki), but has anybody estimated what are the probabilities of getting each of them after Cross + F2L ? I suspect (hope) it's not an uniform distribution, so perhaps the most probable 30 configurations make up for 80% of the probability. That would mean that in 80% of the solves one could do a 1LLL (greatly lowering the turn count), and then the for the rest use classical OLL and PLL. Has this path been investigated too optimize CFOP ?
> Thanks.



They should be evenly distributed.

Remember that permutation and orientation are separate.



guysensei1 said:


> aaaaaaand then there's Bindedsa
> 
> 
> Isn't AntiZBLL the set where corners are oriented but edges not (all) oriented?



I heard the same thing that waffo did about anti-ZBLL; I'm pretty sure anti-ZBLL is no edges oriented but otherwise the same idea as ZBLL.


----------



## cuBerBruce (Jan 6, 2015)

TDM said:


> If the case has rotational symmetry (you can do U2 then the same alg, and it will still solve the same case) then it has either half or a quarter of the chance of a case without symmetry (whether it's a half or a quarter depends on if it's just U2 symmetry, or if it's for all four AUFs i.e. H perm, N perms, skip, and those cases with a 4-edge flip). So for most cases, the probability of getting them is 1/3888. For some it's 1/7776, and for a couple it's 1/15552.



Precisely:

Number of cases with probability 1/3888: 3864
Number of cases with probability 1/7776: 44
Number of cases with probability 1/15552: 8 (including solved case)

Total cases: 3916

When marfer was talking about the number of cases being about 1000, I believe he(?) was thinking of the number of cases if you count mirrors and inverses as the same case. Then the total number of cases is 1212 (or 1211 if you don't count the solved case).


----------



## TDM (Jan 11, 2015)

What's the probability of not having at least one 1x1x2 pair (not calling centre+edge a pair) on a random state 3x3 scramble?


----------



## cmhardw (Jan 11, 2015)

TDM said:


> What's the probability of not having at least one 1x1x2 pair (not calling centre+edge a pair) on a random state 3x3 scramble?



I know I've attempted this one before in this thread. Searching... I used inclusion/exclusion, but perhaps it was argued (Stefan?) that this is not a valid way to count these cases?

--edit--
Lucas answers the question here


----------



## TDM (Jan 11, 2015)

cmhardw said:


> I know I've attempted this one before in this thread. Searching... I used inclusion/exclusion, but perhaps it was argued (Stefan?) that this is not a valid way to count these cases?
> 
> --edit--
> Lucas answers the question here


Thanks! I'm not looking for a very specific answer so Lucas' estimate is fine, I just wanted to know whether I was getting really unlucky in my Roux solves or not


----------



## obelisk477 (Jan 11, 2015)

I remember seeing someone brute force this a while ago, and for some reason it was suspiciously close to 1÷e=.368 . Can't remember where I saw it though


----------



## FJT97 (Jan 15, 2015)

I just had 3 times in a row sure into the same R-Perm.
How is the probability for that?


----------



## obelisk477 (Jan 15, 2015)

Depends on if you're using beginners or edge control or something like that, where you always do OCLL then PLL, or if you do no edge control and standard OLL to PLL


----------



## ryanj92 (Jan 15, 2015)

marfer said:


> Too bad. The initial idea is interesting, but unless the statistical distribution is quite far from the uniform, one will end up needing a few hundred LL patterns to cover the 80% probability mass and that's not practical. I might end up simulating by computer the Cross + F2L stages just too see what exactly is this distribution. I am quite a lazy person though, so don't count on it  Plus I get the feeling that it's not so easy to program, otherwise someone would have already done it.



I'd argue that a simulated distribution isn't that useful - CFOP solvers rarely solve F2L in a 'blind' way and so we would expect the distribution to vary in some way from solver to solver. For instance of one uses basic edge control then all the LL cases with zero edges oriented don't show up...


----------



## FJT97 (Jan 16, 2015)

obelisk477 said:


> Depends on if you're using beginners or edge control or something like that, where you always do OCLL then PLL, or if you do no edge control and standard OLL to PLL



I use zz, so thats in fact with a higher chance..
So, i use zz but no wv and no sun colls...


----------



## marfer (Jan 18, 2015)

cuBerBruce said:


> Precisely:
> 
> Number of cases with probability 1/3888: 3864
> Number of cases with probability 1/7776: 44
> ...



Thanks for the precise numbers. The mirrors I gather they correspond to the four axes of symmetry of a square (Left-Right, Up-Down and diagonals), but what is an inverse of a LL pattern ?

I wrote a small computer simulation to test the probabilities, following strictly the CFOP method: a hand coded cross (not optimal ~average to 10 moves to solve) and then plug in the F2L algorithms from the wiki. Cube solves are started from a scrambled state with 30 random moves. Statistics on 1,000,000 cube solves produce the expected distinct 3916 cases (patterns) after the F2L (any color permutation, U, U2 or U' operatios are considered the same case). To further reduce these I consider equivalent all symmetric patterns on the four symmetry axes of the U face (vertical, horizontal and diagonals). After this step the 3916 cases reduce to 2054, still far from the expected 1212. What am I missing ?

So far, for the 2054 cases, the probability distribution looks depressingly uniform in average (see the attached ). That means there are some patterns that are more probable than others but to reach for example 50 percent probability one need to add about half of the cases. The cumulative distribution function is very close to the main diagonal of the graph (attached ).

For information, here is the probability distribution  and cumulative distribution function  for the 3916 cases without counting symmetrics as the same case. Still very close to uniform distribution.

If someone helps with the missing symmetries, I would like to produce graphs for the fully reduced situation with 1212 cases, so that we can safely conclude that the probabilistic attack on the last layer is doomed to failure because of the uniform probability distribution.


----------



## cuBerBruce (Jan 18, 2015)

marfer said:


> I wrote a small computer simulation to test the probabilities, following strictly the CFOP method: a hand coded cross (not optimal ~average to 10 moves to solve) and then plug in the F2L algorithms from the wiki. Cube solves are started from a scrambled state with 30 random moves. Statistics on 1,000,000 cube solves produce the expected distinct 3916 cases (patterns) after the F2L (any color permutation, U, U2 or U' operatios are considered the same case). To further reduce these I consider equivalent all symmetric patterns on the four symmetry axes of the U face (vertical, horizontal and diagonals). After this step the 3916 cases reduce to 2054, still far from the expected 1212. What am I missing ?



Yes, I agree that 2054 is correct number of cases when mirror symmetry is considered in addition to rotational symmetry.

What you are missing is antisymmetry, or considering inverses to be the same case. Then, the case generated by L F2 R' F' R F' L' and the case generated by L F R' F R F2 L' would be considered the same case, for example. From a speedsolving perspective, I don't know if it's practical to use this, other than rote memorization of what patterns are inverse cases of each other.



marfer said:


> If someone helps with the missing symmetries, I would like to produce graphs for the fully reduced situation with 1212 cases, so that we can safely conclude that the probabilistic attack on the last layer is doomed to failure because of the uniform probability distribution.



I think I can pretty much guarantee that the inclusion of antisymmetry is not going to change the overall look of the graphs very much (except for scaling factors). The 1212 cases are dominated by 801 cases each having a probability of 1/972.


----------



## marfer (Jan 23, 2015)

cuBerBruce said:


> I think I can pretty much guarantee that the inclusion of antisymmetry is not going to change the overall look of the graphs very much (except for scaling factors). The 1212 cases are dominated by 801 cases each having a probability of 1/972.



OK thanks, I think you are right. To complete the situation, here are the probabilities for the 58 OLL cases (57 + the solved OLL case) computed over 1,000,000 cube solves. As expected, close to enough to uniform. See the diagram . If I reduce all symmetric cases (as explained in my previous post) I get 41 different patterns and the uniformity breaks: . There are roughly 3 categories: the most probable 20 cases are still equiprobable, then the next 15, and then the remaining 6 which are less likely than 1%. Anyway, the symmetric cases I suspect cannot be solved by the same algorithm so this will not give a hint in which order to learn the OLL.


----------



## JemFish (Jan 29, 2015)

Got a PLL skip and an OLL skip at the same time this morning, but it was a hand-scramble so I can't reconstruct. What are the chances of that?


----------



## FailCuber (Jan 29, 2015)

I once got 10 gperms in a row before at a comp. 5 was from 1st round and 5was from 2nd what is the probability of that happening?


----------



## cmhardw (Jan 29, 2015)

JemFish said:


> Got a PLL skip and an OLL skip at the same time this morning, but it was a hand-scramble so I can't reconstruct. What are the chances of that?



1/15552



FailCuber said:


> I once got 10 gperms in a row before at a comp. 5 was from 1st round and 5was from 2nd what is the probability of that happening?



If there were only 2 rounds (or three rounds and you made it through 2 of them) then this probability is:
(2/9)^10

If you competed in three rounds, and you're curious about this same situation, then it makes sense to ask a slightly different question:

Given that you competed in three rounds, what's the probability that you get at least 10 consecutive G perms in your 15 official solves?

Turns out that probability is about 7*10^(-7)

Clicky

--edit--

That probability is a little misleading because I think you would have made a post like this if you had gotten 10 of any PLL in a row. The probability that you got any PLL at least 10 times in a row in three rounds (or even in two rounds) would be greater than the probabilities listed, but still on the order of 10^(-6) for the first significant digit


----------



## FailCuber (Jan 29, 2015)

cmhardw said:


> 1/15552
> 
> 
> 
> ...


I don't know that stuff, can you explain it to me more simply etc: 1/X


----------



## JemFish (Jan 29, 2015)

Wow, a 1/15552 change of getting a OLL and an PLL skip at the same time?! I feel special...


----------



## cmhardw (Jan 29, 2015)

FailCuber said:


> I don't know that stuff, can you explain it to me more simply etc: 1/X



(2/9)^10 is approximately 1/3405063

The probability that you get at least 10 consecutive G-perms in 3 rounds of official solves is approximately 1/1525860

The probability in both of those situations for getting any PLL, and not restricting to just G perm, would be on the order of about 10 times larger than each of those numbers.


----------



## qqwref (Jan 29, 2015)

JemFish said:


> Wow, a 1/15552 change of getting a OLL and an PLL skip at the same time?! I feel special...


Considering it was a handscramble, don't feel so special... it's way more likely that you just accidentally gave yourself a position that would be solved after a relatively obvious F2L. The 1/15552 number is for random scrambles (and it assumes you don't do anything at all to affect the LL during F2L).


----------



## JemFish (Jan 29, 2015)

qqwref said:


> Considering it was a handscramble, don't feel so special... it's way more likely that you just accidentally gave yourself a position that would be solved after a relatively obvious F2L. The 1/15552 number is for random scrambles (and it assumes you don't do anything at all to affect the LL during F2L).



I assure you that the scramble was a random scramble, which I achieve by applying random turns to the cube for a long while, and then making sure it's completely scrambled before inspection. Neither did I do anything to affect the last layer, but just normal F2L.


----------



## TDM (Jan 29, 2015)

JemFish said:


> I assure you that the scramble was a random scramble, which I achieve by applying random turns to the cube for a long while, and then making sure it's completely scrambled before inspection. Neither did I do anything to affect the last layer, but just normal F2L.


How do you know that it was completely random?


----------



## JemFish (Jan 29, 2015)

TDM said:


> How do you know that it was completely random?




Either that's a very dumb question or we're using different dictionaries. 

If it's a dumb question then here's my (dumb) answer: the colours were completely messed up.

If we're using different dictionaries, then here's the definition of 'random' according to Oxford dictionaries: "made, done, or happening [or scrambled] without method or conscious decision."


----------



## guysensei1 (Jan 29, 2015)

JemFish said:


> Either that's a very dumb question or we're using different dictionaries.
> 
> If it's a dumb question then here's my (dumb) answer: the colours were completely messed up.
> 
> If we're using different dictionaries, then here's the definition of 'random' according to Oxford dictionaries: "made, done, or happening [or scrambled] without method or conscious decision."


I'm pretty sure you conciously decided which move to make when hand scrambling?


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## AlphaSheep (Jan 29, 2015)

JemFish said:


> Either that's a very dumb question or we're using different dictionaries.



It's certainly not a dumb question. Humans are absolutely terrible judges of whether or not something is random. We normally judge something as random if we see no obvious patterns, whereas something truly random is usually quite likely to have some patterns. A lack of a pattern is a form of order, so when we judge a scramble based on whether or not they "look random", we are actually biasing the scramble to one with some form of order.


----------



## tseitsei (Jan 29, 2015)

JemFish said:


> Either that's a very dumb question or we're using different dictionaries.
> 
> If it's a dumb question then here's my (dumb) answer: the colours were completely messed up.
> 
> If we're using different dictionaries, then here's the definition of 'random' according to Oxford dictionaries: "made, done, or happening [or scrambled] without method or conscious decision."



You obviously don't know what random scrambles are...


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## JemFish (Jan 29, 2015)

tseitsei said:


> You obviously don't know what random scrambles are...



Yep, I have concluded we use different dictionaries. So...enlighten me - what is a random scramble?



AlphaSheep said:


> It's certainly not a dumb question. Humans are absolutely terrible judges of whether or not something is random. We normally judge something as random if we see no obvious patterns, whereas something truly random is usually quite likely to have some patterns. A lack of a pattern is a form of order, so when we judge a scramble based on whether or not they "look random", we are actually biasing the scramble to one with some form of order.



Exactly what are the chances of getting a un-random scramble by hand-scrambling it for 30 seconds at 3 tps?



guysensei1 said:


> I'm pretty sure you conciously decided which move to make when hand scrambling?



And just how could I do that?


----------



## AlphaSheep (Jan 29, 2015)

JemFish said:


> Exactly what are the chances of getting a un-random scramble by hand-scrambling it for 30 seconds at 3 tps?



Your question doesn't make sense. If it's a truly random scramble, then the probability of getting a solved cube is exactly the same as the probability of getting any other scrambled state.



JemFish said:


> Yep, I have concluded we use different dictionaries. So...enlighten me - what is a random scramble?



Random sequences of moves, even ones generated by a computer, are pretty terrible scrambles, because they are far more likely to result in some states, and far less likely to reach some others. The way the official scrambling programs work is that they choose a state at random from all possible states with equal likelihood (basically by placing pieces in random positions and in with random orientations), and then calculate a sequence of moves that will reach that state.


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## JemFish (Jan 29, 2015)

AlphaSheep said:


> Your question doesn't make sense. If it's a truly random scramble, then the probability of getting a solved cube is exactly the same as the probability of getting any other scrambled state.




OK: I scrambled the cube, randomly or not randomly, whichever way you prefer. I had an OLL skip. I then had a PLL skip. The chances of that are less than or equal to 1/15552. Happy?



AlphaSheep said:


> Random sequences of moves, even ones generated by a computer, are pretty terrible scrambles, because they are far more likely to result in some states, and far less likely to reach some others. The way the official scrambling programs work is that they choose a state at random from all possible states with equal likelihood (basically by placing pieces in random positions and in with random orientations), and then calculate a sequence of moves that will reach that state.



So what you're saying, is that if I throw a bunch of sticks on the floor a hundred times within a certain area, it would not be as random as if I put each stick down a hundred times, deliberately, and recorded each 'random' set-up, knowing not to repeat that same set-up in the future? Well this would only be effective if I threw/set-up the sticks at least a billion times.

I haven't even reached a billionth of the total possible number of permutations on a Rubik's Cube, estimated at 43,252,003,274,489,856,000, so a concept of being "far more likely to result in some states, and far less likely to reach some others" is like expecting that you'll somehow get to all the 43 quintillion scrambles possible.


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## tseitsei (Jan 29, 2015)

JemFish said:


> Yep, I have concluded we use different dictionaries. So...enlighten me - what is a random scramble?




let me google that for you

The first link from google explains it quite well already


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## Stefan (Jan 29, 2015)

tseitsei said:


> let me google *something else* for you



ftfy


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## Berd (Jan 29, 2015)

Probability of an Xcross on 4x4 after edge paring using Yau?


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## tseitsei (Jan 29, 2015)

Stefan said:


> ftfy



Ok. Then please enlighten me why random state scramble wouldn't be completely random?


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## Stefan (Jan 29, 2015)

tseitsei said:


> Ok. Then please enlighten me why random state scramble wouldn't be completely random?



Huh? Who said that?


----------



## tseitsei (Jan 29, 2015)

Stefan said:


> Huh? Who said that?



Well I gave him a link that explains what we consider random scrambles and you said that I googled something totally different...


----------



## Robert-Y (Jan 29, 2015)

What's the probability that I get a specific case a specific number of times in a row for a specific number of solves?!

To me it seems that recently some people in this thread have not realized that their questions are actually very oddly specific. Most of these occurrences do not seem very significant or interesting to me at all...

EDIT: Oops. Nevermind, it's just 2 "recent" requests actually...


----------



## Stefan (Jan 29, 2015)

tseitsei said:


> Well I gave him a link that explains what we consider random scrambles and you said that I googled something totally different...



No you claimed to have googled "that", but the talk was about "random scramble" and you instead googled "random state scramble". That's not the same, it's something else. Which is what I said. And in fact, if you take out the word "state" that you added, the first result (at least for me) about cubing is *not* using random state scrambles but random turns.


----------



## tseitsei (Jan 29, 2015)

Stefan said:


> No you claimed to have googled "that", but the talk was about "random scramble" and you instead googled "random state scramble". That's not the same, it's something else. Which is what I said. And in fact, if you take out the word "state" that you added, the first result (at least for me) about cubing is *not* using random state scrambles but random turns.



Using random moves isn't really what (most of the) speedcubers consider being random scrambles...

But yeah I see what you mean now.


----------



## cmhardw (Jan 29, 2015)

Robert-Y said:


> What's the probability that I get a specific case a specific number of times in a row for a specific number of solves?!
> 
> To me it seems that recently some people in this thread have not realized that their questions are actually very oddly specific. Most of these occurrences do not seem very significant or interesting to me at all...
> 
> EDIT: Oops. Nevermind, it's just 2 "recent" requests actually...



I agree that people often ask extraordinarily specific questions when they probably mean a much more general question without realizing.


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## obelisk477 (Jan 29, 2015)

cmhardw said:


> I agree that people often ask extraordinarily specific questions when they probably mean a much more general question without realizing.



I think I'm gonna start responding to this thread only using equivalent probabilities. "What's the odds of getting a U-perm 4 times in a row!?!?!!11?" "Same as the odds of U perm, then an R perm, then a J perm, then an A perm"


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## cmhardw (Jan 29, 2015)

Berd said:


> Probability of an Xcross on 4x4 after edge paring using Yau?



The probability of at least one solved corner edge pair in the F2L after edge pairing, and using Yau, is about 0.01

Clicky

--edit--

Apparently the old way I used to type sums into Wolfram Alpha does not work well anymore. I rewrote the wolfram alpha link using their help files. The link should work now.

--edit 2--

Ugh... I have to include the division by the total number of cases in the summation itself to get it to work properly :\ I'm not a fan of this, but I think the link should finally work.


----------



## TMOY (Jan 30, 2015)

Berd said:


> Probability of an Xcross on 4x4 after edge paring using Yau?



Pleae give a proper definition of "Xcross after edge pairing" or else it's impossible to answer your question. Which pieces have to be where after edge pairing ?
And yes, I know what an Xcross is.


----------



## guysensei1 (Jan 31, 2015)

What's the probability of getting the same OLL and PLL case (but not necessarily from the same angle!) on consecutive solves?


----------



## obelisk477 (Jan 31, 2015)

guysensei1 said:


> What's the probability of getting the same OLL and PLL case (but not necessarily from the same angle!) on consecutive solves?



Depends on the OLL/PLL?


----------



## guysensei1 (Jan 31, 2015)

obelisk477 said:


> Depends on the OLL/PLL?



The average probability?


Also I was referring to getting both at once, not as separate events. Am I making sense?


----------



## obelisk477 (Jan 31, 2015)

guysensei1 said:


> The average probability?
> 
> 
> Also I was referring to getting both at once, not as separate events. Am I making sense?



Well the first OLL/PLL pair is given, so the probability is 1 for it. The probability that any given OLL/PLL 'pair' is then followed by the same 'pair' is 1*(1/58)*(1/22) (if you include skips as possibilities for OLL/PLL) = 1/1276 = .0784%


----------



## Stefan (Jan 31, 2015)

obelisk477 said:


> Well the first OLL/PLL pair is given, so the probability is 1 for it. The probability that any given OLL/PLL 'pair' is then followed by the same 'pair' is 1*(1/58)*(1/22) (if you include skips as possibilities for OLL/PLL) = 1/1276 = .0784%



Not all OLL cases have the same probability. And not even a single one has probability 1/58. Same for PLL.

Imagine one of the OLLs had probability 0.5, and one of the PLLs as well. You'd still say 1/1276, but even just the probability of getting this particular OLL/PLL combination twice in two attempts is already 1/16.


----------



## obelisk477 (Jan 31, 2015)

Stefan said:


> Not all OLL cases have the same probability. And not even a single one has probability 1/58.
> 
> Same for PLL.



He said average probability. The average is just 1/#ofcases right? 57+skip and 21+skip

EDIT: Yes those that would be true for a particular OLL and PLL. But the way I understood the question, he meant an overall average for all OLLs and PLLs in general.


----------



## Stefan (Jan 31, 2015)

obelisk477 said:


> He said average probability.



Just because you had trouble understanding the perfectly fine original question. I don't think he intended to change the meaning.



obelisk477 said:


> The average is just 1/#ofcases right? 57+skip and 21+skip



Well, yeah, but I don't think that's interesting. And it certainly doesn't answer the question for the probability of getting the same OLL/PLL twice in a row.

Also, while the average probability for each such OLL case is 1/58, I'm not sure the average such repetition probability is. And I'm too tired to think about it now it's 6:09am I'm going to bed I don't even feel like using commas anymore


----------



## obelisk477 (Jan 31, 2015)

Stefan said:


> Just because you had trouble understanding the perfectly fine original question. I don't think he intended to change the meaning.
> 
> Well, yeah, but I don't think that's interesting. And it certainly doesn't answer the question for the probability of getting the same OLL/PLL twice in a row.



I didn't misunderstand. I asked for the specific OLL/PLL. And would be willing to calculate it if it were provided!


----------



## Stefan (Jan 31, 2015)

obelisk477 said:


> I didn't misunderstand. I asked for the specific OLL/PLL.



Well you said "Depends on the OLL/PLL?", which is wrong (or rather, since you made it a weird question, the answer is "no").


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## obelisk477 (Jan 31, 2015)

Stefan said:


> Well you said "Depends on the OLL/PLL?", which is wrong (or rather, since you made it a weird question, the answer is "no").



It does depend on it, like you mentioned in your eariler example. It's either a specific OLL/PLL combo, or a general one. If the latter, then my answer is correct (1/12xx); if the former, then I was warranted in asking which OLL and PLL he meant. Also, it would help it OP would weigh in...

EDIT: OP's question could also mean 'what are the odds of getting A oll, then B pll, and then A oll again, and then B pll again.' Then the odds are much smaller


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## Stefan (Jan 31, 2015)

obelisk477 said:


> It does depend on it



No it doesn't.

Imagine there were just two OLL cases, one with 40% chance and one with 60% chance. Now what's the probability of getting the same case in two attempts?

The answer is 52%.

There. A perfectly fine answer to a perfectly fine question. Doesn't depend on the OLL case. Also note it's *not* 50%.


----------



## guysensei1 (Jan 31, 2015)

Umm... Was my question unclear? I'm not understanding this debate.


----------



## Berd (Jan 31, 2015)

guysensei1 said:


> Umm... Was my question unclear? I'm not understanding this debate.


Just let it happen bebe


----------



## Stefan (Jan 31, 2015)

guysensei1 said:


> Umm... Was my question unclear? I'm not understanding this debate.



Your original question was alright (the only problem I see is that you didn't specify whether you're just doing two solves, though that's a reasonable assumption). I'm not sure about your addition of _"The average probability?"_.


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## guysensei1 (Jan 31, 2015)

Stefan said:


> I'm not sure about your addition of _"The average probability?"_.


I'm not sure either. I was slightly confused at obelisk's response. I assumed he was talking about the probability of a single OLL and PLL case so I asked for the average of all of the probabilities but now that I think about it I make no sense


----------



## obelisk477 (Jan 31, 2015)

guysensei1 said:


> I'm not sure either. I was slightly confused at obelisk's response. I assumed he was talking about the probability of a single OLL and PLL case so I asked for the average of all of the probabilities but now that I think about it I make no sense



Yeah, the average of all probabilities is in fact 1/58 and 1/22 for OLL and PLL respectively, but as it turns out those don't really matter for the calculation.

EDIT: Would it still give you a good approximation to do it the way i tried? instead of summing all 57 OLLs or whatever


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## Stefan (Jan 31, 2015)

Ugh, my edit was too slow, so let me move it here:

I think what you wanted is like the 52% of my example. If you understand that calculation, you can do it for real OLL with the probabilities from here. I'm too lazy, plus I'd rather have people do such stuff themselves if I believe they can.


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## guysensei1 (Jan 31, 2015)

Let me confirm this...

I can get the answer by doing
(Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2

And multiply that by

(Probability of PLL1)^2+...+(Probabilty of PLL21)^2+(Probabilty of PLL skip)^2 ?


----------



## cmhardw (Jan 31, 2015)

obelisk477 said:


> Yeah, the average of all probabilities is in fact 1/58 and 1/22 for OLL and PLL respectively, but as it turns out those don't really matter for the calculation.



Oh wow! I never realized this! So:
(P(OLL1)+P(OLL2)+...+P(OLL58))/58 = 1/58

This is true because:


Spoiler



Given a universe with events a_1, a_2, ... , a_n that form a partition of the total space of outcomes then:

1/n * (P(a_1)+P(a_2)+...+P(a_n)) = 1/n

Because P(a_1)+P(a_2)+...+P(a_n)=1 since these events form a partition of the total possible outcomes.

Very cool! I'm excited to have learned something!





guysensei1 said:


> Let me confirm this...
> 
> I can get the answer by doing
> (Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2
> ...



define event e_1 to be OLL1 followed by PLL1
define event e_2 to be OLL1 followed by PLL2
...
define event e_22 to be OLL1 followed by PLL22
define event e_23 to be OLL2 followed by PLL1
etc.

What is the probability of getting event e_1 twice in a row? Of getting event e_2 twice in a row? Of getting event e_3 twice in a row?

Now what is the probability of getting the same OLL and PLL pair twice in a row?

--edit--

Looking at your response I think what you and I have said is equivalent


----------



## Stefan (Jan 31, 2015)

guysensei1 said:


> Let me confirm this...
> 
> I can get the answer by doing
> (Probability of OLL1)^2+(probability of OLL2)^2..........+(Probability of OLL57)^2+(probability of OLL skip)^2
> ...



Yes.


----------



## Stefan (Jan 31, 2015)

cmhardw said:


> Oh wow! I never realized this! So:
> (P(OLL1)+P(OLL2)+...+P(OLL58))/58 = 1/58



I might have never thought about it, either. I did find it interesting as well. Then again, I think it's only useful if you want to lie with statistics  (if you say "average probability" instead of "probability", I suspect most people will ignore the "average", and you can legitimately change the answer that way if you like it better than the real answer)



cmhardw said:


> Looking at your response I think what you and I have said is equivalent



Same idea, though his way is easier since OLL and PLL are independent and he exploits that (especially if you also exploit that many cases share the same probability).


----------



## Antonie faz fan (Apr 22, 2015)

which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)


----------



## Hssandwich (Apr 22, 2015)

Antonie faz fan said:


> which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)



H-perm?


----------



## cmhardw (Apr 23, 2015)

Antonie faz fan said:


> which oll and pll have the biggest propability to occure?? and wich ones the least? ( all edges down all corners up and E perm ?)



OLL
PLL


----------



## JustinTimeCuber (Apr 23, 2015)

OLL-20: 1/216 = 0.046%

PLL-H: 1/72 = 1.389%


OLL/PLL skips have equally low probabilities to the ones listed.


----------



## Suzuha (Apr 23, 2015)

JustinTimeCuber said:


> OLL-20: 1/216 = 0.046%
> 
> PLL-H: 1/72 = 1.389%
> 
> ...



That's interesting. I got that OLL like 6 times but like 2 OLL skips.


----------



## cmhardw (Apr 23, 2015)

GoldenOak said:


> That's interesting. I got that OLL like 6 times but like 2 OLL skips.



Your distribution of X-OLLs and OLL skips is not as rare as it might seem

The probability above is the probability of two or fewer OLL skips in 8 "significant outcomes" where each "significant outcome" is a solve that has either an X-OLL or an OLL skip.

It is often the case that what seems at first to be a strange outcome is more common than you would expect.


----------



## unsolved (Apr 24, 2015)

JustinTimeCuber said:


> OLL-20: 1/216 = 0.046%
> 
> PLL-H: 1/72 = 1.389%
> 
> ...



So what are the solutions to each of those cubes that are the fewest number of moves? In the case of the top one, the last-layer-superflip is the case I am interested in.


----------



## Lucas Garron (Apr 25, 2015)

unsolved said:


> So what are the solutions to each of those cubes that are the fewest number of moves? In the case of the top one, the last-layer-superflip is the case I am interested in.



Visit ACube.js, twist every LL edge, set "ignore position" for every LL piece, then "solve graphical input":
Gets you: F B' R2 D2 R F2 B2 L F B' . U2 R2 // 18q, 12f, 9s

EDIT: Simplified using slice moves: S' U2 R2 U S2 U S' U2 R2


----------



## TheCoolMinxer (Apr 25, 2015)

What is the chance of getting a l3e skip on 7x7, being left with inner parity on the last edge, after realigning the centers?
Happened to me last weekend at a comp, overall PB by 25 sec! (3:58)


----------



## AlexMaass (Apr 26, 2015)

Probability of Feliks getting 3x3 single WR back?


----------



## PenguinsDontFly (Apr 26, 2015)

AlexMaass said:


> Probability of Feliks getting 3x3 single WR back?



Why u ask? Who cares? Congratulate collin instead of wishing it was feliks. I hate it when ppl do this. (ex: cubing world andrew ricci how to get fast at 3x3 and ppl complained about hot it wasnt feliks.) He isnt the only fast cuber by the way. Respect plz for collin.


----------



## AlexMaass (Apr 26, 2015)

PenguinsDontFly said:


> Why u ask? Who cares? Congratulate collin instead of wishing it was feliks. I hate it when ppl do this. (ex: cubing world andrew ricci how to get fast at 3x3 and ppl complained about hot it wasnt feliks.) He isnt the only fast cuber by the way. Respect plz for collin.



Yeah I know Faz isn't the only fast cuber ofc. Collin does deserve it and congrats but I'm rooting for faz to get single wr back.


----------



## cmhardw (Apr 26, 2015)

AlexMaass said:


> Probability of Feliks getting 3x3 single WR back?



After much thought and deliberation, I am at least 95% confident that this probability lies in the closed interval [0,1]


----------



## PenguinsDontFly (Apr 26, 2015)

cmhardw said:


> After much thought and deliberation, I am at least 95% confident that this probability lies in the closed interval [0,1]



Lol thats actually funny.


----------



## penguinz7 (Apr 26, 2015)

AlexMaass said:


> Probability of Feliks getting 3x3 single WR back?



50/50, Either he will or he won't.


----------



## Tempus (Apr 26, 2015)

penguinz7 said:


> 50/50, Either he will or he won't.


Right, just like the odds of the sun rising tomorrow morning. Either it will or it won't, so 50/50.


----------



## JustinTimeCuber (May 8, 2015)

Tempus said:


> Right, just like the odds of the sun rising tomorrow morning. Either it will or it won't, so 50/50.



XD

What is the probability of getting an OLL skip OR (inclusive) a PLL skip (in one solve)?


----------



## TDM (May 8, 2015)

JustinTimeCuber said:


> XD
> 
> What is the probability of getting an OLL skip OR (inclusive) a PLL skip?


OLL: 1/216
PLL: 1/72
OLL or PLL: 1/216 + 1/72 - (1/216)(1/72)
= 287/15552
= 1.85%


----------



## JustinTimeCuber (May 8, 2015)

Couldn't you do it as this: P = 1 - (215/216)(71/72) = 1 - 15265/15552 = 287/15552? Just seems more intuitive to me. 

I just got an OLL skip on my 4x4. Is that a 1/432 chance or am I doing something wrong?


----------



## adimare (May 9, 2015)

Yeah, the probability of not getting parity + getting an OLL skip is 1/432. However, I could argue that getting an OLL skip after parity should also count (this case for most people) bringing the probability back down to 1/216 (this assumes parity is solved by flipping the one miss oriented edge and not just a random one).


----------



## guysensei1 (May 9, 2015)

What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?


----------



## Julian (May 9, 2015)

guysensei1 said:


> What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?


Assuming every solve path goes through EPLL (2-look PLL, COLL/EPLL, etc):

EPLL skip: 1/12
not EPLL skip: 11/12
not EPLL skip 50 times: (11/12)^50 = 0.01289947263

EDIT: misread the question. My answer is for not skipping EPLL in 50 solves.


----------



## guysensei1 (May 9, 2015)

Julian said:


> Assuming every solve path goes through EPLL (2-look PLL, COLL/EPLL, etc):
> 
> EPLL skip: 1/12
> not EPLL skip: 11/12
> ...



I was assuming 1 look PLL.


----------



## Cale S (May 9, 2015)

guysensei1 said:


> What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?



Getting an EPLL is equivalent to having CP solved, which has a 1/6 probability.
The probability of not having solved CP 50 times in a row (assuming no COLL or anything like that) is (5/6)^50 = 0.00010988481... or about 1/9100


----------



## TDM (May 9, 2015)

JustinTimeCuber said:


> Couldn't you do it as this: P = 1 - (215/216)(71/72) = 1 - 15265/15552 = 287/15552? Just seems more intuitive to me.
> 
> I just got an OLL skip on my 4x4. Is that a 1/432 chance or am I doing something wrong?


You could do it that way too.
Yes, it's 1/432.


guysensei1 said:


> What's the probability of not getting EPLL (PLL skips are counted as EPLLS) in an average of 50?


Depends how much OLLCP you know


----------



## Goosly (May 22, 2015)

Since it seems that misscrambles happen way too often, because of scramblers not properly checking the scramble, I was thinking about writing some application to check if a *picture of a scrambled cube matches the scramble*.

One issue I stumbled upon is the fact that people have weird colorscheme's. Therefor, it's only possible to check for "patterns", for example (top side):
[A, A, B]
[C, D, A]
[B, A, E]
Where A, B, etc. could be any color, since we don't know what colors the cube has anyway. But we do know that (UBL, UB, UR, UF) all have the same color.
Since the application needs a picture of a cube, we have 3 sides to check (or, 27 stickers, minus 6 stickers used as "reference")

*My question:*
For a scramble and a "pattern", constructed like described before, what is the probability that the pattern seems to match the scramble, but the cube was actually misscrambled? Or, in other words, how trustworthy is this method? 

Sorry if anything is unclear.


----------



## cashis (May 22, 2015)

What are the odds of two OLL skips in a row?


----------



## JediJupiter (May 22, 2015)

Goosly said:


> *My question:*
> For a scramble and a "pattern", constructed like described before, what is the probability that the pattern seems to match the scramble, but the cube was actually misscrambled? Or, in other words, how trustworthy is this method?
> 
> Sorry if anything is unclear.



I don't understand what's wrong with the current method. You know the orientation you scrambled the cube in, and you know that say, yellow on the scrambled image should be black on the physical cube. 

If this is for a human to check, would they insert the colours into the scrambler beforehand? Or would they have to read it as a bunch of letters? If you're inserting the colours yourself, it'd be hard to implement in competitions and Prisma Puzzle timer already does it if it's for personal use. If you meant for somebody to read it as letters, it's harder to recognise than just knowing that one colour may correspond to a different colour.

Do you mean mis-scrambling by missing or adding one or two moves, or any scrambled state that aren't the one it needs to be? A 3-cycle on the D face might be hard to recognise if a human is just glancing at the cube, but at the same time, it wouldn't be possible to get to that state by only mistaking two moves. Also, you should put an example of how much of the cube is properly checked, since different people check different amounts. Try something like checking the U and F face, since I think most people check those.


----------



## Goosly (May 22, 2015)

JediJupiter said:


> If this is for a human to check, would they insert the colours into the scrambler beforehand? Or would they have to read it as a bunch of letters?



The human doesn't check anything, because clearly humans suck at checking scrambles 
In some further stage, it should be possible to take a picture of the cube with your smartphone and have the app say "Yeah, I think this matches the scramble" or "You misscrambled".


----------



## TDM (May 22, 2015)

cashis said:


> What are the odds of two OLL skips in a row?


(1/216)^2 = 1/46656 = 0.002%


----------



## cashis (May 22, 2015)

TDM said:


> (1/216)^2 = 1/46656 = 0.002%



Thanks. Happened yesterday.


----------



## JediJupiter (May 22, 2015)

Goosly said:


> The human doesn't check anything, because clearly humans suck at checking scrambles
> In some further stage, it should be possible to take a picture of the cube with your smartphone and have the app say "Yeah, I think this matches the scramble" or "You misscrambled".



Okay gotcha. So 3 faces correct, since that's the most with one picture, all adjacent. As a noob at probability calculating I think it's 54/(43*10^15) (I can't remember the actual number of possible states and my internet is too slow right now) which is tiny. But even so, that's from somehow randomly misscrambling the rest, and the amount of those states that could actually be achieved by misscrambling by one or two moves would be a lot smaller, I'd guess for most scrambles it's actually 0. By the way, will you actually make a timer with this function? If so, that's really cool.


----------



## Goosly (May 22, 2015)

JediJupiter said:


> By the way, will you actually make a timer with this function? If so, that's really cool.



I don't think people want to take a picture of their cube after each scramble when practicing at home. What we do want is a more reliable way for checking scrambles at a competition.


----------



## CubesOfTheWorld (Jun 16, 2015)

What is the chance of getting a cross skip on any side? on a specific side?


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## TDM (Jun 16, 2015)

CubesOfTheWorld said:


> What is the chance of getting a cross skip on any side? on a specific side?


On a specific side,
1/(12*11*10*9*2^4) = 1/(11880 * 16)
= 1/190080.
For any side,
30942274/980995276800 = 0.00003154


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## Isaac Lai (Jun 16, 2015)

This is probably around somewhere but I figured it might be easier to ask: What is the chance of a OLL skip?


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## guysensei1 (Jun 16, 2015)

Isaac Lai said:


> This is probably around somewhere but I figured it might be easier to ask: What is the chance of a OLL skip?


1/216 assuming you don't use any LL control during F2L


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## JustinTimeCuber (Jun 16, 2015)

lol your avatars are looking at each other XD


----------



## IRNjuggle28 (Jun 18, 2015)

I just got a LS+LL skip with no AUF and no parity on a 4x4, and no, that's not a typo of any sort. I solved my third pair, and the cube was solved. I'm completely in disbelief, and I feel pretty certain this is the most statistically unlikely thing that will ever happen to me with cubing, no matter how long I cube. I'm absolutely stunned that this happened. It'll be 1/8 as likely as a normal last layer skip, considering that I didn't get either parity or AUF, but I have no idea how unlikely skipping last slot in addition to all that is. 

It was a seemingly normal scramble from a timer I know has decent scrambles, which I solved with Yau. I tried reconstructing, but it was hopeless. There was nothing notable about the solve until the very end; I don't remember anything about the beginning.

After reduction, I had solved the FR and BL pairs, and I did L' U L R' U2 R to solve the back right pair. In doing so, I accidentally solved the FL pair, as well as the last layer. There was no parity. 

*Someone who knows math: what are the odds of skipping last slot, OLL, PLL, OLL parity, PLL parity, and AUF on a 4x4?*


----------



## Logiqx (Jun 18, 2015)

IRNjuggle28 said:


> I just got a LS+LL skip with no AUF and no parity on a 4x4, and no, that's not a typo of any sort.
> 
> *Someone who knows math: what are the odds of skipping last slot, OLL, PLL, OLL parity, PLL parity, and AUF on a 4x4?*



Wow. That's pretty crazy!

The probability of skipping the last slot is 1/150. It's a 1/15 chance of the corner being solved and 1/10 chance of the edge being solved.

The probability of skipping last slot, OLL, PLL, OLL parity, PLL parity and AUF is 1 / (150 * 216 * 72 * 2 * 2 * 4) = 1 / 37,324,800

That's almost 3 times less likely than a lotto win in the UK. All things considered, I think I'd prefer the lotto win.


----------



## FailCuber (Jun 18, 2015)

Lol I just got a LL skip on 4x4. I never got a 3x3 LL skip when I timed my solves, but I got it on 4x4. The chances of getting an LL skip is (LL skip on 3x3) * 4 right?


----------



## Logiqx (Jun 18, 2015)

FailCuber said:


> Lol I just got a LL skip on 4x4. I never got a 3x3 LL skip when I timed my solves, but I got it on 4x4. The chances of getting an LL skip is (LL skip on 3x3) * 4 right?



Correct. 1 / 62,208 if you ignore the possibility of AUF.


----------



## Stefan (Jun 18, 2015)

IRNjuggle28 said:


> I feel pretty certain this is the most statistically unlikely thing that will ever happen to me with cubing



Nah, it's far less likely that you get the 4x4 scramble that you get next, and that will happen to you.


----------



## rybaby (Jun 18, 2015)

What is the probability that, after first 2 blocks of Roux, the F2L is completely solved?
What about if the centers are always oriented in the last move of F2B (i.e. D/U center on U)?


----------



## TDM (Jun 18, 2015)

rybaby said:


> What is the probability that, after first 2 blocks of Roux, the F2L is completely solved?
> What about if the centers are always oriented in the last move of F2B (i.e. D/U center on U)?


First case: 6*5*4 = 120, so 1/120
Second case: 6*5*2 = 60, so 1/60


----------



## Julian (Jun 18, 2015)

TDM said:


> First case: 6*5*4 = 120, so 1/120
> Second case: 6*5*2 = 60, so 1/60


The edges can be flipped, though.


----------



## RomFrta33 (Jun 18, 2015)

Taking orientation into account that'd just be 1/12*10 times two, so 1/60 seems to be good.


----------



## GuRoux (Jun 18, 2015)

rybaby said:


> What is the probability that, after first 2 blocks of Roux, the F2L is completely solved?
> What about if the centers are always oriented in the last move of F2B (i.e. D/U center on U)?



1/480 and 1/240


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## IRNjuggle28 (Jun 18, 2015)

Stefan said:


> Nah, it's far less likely that you get the 4x4 scramble that you get next, and that will happen to you.


Amusingly, I thought of that when writing that post, and assumed nobody would be ridiculous enough to point it out.  Completely true, though.


----------



## Berd (Jun 18, 2015)

Whats the odds of an X-cross skip in Yau? XX-cross!?


----------



## TDM (Jun 18, 2015)

Julian said:


> The edges can be flipped, though.


oh god I'm an idiot

multiply both of those by 1/4


----------



## Tempus (Jun 19, 2015)

Stefan said:


> Nah, it's far less likely that you get the 4x4 scramble that you get next, and that will happen to you.


No, actually. The probability of that outcome is 100%. To conclude that it is less you would have to ignore a fact already stated in the premise.


----------



## Stefan (Jun 20, 2015)

Tempus said:


> No, actually. The probability of that outcome is 100%. To conclude that it is less you would have to ignore a fact already stated in the premise.



I'm talking about that scramble. Not about "the scramble that you get next". Actually getting it is not part of what I'm talking *about*, it's just part of my talking.

But yeah, I'm not sure I have worded or am wording that well.


----------



## Myachii (Jun 20, 2015)

What is the possibility of having the "cross" solved on one face of a clock after scrambling?



IRNjuggle28 said:


> *Someone who knows math: what are the odds of skipping last slot, OLL, PLL, OLL parity, PLL parity, and AUF on a 4x4?*





Logiqx said:


> The probability of skipping last slot, OLL, PLL, OLL parity, PLL parity and AUF is 1 / (150 * 216 * 72 * 2 * 2 * 4) = 1 / 37,324,800
> 
> That's almost 3 times less likely than a lotto win in the UK. All things considered, I think I'd prefer the lotto win.



This means that you could be the only person this has ever happened to and will ever happen to in the cubing community.


----------



## Logiqx (Jun 20, 2015)

Myachii said:


> This means that you could be the only person this has ever happened to and will ever happen to in the cubing community.



Few of us will ever experience this skip but some cubers will be lucky enough to get it.

Tens of thousands of people doing hundreds / thousands of 4x4 solves a year adds up to a large number.


----------



## Myachii (Jun 20, 2015)

Logiqx said:


> Few of us will ever experience this skip but some cubers will be lucky enough to get it.
> 
> Tens of thousands of people doing hundreds / thousands of 4x4 solves a year adds up to a large number.



But 37 million? That number is huge.

Just to put this into context - There is a man who did nothing but count to try and break the World Record for the highest number counted to. He slept, obviously, but from the moment he woke up to the moment he went to sleep he counted. After 3 solid months of counting all day, every day, he broke the world record by reaching 1 million.

There are 12,498 people who have ever solved a 4x4 officially in a WCA competition (I'm going to use this number because it isn't too far off the actual number, because people who have never been to competitions and inactive cubers cancel each other out).

The probability of all the events described occurring is 37,324,800. Take that number and divide it by the number of people who have competed.

This would mean that 12,500 people would each have to do ~3000 (2986.46183 to be exact) solves a year for this to happen once in that year.

Now, how long would this take? The median of all people who have competed is 6249, and the time at that number is 1:24.72.

84.72 seconds multiplied by the 37,324,800 times the cube would have to be solved is 3,162,157,056 seconds, or 878,376.96 hours, or 36,599 days (100.5 years total solving time).

For each person that would be 253,013 seconds, or 70 hours.

tl;dr I would be VERY surprised if I saw another cuber here have the same luck.

EDIT- In other words, what you just did IRNjuggle, you had a 0.0000000267918381 chance of doing

Yeah.


----------



## TDM (Jun 20, 2015)

Myachii said:


> There is a man who did nothing but count to try and break the World Record for the highest number counted to. He slept, obviously, but from the moment he woke up to the moment he went to sleep he counted. After 3 solid months of counting all day, every day, he broke the world record by reaching 1 million.


That seems like a long time to count to a million. Three months is 8 million seconds. Surely, even including sleep, it doesn't take 8 seconds to say each number?


----------



## Logiqx (Jun 20, 2015)

Myachii said:


> But 37 million? That number is huge.
> 
> tl;dr I would be VERY surprised if I saw another cuber here have the same luck.



You're calculation is based on the WCA people doing it in one year. They have already been solving the 4x4 for many years and they will solve it for many years to come.

There are lots of people who cube for a hobby who don't actually compete, I know several amongst my friends. An example would be someone who does 12 solves on the 4x4 each week for an extended period of time... slightly over 600 solves in a year. You only need 60,000 of those guys (five times the WCA number) to do 37 million solves in a year.

Anyhow this logic is flawed because probability doesn't work in such a simple way. Doing 37,324,800 solves doesn't guarantee the outcome... you can equally say there is approximately 50% chance of the outcome by doing half the number of solves.

I think it unlikely for just one person in the community to have this happen.

Edit: There are 27,207 registered members on this forum.


----------



## Isaac Lai (Jun 21, 2015)

What are the chances of getting a 4 mover in 2x2?


----------



## Logiqx (Jun 21, 2015)

I've re-done my calculation because I forgot to consider scramble filtering (i.e. minimum of 4 moves).

4 moves = 1847 / 3673775 = approximately 1 in 1989

Additionally...
5 moves or less = approximately 1 in 310
6 moves or less = approximately 1 in 59
7 moves or less = approximately 1 in 13
8 moves or less = approximately 1 in 3

I got the distribution from Wikipedia - https://en.wikipedia.org/?title=Pocket_Cube


----------



## Myachii (Jun 21, 2015)

Logiqx said:


> You're calculation is based on the WCA people doing it in one year. They have already been solving the 4x4 for many years and they will solve it for many years to come.
> 
> There are lots of people who cube for a hobby who don't actually compete, I know several amongst my friends. An example would be someone who does 12 solves on the 4x4 each week for an extended period of time... slightly over 600 solves in a year. You only need 60,000 of those guys (five times the WCA number) to do 37 million solves in a year.
> 
> ...



1) Probability is always flawed in that case then. This is going off the idea that the last solve of the 37 million would be the desired one. You can never guarantee the outcome, for example you could buy 100 million lottery tickets and never win the jackpot.
2) I highly doubt the number of people who can solve a 4x4 and are currently solving at the required speed is anywhere near 60,000.
3) 27,207 forum members =/= 27,207 4x4 solvers. A lot of them may be inactive too.



TDM said:


> That seems like a long time to count to a million. Three months is 8 million seconds. Surely, even including sleep, it doesn't take 8 seconds to say each number?



June 18th 2007 - September 14th 2007


----------



## TDM (Jun 21, 2015)

Myachii said:


> June 18th 2007 - September 14th 2007


wat

Is there some limit for how slowly you have to count? Because it shouldn't take that long.


----------



## Stefan (Jun 21, 2015)

TDM said:


> wat
> 
> Is there some limit for how slowly you have to count? Because it shouldn't take that long.



Count out loud from let's say 374400 to 374500 and let us know how long it takes you.


----------



## TDM (Jun 21, 2015)

Stefan said:


> Count out loud from let's say 374400 to 374500 and let us know how long it takes you.


Ok, I was wrong then, sorry.


----------



## hkpnkp (Jun 21, 2015)

JustinTimeCuber said:


> lol your avatars are looking at each other XD



LOL


----------



## Myachii (Jun 21, 2015)

TDM said:


> Ok, I was wrong then, sorry.








That's between 999,976 and 1,000,000 by him in the final few minutes. He makes it sound quite musical lol


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## IRNjuggle28 (Jul 4, 2015)

Myachii said:


> But 37 million? That number is huge.
> 
> Just to put this into context - There is a man who did nothing but count to try and break the World Record for the highest number counted to. He slept, obviously, but from the moment he woke up to the moment he went to sleep he counted. After 3 solid months of counting all day, every day, he broke the world record by reaching 1 million.
> 
> ...



That's exactly my thoughts--it felt absolutely surreal. Without thinking about the exact math, it occurred to me that I may be the only person in the world this has ever happened to. It seemed so unbelievable that I wondered if I'd remembered the solve wrong, and done more of the solve than I thought before getting a skip, but I clearly remember the final moves of the solve being L' U L R' U2 R, which guarantees that I'm right about it being a LS+LL skip as it affects more than one F2L slot.

When it comes to massive skips, I've been quite lucky. I've been cubing for less than two years, and I have at least 5 LL skips on 3x3, not even counting this one on 4x4, and not counting the one or two I've gotten on big cubes. I don't get lucky in other ways too often; I don't get an abnormally large number of PLL or OLL skips, but I've had a few incredibly lucky solves. Some people go 5 years without getting a single one of what I must have close to 10 of by now.


----------



## JustinTimeCuber (Jul 5, 2015)

Chances of getting an (x) skip at a competition for 1-4 rounds of 5 solves:



Spoiler: 2x2x2






Spoiler: LL skip



1 round: 3.049%
2 rounds: 6.004%
3 rounds: 8.870%
4 rounds: 11.648%
100 solves: 46.162%





Spoiler: PLL skip



1 round: 59.812%
2 rounds: 83.849%
3 rounds: 93.509%
4 rounds: 97.392%
100 solves: 99.999999%





Spoiler: OLL skip



1 round: 17.197%
2 rounds: 31.436%
3 rounds: 43.227%
4 rounds: 52.990%
100 solves: 97.704%





Spoiler: PBL skip



1 round: 13.138%
2 rounds: 24.551%
3 rounds: 34.464%
4 rounds: 43.074%
100 solves: 94.022%





Spoiler: EG skip



1 round: 0.513%
2 rounds: 1.024%
3 rounds: 1.532%
4 rounds: 2.038%
100 solves: 9.781%








Spoiler: 3x3x3






Spoiler: White cross skip



1 round: 0.042%
2 rounds: 0.084%
3 rounds: 0.126%
4 rounds: 0.168%
100 solves: 0.838%





Spoiler: 1 D-move white cross



1 round: 0.168%
2 rounds: 0.336%
3 rounds: 0.504%
4 rounds: 0.671%
100 solves: 3.311%





Spoiler: PLL skip



1 round: 6.754%
2 rounds: 13.052%
3 rounds: 18.925%
4 rounds: 24.401%
100 solves: 75.306%





Spoiler: OLL skip



1 round: 2.293%
2 rounds: 4.534%
3 rounds: 6.724%
4 rounds: 8.863%
100 solves: 37.126%





Spoiler: CMLL skip because Roux is a thing



1 round: 3.049%
2 rounds: 6.004%
3 rounds: 8.870%
4 rounds: 11.648%
100 solves: 46.162%





Spoiler: LL skip



1 round: 0.032%
2 rounds: 0.064%
3 rounds: 0.096%
4 rounds: 0.128%
100 solves: 0.641%








Spoiler: All even layered cubes






Spoiler: No parity



1 round: 76.270%
2 rounds: 94.369%
3 rounds: 98.664%
4 rounds: 99.683%
100 solves: 99.9999999999%





Spoiler: Every solve has no parity



1 round: 0.097%
2 rounds: 9.537e-5%
3 rounds: 9.313e-8%
4 rounds: 9.094e-11%
100 solves: 6.223e-59%





Spoiler: No double parity



1 round: 99.902%
2 rounds: 99.9999%
3 rounds: 99.9999999%
4 rounds: 99.9999999999%
100 solves: 99.999999999999999999999999999999999999999999999999999999999%





Spoiler: Every solve has no double parity



1 round: 23.730%
2 rounds: 5.631%
3 rounds: 1.336%
4 rounds: 0.317%
100 solves: 3.207e-11%


----------



## cashis (Jul 5, 2015)

Chances of not getting parity in 3 rounds of 4x4?
6x6?
(idk if the chances are different)


----------



## TDM (Jul 7, 2015)

cashis said:


> Chances of not getting parity in 3 rounds of 4x4?
> 6x6?
> (idk if the chances are different)


chance of no parity = 1/4
3 rounds = 15 solves
(1/4)^15 = 9x10^-10 (roughly one in a billion)

6x6 has the same for LL parity, I don't know about edge parity during pairing though


----------



## cashis (Jul 7, 2015)

TDM said:


> chance of no parity = 1/4
> 3 rounds = 15 solves
> (1/4)^15 = 9x10^-10 (roughly one in a billion)
> 
> 6x6 has the same for LL parity, I don't know about edge parity during pairing though



What about just OLL parity?
I guess it would be half?


----------



## TDM (Jul 7, 2015)

cashis said:


> What about just OLL parity?
> I guess it would be half?


The probability of OLL parity on each solve is a half. The probability of no OLL parities (or all OLL parities) in 15 solves is (1/2)^15 = 0.00003 (about 1/33000).


----------



## cuBerBruce (Jul 8, 2015)

cashis said:


> Chances of not getting parity in 3 rounds of ... 6x6?



chance of PLL parity = 1/2 (per solve)
chance of OLL parity in at least 1 orbit = 3/4 (per solve)
3 rounds = 9 solves
no parities in 9 solves:
(1/8)^9 = 1/134217728

(What is the probability of a competition having 3 rounds of 6x6x6?)


----------



## Myachii (Jul 8, 2015)

IRNjuggle28 said:


> That's exactly my thoughts--it felt absolutely surreal. Without thinking about the exact math, it occurred to me that I may be the only person in the world this has ever happened to. It seemed so unbelievable that I wondered if I'd remembered the solve wrong, and done more of the solve than I thought before getting a skip, but I clearly remember the final moves of the solve being L' U L R' U2 R, which guarantees that I'm right about it being a LS+LL skip as it affects more than one F2L slot.
> 
> When it comes to massive skips, I've been quite lucky. I've been cubing for less than two years, and I have at least 5 LL skips on 3x3, not even counting this one on 4x4, and not counting the one or two I've gotten on big cubes. I don't get lucky in other ways too often; I don't get an abnormally large number of PLL or OLL skips, but I've had a few incredibly lucky solves. Some people go 5 years without getting a single one of what I must have close to 10 of by now.



Agreed. I've never had a LL skip and I've been cubing for 2 years and 2 months.


----------



## Berd (Jul 8, 2015)

Myachii said:


> Agreed. I've never had a LL skip and I've been cubing for 2 years and 2 months.



Ive been cubing around a year and a quarter and I'm still LL skipless.


----------



## CubeWizard23 (Jul 8, 2015)

got my pb back in 09 on a AUF LL skip 25 sec


----------



## AlexanderGZH (Aug 9, 2015)

1/190080


----------



## Isaac Lai (Aug 11, 2015)

If you know full OLS, what are the chances of a OLL skip?


----------



## Musicalboy2 (Aug 11, 2015)

Isaac Lai said:


> If you know full OLS, what are the chances of a OLL skip?



....100%?


----------



## Isaac Lai (Aug 11, 2015)

Musicalboy2 said:


> ....100%?



Oh wait ya nvm I was being dumb


----------



## cashis (Aug 16, 2015)

What are the odds of having all but 4 edges solved after hoya stage?
e; WHAT ARE THE ODDS OF THIS HAPPENING TWICE IN A ROW


----------



## PenguinsDontFly (Aug 20, 2015)

What is the chance of having 5 2x2 scrambles in a row with no bar? (by bar I mean a 1/2 face, not a 1/2 layer)


----------



## guysensei1 (Aug 26, 2015)

What is the probability of getting the 'perfect' BLD solve: no parity (this probability is 0.5 right?), no twisted pieces, no cycle breaks?


----------



## not_kevin (Aug 26, 2015)

guysensei1 said:


> What is the probability of getting the 'perfect' BLD solve: no parity (this probability is 0.5 right?), no twisted pieces, no cycle breaks?



No cycle breaks means that there's only one cycle for the piece type in question, which also must include your buffer piece; since twisted pieces can be considered really silly one-piece cycles (eg, FU -> UF), both of these constraints end up being the same. Therefore, the only time you can have this happen is if you have a single cycle for each piece type that also involves your buffer piece, and the length of both cycles is odd (has an even number of targets, or has an odd number of pieces involved).

The number of times this can happen on corners is the sum of the number of n-cycles over the pieces - so 1 (for everything solved) + (7*3) * 6 [you can choose any non-buffer sticker for the first, and the second target has orientation pre-determined to prevent twisted corners] + (7*3) * (6*3) * (5*3) * 4 + (7*3) * (6*3) * (5*3) * (4*3) * (3*3) * 2 = 1247527. Similar logic for edges (basically, start counting down from 11 instead of 7, and there are 2 orientations, not 3) gives 21299668061, if my input to WolframAlpha was correct. Therefore, the number of 'perfect' BLD solves is 1247527 * 21299668061 = 2.657*10^16 (26571910997135147) - since the number of total positions is 43.252*10^19 (43252003274489856000), the probability of this occurring is about 0.061435% - about once every 1627 solves (0.061463%)!


----------



## youSurname (Aug 27, 2015)

Do PLL occurrences change depending on what OLL algs you use. I'm seeing *a lot* of N perms.


----------



## guysensei1 (Aug 27, 2015)

youSurname said:


> Do PLL occurrences change depending on what OLL algs you use. I'm seeing *a lot* of N perms.



Negativity bias?


----------



## obelisk477 (Aug 27, 2015)

youSurname said:


> Do PLL occurrences change depending on what OLL algs you use. I'm seeing *a lot* of N perms.



It depends on how you're running in to them. If you're seeing them occur in full solves (from completely scrambled state to solved), then its just chance, and the OLL you use has no effect on the PLL. If you're using an LL scrambler, then sometimes they only scramble certain ways that affect the probabilities of certain OLLs and PLLs arising.


----------



## guysensei1 (Aug 27, 2015)

obelisk477 said:


> It depends on how you're running in to them. If you're seeing them occur in full solves (from completely scrambled state to solved), then its just chance, and the OLL you use has no effect on the PLL. If you're using an LL scrambler, then sometimes they only scramble certain ways that affect the probabilities of certain OLLs and PLLs arising.



Alternative he learnt OLLCP and uses it to get diagswaps every time


----------



## IRNjuggle28 (Aug 28, 2015)

youSurname said:


> Do PLL occurrences change depending on what OLL algs you use. I'm seeing *a lot* of N perms.


The simplest answer is no, it doesn't affect your PLL. It's possible to use OLL to influence PLL, but you have to do it on purpose. You aren't doing that, and if you were, you would be using OLL to force good cases, in which case you'd get less N perms, not more. It's just bad luck.


----------



## NeilH (Sep 20, 2015)

i just did an ao12 where i got 2 h perms followed by three y perms all in a row

What's the chance of getting the same pll 2 times in a row followed by another pll 3 times in a row?


----------



## RomFrta33 (Sep 20, 2015)

It depends which plls you have twice.. H perm is 1/72 and Y perm is 4/72 (correct if I'm wrong)
So first you get your H perm
Than there's 1/72 chance that you get it once again, then you get another pll, then again so multiply the first 1/72 by 4/72, and again because you got the y perm three times!
So prob(first pll)xprob(second pll)^2
It gives 1/72x4/72x4/72, which is 1/23328


----------



## TheCoolMinxer (Sep 20, 2015)

Probability of getting the same PLL on 4x4 3 times in a row. It was the pure diag corner alg with half of yperm, pll parity, rest of y perm. I dunno if the chance of getting the same PLLs on a 4x4 is lower than on a 3x3...


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## RomFrta33 (Sep 20, 2015)

Idk what you're talkin about, but the probability of a pll on 4^3 is twice lower than on 3^3, because pll parity is 1/2
But if you don't use reduction, like I do sometimes, you can get a looot more differents pll's, that's another story^^


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## cubesp (Sep 20, 2015)

If I well understood the question (I did not read old posts)
I think that the probability of 5 pll in a row is just 
p(1 pll)*p(2 pll)*p(3 pll)*p(4 pll)*p(5 pll) , where p(x pll) is the probability of specific pll for instance p(H)=1/72 (badmephisto.com)
regardless if some pll are the same.
having H cannot influence next pll then there is not a p(H "after a previous H")


----------



## RomFrta33 (Sep 20, 2015)

That's right and wrong 
If i say I'll get 5 h perms in the 5 next cubes, the probability is p(H)^5, which is ridiculously low :3
But if I say i'll get 5 times the same pll, but without saying which one it'll be, the probability is just p(first pll I'll get)^4 ! It depends on the pll you get first, you can get a rough estimate but no more until you know which pll is your first one. The second probability is obviously higher than the first one.
(The last thing you said is obviously right, im talking about your algorithm to calculate the prob of 5 plls)


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## cubesp (Sep 20, 2015)

uhm,
some time passed from my studies, but if I well remember
if you normalize to 72 (p(T)=4/72), 5 pll can form 72^5 combinations.
The event "single pll" is indipendent from the other, then if I want to guess a 5pll combination
and I say UaTTHJb then p(UaTTHJb) should be p(Ua)*p(T)*p(T)*p(H)*p(Jb)
that is also equal to p(TTTTJb) because p(T)=p(Ua)=p(Jb)=4/72 (badmephisto.com)
and is also equal to p(TJbTTT)
Is that right ? Help me remember such rules!!


----------



## RomFrta33 (Sep 20, 2015)

That's right, except that you clearly have not 72^5 combinations of 5 plls! You do if you count AUFs, but actually nobody says "i got a Jb perm with -1/4 auf" 
21 plls equal 21^5 different combinations 
Edit: i love editing
P(Ua,T,T,H,Jb) is not equal to P(TTTTT), because of H perm! The plls you talked about have 1/18 chance to happen, but the H perm has 1/72


----------



## cubesp (Sep 20, 2015)

Right,
otherwise also p(T) should be 1/72 ...


----------



## cubesp (Sep 21, 2015)

I never say that


----------



## joshsailscga (Sep 24, 2015)

The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?


----------



## Ollie (Oct 3, 2015)

Just had 4 consecutive PLL skips occur during a 4x4x4 session.

Usually that's 144^4 = a 1 in 429,981,696 chance, but there were 2 OLLCPs, 1 1LLL case and a normal skip, which a few of have worked out to be 1/82,944. Is this correct?


----------



## Kudz (Oct 16, 2015)

Few days ago I got 2 LL skips in a row on 3x3.
What is propabilty of that? Also I was playing with some OLL skip techniques so, wasn't it double PLL skip?
What I did:
1st: RU'R' F'UF
2nd: case when evrything oriented, skip after using z rotation and J perm insertion.


----------



## Ronxu (Oct 16, 2015)

Kudz said:


> Few days ago I got 2 LL skips in a row on 3x3.
> What is propabilty of that? Also I was playing with some OLL skip techniques so, wasn't it double PLL skip?
> What I did:
> 1st: RU'R' F'UF
> 2nd: case when evrything oriented, skip after using z rotation and J perm insertion.



Neither of them are LL skips.


----------



## ViliusRibinskas (Oct 21, 2015)

What are the chances of getting L4E skip in 4x4+ cubes? I have got only like twice it on 4x4, and never on bigger cubes. So what is the probability for that?



joshsailscga said:


> The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?



I got OLL+PLL skip yesterday, but got PLL parity. For getting OLL+PLL skip chance are of course the same as getting LL skip, but if you mean with parity, Idk then


----------



## shadowslice e (Oct 21, 2015)

joshsailscga said:


> The other day I got a OLL+PLL skip on 4x4, but had double parity. Anyone know the probability for that?



isn't it just chance of an LL skip*parity*parity ((2^3*3!*3^3*3!)*2*2)?


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## Christopher Mowla (Oct 22, 2015)

shadowslice e said:


> isn't it just chance of an LL skip*parity*parity ((2^3*3!*3^3*3!)*2*2)?


The total number of positions of the 4x4x4 last layer with paired dedges is:




​If he is talking about that he got the following case, then there are 2 possible ways he could get it (it or its inverse) with regards to AUF.



​ 
Therefore, the final number is 2/(62,208) = 1/(31,104), which is equivalent to your answer.


----------



## joshsailscga (Oct 22, 2015)

Christopher Mowla said:


> The total number of positions of the 4x4x4 last layer with paired dedges is:
> http://i.imgur.com/BLEjTu3.gif​If he is talking about that he got the following case, then there are 2 possible ways he could get it (it or its inverse) with regards to AUF.
> https://www.speedsolving.com/wiki/images/9/95/DblParity.png​
> Therefore, the final number is 2/(62,208) = 1/(31,104), which is equivalent to your answer.



Sounds cool, thanks for the input. I don't know if it would change any of the numbers but the OLL parity was the speedsolving alg that flips one dedge and swaps two more, as well as flipping + swapping two corners.


----------



## Christopher Mowla (Oct 23, 2015)

joshsailscga said:


> Sounds cool, thanks for the input. I don't know if it would change any of the numbers but the OLL parity was the speedsolving alg that flips one dedge and swaps two more, as well as flipping + swapping two corners.


It sounds like you're describing the following case:





If this is so, this is not double parity, as the double parity case which looks a lot like this does not switch two dedges in addition to flipping one.

So, with this being the case, then since this case is equal to its inverse, there is only one version of it (with respect to AUF). We therefore have 1/(62,208).


----------



## youSurname (Nov 18, 2015)

What is the probability of getting a 3BLD Scramble with no cycle breaks? My Monty Carlo simulations give 736/100130 or roughly 0.74%.


----------



## obelisk477 (Nov 18, 2015)

youSurname said:


> What is the probability of getting a 3BLD Scramble with no cycle breaks? My Monty Carlo simulations give 736/100130 or roughly 0.74%.



That sounds right at least. I mean roughly it should be the odds that your buffer corner and buffer edge are exactly at one spot on the cube (namely, the end of the cycle), or (1/8)*(1/12)=1.04%. That's just an approximation though


----------



## JustinTimeCuber (Nov 18, 2015)

what's the probability that I meet my goal of finally a new PB ao50 by Saturday?
edit: I did the math, it is 110%


----------



## shadowslice e (Nov 18, 2015)

JustinTimeCuber said:


> what's the probability that I meet my goal of finally a new PB ao50 by Saturday?
> edit: I did the math, it is 110%



Nah, you forgot to carry the 4. It's actually 164%


----------



## JustinTimeCuber (Nov 18, 2015)

shadowslice e said:


> Nah, you forgot to carry the 4. It's actually 164%



oops this is why I suck at arithmetic... I should have used my calculator...


----------



## biscuit (Nov 18, 2015)

shadowslice e said:


> Nah, you forgot to carry the 4. It's actually 164%



You forgot the Feliks therom. It should be 16*5*%


----------



## gyroninja (Nov 21, 2015)

What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)
What is the probability of having a 2x1 block (ce pair both with correct orientation) in the last layer solved if all of your edges are preoriented? As with the other question the pieces don't have to be solved relative to the whole cube.

Would also be interesting to see the probabilities without having edges preoriented.


----------



## shadowslice e (Nov 21, 2015)

gyroninja said:


> What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)
> What is the probability of having a 2x1 block (ce pair both with correct orientation) in the last layer solved if all of your edges are preoriented? As with the other question the pieces don't have to be solved relative to the whole cube.
> 
> Would also be interesting to see the probabilities without having edges preoriented.



first question= odds of having phased edges*2 (cause you can have the other two the wrong way around)*4(only 1 position for a corner*3(only one CO allowed/4(AUFS)
= (6*2*4*3)/4= 1/36 (I think)
2) 4*3/4= 1/3


----------



## cuBerBruce (Nov 21, 2015)

gyroninja said:


> What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)



This is what I get:

Start with edges. We have 3 different possibilities for their permutation:
1) correctly permuted: probability 1/6
2) adjacent swap needed: probablility 2/3
3) swap across needed: probability 1/6 (but no possibility of a 2x2 block)

Edge case 1: edges correctly permuted
Corners must have even permutation because edge permutation is even. We have 3 cases:
1) Corners solved (with respect to edges): probability 1/12
2) Corners in a 3-cycle: probability 8/12 = 2/3
3) Corners in a double swap (adjacent or diagonal): probability 3/12 = 1/4

For corner case 3, no corners are in the correct place, so we have no possibility of a 2x2 block.
For corner case 1, all corners are in the correct place, so any oriented corner forms a 2x2 block. Of the 27 total orientation configuarations, 21 have at least one oriented corner. 21/27 = 7/9.
For corner case 2, we need the correctly positioned corner to be correctly oriented: 1 in 3 chance.

Edge case 2:
We can only form a 2x2 block if the corner between the 2 correct edges is correctly placed (1 in 4) and correctly oriented (1 in 3).

So the probability of at least one 2x2 block is (1/6)*(1/12)*(7/9) + (1/6)*(2/3)*(1/3) + (2/3)*(1/4)*(1/3) = 67/648 (or approximately 10.34%).

If we want the probablility of exactly one 2x2 block, then the 21/27 chance must be changed to 8/27. Thus, this probablility is (1/6)*(1/12)*(8/27) + (1/6)*(2/3)*(1/3) + (2/3)*(1/4)*(1/3) = 47/486 (or approximately 9.67%).


----------



## Hssandwich (Nov 21, 2015)

Probability of 3 sub-former WR singles in one day?


----------



## Praetorian (Nov 21, 2015)

Hssandwich said:


> Probability of 3 sub-former WR singles in one day?



NINJA'D


----------



## YouCubing (Nov 21, 2015)

Hssandwich said:


> Probability of *2* sub-former WR singles in one day?



Collin's was a misscramble.


----------



## matty (Nov 21, 2015)

Probability of a world record with misscramble with 2 legit world records all by different people?


----------



## Jbacboy (Nov 21, 2015)

Possibility of 9 parities in an avg of 5? Happened to me at River Hill in 4x4


----------



## Sajwo (Nov 21, 2015)

matty said:


> Probability of a world record with misscramble with 2 legit world records all by different people?



12222 passed since first WR and this happened first time(right?)
so 1/12222 XD


----------



## matty (Nov 21, 2015)

Jbacboy said:


> Possibility of 9 parities in an avg of 5? Happened to me at River Hill in 4x4



(0.5^9)*100 = 0.195%

Happed to me at UKC 2015 in 4x4, I got 8/9 parities.


----------



## Jbacboy (Nov 21, 2015)

matty said:


> (0.5^9)*100 = 0.195%
> 
> Happed to me at UKC 2015 in 4x4, I got 8/9 parities.


Ouch... Thanks!


----------



## Hssandwich (Nov 21, 2015)

At UKC I had no parities in 4x4 in my average of 5. What are the odds of that?


----------



## matty (Nov 21, 2015)

Jbacboy said:


> Ouch... Thanks!



Also 1 in 512 chance :/



Hssandwich said:


> At UKC I had no parities in 4x4 in my average of 5. What are the odds of that?



1 in 1024 chance? <0.1% wut

Pretty sure I calculated right, harry stop getting so lucky....


----------



## Hssandwich (Nov 21, 2015)

matty said:


> 1 in 1024 chance? <0.1% wut
> 
> Pretty sure I calculated right, harry stop getting so lucky....



Sorry man, I was genuinely frustrated since my solves were bad (mostly) for no parity. And also I felt that I was getting an unfair advantage.


----------



## joshsailscga (Nov 22, 2015)

Jbacboy said:


> Possibility of 9 parities in an avg of 5? Happened to me at River Hill in 4x4



LOL I heard you complaining about that


----------



## Jbacboy (Nov 22, 2015)

joshsailscga said:


> LOL I heard you complaining about that


Yeah I was so friggin annoyed, the average could of been really nice.  RIP sub-50


----------



## Tony Fisher (Nov 22, 2015)

Has anyone made a chart showing the probabilities of solve times? So you would for example combine the TPS of the best cubers, the number of official solves they do in a year and the probability of certain skips. From this you could calculate the rough probabilities of specific times occurring during the next year.


----------



## joshsailscga (Nov 22, 2015)

Tony Fisher said:


> Has anyone made a chart showing the probabilities of solve times? So you would for example combine the TPS of the best cubers, the number of official solves they do in a year and the probability of certain skips. From this you could calculate the rough probabilities of specific times occurring during the next year.



Wow, this is actually a really cool idea.


----------



## 4Chan (Dec 8, 2015)

How many last slot cases are there when all edges and corners are permuted?

How do you account for the (several) rotational symmetries?


----------



## shadowslice e (Dec 8, 2015)

4Chan said:


> How many last slot cases are there when all edges and corners are permuted?
> 
> How do you account for the (several) rotational symmetries?



I would guess (3^4)*(2^4)/4(Aufs)/2(reflections)= 162. But how do you have LS when you have correctly permuted all pieces (unless you mean they are misoriented in their slot which is what I assumed)


----------



## Lid (Jan 16, 2016)

Need the probabilities for full PLL with parity including the normal PLLs (44 cases total), couldn't find anything uselful when searching.


----------



## G2013 (Jan 21, 2016)

What are the probabilities to get 5 PLL skips in a row, doesn't matter the AUF? 

Is it 1/440? 22 PLL cases including solved, 4 angles, thus 88, 88*5 is 440. Then why didn't I ever get one? I'm cubing ever since 2006-7.

Where is my fail?

PS: I did get 5 G perms in a row, though, but I know that there are 4 types of G perms and it's more likely...


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## CubesOfTheWorld (Jan 21, 2016)

G2013 said:


> What are the probabilities to get 5 PLL skips in a row, doesn't matter the AUF?
> 
> Is it 1/440? 22 PLL cases including solved, 4 angles, thus 88, 88*5 is 440. Then why don't I get it so often? Actually I never got that.
> 
> Where is my fail?



1/72 chance to get a PLL skip (not all cases have the same probability).
(1/72)^5 = 1/1934917632 which is about 0.0000000517%


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## ViliusRibinskas (Mar 9, 2016)

In 3 solves 2/3 OLLs were the least common OLL (1/216 chance). What is the probability of getting that in 3 solves?


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## shadowslice e (Mar 9, 2016)

ViliusRibinskas said:


> In 3 solves 2/3 OLLs were the least common OLL (1/216 chance). What is the probability of getting that in 3 solves?



645/10077696


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## Isaac Lai (Mar 9, 2016)

What is the probability of getting 4 4 movers in a 2x2 ao5?


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## IQubic (Mar 22, 2016)

So, let's assume I have solved all of F2L, except for a single edge, which I place in its position, but flipped. What are the chances that all of the remaining pieces are, save for one edge, correctly oriented?


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## Cale S (Mar 22, 2016)

IQubic said:


> So, let's assume I have solved all of F2L, except for a single edge, which I place in its position, but flipped. What are the chances that all of the remaining pieces are, save for one edge, correctly oriented?



CO skip = 1/27
3 edges oriented = 1/2

1/54


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## IQubic (Mar 23, 2016)

How likely is it to get an all edges oriented OLL? I am not using any sort of edge control or other such tricks.


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## CubesOfTheWorld (Mar 23, 2016)

IQubic said:


> How likely is it to get an all edges oriented OLL? I am not using any sort of edge control or other such tricks.



The fourth edge's orientation is determined by the other three:
(1/2)^3 = 1/8 chance of getting an OLL with the top cross already solved.


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## PurpleBanana (Apr 13, 2016)

Given a random state scramble, what is the probability that at least one edge of the CFOP white cross is already solved? By "solved" I mean it has to be next to the white center and oriented correctly, but not necessarily matched up with its corresponding center. 

For any individual edge, the probability would be 1/6. 1/3 = probability the edge is in the cross layer, 1/2 = probability the edge is oriented correctly, 1/3 x 1/2 = 1/6. So the probability that at least one of the edges is solved would be 1/6 + 1/6 + 1/6 + 1/6 = 2/3. That seems a bit high. Am I missing something? 

Sorry if this question has already been answered but I did a search and couldn't find anything.


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## YouCubing (Apr 13, 2016)

Probability of a solved center on 4x4?


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## obelisk477 (Apr 13, 2016)

PurpleBanana said:


> Given a random state scramble, what is the probability that at least one edge of the CFOP white cross is already solved? By "solved" I mean it has to be next to the white center and oriented correctly, but not necessarily matched up with its corresponding center.
> 
> For any individual edge, the probability would be 1/6. 1/3 = probability the edge is in the cross layer, 1/2 = probability the edge is oriented correctly, 1/3 x 1/2 = 1/6. So the probability that at least one of the edges is solved would be 1/6 + 1/6 + 1/6 + 1/6 = 2/3. That seems a bit high. Am I missing something?
> 
> Sorry if this question has already been answered but I did a search and couldn't find anything.



What you're looking for is the odds of the first edge being 'solved' plus the odds of the second cross edge being 'solved', etc etc. However, you run up against the problem that the second, third, and fourth cross edges actually have to be really 'solved' relative to the first cross edge. So while odds for correct permutation for the first cross edge are indeed 4/12=1/3, the odds for the second cross edge's permutation being correct go down to 1/11, being that there is only one spot relative to the first 'solved' cross edge that it can go. and then down to 1/10 for the third cross edge, etc. It ends ups being 1/6+1/22+1/20+1/18=31.77%


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## PurpleBanana (Apr 13, 2016)

obelisk477 said:


> What you're looking for is the odds of the first edge being 'solved' plus the odds of the second cross edge being 'solved', etc etc. However, you run up against the problem that the second, third, and fourth cross edges actually have to be really 'solved' relative to the first cross edge.


It wouldn't matter where the other cross edges were, all that matters is that there is at least one correctly oriented edge in that layer. There could be more than one and they don't have to be correctly permuted relative to each other.

Edit: For what it's worth, I just did an experimental test of 200 scrambles and 110 of them had a solved edge = 55%. This isn't a very large sample size so the value isn't going to be accurate but should be in the ballpark of the actual figure.

Edit2: I expanded the experiment sample size to 500 scrambles. Number with the solved edge = 262/500 = 52.4%.

Edit3: Expanded the sample size to 1000. 531/1000 = 53.1% with solved edge. This value should be pretty close to the actual probability. Right now I'm working on a different way of calculating it.


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## Herbert Kociemba (Apr 13, 2016)

My suggestion ist
(Binomial[4,1]*Binomial[8,3]*1*2^3+Binomial[4,2]*Binomial[8,2]*3*2^2+Binomial[4,3]*Binomial[8,1]*7*2^1+Binomial[4,4]*Binomial[8,0]*15*2^0)/(Binomial[12,4]*2^4)= 4271/7920 = 0.539268


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## PurpleBanana (Apr 14, 2016)

Herbert Kociemba said:


> My suggestion ist
> (Binomial[4,1]*Binomial[8,3]*1*2^3+Binomial[4,2]*Binomial[8,2]*3*2^2+Binomial[4,3]*Binomial[8,1]*7*2^1+Binomial[4,4]*Binomial[8,0]*15*2^0)/(Binomial[12,4]*2^4)= 4271/7920 = 0.539268


I don't understand the math you did there, but if you're correct then my experimental value had a percent error of (0.539268-0.531)/0.539268 = 1.5%, not too bad given the sample size.


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## Herbert Kociemba (Apr 14, 2016)

PurpleBanana said:


> I don't understand the math you did there, but if you're correct then my experimental value had a percent error of (0.539268-0.531)/0.539268 = 1.5%, not too bad given the sample size.


Consider the 4 edges with one face being white as indistinguishable. We can do this because the second colour does not matter. Then there are Binomial[12,4] possibilities to set these 4 edges on the 12 possible places. Because there are 2 orientations for each edge, we have to multiply by 2^4. This explains the denominator.

Now take for example the term Binomial[4,3]*Binomial[8,1]*7*2^1.
Binomial[4,3]*Binomial[8,1] describes the number of possiblities to have exactly 3 edges in the correct layer. You have Binomial[4,3] possibilites to put 3 edges in the correct layer and Binomial[8,1] possiblities to put the remaining edge in the two other layers. There are 2^3 ways to orient the 3 edges in the correct layer. Only in 1 case all orientations are wrong, in 2^3-1=7 cases at least one orientation is correct. This explains the factor 7. The remaining edge in the "wrong" layers has 2^1 possiblities. This explains the factor 2^1.

The other terms in the nominator describe the number of possiblities, that exactly 1, 2 or 4 edges are in the correct layer with at least one edge correctly orientated.


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## PurpleBanana (Apr 14, 2016)

Herbert Kociemba said:


> Consider the 4 edges with one face being white as indistinguishable. We can do this because the second colour does not matter. Then there are Binomial[12,4] possibilities to set these 4 edges on the 12 possible places. Because there are 2 orientations for each edge, we have to multiply by 2^4. This explains the denominator.
> 
> Now take for example the term Binomial[4,3]*Binomial[8,1]*7*2^1.
> Binomial[4,3]*Binomial[8,1] describes the number of possiblities to have exactly 3 edges in the correct layer. You have Binomial[4,3] possibilites to put 3 edges in the correct layer and Binomial[8,1] possiblities to put the remaining edge in the two other layers. There are 2^3 ways to orient the 3 edges in the correct layer. Only in 1 case all orientations are wrong, in 2^3-1=7 cases at least one orientation is correct. This explains the factor 7. The remaining edge in the "wrong" layers has 2^1 possiblities. This explains the factor 2^1.
> ...


It seems you're right. Your reasoning all seems sound. The value you obtained also seems reasonable and is corroborated by my experimental evidence. 

Thanks for working on this problem.


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## PurpleBanana (Apr 14, 2016)

YouCubing said:


> Probability of a solved center on 4x4?


https://www.speedsolving.com/forum/threads/probability-thread.20384/page-4#post-396547


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## YouCubing (Apr 14, 2016)

PurpleBanana said:


> https://www.speedsolving.com/forum/threads/probability-thread.20384/page-4#post-396547


huh. That's a lot more common than I expected xD


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## PurpleBanana (Apr 14, 2016)

YouCubing said:


> huh. That's a lot more common than I expected xD


Really? It's less common than I expected.


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## PurpleBanana (Apr 20, 2016)

Probability of having any solved 2x1x1 block given a random 2x2 scramble?


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## unsolved (Apr 20, 2016)

The answer will be the total number of 2x1x1 blocks possible, divided by the total number of arrangements.

The total arrangements is:







I don't know the total number of 2x1x1 blocks though.


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## cmhardw (Apr 20, 2016)

unsolved said:


> The answer will be the total number of 2x1x1 blocks possible, divided by the total number of arrangements.
> 
> The total arrangements is:
> 
> ...



You'd have to consider cases where there are more than one 2x1x1 blocks as well.


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## Joel Banks (May 3, 2016)

Odds of all 4 f2l cases already being made?


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## PurpleBanana (May 3, 2016)

Joel Banks said:


> Odds of all 4 f2l cases already being made?


About 1 in 9.6 billion, I think.


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## Lucas Garron (May 3, 2016)

PurpleBanana said:


> About 1 in 9.6 billion, I think.



What's your source? I get only one in 3.6 billion:

 1/(corners in place * corners oriented * edges in place * edges oriented)
= 1/((8*7*6*5)*(3^4)*(8*7*6*5)*(2^4))
= 1/3657830400


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## PurpleBanana (May 3, 2016)

You're right, I just realized the error I made.


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## teboecubes (May 4, 2016)

What's the probability of this:

I was doing a solve on the iitimer app, and I came across this one scramble that gave me a cross skip and pair built but not placed no joke!!! I just had to do an AUF, or ADF since it was the cross, and do R U' R. Surprisingly it was not a PB (I was a bit slow on the other pairs)


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## PurpleBanana (May 5, 2016)

teboecubes said:


> What's the probability of this:
> 
> I was doing a solve on the iitimer app, and I came across this one scramble that gave me a cross skip and pair built but not placed no joke!!! I just had to do an AUF, or ADF since it was the cross, and do R U' R. Surprisingly it was not a PB (I was a bit slow on the other pairs)


1/7920


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## TheSilverBeluga (May 9, 2016)

This actually happened to me a few years back. I was still solving beginner's method, and I had one solve where I finished putting in the last corner of the first layer, and all I had left was AUF. Basically, I skipped having to solve the last twelve pieces.
If my math is right, the chance of this happening is 4*(1/8!)*2*(1/2)^7*(1/4!)*2*(1/3)^3 = 4*2*2*1/40320*1/128*1/24*1/27 = 1/209,108,880
(I think I broke the universe.)


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## Joel Banks (May 9, 2016)

How many combinations are there on a Rubik's cube if you can disassemble it and put it back together? (One edge or corner COULD be flipped)


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## TheSilverBeluga (May 9, 2016)

Joel Banks said:


> How many combinations are there on a Rubik's cube if you can disassemble it and put it back together? (One edge or corner COULD be flipped)


There are eight corners with three possible orientations = 1/8! * (1/3)^8
There are twelve edges with two possible orientations = 1/12! * (1/2)^12
1/40320 * 1/6561 * 1/479001600 * 1/4096 = 5.19 x 10^20 = about 519000000000000000000. (The calculator I was using didn't want to use exact numbers because it was too big.)


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## Joel Banks (May 9, 2016)

Same question as last... Just with 4x4


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## TheSilverBeluga (May 9, 2016)

Joel Banks said:


> Same question as last... Just with 4x4


Since you're talking about the number of ways it could be put back together, I'm going to count two scrambles that are technically the same but have swapped center pieces as different. (I hope that makes sense.)

There are 8 corner pieces with 3 possible orientations = 1/8! * (1/3)^8
There are 24 edge pieces with 1 possible orientation = 1/24!
There are 24 center pieces with 1 possible orientation = 1/24!

1/8! * (1/3)^8 * 1/24! * 1/24! = 101,836,133,451,709,433,602,625,095,475,784,887,762,411,323,392,000,000,000

If you don't want to count swapped centers as different:

Each center color can be swapped with other center pieces of the same color 4! = 24 different ways. There are six colors, so 24 * 6 = 144.

101,836,133,451,709,433,602,625,095,475,784,887,762,411,323,392,000,000,000 / 144 = 707,195,371,192,426,622,240,452,051,915,172,831,683,411,968,000,000,000


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## Joel Banks (May 10, 2016)

Wouldn't the edge pieces have 2 orientations? Also your good at this, want more things to find out or should I stop now.


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## TheSilverBeluga (May 10, 2016)

Joel Banks said:


> Wouldn't the edge pieces have 2 orientations? Also your good at this, want more things to find out or should I stop now.


On a 4x4, the edges aren't centered. If you take out an edge piece and flip it around, you won't be able to put it back in because the hook that connects it to the rest of the puzzle will be pointing the wrong way. On a 5x5, for example, the very center edge has two orientations, but the two edges on either side both only have one, for the same reason.


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## gyroninja (May 10, 2016)

Joel Banks said:


> Wouldn't the edge pieces have 2 orientations? Also your good at this, want more things to find out or should I stop now.


Edge pieces on the 4x4 only have permutation.


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## Herbert Kociemba (May 10, 2016)

TheSilverBeluga said:


> Each center color can be swapped with other center pieces of the same color 4! = 24 different ways. There are six colors, so 24 * 6 = 144.



It is 4!^6, not 4!*6
So the number of possibilites is 24!*24!/(4!^6)*8!*3^8.
And I would divide this number by 24 because of the 24 different possible orientations of the cube in space.

For an nxnxn with n even we have (n-2)*12 edges which make (n-2)/2 edge clusters with 24 edges each. We have (n-2)^2*6 centers which make (n-2)^2/4 center clusters with 24 centers each. So there are

24!^((n-2)/2)*(24!/(4!^6))^((n-2)^2/4)*8!*3^8 /24 

possibilites to build an nxnxn cube with even n.


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## Joel Banks (May 10, 2016)

gyroninja said:


> Edge pieces on the 4x4 only have permutation.


Okay. That makes sense


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## 4Chan (May 16, 2016)

Hay ya'll, I'm having a brain fart because I'm studying for finals.

What's the probability that all 5 corners are oriented during Last Slot?
I don't know how to account for redundant cases correctly due to AUF because I'm a retard who is bad at maths.

EDIT: The answer is 1/81, thanks notKevin!!


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## sneze2r (May 16, 2016)

Hello,
I've digged out an old thread from polish forum where I estimated probablilities of getting an specific bld case(number of targets & number of twists for each edges and corners) by doing simulation in R. Since I had lazy weekend I've decided to do write this code again. Here is the table with results of this simulation:



Spoiler: R code



k=c()
twists=c()
for(j in 1:300000){
a=sample(1:12)
h=0
i=1
obr=sample(c(0,1),11,replace=T)
if(sum(obr)%%2==1){obr[12]=1}else{obr[12]=0}
obr=obr+1
twists[j]=sum(as.numeric(a==c(1:12))==obr)
while(a_!=i&a!=0){
h=h+1
ind=a
 a=a[a]
a[ind]=0}
for(i in 2:12){
if(a!=i&a!=0){h=h+2}
while(a!=i&a!=0){
h=h+1
ind=a
 a=a[a]
a[ind]=0
}
}
k[j]=h
}

results=paste(k,twists,sep=',')
wyniki=c(table(results)[27:length(table(results))],table(results)[1:26])
#^here i had to "re-sort" this table results because it was previously#
#sorted automaticly like this:14,4 15,1 15,2 <-(!)-> 4,3 4,4 4,5... etc#
barplot(height=wyniki/length(k),main="probability barplot of (targets,twist) cases for edges")
tablica=table(targets=k,twists)
dd <- addmargins(tablica, FUN = list(Total = sum), quiet = TRUE)
write.table(dd,file="dane.csv",sep=",",row.names=T)
}
_


_
*In this table first column means number of targets, first row- numbers of twists. For instance, if i want calculate probability of getting 7-9 targets with maximum 2 twists, i read records from a table: (70+314+471+627+2067+2229+4131+8941+6042)/300000=8,297333%*


Spoiler: Table with simulation result for edge cases(csv)



"targets-twists","0","1","2","3","4","5","6","7","Total"
"3",0,0,0,0,1,0,0,0,1
"4",0,0,2,0,0,0,0,0,2
"5",1,7,11,12,11,5,1,0,48
"6",6,44,67,86,42,18,0,1,264
"7",70,314,471,345,110,7,0,0,1317
"8",627,2067,2229,999,136,10,0,0,6068
"9",4131,8941,6042,1160,97,5,0,0,20376
"10",18423,25064,7096,709,51,0,0,0,51343
"11",51388,30150,4174,365,17,0,0,0,86094
"12",63957,17032,1846,114,0,0,0,0,82949
"13",34568,5636,583,6,0,0,0,0,40793
"14",8278,1385,69,0,0,0,0,0,9732
"15",752,237,1,0,0,0,0,0,990
"16",13,10,0,0,0,0,0,0,23
"Total",182214,90887,22591,3796,465,45,1,1,300000





Spoiler: Table with simulation result for corner cases(csv)



"targets-twists","0","1","2","3","4","5","Total"
"0",1,2,3,5,3,1,15
"1",6,18,16,13,6,0,59
"2",44,111,95,63,10,1,324
"3",320,658,543,177,34,4,1736
"4",2096,3217,1914,394,43,2,7666
"5",10207,11040,3736,323,18,0,25324
"6",35277,23422,3306,226,8,0,62239
"7",74702,21358,1665,125,0,0,97850
"8",67991,8612,759,0,0,0,77362
"9",22453,2578,82,0,0,0,25113
"10",1815,497,0,0,0,0,2312
"Total",214912,71513,12119,1326,122,8,300000





If You want to calculate probability of getting certain case(for instance 10 or 12 targets for edges with maximum 2 twist AND 6-8 targets for corners with maximum one corner twist, You calculate probablilites separate for edges and corners, than multiply those and multiply by 2. P(A)*P(B)*2
_


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## Joel Banks (May 22, 2016)

If you take apart a megaminx and put it back together, what are the odds it would be solveable?


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## xyzzy (May 22, 2016)

Joel Banks said:


> If you take apart a megaminx and put it back together, what are the odds it would be solveable?



There's a 1/3 chance of the corner orientation being solvable, 1/2 chance of the edge orientation being solvable, 1/2 chance of corner permutation being solvable, and 1/2 chance of edge permutation being solvable, so multiplying them together gives 1 in 24.


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## nathannathan1232 (May 23, 2016)

xyzzy said:


> There's a 1/3 chance of the corner orientation being solvable, 1/2 chance of the edge orientation being solvable, 1/2 chance of corner permutation being solvable, and 1/2 chance of edge permutation being solvable, so multiplying them together gives 1 in 24.


wouldnt permutation be betweet both edges and corners? a 2-cycle would make the corners theoreticly unsolvable, but having the edges also need a 2-cycle would fix it.


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## bobthegiraffemonkey (May 23, 2016)

nathannathan1232 said:


> wouldnt permutation be betweet both edges and corners? a 2-cycle would make the corners theoreticly unsolvable, but having the edges also need a 2-cycle would fix it.


Nope, megaminx works a little differently from a 3x3. Every twist is an even permutation of corners and an even permutation of edges, so you can't do a 2-cycle of edges even when ignoring corners and vice versa. 1/24 is the correct answer to the original question.


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## Drad (May 23, 2016)

What are the chances of getting a 3x3 skip on 4x4 ?


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## gyroninja (May 23, 2016)

Drad said:


> What are the chances of getting a 3x3 skip on 4x4 ?


I think it would just be
1/(43,252,003,274,489,856,000 * 2 * 2)

=1/173,008,013,097,959,424,000

Chance of getting a solved 3x3 stage with oll / pll parity. (using like redux)


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## FJT97 (May 26, 2016)

Ok, i think this might be a little complicated and hard to understand now..

Note: I count y2 as two rotations.

Its about 4x4. I'm using Yau. Its about edge pairing after i placed the last cross edge.
What i would have done normally:
1: Uw. 2: looking at the LFU edge piece. 3: placing the matching edge in RFD. 4: Rotating y. Repeating step 2 and 3. Repeating step 4. repeating step 2 and 3. 5: Uw'. 6: Rotating y2 . Now i have 3 solved dedges and a unsolved in LF and can continue.
Of course, that method is kinda bad, too many rotations.

Now i have two alternatives on which i want to know, which one requires less cube rotations on average.

First variant:

step 1 and 2 and 3 are the same as above. 4: placing the BRD edge without rotating if that is possible with just R and U moves. If not, then rotate and place it "normally" as i did above. Then i'd place the next edges "normally" again. Just rotating if its not doable with just R and U or L and U moves. i get to the position then, where i just have a unsolved dedge in LF then. This method requires 0 to 4 rotations.

Second variant:

1 Uw. 2: This is really the only difference: I look at the LBU edge now, not at the LFU. Then i just go on with the steps of the first variant. The advantage is, that i can place the first two edges without any cube rotations as i allow myself to do F moves. If i don't need a rotation then for the last edge, i would need one rotation (y') to get the unsolved dedge to LF.

Hope you got my point. In the first variant i don't even need one cube rotation if anythings perfect. i the second variant, i need at least one rotation (the y' at the end) but i don't do any rotations for the second edge.

So with which of the two variants i end up with less rotations?

P:S.: I know that i could just do B moves in both variants and safe the rotations completely and i know, that its probably best to choose between the variants if i see the corresponding edges in a real solve, but thats not the point. Its about the pure probability


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## shadowslice e (Jun 3, 2016)

So how many cases are there for pseudo-2-gen LL {R,U,r}? I get ((2^3)(3^3)(4!))/2/4/2=81 (so it should be around there). Does anyone know the exact answer?


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## gyroninja (Jun 3, 2016)

shadowslice e said:


> So how many cases are there for pseudo-2-gen LL {R,U,r}? I get ((2^3)(3^3)(4!))/2/4/2=81 (so it should be around there). Does anyone know the exact answer?


Your way off. Rough approximation would be 8 eo * 84 2gll = 672.


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## shadowslice e (Jun 3, 2016)

gyroninja said:


> Your way off. Rough approximation would be 8 eo * 84 2gll = 672.


oops I can't type; my esimate is around 648 or 324 if inverses are not opposite


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## xyzzy (Jun 4, 2016)

shadowslice e said:


> So how many cases are there for pseudo-2-gen LL {R,U,r}? I get ((2^3)(3^3)(4!))/2/4/2=81 (so it should be around there). Does anyone know the exact answer?



226 if we don't count rotations, mirrors or inverses as distinct cases.


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## CLL Smooth (Jul 4, 2016)

Anyone know how many CLL cases a megaminx has?


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## xyzzy (Jul 4, 2016)

CLL Smooth said:


> Anyone know how many CLL cases a megaminx has?



Let's say we count mirrors and inverses as distinct cases, and assume all the edges are already solved. The only cases with rotational symmetry are the skip case, the two O PLLs and the two "star" PLLs. (Not sure what the common names for these PLLs are; I don't do megaminx much.) Thus there are ((5!/2)(3^4)−5)/5 + 5 = 976 CLL cases in total.

If we don't assume the edges are solved, then we may fix (say) the UFR corner as a reference; this gives (4!/2)(3^4) = 972 cases without accounting for rotationally-equivalent cases. In addition to the CLL-skip case, there's now one more case with rotational symmetry, so there're (972−2)/5 + 2 = 196 CLL cases if edges aren't solved.

If we do count mirrored cases as the same, then the numbers drop to 506 (edges solved) and 116 (edges not solved); if we also count inverses as the same, then they further drop to 314 (edges solved) and 76 (edges not solved).

Edit: The numbers in the previous paragraph were wrong (there was a bug in the code I was using to check) and have been fixed.


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## CLL Smooth (Jul 4, 2016)

xyzzy said:


> Let's say we count mirrors and inverses as distinct cases, and assume all the edges are already solved. The only cases with rotational symmetry are the skip case, the two O PLLs and the two "star" PLLs. (Not sure what the common names for these PLLs are; I don't do megaminx much.) Thus there are ((5!/2)(3^4)−5)/5 + 5 = 976 CLL cases in total.
> 
> If we don't assume the edges are solved, then we may fix (say) the UFR corner as a reference; this gives (4!/2)(3^4) = 972 cases without accounting for rotationally-equivalent cases. In addition to the CLL-skip case, there's now one more case with rotational symmetry, so there're (972−2)/5 + 2 = 196 CLL cases if edges aren't solved.
> 
> If we do count mirrored cases as the same, then the numbers drop to 490 (edges solved) and 100 (edges not solved); if we also count inverses as the same, then they further drop to 252 (edges solved) and 52 (edges not solved).



Wow! That's a lot to digest.
If I understand you correctly, which I'm unsure of, this means there are 196 distinct LL cases on the Shengshou 2x2 megaminx (kilominx). Sound right?


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## xyzzy (Jul 4, 2016)

CLL Smooth said:


> If I understand you correctly, which I'm unsure of, this means there are 196 distinct LL cases on the Shengshou 2x2 megaminx (kilominx). Sound right?



Yup.


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## EetuK00 (Jul 15, 2016)

What's the probability for no cycle breaks in 3x3 blind memo?

And bonus: How many different memos are there?


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## tseitsei (Jul 15, 2016)

EetuK00 said:


> What's the probability for no cycle breaks in 3x3 blind memo?
> 
> And bonus: How many different memos are there?


I'm working on the cycle break question but isn't the bonus question quite obvious or am I being stupid?
Obviously there must be as many different memos as there are scrambles. Otherwise it would be impossible to know which scramble you are currently solving...

So bonus question answer for 3x3 is
{\displaystyle (8!\cdot 3^{8-1})\cdot \left({\frac {12!}{2}}\cdot 2^{12-1}\right)=43\,252\,003\,274\,489\,856\,000\approx 43\cdot 10^{18}}






But then again it depends what you mean by memo. Of course you can artificially create memos that don't correspond to any possible 3x3 scramble but if that is the case there are infinite amount of different memos. But to be able to solve every 3x3 scramble you need as many different memos as there are scrambles.


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## EetuK00 (Jul 15, 2016)

tseitsei said:


> I'm working on the cycle break question but isn't the bonus question quite obvious or am I being stupid?
> Obviously there must be as many different memos as there are scrambles. Otherwise it would be impossible to know which scramble you are currently solving...
> 
> So bonus question answer for 3x3 is
> {\displaystyle (8!\cdot 3^{8-1})\cdot \left({\frac {12!}{2}}\cdot 2^{12-1}\right)=43\,252\,003\,274\,489\,856\,000\approx 43\cdot 10^{18}}



Oh yeah the bonus question is actually quite obvious now that I think about it


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## xyzzy (Jul 15, 2016)

EetuK00 said:


> What's the probability for no cycle breaks in 3x3 blind memo?



One could "break cycles" in order to flip edges or twist corners, but (I think?) most people just use algs for that, so assuming that, we can ignore the orientation of the pieces entirely. In other words, we're asking for the probability that we have only one corner cycle and only one edge cycle. (Please correct me if this is wrong and not what you're asking for, because I have no idea how people do BLD.)

Consider the two cases based on permutation parity. Suppose the corner/edge permutation parities are even. (They must be either both even or both odd.)

P(one corner cycle | even parity) = (P(corner 7-cycle) + P(corner 5-cycle) + P(corner 3-cycle) + P(corners solved)) / P(even parity)
= [(C(8,7)6! + C(8,5)4! + C(8,3)2! + C(8,1)0!)/8!] ⋅ 2
= [1/(7⋅1!) + 1/(5⋅3!) + 1/(3⋅5!) + 1/(1⋅7!)] ⋅ 2
= 43/120

By the same reasoning:
P(one edge cycle | even parity) = [1/(11⋅1!) + 1/(9⋅3!) + 1/(7⋅5!) + 1/(5⋅7!) + 1/(3⋅9!) + 1/(1⋅11!)] ⋅ 2
= 48187/217728

As the corners and edges are conditionally independent once we know the parity, we can conclude that the probability of not having to break cycles when the parities are even is:
P(one corner cycle, one edge cycle | even parity) = P(one corner cycle | even parity) P(one edge cycle | even parity)
= 2072041/26127360
≈ 7.93%

Then we can do the same computation for the odd parity case:
P(one corner cycle, one edge cycle | odd parity) = 39784637/326592000
≈ 12.18%

And conclude:
P(one corner cycle, one edge cycle) = (1/2)(2072041/26127360) + (1/2)(39784637/326592000)
= 131370299/1306368000
≈ 10.06%


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## EetuK00 (Jul 15, 2016)

Cool! That answers my question perfectly


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## tseitsei (Jul 15, 2016)

No for the real question. I hope someone can verify my result but I think this should be correct 

For edges I get 22.65% chance for no cycle breaks.
And for corners I get 33.98% chance for no cycle breaks.
And together 22.65% * 33.98% = 7.70% for no cycle breaks in whole solve
NOTE: I'M NOT COUNTING FLIPPED/TWISTED PIECES AS CYCLE BREAKS HERE!
Note2: Should I do something for the total propability because of parity? Not sure about this one...

How I got hose numbers:


Spoiler



for edges:
First I have 1/12 of a chance that the buffer piece is in its correct location to start with.
In this case the only way I will have no cycle breaks is if all other edges are in their correct positions already (flipped or not). And possibility for that is quite obviously 1/11!.

Then next case: I have 11/12 * 1/11 (which obviously =1/12 that I only noticed a while later) chance that the buffer piece is my second target (the one right after the buffer piece) since 11/12 that it is not in the buffer position and then 1/11 that it is in the next location we go. And now I have solved 2 edges already so in order to have only 1 cycle the remaining edges must be in their correct positions. Propability for that is 1/10!.

Once more to make the idea clear: I have 11/12 * 10/11 * 1/10 =1/12 chance that the buffer piece is my 3rd target. And again the remaining 9 edges need to have correct permutation for only 1 cycle to exist so propability is 1/9!.

We continue this until we have covered all cycle lengths so up to cycle length of 11 where we have 1/12 chance for the buffer to be there and 0 edges remaining after that so we get 1/0! for propability for our remaining edges.

so now we put it all together and get a neat sum (don't know the correct format for this forum but...):

sum(n goes 0...11)[ (1/12)*(1/n!) = 22.65%

and similarly for corners we get a sum:
sum(n goes 0...7)[(1/8)*(1/n!) = 33.98%



EDIT: ninja'd and he did it better  So I should have done something with parity... Well my numbers were approximately correct anyway


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## DGCubes (Jul 15, 2016)

EetuK00 said:


> Oh yeah the bonus question is actually quite obvious now that I think about it



Technically multiple memos could correspond to the same scramble. For example, doing cycles in different orders makes a difference, as does how you memorize flipped edges and twisted corners. I think a lot of it also depends on a person's particular memo method. So, I don't think you could really get a true answer to the bonus question.


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## tseitsei (Jul 15, 2016)

DGCubes said:


> Technically multiple memos could correspond to the same scramble. For example, doing cycles in different orders makes a difference, as does how you memorize flipped edges and twisted corners. I think a lot of it also depends on a person's particular memo method. So, I don't think you could really get a true answer to the bonus question.


Yep there are really 2 possible answers:
Q: How many memos do you NEED to be able to solve any 3x3 scramble?
A: The amount of different scrambled states possible.
Q: How many different memos could be made?
A: Infinite amount.


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## EetuK00 (Jul 15, 2016)

tseitsei said:


> Yep there are really 2 possible answers:
> Q: How many memos do you NEED to be able to solve any 3x3 scramble?
> A: The amount of different scrambled states possible.
> Q: How many different memos could be made?
> A: Infinite amount.




Wait what? Infinite? If we give every sticker its own letter, you cant write out infinite amount of memos(?) Am I missing something here?


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## tseitsei (Jul 15, 2016)

EetuK00 said:


> Wait what? Infinite? If we give every sticker its own letter, you cant write out infinite amount of memos(?) Am I missing something here?


Well depends on how we define memo. If I have to memo letters WR I could memo WalRus or WaR or World Record... would those be the same memo or different memo.

Also even if we define 'memo' to be just the string of letters we can always just lengthen the memo (there is no practical reason to do so but we CAN do it.). Trivial example: I have a 3 cycle buffer -> A -> B -> buffer. Normally you would memo AB But you could also memo BABA for the same outcome. Or something like IB IA BI. As I said stupid but possible...


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## EetuK00 (Jul 15, 2016)

tseitsei said:


> Well depends on how we define memo. If I have to memo letters WR I could memo WalRus or WaR or World Record... would those be the same memo or different memo.
> 
> Also even if we define 'memo' to be just the string of letters we can always just lengthen the memo (there is no practical reason to do so but we CAN do it.). Trivial example: I have a 3 cycle buffer -> A -> B -> buffer. Normally you would memo AB But you could also memo BABA for the same outcome. Or something like IB IA BI. As I said stupid but possible...



Ahh I see


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## EetuK00 (Jul 16, 2016)

Well what's the probability for not having to orientate centers in 3x3 supercube?
(If you completely ignore center orientarion during the solve)


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## xyzzy (Jul 18, 2016)

EetuK00 said:


> Well what's the probability for not having to orientate centers in 3x3 supercube?



There're 4 possible orientations for each centre and there're 6 centres, along with a parity constraint (you can't rotate a single centre 90°), so that's \( 4^6/2=2048 \) centre orientations and a 1/2048 chance of not having to orient centres.


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## PhillipEspinoza (Aug 25, 2016)

Can anyone tell me:

1) What are the chances, after EO Line, that any F2L edge (4/12) will be paired up with a correctly oriented 3rd layer corner (4/8)? 

2) What about the chances that a top layer edge piece (non EO Line, so 2/12) will be connected to a correctly oriented 3rd layer corner (as if it were a 1st layer corner)? [You have 4 correct positions with 4 3rd layer corners that can join either side of the 2 edge pieces. There are 8 corners and they each have 3 different ways to be oriented. Hope this helps.]

3) What is the chance that either 1 OR 2 will randomly occur after EO Line during block building?

4) We all know the probability of an OLL skip with ZZ is 1/27. But what is the probability of an OLL skip after ELS? In other words, if you just randomly insert the edge piece into LS, without regard to corner orientation of the corner it's paired with, does that decrease the chances of an OLL skip to 1/81?

5) If you do ELS with a correctly oriented LL corner, your chances of ending up with all pieces in LL oriented is 1/27 right? And if I do not get an orientation skip, and I do a CLS alg, my chances of skipping PLL are 1/72. Given this, what are my chances that I will either get an OLL skip or a PLL skip (for ZZ)? In other words, what are the chances of a LL skip after LS using these combined methods?

THANKS IN ADVANCE TO ANYONE WHO CAN MAKE SENSE OF WHAT I'M SAYING IN MY SLEEP DEPRIVED STATE. MUCH APPRECIATED!


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## AlphaSheep (Aug 25, 2016)

PhillipEspinoza said:


> Can anyone tell me:
> 
> 1) What are the chances, after EO Line, that any F2L edge (4/12) will be paired up with a correctly oriented 3rd layer corner (4/8)?
> 
> ...


1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%.

2) Not sure I understand. Can you give an example?

3) Not sure how to work this out, because the state after 1 move is not completely independent of the state before that move. I think that the answer depends on how inefficient your solution is. The more moves you take, the more likely it is to happen, I'm sure. 

4) Yes 1/81 is correct. 

5) Yes, if the corner you insert is oriented then the chance of the rest also being oriented is 1/27. And if you don't influence permutation of the other corners during CLS, then yes, theres a 1/72 chance of a PLL skip. That means there's a 49/972 (about 5%) chance of skipping at least one of OLL or PLL and 1/1944 chance of skipping both. Same as nomal OLL+PLL for ZZ.


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## guysensei1 (Aug 25, 2016)

tseitsei said:


> but isn't the bonus question quite obvious or am I being stupid?


I think you're missing something here, there's more than 1 valid way to memorise scrambles with cycle breaks. You can choose any unsolved piece to shoot to after all, so there exist many many scrambles with multiple valid memoes.

Example: assume UB and UF are swapped, and your buffer is somewhere else. You can memo UB UF UB or UF UB UF or BU FU BU or FU BU FU (convert those to whatever scheme you're using.)


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## xyzzy (Aug 25, 2016)

AlphaSheep said:


> 1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%



The probabilities aren't independent (e.g. if you have FR in the UB position and BR in the UR position, the matching corner position for both of those edges is UBR), so this is not exactly accurate, but it's pretty close to the correct value. I'm too tired to exert my brain, so I just did a simulation on a million random states and got this:

No pairs: 453642
1 pair: 433056
2 pairs: 106154
3 pairs: 7070
4 pairs: 78


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## SiTeMaRo (Aug 25, 2016)

cmhardw said:


> I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.
> 
> The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:
> 
> ...


And I still had 3 LL skips in just an year.


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## AlphaSheep (Aug 25, 2016)

xyzzy said:


> The probabilities aren't independent (e.g. if you have FR in the UB position and BR in the UR position, the matching corner position for both of those edges is UBR), so this is not exactly accurate, but it's pretty close to the correct value. I'm too tired to exert my brain, so I just did a simulation on a million random states and got this:
> 
> No pairs: 453642
> 1 pair: 433056
> ...


Yeah, I just realised that I was thinking that the corner can only connect on one side, so I was completely wrong on that too. I'll have to rethink it. It's cool to have simulation results though because it means I know what the answer needs to be


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## hagner (Aug 25, 2016)

back when i was doing lbl i got a l2l skip (last two layers) except for oll 23 otherwise everything was permutatet and oriented after first layer and i just did sexy moves to orient oll 23 and auf. what are the odds for this?

my times back then where around 1 minute, maybe even 1 and a half and i got a 33 second solve, but of course i didn't count as pb.


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## PhillipEspinoza (Aug 26, 2016)

AlphaSheep said:


> 1) I'd prefer someone else to confirm my answers but I think the chances of one specific edge being paired with any 3rd layer corner in a specific orientation is 1/6 (1/3 chance of being oriented and 1/2 chance of being from the last layer). That means that the odds of at least one out of two specific edges being like that are 11/36, and for at least one out of 3 is 91/216 and for at least one of four specific edges, its 671/1296 or 51.77%.
> 
> 2) Not sure I understand. Can you give an example?
> 
> ...



Thanks for your response. Surely, in your #5 response, the skipping of both CLS then PLL is not possible if the corner piece in DFR is a fixed LL corner. In other words, an LL skip after building the 2nd to last slot is not possible, even though it is possible to skip orientation, and be left with a TTLL case. Or you can be left with a CLS case after which you might get a PLL skip. But I don't think you can have both. But in this scenario, ~1/20 solves will have the LL done in one LS step via skip after CLS or execution of TTLL. That's useful to know THANKS!


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## CyanSandwich (Oct 6, 2016)

What are the chances that your last slot+LL is simply 2 twisted corners?

I got this, but I'm no probability expert.
(5! * (3^4) * (5!/2) * (2^4)) / 20 = 466560


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## wowitsbryce (Oct 13, 2016)

I just finished learning ELL. So, I started thinking, what are the chances of having the corners of the last layer solved? In other words, what are the chances of solving the first two layers and immediately get an ELL case? Does the probability change if partial edge control is used?


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## Berd (Oct 13, 2016)

wowitsbryce said:


> I just finished learning ELL. So, I started thinking, what are the chances of having the corners of the last layer solved? In other words, what are the chances of solving the first two layers and immediately get an ELL case? Does the probability change if partial edge control is used?


Edge control won't affect the probablilliy of an CLL skip.


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## AlphaSheep (Oct 14, 2016)

wowitsbryce said:


> I just finished learning ELL. So, I started thinking, what are the chances of having the corners of the last layer solved? In other words, what are the chances of solving the first two layers and immediately get an ELL case? Does the probability change if partial edge control is used?


There's a 1 in 162 chance, which is 0.62%.


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## Altha (Oct 14, 2016)

What's the chance of any scrambled cube having at least 1 twisted corner? (or 1-[no twisted corners])


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## Youds edy (Oct 22, 2016)

Just read, it's amazing.


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## ViliusRibinskas (Nov 2, 2016)

What is the chance to get LL skip on square-1 and LL skip on 2x2?


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## guysensei1 (Nov 2, 2016)

ViliusRibinskas said:


> What is the chance to get LL skip on square-1 and LL skip on 2x2?


Sq1: (1/72)/2=1/144 which is the same as a PLL skip on even layered cubes.
2x2: 1/162


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## Torch (Nov 16, 2016)

I have an interesting problem that someone might be able to help with; it's really more of a combinatoric problem than a probability problem, but here it is anyway:

If you have an icosahedron and and color each of the 30 edges red or blue randomly, how many possible unique colorings are there? This counts all rotationally symmetric positions as identical. This seems pretty difficult to me, but if someone could at least give me an estimate that would be great.


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## xyzzy (Nov 17, 2016)

Torch said:


> If you have an icosahedron and and color each of the 30 edges red or blue randomly, how many possible unique colorings are there? This counts all rotationally symmetric positions as identical. This seems pretty difficult to me, but if someone could at least give me an estimate that would be great.



Such computations are usually done with the lemma that is not Burnside's.



Spoiler: math



We first need to list all the rotational symmetries of the icosahedron. There is 1 order-1 rotation (the identity, or "do nothing", rotation), 24 order-5 rotations (rotate around a vertex either 1/5 or 2/5 of a revolution), 20 order-3 rotations (rotate around a face) and 15 order-2 rotations (rotate around an edge).

Then for each rotation, we count how many ways there are to edge-colour the icosahedron such that applying the rotation doesn't change the colouring. Rather than just two colours, let's say we use n colours. For the order-1 rotation, that gives n^30 colourings; for the order-5 rotations, that gives n^6 colourings each; for the order-3 rotations, that gives n^10 colourings each; for the order-2 rotations, that gives n^14 colourings each. Adding these up, we get n^30 + 24n^6 + 20n^10 + 15n^14, which not-Burnside's-lemma says is exactly equal to the total number of rotations (60) multiplied by the number of rotationally distinct colourings.

Now we can substitute n=2 and get punch our calculator to get 17900160 as our answer, but also notice that the largest contributor to the above expression is by far the n^30 term, so we get a pretty good approximation just from dividing 2^30 (the number of colourings) by 60 (the number of rotational symmetries), which equals 17895697.067.



(I may or may not have made silly mistakes above, but you get the idea.)


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## ViliusRibinskas (Nov 17, 2016)

So how many cases are there on 2x2 when you have all of your yellow and white corners oriented, 3/4 of white layer is solved and the 4th piece of the white layer is on the top?

P.S. I don't know if this is the correct thread or not.


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## shadowslice e (Nov 17, 2016)

ViliusRibinskas said:


> So how many cases are there on 2x2 when you have all of your yellow and white corners oriented, 3/4 of white layer is solved and the 4th piece of the white layer is on the top?
> 
> P.S. I don't know if this is the correct thread or not.


According to this page, there are 9 cases (one solved) when considering AUFs.


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## Cale S (Nov 17, 2016)

shadowslice e said:


> According to this page, there are 9 cases (one solved) when considering AUFs.



3 PLLs and 6 cases for CP after inserting


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## shadowslice e (Nov 17, 2016)

Cale S said:


> 3 PLLs and 6 cases for CP after inserting


*1 skip, 2 PLLs, 6 CPLS cases


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## Daniel Lin (Nov 17, 2016)

Altha said:


> What's the chance of any scrambled cube having at least 1 twisted corner? (or 1-[no twisted corners])


ez pz 

7*2/24-21*4/24/21+35*8/24/21/18-35*16/24/21/18/15+21*32/24/21/18/15/12-7*64/24/21/18/15/12/9+1*128/24/21/18/15/12/9/6

=44.34%


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## Cale S (Nov 17, 2016)

shadowslice e said:


> *1 skip, 2 PLLs, 6 CPLS cases



oh lol why did I think skip was a PLL, I guess I was thinking of 3x3 PLLs with adj, diag, or no swap


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## Altha (Nov 18, 2016)

Daniel Lin said:


> ez pz
> 
> 7*2/24-21*4/24/21+35*8/24/21/18-35*16/24/21/18/15+21*32/24/21/18/15/12-7*64/24/21/18/15/12/9+1*128/24/21/18/15/12/9/6
> 
> =44.34%


Excluding the buffer would it be that percentage multiplied by 7/8 or do I suck at maths


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## Daniel Lin (Nov 18, 2016)

Altha said:


> Excluding the buffer would it be that percentage multiplied by 7/8 or do I suck at maths


You can't have just the the buffer twisted, then it would be unsolvable!
I just ignored the buffer, since it's determined by the rest of the 7 corners.

Basically, there are 7 corners. The probability for each corner of being twisted is 2/24. To find the probability of any one of them being twisted, you add them all up. But then you would be overcounting, because two corners could be twisted. So you subtract the probablity of any two combos of corners being twisted. But then you'd be undercounting, since three could be twisted. So you add the prob of three corner twists and so on...

EDIT: o wait i'm wrong. the buffer does matter. dangit why do i always mess up XD


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## Cale S (Nov 18, 2016)

Altha said:


> What's the chance of any scrambled cube having at least 1 twisted corner? (or 1-[no twisted corners])



I tried it including the buffer and got an answer a little higher than Daniel's which ignored buffer


Spoiler



(!8)*(3^7) + 8*(!7)*(3^6) + (8 choose 2)*(!6)*(3^5) + (8 choose 3)*(!5)*(3^4) + (8 choose 4)*(!4)*(3^3) + (8 choose 5)*(!3)*(3^2) + (8 choose 6)*3 + 1

45,273,046 positions with no twisted corners

total positions = 3^7*8! = 88,179,840

so (88179840 - 45273046)/88179840

which is 48.658...%


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## Goosly (Nov 18, 2016)

Daniel Lin said:


> The probability for each corner of being twisted is 2/24.



Why 2/24?
Maybe I didn't understand the question but... why isn't the chance of at least one twisted corner (1 - (1/3)^7) ?

Also how could it ever be 44% or 48%? Almost half of the scrambles would have no twisted corners, which is clearly wrong.


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## Cale S (Nov 18, 2016)

Goosly said:


> Why 2/24?
> Maybe I didn't understand the question but... why isn't the chance of at least one twisted corner (1 - (1/3)^7) ?
> 
> .



That's the probability of at least one corner unoriented, which is different from twisted in place


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## gateway cuber (Nov 18, 2016)

CLL Smooth said:


> Wow! That's a lot to digest.
> If I understand you correctly, which I'm unsure of, this means there are 196 distinct LL cases on the Shengshou 2x2 megaminx (kilominx). Sound right?


what!?!?!?!? When I finally buy a Kilo I am sooooooo learning 1LLL


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## CLL Smooth (Nov 18, 2016)

gateway cuber said:


> what!?!?!?!? When I finally buy a Kilo I am sooooooo learning 1LLL


Recognition looks atrocious. Plus only 3 PLLs after OCLL, it hardly seems worth it.


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## Daniel Lin (Nov 19, 2016)

Goosly said:


> Why 2/24?
> Maybe I didn't understand the question but... why isn't the chance of at least one twisted corner (1 - (1/3)^7) ?
> 
> Also how could it ever be 44% or 48%? Almost half of the scrambles would have no twisted corners, which is clearly wrong.


Just pick a random sticker like UBR. There are 24 corners stickers, and UBR would be twisted if it were at RUB or BUR


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## gateway cuber (Nov 20, 2016)

CLL Smooth said:


> Recognition looks atrocious. Plus only 3 PLLs after OCLL, it hardly seems worth it.


yeah see yur point... then how many algs is ocll?


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## CLL Smooth (Nov 21, 2016)

gateway cuber said:


> yeah see yur point... then how many algs is ocll?


I think it's 16 cases. Recognition is same as megaminx OCLL but most algs are easier.


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## DELToS (Dec 6, 2016)

What is the probability of having an ELL case using CFOP after finishing F2L?


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## ViliusRibinskas (Dec 6, 2016)

What is the chance of getting LL and OLL skip doing EO before inserting 4th f2l pair?


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## shadowslice e (Dec 6, 2016)

ViliusRibinskas said:


> What is the chance of getting LL and OLL skip doing EO before inserting 4th f2l pair?


0?

Not really sure what you mean...


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## xyzzy (Dec 6, 2016)

DELToS said:


> What is the probability of having an ELL case using CFOP after finishing F2L?



1/162. (1/27 for corners to be oriented, 1/6 for corners to be permuted.)


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## ViliusRibinskas (Dec 7, 2016)

shadowslice e said:


> 0?
> 
> Not really sure what you mean...


You do EO when iserting 4th f2l pair and all the edges are oriented. So what is the chance to get OLL or LL skip?


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## AlphaSheep (Dec 7, 2016)

ViliusRibinskas said:


> You do EO when iserting 4th f2l pair and all the edges are oriented. So what is the chance to get OLL or LL skip?


Chances of OLL skip are 1 in 27 and chances of LL skip are 1 in 1944.


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## DELToS (Dec 10, 2016)

What is the probability of using a random PLL generator (like the PLL option on csTimer) and not getting an E-Perm until the 46th PLL, and then getting the E-Perm 3 times in a row?


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## obelisk477 (Dec 10, 2016)

DELToS said:


> What is the probability of using a random PLL generator (like the PLL option on csTimer) and not getting an E-Perm until the 46th PLL, and then getting the E-Perm 3 times in a row?



Roughly (21/22)^45 * (1/36)^3, or .0002642%. So about to put it another way, you would get one of these occurences for every 20 LL skips

EDIT: I say roughly because I don't want to calculate the average PLL probability excluding E perm. 21/22 is close enough


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## Daniel Lin (Dec 10, 2016)

obelisk477 said:


> EDIT: I say roughly because I don't want to calculate the average PLL probability excluding E perm. 21/22 is close enough



yeah it's 68/72


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## DELToS (Dec 10, 2016)

obelisk477 said:


> Roughly (21/22)^45 * (1/36)^3, or .0002642%. So about to put it another way, you would get one of these occurences for every 20 LL skips
> 
> EDIT: I say roughly because I don't want to calculate the average PLL probability excluding E perm. 21/22 is close enough


I've probably done over ten to fifteen thousand solves (very rough estimate, cubing for 2.5 years) and I've never gotten a LL skip...


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## CubingGenius (Dec 13, 2016)

What is the probability of doing a 2 gen scramble by hand randomly and getting a 1 move solutions? 1 out of 146966400 if I'm correct?


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## qbtronix (Dec 21, 2016)

What is the probability of a Roux CMLL + L6E Skip? I.e. a solved cube after just the 2 left and right blocks.


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## Daniel Lin (Dec 21, 2016)

qbtronix said:


> What is the probability of a Roux CMLL + L6E Skip? I.e. a solved cube after just the 2 left and right blocks.


if you mean COMPLETELY solved 
then 
1/(3^3*4!*2^5*6!/2)=1.33959191e-7


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## shadowslice e (Dec 21, 2016)

Daniel Lin said:


> if you mean COMPLETELY solved
> then
> 1/(3^3*4!*2^5*6!/2)=1.33959191e-7


You need to factor in the fact you can have centres in any orientation (or yellow top, white bottom) so the number needs to be divided by 4 which is 3.34897977e-8


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## qbtronix (Dec 23, 2016)

shadowslice e said:


> You need to factor in the fact you can have centres in any orientation (or yellow top, white bottom) so the number needs to be divided by 4 which is 3.34897977e-8


Dang. I'm not a Roux solver, (or even a fast CFOP solver) and, believe it or not, had this while fooling around on a hand-scramble. I've had a few CFOP LL skips, most with AUF, but being orders of magnitude less likely my only conclusion is that I must have scrambled it exceedingly poorly. I had had two beers in me so perhaps my mind was playing tricks on me.


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## Daniel Lin (Dec 29, 2016)

how many total double 2cycles of edges are there on a 3x3?

like S' U2 M' U' M S U'

cases from different angles count as the same

EDIT: this is more of a combinatorics question


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## xyzzy (Dec 29, 2016)

Daniel Lin said:


> how many total double 2cycles of edges are there on a 3x3?
> 
> like S' U2 M' U' M S U'
> 
> cases from different angles count as the same



I'm assuming you don't count stuff like (M' U M U)3 U2 as a double 2-cycle because that would be too annoying to count. There are 289 cases up to rotational symmetry, or only 170 up to rotational and mirror symmetry.



Spoiler: Burnside's lemma stuff



There are five conjugacy classes of rotational symmetries: identity (1), 90° rotation about a face (6), 180° rotation about a face (3), rotation about an edge (6) and rotation about a corner (8). The numbers of 2+2-cycles preserved by any representative of each of these conjugacy classes are 5940, 6, 120, 100 and 0, respectively, so we can just plug this into Burnside's lemma to get (5940×1+6×6+120×3+100×6+0×8)/(1+6+3+6+8) = 289.

If you want to exclude mirror-identical cases, then you also have to consider the conjugacy classes of reflection symmetries: point reflection (1), axis-aligned plane reflection (3), 90° rotoreflection (6), face-diagonal-aligned plane reflection (6) and 60° rotoreflection (8). The numbers of 2+2-cycles preserved for these classes are 120, 156, 6, 100 and 0, respectively. Plugging the data from all ten conjugacy classes into Burnside's lemma, we get (5940×1+6×6+120×3+100×6+0×8+120×1+156×3+6×6+100×6+0×8)/(1+6+3+6+8+1+3+6+6+8) = 170.


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## Daniel Lin (Dec 29, 2016)

xyzzy said:


> I'm assuming you don't count stuff like (M' U M U)3 U2 as a double 2-cycle because that would be too annoying to count. There are 289 cases up to rotational symmetry, or only 170 up to rotational and mirror symmetry.
> 
> 
> 
> ...


thanks!
you are awesome

if you counted the annoying stuff, would it be approximately double?


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## xyzzy (Dec 29, 2016)

Daniel Lin said:


> if you counted the annoying stuff, would it be approximately double?



The annoying stuff turned out to actually not be so annoying to count once I changed how I was counting the number of cases. There are 265 order-4 cases up to rotational symmetry, or total 289+265 = 554 cases including normal 2+2-cycle cases, so it's a bit less than double.


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## Daniel Lin (Dec 29, 2016)

how many double 2cycles of corners are there?

less than edges I'm guessing?


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## GenTheThief (Dec 29, 2016)

Daniel Lin said:


> how many total double 2cycles of edges are there on a 3x3?
> 
> like S' U2 M' U' M S U'
> 
> ...





Daniel Lin said:


> how many double 2cycles of corners are there?
> 
> less than edges I'm guessing?


Deciding if they are worth learning for BLD? Or just curious?


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## Daniel Lin (Dec 29, 2016)

GenTheThief said:


> Deciding if they are worth learning for BLD? Or just curious?


I'm planning on making a list for BLD

The problem is, i don't have a way of organizing all 289 cases or even listing them all.

idk if it's worth learning all of them. But it saves you from having to do 3 comms, so learning at least a few would definitely be worth it (I know like 10 cases).


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## xyzzy (Dec 30, 2016)

Daniel Lin said:


> how many double 2cycles of corners are there?
> 
> less than edges I'm guessing?



Probably around 80 to 90 not counting the twisted cycle cases (e.g. [R', U' D L' U D']), roughly triple that if you include the twisted cycle cases.

(Not accounting for symmetry, there are 1890 cases, and since there are 24 rotational symmetries, we just divide 1890 by 24 to get an estimate of 78.75.)


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## YouCubing (Dec 30, 2016)

probability of a cross solved on clock? I remember there was a scramble on the cubingtime weekly comp that had that


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## Ronxu (Dec 31, 2016)

YouCubing said:


> probability of a cross solved on clock? I remember there was a scramble on the cubingtime weekly comp that had that


2*(1/12)^5


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## Mastermind2368 (Dec 31, 2016)

What are the chances of ls and LL skip on ZZ?


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## shadowslice e (Dec 31, 2016)

Mastermind2368 said:


> What are the chances of ls and LL skip on ZZ?


(2*4)/((3^3)(4!)(4!))=1/1944 (this is including ones which are an AUF from solved)
Not sure what you mean by ls and LL skip though but if you mean build F2L-1 and the rest is solved,
2/((3^4)(5!)(5!))= 1/583200


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## xyzzy (Jan 1, 2017)

Mastermind2368 said:


> What are the chances of ls and LL skip on ZZ?



Yeah, this is too vague to answer without further information. Obviously, once you have three "pairs" done, you're not going to count whatever's left as an LS skip unless the F2L happens to be solved too, so for a _real_ LS skip, the fourth pair has to be coincidentally solved together with the third. This happens _much more rarely_ than doing three slots in fixed order (e.g. always doing LB-LF-RB) and finding that the fourth is also solved (which has probability 1/75).

Most common F2L algs for solving a single slot don't affect other slots, unless a piece you need for the pair is stuck in said other slot. Not only this, you'd normally prefer to solve the other pair first, unless it _also_ has stuck pieces. This is a fairly rare situation, and even rarer is when you have something like this _and_ you can't use keyhole-style multislotting to fix it.

Another way of thinking about it is that you're asking about the probability of finding an F2L solution as short as an F2L-1 solution. You have _four_ choices of F2L-1, and each of them requires at most as many moves as solving the full F2L, so clearly, the chances of full F2L being as easy as all four of these choices has to be pretty tiny.

Of course human solvers exhibit a bias in the order in which the slots are solved, which complicates matters even further.

tl;dr: No easy closed form solution; do an experiment over a few thousand solves and count how many of them have "LS skips". That would be your answer for the probability of an LS skip, which you can divide by 1944 to get the approximate probability of an LSLL skip.


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## Mastermind2368 (Jan 2, 2017)

xyzzy said:


> Yeah, this is too vague to answer without further information.[/QUOTE
> 
> I ment if you had your EO-Line, 2/4 pairs and when you put in the second last pair, you solved the last pair and the LL


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## xyzzy (Jan 2, 2017)

The said "other information" wasn't pertaining to what you meant by "LS skip" (which I think is fairly universal), but to what F2L algs you use and how you solve F2L in general, because the answer to your question of the probability of an LS skip depends on that.

Take a look at this F2L case. If you use R2 U2 R' U2 R2 to insert the last two F2L edges at the same time, does that count as an LS skip? Even though there's no distinct "last slot"? Also note that if you solve the "third slot" differently, for example with R2 U' R' U R2, then there might no longer be an LS skip.

Multislotting and freestyle blockbuilding tricks severely complicate the notion of a "last slot", but even without these, the key consideration here really is whether you consider a certain fixed slot to be the "last slot", or if the "last slot" is always the last unsolved slot. In the former case, then skipping LSLL is as shadowslice answered: 1/75 multiplied by the probability of a ZBLL skip.

In the latter case, that's what I was trying to explain before, and there's no easy answer that can be obtained just from multiplying a bunch of small numbers together.


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## CubingGenius (Jan 3, 2017)

What is the chance of having zero misoriented edges on two axis while the third one has at least two misoriented edges on a cube that is in a random, solvable state?


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## xyzzy (Jan 3, 2017)

CubingGenius said:


> What is the chance of having zero misoriented edges on two axis while the third one has at least two misoriented edges on a cube that is in a random, solvable state?



P(edges oriented on all three axes) = 1/(2^11 ⋅ 12!/4!^3) = 1/70963200
P(edges oriented on F-B and L-R axes) = 1/(2^11 ⋅ 12!/(4!8!)) = 1/1013760
P(edges oriented on F-B and L-R, but not on U-D) = 1/1013760 − 1/70963200 = 23/23654400
P(edges oriented on exactly two axes ) = 3 ⋅ 23/23654400 = 23/7884800

Roughly 1 in 340000, in other words.


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## guysensei1 (Jan 3, 2017)

In a 4BLD solve, what is the least number of solved pieces you can get, assuming you can always choose the orientation with the most solved pieces?


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## CubingGenius (Jan 3, 2017)

xyzzy said:


> P(edges oriented on all three axes) = 1/(2^11 ⋅ 12!/4!^3) = 1/70963200
> P(edges oriented on F-B and L-R axes) = 1/(2^11 ⋅ 12!/(4!8!)) = 1/1013760
> P(edges oriented on F-B and L-R, but not on U-D) = 1/1013760 − 1/70963200 = 23/23654400
> P(edges oriented on exactly two axes ) = 3 ⋅ 23/23654400 = 23/7884800
> ...



Thank you very much!


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## Cale S (Jan 3, 2017)

guysensei1 said:


> In a 4BLD solve, what is the least number of solved pieces you can get, assuming you can always choose the orientation with the most solved pieces?



There was discussion of this a few years ago, I don't remember if an exact answer was reached
I think something like 6 was the lowest after several trials 

found the thread: https://www.speedsolving.com/forum/threads/most-common-best-of-4bld-centers-solved-problem.44081/


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## CubingGenius (Feb 6, 2017)

If you take all the solvable positions on the rubik's cube, count the EO from all 3 angles and then choose the EO with the lowest number of misoriented edges, (if multiple orientations have the same number, green front/red front/white front is the preference) what percent of the time wil each EO use rounded to at least 2 decimal places? (eg. green front has 6, red front has 4 and white front has 4, so red is chosen because red and white have the same number, but red is chosen before white because of the preference mentioned above.)

I hope someone is able to answer this challenging question. If you need more clarity, I will do my best to explain it better.


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## xyzzy (Feb 6, 2017)

CubingGenius said:


> If you take all the solvable positions on the rubik's cube, count the EO from all 3 angles and then choose the EO with the lowest number of misoriented edges, (if multiple orientations have the same number, green front/red front/white front is the preference) what percent of the time wil each EO use rounded to at least 2 decimal places? (eg. green front has 6, red front has 4 and white front has 4, so red is chosen because red and white have the same number, but red is chosen before white because of the preference mentioned above.)
> 
> I hope someone is able to answer this challenging question. If you need more clarity, I will do my best to explain it better.



At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.

Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.


```
from itertools import product
from collections import Counter

combs = Counter(tuple(sorted(a[:4]) + sorted(a[4:8]) + sorted(a[8:])) for a in product(range(3), repeat=12) if 4 == a.count(0) == a.count(1))
orients = [a for a in product(range(2), repeat=12) if sum(a) % 2 == 0]

axis_counts = [0, 0, 0]

for comb in combs:
    for orient in orients:
        flip_counts = [0, 0, 0]
        for i in range(3):
            flip_counts[ i] = sum((orient[j] + (j // 4 == i) + (comb[j] == i)) % 2 for j in range(12))
        axis_counts[flip_counts.index(min(flip_counts))] += combs[comb]

print(axis_counts)
```

The output is [37365070, 20481060, 13117070], _id est_ the first choice gets chosen 52.654% of the time, the second choice 28.862%, and the third choice 18.484%.


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## CubingGenius (Feb 6, 2017)

xyzzy said:


> At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.
> 
> Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.
> 
> ...



Thank you very much for the answer. You appear to have an amazing knowledge of cube theory.


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## TDM (Feb 8, 2017)

xyzzy said:


> At first I was going to say this wasn't challenging, then I got stumped on how to count flipped edges in multiple orientations.
> 
> Define a reference orientation by assuming every quarter turn flips four edges. To count how many flipped edges there are on, say, the standard ZZ EO reference, "flip" the edges that are on the S slice, "flip" the edges that would be on the S slice on a solved cube, and then count. Repeat this for the M and E slices to get the numbers of misoriented edges on the three possible ZZ EO references.
> 
> ...


How about with two orientations instead of three?


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## xyzzy (Feb 9, 2017)

TDM said:


> How about with two orientations instead of three?



Good thing I posted the code instead of letting it get buried in my command history!

Just change flip_counts to be a two-element list and the range(3) to range(2); running the modified code snippet gives 67.1% for first choice and 32.9% for second choice.

(If you want to calculate this by hand, unlike the three-axis calculation, the number of cases here is small enough to be doable in maybe a few hours, but thank goodness computers exist.)


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## obelisk477 (Mar 2, 2017)

How many LL cases are there if there is a formed 2x1x1 block on U, either flipped or not?


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## shadowslice e (Mar 3, 2017)

obelisk477 said:


> How many LL cases are there if there is a formed 2x1x1 block on U, either flipped or not?


According to the wiki, there are 58 unique cases. https://www.speedsolving.com/wiki/index.php/Tripod_Method


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## xyzzy (Mar 3, 2017)

shadowslice e said:


> According to the wiki, there are 58 unique cases. https://www.speedsolving.com/wiki/index.php/Tripod_Method



It's only a 2×1×1 block (i.e. a pair), not a square, so there are way more cases than tripod LL. I'd guess that there are around 2000 different cases.


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## shadowslice e (Mar 3, 2017)

xyzzy said:


> It's only a 2×1×1 block (i.e. a pair), not a square, so there are way more cases than tripod LL. I'd guess that there are around 2000 different cases.


Oh ok about 2^2*3^2*3!*3!/2/4=162 cases then I think.


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## AlphaSheep (Mar 3, 2017)

obelisk477 said:


> How many LL cases are there if there is a formed 2x1x1 block on U, either flipped or not?


I guess more than half of full LL.

Consider any particular edge. There are 2 possible corners that can connect to it. The chance of the first corner being on the correct side of the edge and oriented is 1 in 12. There's also a 1 in 2 chance that the first corner won't be touching the edge at all and 1 in 6 chance that the first corner is on the correct side but not oriented. In both of these cases there's a 1 in 9 chance of the second corner being on the correct side and oriented. That means the probability of having a 2x1x1 block involving the specific edge is 1/12+(1/2+1/6)*(1/9) = 17/108, or 15.74% of LL cases. Then you need to account for the fact that one of the other edges could be part of a 2x1x1 block, but my attention span ran out at about this point.


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## Teoidus (Mar 8, 2017)

What is the probability that, in a random scramble, at least 1 corner edge pair is solved?

What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?


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## Cale S (Mar 8, 2017)

Teoidus said:


> What is the probability that, in a random scramble, at least 1 corner edge pair is solved?
> 
> What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?



By solved do you mean both pieces in place? 

I think the probability of a corner edge pair being formed is close to 1 - 1/e


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## obelisk477 (Mar 8, 2017)

Teoidus said:


> What is the probability that, in a random scramble, at least 1 corner edge pair is solved?
> 
> What is the probability that, in a random scramble, at least 1 corner edge pair is 1 move from solved?



The 1 - 1/e figure I think is right as well, so like 63.2% ish.

1 move from solved *or* solved has to be pretty close to 100%, but I'm speculating.


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## FJT97 (Mar 8, 2017)

Its 8 moves max for a cfop cross. How many moves for an eoline? And how many on average?

Thanks


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## sqAree (Mar 8, 2017)

FJT97 said:


> Its 8 moves max for a cfop cross. How many moves for an eoline? And how many on average?
> 
> Thanks



EOLine takes an average of ~6.127 moves and a maximum of 9 moves

From http://cube.crider.co.uk/zz.php?p=eoline


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## Teoidus (Mar 9, 2017)

Cale S said:


> By solved do you mean both pieces in place?
> 
> I think the probability of a corner edge pair being formed is close to 1 - 1/e



Sorry I meant formed, not necessarily solved.

Do you happen to how people arrived at that figure?


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## Cale S (Mar 9, 2017)

Teoidus said:


> Sorry I meant formed, not necessarily solved.
> 
> Do you happen to how people arrived at that figure?



I'm not sure but I think it's mathematically similar to how the fraction of permutations that are a derangement approaches 1/e


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## xyzzy (Mar 9, 2017)

Teoidus said:


> Sorry I meant formed, not necessarily solved.
> 
> Do you happen to how people arrived at that figure?



There are 24 possible corner/edge pairs, and each one has (exactly) 1/24 chance of being formed. Assuming that the probabilities are approximately independent, this means that we can approximate the distribution of the number of pairs formed as a Poisson distribution with mean 1; under this approximation, the probability that there are no pairs at all is approximately 1/e.

Or, if we don't want to use a Poisson approximation, then under the same simplifying assumption that the probabilities are approximately independent, P(no pairs) ~ \( (1-1/24)^{24}\approx36\% \).


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## Daniel Lin (Mar 25, 2017)

ok ishaan asked me what the probablity is that you get at least one flipped edge for 3bld (excluding the buffer piece). check my work if you want lol

i got 36.82 %



Spoiler



(11 choose 1)/24-(11 choose 2)/24/22+(11 choose 3)/24/22/20-(11 choose 4)/24/22/20/18+(11 choose 5)/24/22/20/18/16-(11 choose 6)/24/22/20/18/16/14+(11 choose 7)/24/22/20/18/16/14/12-(11 choose 8)/24/22/20/18/16/14/12/10+(11 choose 9)/24/22/20/18/16/14/12/10/8-(11 choose 10)/24/22/20/18/16/14/12/10/8/6+(11 choose 11)/24/22/20/18/16/14/12/10/8/6/4


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## Hazel (Apr 4, 2017)

I think this is the right thread for this, maybe?
Hoe many cases would there be for this Last Slot + Last Layer thing?:

- LL Corners are already solved (both oriented and permuted)
- FRD corner is already solved
- All the Last Layer edges on the U layer are oriented, but not necessarily the LL edge in FR

For a better visual, here's a setup for a case of what I'm talking about: F' U2 F U' F' U' F U2 F' U F U'

Also, how many cases would there be if the LL edge in FR was oriented?


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## Cale S (Apr 5, 2017)

Aerma said:


> I think this is the right thread for this, maybe?
> Hoe many cases would there be for this Last Slot + Last Layer thing?:
> 
> - LL Corners are already solved (both oriented and permuted)
> ...



12 cases for when the edge in FR is oriented, 12 more for when it is unoriented (just the mirrors)


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## Joel2274 (Apr 5, 2017)

Aerma said:


> I think this is the right thread for this, maybe?
> Hoe many cases would there be for this Last Slot + Last Layer thing?:
> 
> - LL Corners are already solved (both oriented and permuted)
> ...


I believe you are talking about OLS. If so, there are 17,712 cases I believe. (I didn't use math I just saw that stat somewhere)


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## obelisk477 (Apr 5, 2017)

Joel2274 said:


> I believe you are talking about OLS. If so, there are 17,712 cases I believe. (I didn't use math I just saw that stat somewhere)



Full OLS is more like 9000 IIRC. I think I remember @Cale S worked it out once


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## Hazel (Apr 5, 2017)

Again, I don't know if this is the right thread or not, but here's another question:
What would the maximum movecount be for creating a solved "block" of all pieces except the last layer, FD and FR edge, and FDR corner?


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## Joel2274 (Apr 5, 2017)

obelisk477 said:


> Full OLS is more like 9000 IIRC. I think I remember @Cale S worked it out once


Here's where I found the number at. It says somewhere at the right, but maybe it hasn't been updated or something.


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## xyzzy (Apr 5, 2017)

Aerma said:


> Again, I don't know if this is the right thread or not, but here's another question:
> What would the maximum movecount be for creating a solved "block" of all pieces except the last layer, FD and FR edge, and FDR corner?



Determining the exact value would require going through almost 28 billion cases, so I don't think you'll be getting an answer to this question from someone else, ever.

If you really want to know what the worst optimal move count is, there isn't really much of a choice but to write some code and determine how many moves each of those 28 billion cases need. See also: https://www.jaapsch.net/puzzles/compcube.htm


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## Hazel (Apr 7, 2017)

How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
Example setup for one of the cases, in case I'm not making sense:
D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'


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## obelisk477 (Apr 7, 2017)

Aerma said:


> How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
> Example setup for one of the cases, in case I'm not making sense:
> D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'



And everything else is solved?


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## lucarubik (Apr 7, 2017)

i asked this 6 years ago but i didnt get an answer, ive been looking for it and the answer seems to be 10.xx, but i cant trust myself
whats the average number of targets if you are using 2 fixed buffers? (dont mind orientations; passive pieces unoriented count as 0 targets, parity edge and corner count as 1 each and breaking into a new n cycle counts as n+1 targets)
also whats the average of passive pieces to twist? not counting the buffers i guess, or whatever is easier to calculate


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## AlphaSheep (Apr 7, 2017)

Aerma said:


> How many cases would there be if 6 edges on the 3x3 were to be scrambled (4 on U and 2 on E), and all the yellow edges on the U layer were oriented correctly (assuming the U side is yellow)?
> Example setup for one of the cases, in case I'm not making sense:
> D' F2 D2 B' R2 B D' U R' F2 R' U2 R2 U'


Assuming the 2 F2L slots are adjacent, there's are 403 cases, excuding solved, I think.

Someone else please check my math :
Both E edges in E: (4 EPLLs + 5 parity EPLLs) * 2 orientations + 2-flip = 19 cases
First E edge in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(3 edges that can be flipped - 1 last edge orientation determined by others)) = 96 cases
Mirrors of above cases: 96 cases (no symmetries possible)
Both E edges in U: (4! ways to arrange U edges * 1 (E edges determined by parity)) * (2 possible orientations for edges^(4 edges that can be flipped - 1 last edge orientation determined by others)) = 192 cases.


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## Hazel (Apr 7, 2017)

AlphaSheep said:


> Assuming the 2 F2L slots are adjacent, there's are 403 cases, excuding solved, I think.
> 
> Someone else please check my math :
> Both E edges in E: (4 EPLLs + 5 parity EPLLs) * 2 orientations + 2-flip = 19 cases
> ...


Well, I was experimenting with a method idea but this would be too much for that :/


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## Hazel (Apr 11, 2017)

The standard likelihood of getting a PLL skip is 1/72, but what is it if you know full COLL?


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## sqAree (Apr 11, 2017)

Aerma said:


> The standard likelihood of getting a PLL skip is 1/72, but what is it if you know full COLL?



There's a 1/8 chance to have EO done when arriving at LL. However, if you skip OLL you won't be able to use COLL. The chance of an OLL skip is 1/216. The chance for an EPLL skip is 1/12. Hence it's (1/8 - 1/216) * 1/12 + (1 - 1/8 + 1/216) * 1/72 = 173/7776 which is approximately 2.2% so around 1.6 times as likely as without COLL.


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## Mastermind2368 (Apr 14, 2017)

How many cases are their on a SQ1 when you have two edges in the bottom layer and the other 4 in the U layer?


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## Hazel (Apr 22, 2017)

How many L8E cases are there if all the edges are oriented?


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## xyzzy (Apr 22, 2017)

Aerma said:


> How many L8E cases are there if all the edges are oriented?



Which eight edges are you considering? (Also, regardless of which eight edges you choose, the answer is likely to be in the thousands, since 8!/2 = 20160.)


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## Bas Verseveldt (Apr 22, 2017)

lucarubik said:


> i asked this 6 years ago but i didnt get an answer, ive been looking for it and the answer seems to be 10.xx, but i cant trust myself
> whats the average number of targets if you are using 2 fixed buffers? (dont mind orientations; passive pieces unoriented count as 0 targets, parity edge and corner count as 1 each and breaking into a new n cycle counts as n+1 targets)
> also whats the average of passive pieces to twist? not counting the buffers i guess, or whatever is easier to calculate


Note that this is not a very accurate approximation as you do need to memorize misoriented pieces, and you do need to execute algorithms to solve them, hence they affect both memorization and execution time. Parity will only affect execution time (but as the chance of parity is 1/2, we can easily deal with it). 
Anyway, I brute forced this problem by varying the number of pieces P from 3 to 10 (unfortunately my computer couldn't handle 12 due to there being many permutations) and got a very clear linear trend T = 1.0916P - 1.7731 for the average number of targets T. 
For P = 8 (corners) the exact average is 6.968 while the model predicts 6.959 targets, so it is clearly very accurate.
For P = 12 (3x3 edges) this model gives an estimated 11.32 targets with an estimated error smaller than 0.02. So it is definitely not 10.xx. 
For P = 24 (big cube wings) the estimate is 24.4 targets (altough I'm not sure how accurate that estimate is as it is extrapolated so far)
This formula does not apply for big cube centers, because there are identical pieces which this formula does not take into account. For those, I would use the approximation that it is possible to avoid cycle breaks altogether (which is practically always the case) and that we cannot choose our orientation (like on a 5x5). In that approximation, the average number of solved pieces can be computed exactly. Every piece has a chance of 1/6 to be solved, so an average of 23 * 1/6 = 3.83 pieces is solved (remember: not 24, because we never don't consider the buffer to be solved). Neglecting cycle breaks this gives us an average of 23 - 3.83 = 19.17 targets. 
On a 4x4 we can choose the orientation based on centers, and I don't know a nice way of handling the problem then, as we cannot consider the rotations to be independent positions. From experience, I guess that the average number of targets is about 15 if we choose the optimal orientation.


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## lucarubik (Apr 22, 2017)

Bas Verseveldt said:


> Note that this is not a very accurate approximation as you do need to memorize misoriented pieces, and you do need to execute algorithms to solve them, hence they affect both memorization and execution time. Parity will only affect execution time (but as the chance of parity is 1/2, we can easily deal with it).
> Anyway, I brute forced this problem by varying the number of pieces P from 3 to 10 (unfortunately my computer couldn't handle 12 due to there being many permutations) and got a very clear linear trend T = 1.0916P - 1.7731 for the average number of targets T.
> For P = 8 (corners) the exact average is 6.968 while the model predicts 6.959 targets, so it is clearly very accurate.
> For P = 12 (3x3 edges) this model gives an estimated 11.32 targets with an estimated error smaller than 0.02. So it is definitely not 10.xx.
> ...


so for non parity 3x3 scrambles the average is 9,144, thats a bit less than I expected, not too much tho
what about parity scrambles, i would asume you dont just add one (two targets) do you, i wouldnt know where to start to calculate it, ive never been taught probability, i would defenetly f*** up at 5 points becouse i dont have any experience
what about passive pieces to twist, every corner has 2/24 chances to be on its place but twisted, right? so you have 14/24 chances of having at least one, exclouding the buffer, but this doesnt take into consideration that if 7 corners are in the correct orientation the 8th has to be oriented too, god im so bad at this... does anyone know about a tool i can use to "calculate" this, just by simulation?
thanks for the answer, thats 1/3 of the problem solved


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## Bas Verseveldt (Apr 22, 2017)

lucarubik said:


> so for non parity 3x3 scrambles the average is 9,144, thats a bit less than I expected, not too much tho
> what about parity scrambles, i would asume you dont just add one (two targets) do you, i wouldnt know where to start to calculate it, ive never been taught probability, i would defenetly f*** up at 5 points becouse i dont have any experience
> what about passive pieces to twist, every corner has 2/24 chances to be on its place but twisted, right? so you have 14/24 chances of having at least one, exclouding the buffer, but this doesnt take into consideration that if 7 corners are in the correct orientation the 8th has to be oriented too, god im so bad at this... does anyone know about a tool i can use to "calculate" this, just by simulation?
> thanks for the answer, thats 1/3 of the problem solved


On average, there is one piece in its correct place (that is a well known problem in probability theory) _regardless_ of the number of pieces we consider. This can be seen as follows. Suppose we have P pieces. There is a chance of 1/P that piece 1 is in its correct place, so the expected contribution of this piece (averaged over all permutations) is 1/P. This argumentation also holds for all other pieces so the expected number of pieces in solved position is 1/P * P = 1. For corners, this means that the average number of twisted corners is 1 * 7/8 * 2/3 = 7/12 = 0.583 (here 1 is the number of solved pieces, 7/8 because we do not consider the buffer, and 2/3 the chance that is it oriented incorrectly). The fact that the orientation of the 8th corner depends on the orientation of the other 7 corners is irrelevant, as we can choose the buffer corner as 8th and its orientation is irrelevant. For edges, the same argumentation goes, giving an expected number of 1 * 11/12 * 1/2 = 11/24 = 0.458 flipped edges.
If you wish to brute force a question like this, you will need either a programming code (more difficult to learn) or a commando-based mathematical program (easier, i use Mathematica but you will need a license to use it).
The chance of at least one corner being twisted is NOT 14/24. By that argument, the chance of throwing at least one six with a dice in six throws would be 6 * 1/6 = 1, which is false as it is perfectly possible for there not to be any sixes. The _expected value_, however, is equal to 14/24 = 7/12. However, it is possible that two or more corners are twisted. Those cases are counted multiple times in this average. Therefore, the actual probability will be lower than 14/24 = 0.583
Computing this is a well-known problem in mathematics: _a group of people have a hat, they take it off, throw it in a bin, and everybody picks a hat at random. What is the chance that nobody picks his own hat? _
This chance turns out to be 1/e for a sufficiently large (>=5 people) group, where e is the exponential constant. The chance of at least somebody picking up the right one is therefore 1 - 1/e. 
In this case, the corners are the hats and the positions where they belong are the men, and by scrambling the hats are distributed. 
Therefore, the chance of at least one misoriented corner is approximately (1 - 1/e) * 2/3 = 0.421. 
Similarly, for edges, the chance is (1 - 1/e) * 1/2 = 0.316. 
In my calculation of the average number of targets of 11.32 I did not consider parity. The total number of targets averaged over _both parity and non-parity _scrambles is therefore equal to 6.97 + 11.32 = 18.29. I don't know if there is a difference between parity and non-parity scrambles here, but I could easily calculate that using my algorithm if you're curious.


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## lucarubik (Apr 22, 2017)

if a corner has 2/24 chances to be in its place but twisted that doesnt mean if i scramble 24 times its gonna be on its place but twisted two times, so i was right till that point
if every corner has 1/12 chances then given 12 there will be 1 twisted on average so given 7 there will be 7/12, cool, it makes sense doesnt it! im horrible at this, i struggled to come up with this logic proccess, im horrible at intelligence, feels bad.
did your brute force experiment break cycles as we blders do? would it be such a hard ecuation to formulate that you, that seem to know about odds and stuff, rather brute forces it?
it really surprises me the average of cycles, if you optimize parity and count it as a cylce is 9,114, i defently wouldve said more
i got the 10.xx off a post i read a couple of pages ago in this thread, where instead of directly taking the average of targets it counted the odds of each number of taregts, till 10, and the most probable were 4 corners and 6 edges i believe
thanks again by the way its really cool for me to know this stuff, damn i cant beleive i almost forgot to say thanks


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## lucarubik (Apr 22, 2017)

sneze2r said:


> Hello,
> I've digged out an old thread from polish forum where I estimated probablilities of getting an specific bld case(number of targets & number of twists for each edges and corners) by doing simulation in R. Since I had lazy weekend I've decided to do write this code again. Here is the table with results of this simulation:
> 
> 
> ...


this is the post i was refering, it counts the odds of having x targets, being x all possible values not just till 10, and the most probable are 7 corners and 12 edges then i saw 8 corners was way more probable than 6 so i rounded to 10.xx average instead of 9.5 wich i guess was a mistake if both posts are correct (or if both are proportionally incorrect lol)


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## WACWCA (Apr 24, 2017)

Probably out there somewhere but What is the probability there are 2 oriented sides on 2x2 with any combination of opposite colors, like in Guimmond


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## xyzzy (Apr 24, 2017)

WACWCA said:


> Probably out there somewhere but What is the probability there are 2 oriented sides on 2x2 with any combination of opposite colors, like in Guimmond



Fix the DBL corner. This leaves seven corners to be oriented relative to that fixed corner.

P(U and D oriented) = P(L and R oriented) = P(F and B oriented) = \( 3^{-6}=1/729 \)
P(oriented on all axes) = P(U and D oriented) P(corners separated into tetrads) = \( 3^{-6}\times\binom84^{-1}=1/51030 \)
P(oriented on at least one axis) = P(U and D oriented) + P(L and R oriented) + P(F and B oriented) − 2 P(oriented on all axes) = \( 104/25515\approx0.41\% \)


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## Hazel (Apr 26, 2017)

How many L8E cases are there where the equator edges and U layer edges are scrambled but the E layer edges are all in the E layer and all the U layer edges are in the U layer, *and *all the U layer edges are properly oriented?


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## Teoidus (Apr 26, 2017)

Without counting AUFs: 4!/2 in U * 4! in E * 2^4/2 EO = 12 * 24 * 8
With AUFs: (4 + 4) in U * (2 + 5 + 5) in E = 8 * 12? I think?


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## AlphaSheep (Apr 26, 2017)

Aerma said:


> How many L8E cases are there where the equator edges and U layer edges are scrambled but the E layer edges are all in the E layer and all the U layer edges are in the U layer, *and *all the U layer edges are properly oriented?


In the top layer, you have 5 EPLLs (1 is solved) and 5 parity EPLLs.

For each of those you have 12 permutationsof the E layer edges for parity and 12 for no parity, but if they are oriented, you can reduce it to 5 by symmetry (for no parity its solved, two 3 cycles, adjacent double 2 swap, and diagonal double 2 swap) You can y rotate to get specific cases. I think it's also 5 for parity (adjacent swap, diagonal swap, clockwise and anti clockwise cycles, and the weird case that looks like a T).

That gives 5*5 + 5*5 - 1 (solved) = 49 cases _if all E edges are oriented_.

If you allow E edges to not be oriented then I think it increases to (5*5 + 5*5)*(2^3) - 1 = 399 cases. It's too early in the morning to try thinking of symmetry with bad edges.


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## WACWCA (Apr 27, 2017)

I know the gods number for 2x2 is 11, but what is the average optimal move count?


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## xyzzy (Apr 27, 2017)

WACWCA said:


> I know the gods number for 2x2 is 11, but what is the average optimal move count?



8042347/918540 ~ 8.756 moves.

The distribution is available on Jaap's Puzzle Page.


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## LexCubing (Apr 29, 2017)

There 72 permutations for any F2L case. 7 F2L cases: Solved, Corner in, Edge in, 4 cases when they're both at the U layer. This last slot by the way.

So 504 permutations?

Also if I want to know how many LSLL cases would a I do this:

504 x 27 

There are 27 unique CO cases. Correct me if I am wrong.

Or:

504 x 108

108 CO cases since I took into account permutation?


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## shadowslice e (Apr 29, 2017)

If you want to solve the whole cube after F2L-1, you would have roughly (5!*5!*3^4*2^4)/2/4/4=583200 though likely slightly more as many cases would have multiple rotation symmetries (like the h perm is the same from multiple angles) (see the lemma that is not Burnside’s)

This is because you have
5! permutations of edges,
5! permutations of corners
3^4 orientations of corners (last corner orientation is dependant on previous 4)
2^4 orientations of edges (same reasoning as above)
1/2 parity
1/4 preauf
1/4 postauf


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## LexCubing (May 1, 2017)

When calculating let's say ZBLL.You don't 4!^2 x 3^3 because that doesn't give us the unique cases. We times it by 8 since there's just really the Solved, H, Pi, S, As, L, T, and U.

So for LSLL do we

5!^ x 108 or 5! x 27!

There are 27 unique CO for LSLL but when I saw TSLE there's 108 cases?


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## AlphaSheep (May 1, 2017)

LexCubing said:


> When calculating let's say ZBLL.You don't 4!^2 x 3^3 because that doesn't give us the unique cases. We times it by 8 since there's just really the Solved, H, Pi, S, As, L, T, and U.
> 
> So for LSLL do we
> 
> ...


That's not quite right. With ZBLL, you do 
4! (corner permutation)
* 4! (edge permutation)
* 3^3 (corner orientation)
/ 2 (parity)
/ 4 (AUF afterwards doesn't matter)
= 1944 states. 
You can't just times by 8 because solved and H appear in different proportions to the others. This doesn't account for symmetry, and as has been pointed out a few times, the easiest way to deal with symmetry is to list the symmetric cases. Listing these and removing the symmetry cases gives 494 cases (493 excluding solved).

For LSLL, you've also got it wrong. 5! x 27! is around 30 billion times larger than the number of cube states (43 quintillion).
The number of LSLL states (with EO done) is
5! (corner permutation)
* 5! (edge permutation)
* 3^4 (corner orientation)
/ 2 (parity)
/ 4 (AUF afterwards doesn't matter)
= 145800 states.
Counting out the symmetric cases is actually easier than you'd think. The trick is that a case can only have rotational symmetry if *both* the LS corner and edge are correctly placed. If either one of them is in the last layer, then rotational symmetry is impossible.

The same trick can be used in TSLE. So for TSLE, we are concerned with the orientation of 4 corners (the 5th one is determined by the other 4), and the position of 1 edge. So the number of states is
5 (number of positions for the edge)
* 3^4 (corner orientations)
= 405 states.
Note in this case there is no parity (because we're not completely solving permutation) and there is no correction for final AUFs because AUFs afterwards do not change the goal state.

Now we need to count symmetries. Remember what I said about states with a LS edge in LL not having any symmetries? We can use that here. Of the 405 states, there are 81 with the LS edge in place and 324 with it in LL, so those 324 states can't have any symmetry. Therefore, we can do an AUF before these cases to transform these into another case, so just by adding AUFs in front, these 324 states represent (324 / 4) = 81 unique cases.

That leaves the 81 remaining cases with the LS edge placed. Of these, there are three possible twists for the LS corner. If it's solved, we list out all the possible arrangements of the top layer - there are 27 of them. Of these, we note that H has a symmetry in U2 (U2 doesn't change the H case) and the solved state has symmetry in U. So the solved case can't be reduced, and only 2 states can be reduced to H. So, of the 27, 1 is solved, 2 are H, leaving (24 / 4) = 6 other cases, for 8 in total (or 7 excluding the solved case).

If the corner is twisted, then there are similarly 27 states for each twist direction, but one of them has symmetry in U (the one where all for corners in LL are twisted the same direction), and one state has symmetry in U2 (the one where two diagonal corners are twisted in the same direction and the other two are solved). That gives again (1+2/2+24/4) = 8 cases. for each twist direction.

Therefore the total number of unique TSLE cases taking symmetry into account is 81 + 7 + 8 + 8 = 104 cases (or 105 if you include the solved case).

If you want to work out the unique cases for LSLL, then you have the same symmetries to deal with. It's not nearly as difficult as it seems. Why don't you give it a try?


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## LexCubing (May 2, 2017)

Thanks man and also that was supposed to be 27 not 27!  Oops.


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## Malkom (May 2, 2017)

What's the probability 5/10 solves end up in Vperms? Is it as simple as (4/74)^5 x (70/74)^5?


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## xyzzy (May 2, 2017)

Malkom said:


> What's the probability 5/10 solves end up in Vperms? Is it as simple as (4/74)^5 x (70/74)^5?



For a fixed set of 10 solves, assuming you don't do OLLCP in any form (this includes not using COLL), a V perm would show up with probability 1/18. Getting five V perms in these ten solves would then have a probability \( \binom{10}5(1/18)^5(17/18)^5\approx0.01\% \). (The probability doesn't change substantially if you ask about having at least 5 V perms, rather than exactly 5.)

But of course you're not doing only 10 solves, so if you're looking for the probability of 10 consecutive solves in a large session having 5 V perms, this probability tends to 1 (exponentially) as the session gets larger.


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## LexCubing (May 6, 2017)

LL skip is 1/72 and H PLL is 1/108. Why? How do you do this?

How do you know a step is N moves efficient on average? How do you say 1LLLL is 15 moves on avg and Cross is 4.9 moves efficient?

How do you calc that?


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## shadowslice e (May 6, 2017)

1) The skip chance us based on how many cases there are and how many cases you want. The study of it is combinatorics and there are some basic techniques which can get a rough estimate (and LL skip is not 1/72 it's more like 1/4000 off the top of my head).

2) The average movecount is done by finding the movecount for each case and then averaging no real quick way to do it.


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## Rcuber123 (May 6, 2017)

LL skip is 1 in 15552


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## xyzzy (May 6, 2017)

LexCubing said:


> LL skip is 1/72 and H PLL is 1/108. Why? How do you do this?



Step 1: Enumerate the cases in a way such that the cases are equiprobable.
Step 2: Form equivalence classes.
Step 3: Get the size of an equivalence class.
Step 4: There is no step 4.

For PLL, there are 4! 4! / 2 = 288 permutations in total. You consider some of them to be the same (e.g. up to rotation and AUF), so you form equivalence classes. The PLL skip case is an equivalence class with four elements (solved, U, U2, U'), so the probability of getting that is 4/288 = 1/72. Likewise, H perm is an equivalence class with four elements, so the probability of getting that is also 1/72. Z perm is an equivalence class with eight elements, so it has a probability 1/36, and so on.

You _can_ argue about this in terms of symmetries, but it gets a bit confusing to think about why the two N perms each have probability 1/72 unless you have a good grasp of what's going on.



LexCubing said:


> How do you know a step is N moves efficient on average? How do you say 1LLLL is 15 moves on avg and Cross is 4.9 moves efficient?



What shadowslice said, but specifically for anything that doesn't temporarily break up already existent progress (e.g. cross or EOline), we normally use a program to do a breadth first search starting from the solved case and applying moves from there. This avoids the need to explicitly solve every case (or indeed, to explicitly solve any case at all).


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## LexCubing (May 6, 2017)

Rcuber123 said:


> LL skip is 1 in 15552


Oops that was PLL skip when using COLL.



xyzzy said:


> Step 1: Enumerate the cases in a way such that the cases are equiprobable.
> Step 2: Form equivalence classes.
> Step 3: Get the size of an equivalence class.
> Step 4: There is no step 4.
> ...



So how does symmetry affect probabilities and what step 4? Why did you even write it? What did you mean no step 4?


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## xyzzy (May 6, 2017)

hello????

Anyway, cases that are more symmetric occur less frequently, with the ratio being determined by the size of the stabiliser. To understand what "symmetry" even means, of course you'll need some group theory for that, and I'm _not_ going to compress a one-semester undergraduate course into a single forum post, because that's obviously impossible.

(Almost every "intro-level" video/text I've seen about group theory and Rubik's cubes is outright wrong in some subtle way, even if they get the correct result. This is clearly not easy to get right, so you might as well just use a different approach. That's why I gave a combinatorial approach rather than a group theoretic one.)


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## LexCubing (May 6, 2017)

xyzzy said:


> hello????
> 
> Anyway, cases that are more symmetric occur less frequently, with the ratio being determined by the size of the stabiliser.



I meant can you talk more about the confusing N perms. You say something about arguing in terms of symmetries?

Just to check you say Z PLL has 8 classes? So the Solved, U2, U' and another Solved, U, U2, U' because there's 2 Z PLLs? Is that what you meant by 8 since there's 2 Z PLLs, one going left and the other to the right?

What do you do when you combine both orientation and permutation?

Okay let's try this. 288 x 27 = 7776
A H pure flip has solved and U2 so 2/7776 so 1/3338?


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## xyzzy (May 6, 2017)

LexCubing said:


> Just to check you say Z PLL has 8 classes? So the Solved, U2, U' and another Solved, U, U2, U' because there's 2 Z PLLs? Is that what you meant by 8 since there's 2 Z PLLs, one going left and the other to the right?



The eight ways to have a Z perm: 1 2 3 4 5 6 7 8

"Z perm" is the equivalence class (_singular_ "class", not plural "classes") consisting of these eight permutations of last layer pieces.


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## LexCubing (May 6, 2017)

xyzzy said:


> The eight ways to have a Z perm: 1 2 3 4 5 6 7 8
> 
> "Z perm" is the equivalence class (_singular_ "class", not plural "classes") consisting of these eight permutations of last layer pieces.



What about when you start combing orientation and permutation?

Okay let's say a pure twist H ZBLL is

288 x 27 = 7776

2/7776 since we have solved and U2?


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## xyzzy (May 6, 2017)

LexCubing said:


> Okay let's say a pure twist H ZBLL is
> 
> 288 x 27 = 7776
> 
> 2/7776 since we have solved and U2?



The 7776 is correct; the 2 is wrong.

Pick any alg to generate the pure twist H ZBLL and call it X. Now look at these 16 possibilities: X, U X, U2 X, U' X, X U, U X U, U2 X U, U' X U, …, U' X U'. These give rise to exactly 8 distinct permutations/orientations of the last layer (try it out for yourself), so the probability is 8/7776 = 1/972.


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## LexCubing (May 7, 2017)

xyzzy said:


> The 7776 is correct; the 2 is wrong.
> 
> Pick any alg to generate the pure twist H ZBLL and call it X. Now look at these 16 possibilities: X, U X, U2 X, U' X, X U, U X U, U2 X U, U' X U, …, U' X U'. These give rise to exactly 8 distinct permutations/orientations of the last layer (try it out for yourself), so the probability is 8/7776 = 1/972.



Okay let me get this straight. First I calculate every possible valid cases. Look at the 16 possibilties and reduce it to the # of distinct cases.

Last question why that 16 possibilities? Any reason for that? I get the U X, ... X U since they're the AUFs but why do U X U, etc.? Won't U X U = U2 X?


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## xyzzy (May 7, 2017)

LexCubing said:


> Last question why that 16 possibilities? Any reason for that? I get the U X, ... X U since they're the AUFs but why do U X U, etc.? Won't U X U = U2 X?



The order of X and U cannot be switched. Compare: U X and X U.


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## LexCubing (May 7, 2017)

xyzzy said:


> The order of X and U cannot be switched. Compare: U X and X U.



I'm not asking U X vs X U.

I'm asking about U X U vs U2 X.
Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?


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## xyzzy (May 7, 2017)

LexCubing said:


> I'm not asking U X vs X U.
> 
> I'm asking about U X U vs U2 X.
> Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?



No!!!!! They do not commute! Have you _actually_ tried out both U X U and U2 X to see how they differ?

You get the same ZBLL case, but at a different angle. _The angle matters._


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## TDM (May 7, 2017)

LexCubing said:


> I'm not asking U X vs X U.
> 
> I'm asking about U X U vs U2 X.


isn't that literally the same thing but with an extra U move added to the front of both...?


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## AlphaSheep (May 7, 2017)

LexCubing said:


> I'm not asking U X vs X U.
> 
> I'm asking about U X U vs U2 X.
> Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?


An AUF before depends on the angle of the case, for example, a U perm with the bar on the left vs the bar in front. An AUF afterwards depends on the colours, so for example an orange bar vs blue bar.

For a Ua perm, there are 4 possible colours for the bar, and 4 possible locations for the bar, which give the 16 possible Ua cases.


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## LexCubing (May 7, 2017)

xyzzy said:


> No!!!!! They do not commute! Have you _actually_ tried out both U X U and U2 X to see how they differ?
> 
> You get the same ZBLL case, but at a different angle. _The angle matters._


Okay thanks for the help.


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## obelisk477 (May 14, 2017)

What are the odds, on a cube with only 6 scrambled corners (they could be twisted as well), that the corners will be a set of two 3 cycles? 

Asking for FMC


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## xyzzy (May 14, 2017)

obelisk477 said:


> What are the odds, on a cube with only 6 scrambled corners (they could be twisted as well), that the corners will be a set of two 3 cycles?


Consider the possible cycles on the six unsolved corners.

(5, 1):
Possible permutations with this cycle structure: \( \binom6{5,1}\cdot(5-1)!=144 \)
Orientations: \( 2\cdot3^4=162 \)

(4, 2):
Permutations: \( \binom6{4,2}\cdot(4-1)!(2-1)!=90 \)
Orientations: \( 3^5=243 \)

(3, 3):
Permutations: \( \binom6{3,3}\frac1{2!}\cdot(3-1)!^2=40 \)
Orientations: \( 3^5=243 \)

(3, 1, 1, 1):
Permutations: \( \binom6{3,1,1,1}\frac1{3!}\cdot(3-1)!=40 \)
Orientations: \( 2^3\cdot3^2=72 \)

(2, 2, 1, 1):
Permutations: \( \binom6{2,2,1,1}\frac1{2!2!}\cdot(2-1)!^2=45 \)
Orientations: \( 2^2\cdot3^3=108 \)

(1, 1, 1, 1, 1, 1):
Permutations: \( 1 \)
Orientations: 1 for all twisted cw, 1 for all twisted ccw, \( \binom63=20 \) for three cw and three ccw; total 22.

Conditioned on the last six corners all being unsolved, the probability of getting a "good" (3, 3) case (i.e. both cycles aren't twisted) is \( 40\cdot3^4/(\text{sum of the stuff above})=3240/62680\approx5.17\% \).


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## FastCubeMaster (May 15, 2017)

Does anyone know where I can find or know themselves the chances for each of the UF VLS cases occurring?
Just finished learning and curious to know


----------



## TDM (May 15, 2017)

FastCubeMaster said:


> Does anyone know where I can find or know themselves the chances for each of the UF VLS cases occurring?
> Just finished learning and curious to know


Since you can't reduce any cases to another cases by AUF, they are all equally likely.

So 1/216 given the pair is solved, or 1/27 given you have a UF case.


----------



## FastCubeMaster (May 15, 2017)

TDM said:


> Since you can't reduce any cases to another cases by AUF, they are all equally likely.
> 
> So 1/216 given the pair is solved, or 1/27 given you have a UF case.



O, definitely makes sense but seems a bit weird for me cos a few cases keep occuring, more than others. I guess it's just chance


----------



## Daniel Lin (May 21, 2017)

given a 2 random edges and 2 random corners being swapped on the cube

what's the probability of a least one corner and edge being adjacent?

examples of an adj case
R' F R U2 r2 F r U' r U2 (UFR and UR are next to each other)

non adj

u R' U' R' F R2 U' R' U' R U R' F' R E


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## Cale S (May 21, 2017)

Daniel Lin said:


> given a 2 random edges and 2 random corners being swapped on the cube
> 
> what's the probability of a least one corner and edge being adjacent?
> 
> ...



The corner pieces can be adjacent (3/7 chance), diagonal on a layer (3/7 chance), or on opposite corners of the cube (1/7 chance)

Finding probability they aren't adjacent: 

If the corners are adjacent, first edge has a 7/12 chance, and second has 6/11 chance

If the corners are diagonal, first edge has 6/12 = 1/2 chance, second has 5/11

If the corners are completely opposite, the numbers are the same as diagonal

1 - ((3/7)(7/12)(6/11) + (3/7)(1/2)(5/11) + (1/7)(1/2)(5/11))

*73.38%* chance of having adjacent corner and edge


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## Daniel Lin (May 29, 2017)

ok 1 more question

how many total 2e2c cases are there if you count U/D, F/B, and L/R mirrors as the same?

so that
R U R' F' R U R' U' R' F R2 U' R' U'=
R' D' R F R' D' R D R F' R2 D R D=
L D L' F' L D L' D' L' F L2 D' L' D'

can someone do the burnsides lemma thingy?


----------



## xyzzy (May 29, 2017)

Daniel Lin said:


> how many total 2e2c cases are there if you count U/D, F/B, and L/R mirrors as the same?
> 
> can someone do the burnsides lemma thingy?



Identity symmetry: \( (\binom82\cdot3)\cdot(\binom{12}2\cdot2)=11088 \)
Mirror on one axis: \( (4\cdot3)\cdot(12+8)=240 \) (12 for edges in the slice preserved under mirroring, 8 for edges outside of that slice)
Mirror on two axes: \( (4)\cdot(12)=48 \)
Mirror on all axes: \( (4\cdot3)\cdot(12/2\cdot2)=144 \)
Burnside's lemma magic: \( \frac18(11088+3\cdot240+3\cdot48+144)=1512 \)


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## Herbert Kociemba (May 30, 2017)

Daniel Lin said:


> R U R' F' R U R' U' R' F R2 U' R' U'=
> R' D' R F R' D' R D R F' R2 D R D=
> L D L' F' L D L' D' L' F L2 D' L' D'


Is there a particular reason that e.g. F U F' L' F U F' U' F' L F2 U' F' U' should be counted as different? This belongs to a 90° rotation which cannot be generated by the three mirrors.


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## Abram Lookadoo (Jun 25, 2017)

can someone find the probability of...

any 1x1x3 block formed after scramble
being 1 move away from a 1x1x3 block after scramble (in axis turn metric)
having at least one F2L pair anywhere on the cube after making a 2x2x3 block
having both F2L pairs anywhere on the cube after making a 2x2x3 block
2gr skip with phasing (and average movecount if able(htm))
2gr skip with 2 corners oriented (and average movecount if able(htm))


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## WACWCA (Jun 28, 2017)

Probability of solved 2x2 face, any kind


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## Liquorice (Jul 9, 2017)

What is the probability of parity in 3x3 blindfolded?

What is the exact probability of getting an edges memo of length 11 (11 edges need to be solved, buffer does not count)? If buffer piece is solved, it adds 1 to the length. There may be a permuted flipped edge and a solved edge (11 +1 -1 = 11).
Length 12 (one edge is flipped and permuted or two edges are flipped/permuted and one is solved etc.)? Length 1, 2, 3 ... 20, 21, 22?
Corners memo length?


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## Ronxu (Jul 9, 2017)

Liquorice said:


> What is the probability of parity in 3x3 blindfolded?


0.5


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## Liquorice (Jul 9, 2017)

Ronxu said:


> 0.5


Why? How do you know?


----------



## Ronxu (Jul 9, 2017)

Liquorice said:


> Why? How do you know?


If the scramble has an odd number of quarter turns you have parity. This happens because a quarter turn does an odd number of both edge and corner 2-cycles. if you have an even number of quarter turns, then the total number of 2-cycles is even on both edges and corners and there's no parity. Both cases are equally likely.


----------



## Underwatercuber (Jul 17, 2017)

Probabilities of having a 3/4/5bld solve without parity or any cycle breaks.


----------



## xyzzy (Jul 17, 2017)

Underwatercuber said:


> Probabilities of having a 3/4/5bld solve without parity or any cycle breaks.



For 3bld, it's around 3.97%:



xyzzy said:


> One could "break cycles" in order to flip edges or twist corners, but (I think?) most people just use algs for that, so assuming that, we can ignore the orientation of the pieces entirely. In other words, we're asking for the probability that we have only one corner cycle and only one edge cycle. (Please correct me if this is wrong and not what you're asking for, because I have no idea how people do BLD.)
> 
> Consider the two cases based on permutation parity. Suppose the corner/edge permutation parities are even. (They must be either both even or both odd.)
> 
> ...



4bld is too hard to exactly calculate because it depends on how you orient the cube. Too lazy to calculate for 5bld now, but I think you can reuse the above calculation and replace 8 or 12 with 24 as appropriate for the wings, and for the centres you don't really need cycle breaks at all, I think.


----------



## greentgoatgal (Jul 23, 2017)

DT546 said:


> not cubing related, but interesting
> 
> if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday



You are also more likely to get struck by lightning on your way to get the lottery ticket than you are to win the lottery.


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## obelisk477 (Jul 23, 2017)

greentgoatgal said:


> You are also more likely to get struck by lightning on your way to get the lottery ticket than you are to win the lottery.



If you drove to the gas station each time to buy a lottery ticket for the expected number of tickets you would need to win the lottery, you would be killed in a car wreck 6 or 7 times before actually getting a winning ticket.


----------



## applezfall (Aug 2, 2017)

What is the probability of an 4 move 2x2 solve?
And what's the rarest and least rare pll?


----------



## KAINOS (Aug 3, 2017)

applezfall said:


> What is the probability of an 4 move 2x2 solve?
> And what's the rarest and least rare pll?


For the first question check out this website: https://www.jaapsch.net/puzzles/cube2.htm
The rarest PLLs are H and N perms, (probability=1/72) and other than E and Z (1/36) the rest of them have the same chance of happening(1/18). If you group variants as a single case the most common one would be G perm, though. (1/18*4=2/9)


----------



## RubixKid (Aug 13, 2017)

The chance of a last 4 centers skip on Skewb is 1/8

I got 4 in a row once which is a 1/4096 probability!


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## Cale S (Aug 13, 2017)

RubixKid said:


> The chance of a last 4 centers skip on Skewb is 1/8
> 
> I got 4 in a row once which is a 1/4096 probability!



it's 1/12


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## Daniel Lin (Aug 21, 2017)

probability of having a face of opposite colors solved already on a 2x2?

or if someone has a frequency table for the number of states for each movecount that would be nice


----------



## FastCubeMaster (Aug 21, 2017)

Can someone find the highest ratios of cubers (competed officially) in a country to the country's population.


----------



## Dr_Detonation (Aug 21, 2017)

There are 15,403 Americans who have competed officially. The U.S. population was estimated at 323 million in 2016. So, it's a little less than 1:20,970 for the USA


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## Oliver1010 (Aug 27, 2017)

I just got a cross skip this is the scramble
D' L2 F' U2 L2 B2 F2 U2 R2 B R2 L F D F L B U

I also have gotten two OLL skips in a row before


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## Rubix Cubix (Aug 27, 2017)

FastCubeMaster said:


> Can someone find the highest ratios of cubers (competed officially) in a country to the country's population.



Haven't got anything to back it up with evidence, but somebody told me the other day it was Poland.


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## Oliver1010 (Aug 29, 2017)

Does anyone know the chances of getting a first layer skip?
I got two first layer skips on a 2x2
U F' U R F' R F U2 R' U'
R F2 R2 U2 F' U R F' U F'


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## AlphaSheep (Aug 29, 2017)

Oliver1010 said:


> Does anyone know the chances of getting a first layer skip?
> I got a first layer skip on a 2x2
> U F' U R F' R F U2 R' U'


You can always solve at least one corner by rotating. Then there are 7 places for the 2nd corner, 6 for the 3rd, and 5 for the 4th. Each of those 3 corners has a 1/3 chance of being oriented correctly, so the chance of a layer skip is 1 in 7*6*5*3*3*3, which is 1 in 5670.

Edit: this is the chance of a first layer skip of a specific colour, eg white.


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## Oliver1010 (Aug 30, 2017)

AlphaSheep said:


> You can always solve at least one corner by rotating. Then there are 7 places for the 2nd corner, 6 for the 3rd, and 5 for the 4th. Each of those 3 corners has a 1/3 chance of being oriented correctly, so the chance of a layer skip is 1 in 7*6*5*3*3*3, which is 1 in 5670.
> 
> Edit: this is the chance of a first layer skip of a specific colour, eg white.



What are the chances of a first layer skip on other size cubes?


----------



## teboecubes (Sep 16, 2017)

One I got a 2x2 first layer skip, did OLL, then PLL skip


----------



## Oliver1010 (Sep 16, 2017)

What is the probability of a T Perm during blindfolded?


----------



## obelisk477 (Sep 16, 2017)

Cale S said:


> That has a 1 in 6 chance, not unlikely at all



? 1/6 is just for PLL skip, not first layer skip as well. Not sure what the odds for that are though. Guess it depends on color neutrality


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## Underwatercuber (Sep 16, 2017)

What's the probability of having a clock cross correctly solved and oriented? Solved but unoriented? 1 move from being unsolved?


----------



## Cale S (Sep 16, 2017)

obelisk477 said:


> ? 1/6 is just for PLL skip, not first layer skip as well. Not sure what the odds for that are though. Guess it depends on color neutrality



oops read the post wrong lol


----------



## guysensei1 (Sep 16, 2017)

Asked before but lost the post so i'll ask again, whats the probability that a 3x3 LL case can be AUFed into a 3C3E?


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## TipsterTrickster (Sep 19, 2017)

I'm guessing the probibility of getting a 3x3 stage skip on big cube is 1/43 quintillion (correct me if I'm wrong) but what about a f2l skip?


----------



## obelisk477 (Sep 19, 2017)

TipsterTrickster said:


> I'm guessing the probibility of getting a 3x3 stage skip on big cube is 1/43 quintillion (correct me if I'm wrong) but what about a f2l skip?



It's actually 4 times that because of parities


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## xyzzy (Sep 19, 2017)

obelisk477 said:


> It's actually 4 times that because of parities


A quarter, not four times, but only on even-order cubes.


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## obelisk477 (Sep 19, 2017)

xyzzy said:


> A quarter, not four times, but only on even-order cubes.



Right. I mean 4 times worse. And I guess it would be half on odd numbered cubes


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## Cale S (Sep 19, 2017)

obelisk477 said:


> Right. I mean 4 times worse. And I guess it would be half on odd numbered cubes



Why half on odd numbered cubes?


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## obelisk477 (Sep 19, 2017)

Cale S said:


> Why half on odd numbered cubes?



No PLL parity?

EDIT: Oh, I see. I was equating edge pairing parity on odd cubes with OLL parity on even cubes. Which is valid if you say your 3x3 stage is 'done' before the edge pairing parity stage.


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## TipsterTrickster (Sep 19, 2017)

obelisk477 said:


> It's actually 4 times that because of parities


 oh yea good point


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## CuberFles (Sep 19, 2017)

What are the odds for hitting edge parity on a 4 x 4? I'd say 40 %, since there are 5 possible situations after F3L (0, 1, 2, 3 or 4 dedges flipped) and only 2 of these are "wrong" (1 and 3). However, after several dozens of solves I can't seem to get any less than a 60 % chance to hit edge parity. I'm using the 3 x 3 reduction method (centers first, then pair the edges, then solve like a 3 x 3). Is my method invoking unfair amounts of parity cases? Is there anything I can do to hit parity less frequently?


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## obelisk477 (Sep 19, 2017)

CuberFles said:


> What are the odds for hitting edge parity on a 4 x 4? I'd say 40 %, since there are 5 possible situations after F3L (0, 1, 2, 3 or 4 dedges flipped) and only 2 of these are "wrong" (1 and 3). However, after several dozens of solves I can't seem to get any less than a 60 % chance to hit edge parity. I'm using the 3 x 3 reduction method (centers first, then pair the edges, then solve like a 3 x 3). Is my method invoking unfair amounts of parity cases? Is there anything I can do to hit parity less frequently?



Call the UB edge 1, the UR edge 2, the UF edge 3, and the UL edge 4, for the sake of this discussion. Now let us consider the ways that you can have parity.

*0 Edges*
1,2,3,4 are flipped - NO PARITY

*1 Edge*
1 is flipped - PARITY
2 is flipped - PARITY
3 is flipped - PARITY
4 is flipped - PARITY

*2 Edges*

1 and 2 are flipped - NO PARITY
2 and 3 are flipped - NO PARITY
3 and 4 are flipped - NO PARITY
4 and 1 are flipped - NO PARITY
1 and 3 are flipped - NO PARITY
2 and 4 are flipped - NO PARITY

*3 Edges*

1 is not flipped - PARITY
2 is not flipped - PARITY
3 is not flipped - PARITY
4 is not flipped - PARITY

*4 ֵEdges*

All four are not flipped - NO PARITY

If you count up each of the 'Parity' and 'No parity', you'll see that the split is 50%. The fact that you're seeing 60% is just unfortunate, and over a long number of solves, you'd see it settle to 50%.


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## CuberFles (Sep 20, 2017)

Aha, thanks for pointing that out


----------



## Oliver1010 (Sep 23, 2017)

TipsterTrickster said:


> I'm guessing the probibility of getting a 3x3 stage skip on big cube is 1/43 quintillion (correct me if I'm wrong) but what about a f2l skip?



https: //www.youtube.com/watch?v=v6s9l7kgwRM
About 1 in 3.66 billion


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## Oliver1010 (Sep 23, 2017)

What are the odds of a last layer skip, oll skip, or pll skip on a 4x4 or even layered cube


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## xyzzy (Sep 23, 2017)

Oliver1010 said:


> What are the odds of a last layer skip, oll skip, or pll skip on a 4x4 or even layered cube



OLL skip: 1 / (2^4 3^3) = 1/432
Full PLL skip: 1 / (4! 4!) = 1/576
PLL skip (allowing for AUF): 4 / (4! 4!) = 1/144
Full LL skip: (1/432) (1/576) = 1/248832
LL skip (allowing for AUF): (1/432) (1/144) = 1/62208


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## 94matt (Oct 5, 2017)

HI guys just want to know that if you have all the corners and one side completed, how many possibilities do you have for the last 8 egdes?


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## TDM (Oct 5, 2017)

94matt said:


> HI guys just want to know that if you have all the corners and one side completed, how many possibilities do you have for the last 8 egdes?


There are 8!/2 possible permutations and 2^7 possible orientations of the last eight edges. Multiplying those gives a total of 2,580,480 cases.


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## guysensei1 (Oct 6, 2017)

TDM said:


> There are 8!/2 possible permutations and 2^7 possible orientations of the last eight edges. Multiplying those gives a total of 2,580,480 cases.


This only works if you are considering rotated versions of the same case as distinct. For example there are 8 'ways' for a clockwise U perm to appear that fits OP's description.


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## Duncan Bannon (Oct 9, 2017)

Most of the people here probably know this, but the chance of a PLL skip is the same as getting a H perm. 1/72


Also, how many algs would there be for FULL 1LLL


----------



## Rubix Cubix (Oct 9, 2017)

Duncan Bannon said:


> Most of the people here probably know this, but the chance of a PLL skip is the same as getting a H perm. 1/72
> 
> 
> Also, how many algs would there be for FULL 1LLL



If you want to do 1LLL, then you could learn ZBLL, only 493 cases I think


----------



## ErwinOlie (Oct 9, 2017)

Duncan Bannon said:


> Most of the people here probably know this, but the chance of a PLL skip is the same as getting a H perm. 1/72
> 
> 
> Also, how many algs would there be for FULL 1LLL



The wiki gives the number *3915* with full explanation where that number comes from.
https://www.speedsolving.com/wiki/index.php/1LLL


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## Duncan Bannon (Oct 9, 2017)

Okay, Thanks!


----------



## Duncan Bannon (Oct 19, 2017)

What are the chances of a no AUF PLL skip for 2x2? Thanks


----------



## TDM (Oct 19, 2017)

Duncan Bannon said:


> What are the chances of a no AUF PLL skip for 2x2? Thanks


Probability of CP skip: 1/6
Probability of AUF skip: 1/4
Total probability of AUFless PLL skip: 1/24.


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## Duncan Bannon (Oct 19, 2017)

TDM said:


> Probability of CP skip: 1/6
> Probability of AUF skip: 1/4
> Total probability of AUFless PLL skip: 1/24.



Sweet, You the man, thanks.

Edit- that was really easy math Duncan, LOL
Edit2- Is the chance of PLL skip really 1/6? Because of the H cases wouldn’t it be 6/30? 
Edit3- What are chances of full No AUF LL skip on 2x2? Thanks


----------



## Username (Oct 19, 2017)

Duncan Bannon said:


> Edit2- Is the chance of PLL skip really 1/6? Because of the H cases wouldn’t it be 6/30?



Yes, there are 4! ways to permute the four corners. Four of them are "solved" (disregarding AUF). 4!/4 = 3! = 6, hence 1/6

Similarly, there are 4! different ways to arrange four corners when they are oriented. Only one of them is completely solved, thus PLL skip with no AUF is 1/24


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## Duncan Bannon (Oct 19, 2017)

Oh, very cool. Thanks.


----------



## CornerCutter (Oct 19, 2017)

What is the probability of getting a 6 move Pyraminx scramble?


----------



## xyzzy (Oct 20, 2017)

CornerCutter said:


> What is the probability of getting a 6 move Pyraminx scramble?


Jaap's Puzzle Page has complete move count distributions for the smaller puzzles, including pyraminx.


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## Underwatercuber (Oct 21, 2017)

Oliver1010 said:


> I have gotten
> 2 oll skips in a row
> 2 pll skips in a row
> 2 same ll cases in an ao100
> ...


1. What do you mean by first layer skip
2. what to you mean by a ZBLL skip
3. Skips are unrelated to how long you have been cubing


----------



## Oliver1010 (Oct 21, 2017)

Duncan Bannon said:


> Edit3- What are chances of full No AUF LL skip on 2x2? Thanks



There are 4!=24 ways to place the corners in the top layer, ignoring orientation. 
Each of the corners has 3 different ways it can be twisted. The 4th corner's orientation depends on the orientation of the other corners. We take 1/(24*(3^3))= 1/648 chance of a LL skip no AUF


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## obelisk477 (Oct 22, 2017)

With TNoodle scrambles, what is the probability of any given tip on a pyraminx being scrambled? Is it 1/3 or 1/2?


----------



## whatshisbucket (Oct 23, 2017)

obelisk477 said:


> With TNoodle scrambles, what is the probability of any given tip on a pyraminx being scrambled? Is it 1/3 or 1/2?


I had tnoodle generate 50 pyraminx scrambles and there were 129/200 tips scrampled, much closer to 2/3 than 1/2, so there is probably an equal chance of any given tip being in each of the three positions.


----------



## Hazel (Oct 23, 2017)

is there a movecount distribution for 3x3 last layer?


----------



## xyzzy (Oct 23, 2017)

Aerma said:


> is there a movecount distribution for 3x3 last layer?


http://cubezzz.dyndns.org/drupal/?q=node/view/137


----------



## CLL Smooth (Nov 11, 2017)

How many cases are there if you have F2L-1 corner+ a tripod (1x2x2) in the LL? Does it matter the orientation of the tripod?


----------



## teboecubes (Nov 11, 2017)

Underwatercuber said:


> 3. Skips are unrelated to how long you have been cubing



Sort of, because a cuber who got has only done 500 solves in month has a lesser chance than a cuber who has done 100,000 solves over 5 years.


----------



## Sue Doenim (Nov 11, 2017)

CLL Smooth said:


> How many cases are there if you have F2L-1 corner+ a tripod (1x2x2) in the LL? Does it matter the orientation of the tripod?


4!×27×2=1296 cases. Only 648 with EO. 162 if you do EO and force the F2L corner into place, not necessarily oriented.

Could anyone get me:
1- 2x2 face average optimal movecount
and
2- 2x2 orientation optimal average movecount?

EDIT: Never mind about the second one.


----------



## dnguyen2204 (Dec 4, 2017)

What are the chances of getting a half oll skip (corners or edges) or half pll skip (corners or edges) using CFOP 4LLL?


----------



## xyzzy (Dec 4, 2017)

dnguyen2204 said:


> What are the chances of getting a half oll skip (corners or edges) or half pll skip (corners or edges) using CFOP 4LLL?



11/72 and 2/9 respectively, if full skips are excluded.


----------



## dnguyen2204 (Dec 4, 2017)

xyzzy said:


> 11/72 and 2/9 respectively, if full skips are excluded.


Nice, not too rare.

That's the combined probability? Can you separate edges and corners?


----------



## Rpotts (Dec 4, 2017)

dnguyen2204 said:


> What are the chances of getting a half oll skip (corners or edges) or half pll skip (corners or edges) using CFOP 4LLL?



1/8 to skip Edge Orientation 
1/27 to skip Corner Orientation
1/6 to skip Corner Permutation 
1/12 to skip Edge Permutation 

So about 16% to skip half or both steps of OLL, and about 25% to skip half or both steps of PLL.


----------



## Hazel (Dec 5, 2017)

What are the chances of getting a solved 2x2x1 block with white as one of the 1x2 parts 2 scrambles in a row? I got that


----------



## teboecubes (Dec 6, 2017)

Aerma said:


> What are the chances of getting a solved 2x2x1 block with white as one of the 1x2 parts 2 scrambles in a row? I got that


Dont know exactly, but I think it would be ((P Edge Solved)^2 x (P Corner Solved)) / 12. Im not sure though


----------



## FastCubeMaster (Dec 21, 2017)

May have been asked before, but what is the probability of getting a 4 mover in 2x2?
I assume this is the same as getting one officially. (- the 3 moves or less cases, of course)


----------



## AlphaSheep (Dec 21, 2017)

FastCubeMaster said:


> May have been asked before, but what is the probability of getting a 4 mover in 2x2?
> I assume this is the same as getting one officially.


The probability is 1 in 1989.

There are 3 674 160 states, 1 of which is solved, and 384 of which can be solved in 3 moves or less (these are excluded in official scrambles). 1847 states can be solved in 4 moves.
(3674160 - 385) / 1847 = 1989.05


----------



## Mellis Ferton (Dec 28, 2017)

Chance to get an xx-cross with an LL skip on 3x3?


----------



## Duncan Bannon (Dec 28, 2017)

@AlphaSheep This is INSANE math. I’m pretty good in math, how can I do this? And thats about right as I’ve got a 4 mover and 5 mover in about 2k solves.


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## whatshisbucket (Dec 28, 2017)

Mellis Ferton said:


> Chance to get an xx-cross with an LL skip on 3x3?


Depends on how you define the probability of getting an xx-cross. The chance of getting an LL skip with no influence is 1 in 15552. If you define getting an xx-cross as having 2 F2L pairs solved after you complete the cross, assuming that you didn't try to influence the pairs at all, the probability is a little less than 1 in 18816, but surely there is some influence here that significantly increases the probability. This would make the desired probability about 1 in 300 million.



Duncan Bannon said:


> @AlphaSheep This is INSANE math. I’m pretty good in math, how can I do this? And thats about right as I’ve got a 4 mover and 5 mover in about 2k solves.



It's not complex math; it's right there in the post.


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## Duncan Bannon (Dec 28, 2017)

I know, but how does he know the that X amount can be solve in 4 moves and such?


----------



## whatshisbucket (Dec 28, 2017)

Duncan Bannon said:


> I know, but how does he know the that X amount can be solve in 4 moves and such?


https://en.wikipedia.org/wiki/Pocket_Cube


----------



## Duncan Bannon (Dec 28, 2017)

Thanks You! That’s useful.


----------



## Franklin (Jan 22, 2018)

When I was in grade 3, I solved a layer of a rubik's cube... AND IT WAS COMPLETELY SOLVED. OK OK I kind of used a method, the steps were: 1.Solve a layer 2. Solve a different layer 3. Repeat step 2 until entire cube is solved. Kind of how you would solve a pyraminx as a beginner. But that is still incredible, probably one of the luckiest solves ever. What's the chance of something like that happening? How much layers would it take to usually solve that?


----------



## DGCubes (Jan 22, 2018)

Sounds almost like CFinity (but even less likely to work).

The chances of all 8 other edges and all 4 other corners being solved is:
1 / [(8! * 2^8 * 4! * 3^4) / (2 * 3 * 2)] = 1/1,672,151,040 = 0.000000000598 = 0.0000000598%

However, I think we can assume that you would be able to solve the cube if it were only a turn away (such as if the second or third layers were off by a single move). There are 16 cases like this, reducing your chances of reaching one of these to:
16/1,672,151,040 = 1/104,509,440 = 0.00000000957 = 0.000000957%
Basically, one in a hundred million.

If you wanted a 50% chance of having a solved cube (computed as a 50% chance of not having an unsolved cube, and assuming each layer is independent of each previous layer), you would need to do:
0.9999999904315 ^ n = 0.5
n ≈ 72,440,526 layers

(Although, since you had done some layers already, this might have somewhat affected the outcome of each trial, meaning they are not entirely independent. The average would likely be slightly less, but I'd guess it'd still hover around 70 million. Someone can correct me on that; it's probably a bit off. If anyone sees anything wrong with any of my math, feel free to point it out.)

In other words, you got VERY LUCKY!


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## Hazel (Feb 7, 2018)

What is the probablity of having two 2x2 scrambles in a row that both have a completed adjacent-swap side?


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## teboecubes (Feb 7, 2018)

Aerma said:


> What is the probablity of having two 2x2 scrambles in a row that both have a completed adjacent-swap side?



Let's see...


So,
I have to calculate the probabilty of having two pieces next to each other, so around 1/7, then the same orientation to make a bar, so 1/3 for each piece, or 1/9, times the 1/7 is already 1/63 for a bar, 

to have the next piece in for the adjacent swap would make it 1/1134,

then the last one would make it 1/17010

there are 4 different adjacent swaps for 1 face, which increases the chances to 4/17010, or 2/8505

having two of these in a row would make it 1/72335025, or 1/17010 ^ 2

so the answer is 1/72335025, or 0.000000138245615, or 0.0000138245615%


Remember, I might not be completely accurate, but this was my thought process, and correct me if I made any mistake.


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## whatshisbucket (Feb 7, 2018)

hm
6 colors to make the face
4 possible colors for the bar
24 possible permutations of the other corners
27 possible orientations of the other corners
that's 15,552 scrambles (funny, 1/15552 is the probability of LL skip on 3x3)

but that overcounts, since there are scrambles that have two completed opposite faces:
3 pairs of colors to make the faces
4 possible colors for each of the two bars
4 cuz AUF
that's 192

thus 15,360 scrambles possible.
total of 3,674,160 2x2 scrambles
(15,360/3,674,160)^2=(64/15309)^2=4096/234,365,481= about 1/57,218.


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## Hazel (Feb 8, 2018)

Oh wow, that's a lot lower chance than I expected! I guess I got pretty lucky then haha


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## Kumato (Feb 19, 2018)

Hello everyone! 

I was wondering if anybody knew if there are PLLs or OLLs that were more common than others. For instance, I've been noticing quite more J perms than N perms, or more U perms than Z or H.

The OLL range is bigger, so if someone has noticed a variation on chance between different OLLs, I would like you to write it as a reply.

To make it clear, what I mean by this, is that a certain case as a bigger percentage chance of appearing.

Also I made a poll in whether you think change may vary between different cases.

Thank you and have a good day! 

(PS: I'm sorry if someone made this thread before, but I could not find it.)


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## DGCubes (Feb 19, 2018)

Yes, the probabilities of each one are known. It has to do with symmetry; there's a pretty good explanation in this Reddit thread. Your findings make sense (especially if you consider a U-perm either of the two U-perms; same for a J-perm). You're 8 times more likely to get either U-perm than an H-perm, and 4 times more likely to get either U-perm than a Z-perm.

The Reddit thread links this page for OLL probabilities and this one for PLL probabilities.


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## Kumato (Feb 19, 2018)

OK, first I can't believe DG answered this...
And secondly, Thanks!!!


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## Cale S (Feb 19, 2018)

PLLs with less rotational symmetry have higher probabilities mathematically, but in real solves people influence steps to get more solves with oriented edges or permuted corners


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## cubeshepherd (Mar 4, 2018)

I am not positive if this has been asked yet or not, but what are the odds of your Ao5 and Ao12 being the exact same time. I ask because for the third time this week I have gotten the same average in Ao5 and Ao12 in three different events. One is 3x3 at 12.42, one is Pyraminx at 4.54 and the other was Skewb at 5.40. Thank you in advance for your help and answers.


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## AlphaSheep (Mar 5, 2018)

cubeshepherd said:


> I am positive if this has bee asked yet or not, but what are the odds of your Ao5 and Ao12 being the exact same time. I ask because for the third time this week I have gotten the same average in Ao5 and Ao12 in three different events. One is 3x3 at 12.42, one is Pyraminx at 4.54 and the other was Skewb at 5.40. Thank you in advance for your help and answers.


Depends. This is impossible to calculate since its impossible to know what the underlying distribution of your solves are. If you are very consistent then the chances of this happening are really high. If you are so consistent that you usually get the same times every solve, then the chances of this happening are almost 100%. If you're very inconsistent, then the odds are still higher than you'd expect, because your times are distributed around your global average.


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## hotufos (Mar 6, 2018)

What is the probability of a cross skip?
TL;DR: I think 1/31,680. I'm not positive.

Many of us know that the chance you'll get an OLL skip in any given solve is 1/216; for PLL skip, 1/72; and for a full LL skip, 1/15,552. But what is the chance you will get a cross skip in a solve? I've never heard of anyone ever getting one. I tried to figure out the chance. Here's what I came up with.

The cube has 43,252,003,274,489,856,000 positions (12! x 8! x 2^11 x 3^7 / 2). How many of these positions have a cross on them? First of all, in a cross, all of the corners are completely irrelevant. We can start our calculation by using the number of positions of the corners, which is 88,179,840 positions (8! x 3^7). For there to be a cross, four edges matter and eight don't. For the four edges that matter, the cross edges, there are six options, because the cross could be on any of the six faces. The eight edges that don't matter could be permuted 8!/2 ways. This is because there are eight places to put the first irrelevant edge (because four spots are already occupied by the cross edges), seven places to put the second (because now there is a spot occupied by an irrelevant edge), six places for the third, and so on. The /2 at the end is because we have already chosen a position for the corners, and it's impossible to switch just two pieces on a cube. There are 2^7 ways that the eight irrelevant edges could be oriented, since the orientation of the eighth edge depends of the orientation of the other seven.

This means that there are, in total, 8! x 3^7 x 6 x 8! x 2^7 / 2 = 1,365,277,881,139,200 cube states that have a cross on them. Dividing this number by the total number of 3x3 positions gets us the chance that any random scramble will have a cross on it, thus, a cross skip. Ultimately this comes to 1/31,680, which is very close to half as likely as a last layer skip.

All of this is my own work (except figuring out 43 quintillion). I'm not positive that this is correct. I think my formula may not properly address cube states with multiple crosses on them. Any thoughts are welcome.


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## efattah (Mar 6, 2018)

Your calculation is very interesting, and I feel it is quite accurate. It does make me think though, since I use LMCF instead of CFOP, I too am looking for 4 pieces in the correct position as the first step (a full EG1 face, four corners with the color, any permutation). However, according to my calculation the probability of skipping the EG1 face is 1544 - 1. I have had it happen many times. In both cases we need 4 pieces pre-solved. In one case the odds are 30,000+ against, in the other case 1500+ against. The same thing happens when we look at F2L vs. E2L pairs, in terms of the chance of skipping the first pair. In CFOP you must pair an edge with its exact corner. In E2L, any two solved edges count as a pair and any two edges can be solved as a pair. So again the chance of a skip is way higher.


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## whatshisbucket (Mar 6, 2018)

Consider only permutation and orientation of the edges (we need not worry about the parity of their permutation). There are simply 12!*2^11=980995276800 possible solvable states of the edges. A cross necessitates having 4 edges solved, leaving 8!*2^7 possibilities. With 6 colors, this is 30965760, 1/31680 of our sample set. Of course, as you mentioned, this does not account for the unlikely possibility that there are multiple crosses. 

Let's do this:
2 crosses:
Adjacent crosses (RU): 12 pairs of faces*(5 unimportant edges)!*2^4=23040
Opposite crosses (UD): 3 pairs of faces*(4 unimportant edges)!*2^3=576

3 crosses:
(RUD): 12 sets of faces*(2 unimportant edges)!*2^1=48
(RUF): 8 sets of faces *(3 unimportant edges)!*2^2=192

4 crosses:
(RUFD): 12 sets of faces*(1 unimportant edge)!*2^0=12
(RULD): 3 sets of faces*(0 unimportant edges)!*1 (we are guaranteed that EO is valid here since all of the edges are solved)=3

5 crosses:
6 sets of faces*0!*1=6

6 crosses:
1.

Note that every case with at least 4 crosses has 6, but we ignore this fact for the purposes of counting correctly. 

There are (sort of) 30965760 states with 1 cross. We have double counted the states with 2 crosses, so we subtract these. However, that means that each case with 3 crosses has been counted 3-3=0 times, so we add these back. Etc.

Our final count is 30965760-(23040+576)+(48+192)-15+6-1=30942374. This is roughly 1/31704 of all possible states.


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## whatshisbucket (Mar 6, 2018)

efattah said:


> Your calculation is very interesting, and I feel it is quite accurate. It does make me think though, since I use LMCF instead of CFOP, I too am looking for 4 pieces in the correct position as the first step (a full EG1 face, four corners with the color, any permutation). However, according to my calculation the probability of skipping the EG1 face is 1544 - 1. I have had it happen many times. In both cases we need 4 pieces pre-solved. In one case the odds are 30,000+ against, in the other case 1500+ against. The same thing happens when we look at F2L vs. E2L pairs, in terms of the chance of skipping the first pair. In CFOP you must pair an edge with its exact corner. In E2L, any two solved edges count as a pair and any two edges can be solved as a pair. So again the chance of a skip is way higher.


The reason for the higher probability of a skip is that the pieces of the EG face (not necessarily EG-1, correct?) need only be solved relative to each other (and not even that since the permutation doesn't matter). Let's compute the probability of that for fun:



Spoiler: calculations



1 face:
6 colors*6 permutations of solved layer*27 orientations*24 permutations=23328

2 faces:
adjacent: 12 pairs of faces*4 permutations of pieces in solved layers*3 orientations*2 permutations=288
opposite: 3 pairs of faces*6^2 permutations of pieces within layers*4 AUF=432

3 faces:
(RUF): 8 sets*1 (cuz the cube must be solved)=8
(RUD): 12 sets*4 permutations=48

4 faces:
15 sets of 4*1 (cuz the cube is solved)=15

5 faces:
6

6 faces:
1

Total=23328-(288+432)+56-15+6-1=22654
This is roughly 1/162 of the 3674160 possible states.


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## 1001010101001 (Mar 6, 2018)

What about ZZ users ( EO skip, EOLINe skip)


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## 1001010101001 (Mar 6, 2018)

What about ZZ users ( EO skip, EOLINe skip)


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## Sue Doenim (Mar 6, 2018)

EO skip is 1/2048, EOline is 1/270336.


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## dnguyen2204 (Mar 7, 2018)

What's the probability of having a "bad" cross case (optimal solution >6 moves) on white? On white and yellow?


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## xyzzy (Mar 7, 2018)

dnguyen2204 said:


> What's the probability of having a "bad" cross case (optimal solution >6 moves) on white? On white and yellow?


18.45% and 3.44% respectively. (And you didn't ask about it, but for full CN it drops to 0.03%.)

Source: http://www.cubezone.be/crossstudy.html


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## Kumato (Mar 11, 2018)

hotufos said:


> What is the probability of a cross skip?
> TL;DR: I think 1/31,680. I'm not positive.
> 
> Many of us know that the chance you'll get an OLL skip in any given solve is 1/216; for PLL skip, 1/72; and for a full LL skip, 1/15,552. But what is the chance you will get a cross skip in a solve? I've never heard of anyone ever getting one. I tried to figure out the chance. Here's what I came up with.
> ...



I have done around less than 12k solves on 3x3, and I have gotten both a LL skip, and a cross skip.


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## shadowslice e (Mar 11, 2018)

Kumato said:


> I have done around less than 12k solves on 3x3, and I have gotten both a LL skip, and a cross skip.


That's not unusual


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## teboecubes (Mar 26, 2018)

How many 2x2 first layer cases are there?


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## alwin5b (Mar 26, 2018)

teboecubes said:


> How many 2x2 first layer cases are there?



without regarding symmetry/mirrors, it is (3^4) * [ 8! / (4! * 4!) ] = 5670


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## Duncan Bannon (Mar 29, 2018)

What I saw chance of OLL skip with Corner Permutation?


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## Sue Doenim (Mar 29, 2018)

1/1296, or about 0.07%.


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## teboecubes (Apr 12, 2018)

What are, or how can I figure out, the probabilities of skewb OLLs (sarah intermediate)?


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## whatshisbucket (Apr 12, 2018)

I believe each of the 11 unsolved cases has an equal 4/45 chance of occurring, and a skip occurs 1/45 of the time.


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## obelisk477 (May 19, 2018)

Average # of cycles for 4x4 wings?


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## xyzzy (May 20, 2018)

obelisk477 said:


> Average # of cycles for 4x4 wings?


1/1 + 1/2 + 1/3 + … + 1/24 − 1 ≈ 2.776 if you don't count 1-cycles (pieces that are already solved).

Quick proof: If you start with an empty set, there'll obviously be 0 cycles. The kth element added will introduce a new cycle with probability 1/k, so this gives the harmonic number H_24 ≈ 3.776. To exclude the solved pieces, note that each element is a fixed point (i.e. the piece is solved) with probability 1/24, so the expected number of solved pieces is 1 and we subtract that from H_24.

A caveat is that this is assuming you don't change your orientation to influence the permutation, because that would make the problem intractable. Also, keep in mind stuff like the gambler's fallacy—just because you have three cycles already doesn't make it any more or less likely that you'll encounter a cycle break later. The probability of a cycle break only depends on how many pieces are left to trace, not on how many cycle breaks there already are.


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## xyzzy (May 20, 2018)

M0r0 said:


> Chances of me solving a Skewb in four moves (It happened to me at a contest this morning.)
> 
> Basically when solving the white side, I accidentally solved the whole thing.


0 with WCA-compliant scrambles:

4b3c) Skewb: The (random) state must require at least 7 moves to solve.

But this is the probability thread, not the regulations thread, so check out Jaap's Puzzle Page for statistics on how many moves are needed. 2073 out of 3149280 states can be solved in four moves or less, so the probability is 2073/3149280 ~ 0.07%.


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## whatshisbucket (May 21, 2018)

M0r0 said:


> But it literally happened to me at a contest today...
> It was only around 4 or 5 moves.


Then it was a misscramble and you should contact someone to get that changed to a DNF (assuming it's too late to get an extra attempt).


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## TipsterTrickster (May 21, 2018)

i have had a 4 move skewb scramble before, it was a hedge with the first move the other way


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## TipsterTrickster (May 21, 2018)

M0r0 said:


> But it literally happened to me at a contest today...
> It was only around 4 or 5 moves.


Wca?
(ps sorry for the double post i couldn't figure out how to edit in quotes)


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## Duncan Bannon (May 21, 2018)

TipsterTrickster said:


> Wca?
> (ps sorry for the double post i couldn't figure out how to edit in quotes)



The way I would do it would be to edit it. Once you click edit, type left bracket, then say the word quote then right bracket then copy in the message they say then type the same thing as the quote thing but add a forward slash between the [ and the Q

It’s hard to explain in words without it making it a quote  So feel free to ask questions.


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## Duncan Bannon (Jun 4, 2018)

Probability of doing OLL parity, it solving OLL and U2 Auf PLL skip?


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## Sue Doenim (Jun 4, 2018)

1 in 62208, or 0.00160751%.


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## Duncan Bannon (Jun 4, 2018)

That's neat. Now just have to get that LL skip. Thanks!


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## Hazel (Jun 5, 2018)

I just got two PLL skips in a row, then four full-step solves, then another PLL skip aided by a COLL.
The chances of a PLL skip is 1/72, so I'll start with 1/(72^2). The chance of not getting a PLL skip is 71/72, so I'll change to expression to (1/(72^2))*((71/71)^4)). The chance of getting a COLL case that leads to a PLL skip is 173/7776, so the expression turns to (1/(72^2))*((71/71)^4))*(173/7776), which is equal to 4396220813/(1.0833062*10^15), which is equal to about 1 in 246,417 or about a 0.00041% chance of occurring!
Someone tell me if this math is wrong


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## obelisk477 (Jun 5, 2018)

Aerma said:


> I just got two PLL skips in a row, then four full-step solves, then another PLL skip aided by a COLL.
> The chances of a PLL skip is 1/72, so I'll start with 1/(72^2). The chance of not getting a PLL skip is 71/72, so I'll change to expression to (1/(72^2))*((71/71)^4)). The chance of getting a COLL case that leads to a PLL skip is 173/7776, so the expression turns to (1/(72^2))*((71/71)^4))*(173/7776), which is equal to 4396220813/(1.0833062*10^15), which is equal to about 1 in 246,417 or about a 0.00041% chance of occurring!
> Someone tell me if this math is wrong



The math sounds right, but I've always found stats like these particularly unremarkable. For example:

I got an Ra perm, followed by a T perm, followed by a Gb, followed by an E-perm! The chances of this occuring are (1/18)^3 * (1/36) = 1/209,952. How rare! Isn't that crazy?

But it really isn't that remarkable in the sense that four random PLLs in a row isn't particularly exciting, even though it has roughly the same probability of occuring as your two PLL skips, four full step, then EPLL skip.


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## Hazel (Jun 5, 2018)

obelisk477 said:


> The math sounds right, but I've always found stats like these particularly unremarkable. For example:
> 
> I got an Ra perm, followed by a T perm, followed by a Gb, followed by an E-perm! The chances of this occuring are (1/18)^3 * (1/36) = 1/209,952. How rare! Isn't that crazy?
> 
> But it really isn't that remarkable in the sense that four random PLLs in a row isn't particularly exciting, even though it has roughly the same probability of occuring as your two PLL skips, four full step, then EPLL skip.


That crossed my mind and you are correct, but it was more remarkable to me that I got 3 PLL skips in such a short span of time (one with some control).


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## whatshisbucket (Jun 5, 2018)

this is a little more remarkable: A few weeks ago I had an accidental XXstar on megaminx. 
That's (1/60)(1/57) for the corners and (1/50)(1/48) for the edges, with 10 possible pairs of F2L pairs, for 1/820,800. 
This overcounts a little bit (because of XXXstars) but who cares. For reference, a LL skip comes 1 in every 933120 solves on megaminx.


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## whatshisbucket (Jun 5, 2018)

Aerma said:


> That crossed my mind and you are correct, but it was more remarkable to me that I got 3 PLL skips in such a short span of time (one with some control).



Well the chance of getting exactly 3 PLL skips in 7 solves (ignoring COLL, obviously that significantly increases the probability) is 21(71^4)/(72^7), which is about 1/11280, already more likely than an LL skip.


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## Hazel (Jun 6, 2018)

whatshisbucket said:


> Well the chance of getting exactly 3 PLL skips in 7 solves (ignoring COLL, obviously that significantly increases the probability) is 21(71^4)/(72^7), which is about 1/11280, already more likely than an LL skip.


How about what happened to me a few months ago, I was doing PLL training on csTimer and I got 71 solves in a row with not a single E-perm, but then the 72nd, 73rd, and 74th solves were all E-perms!


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## Hazel (Jun 13, 2018)

What's the probability of getting the same VLS case twice in a row (without worrying about angle)? Specifically this one:


Spoiler


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## Thom S. (Jun 24, 2018)

Aerma said:


> What's the probability of getting the same VLS case twice in a row (without worrying about angle)?



There are 216 VLS cases and becsude they have the Slot in them I believe there is no symmetry - every case has a probability of 1/216.
That means your probability is 1/46,656


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## xyzzy (Jun 24, 2018)

Thom S. said:


> There are 216 VLS cases and becsude they have the Slot in them I believe there is no symmetry - every case has a probability of 1/216.
> That means your probability is 1/46,656


Wrong for the specific part of the question you quoted. (Hint: If you flip a coin twice, what're the chances of getting the same outcome twice in a row? For bonus points: what if you get to flip the coin more than twice?)

There are often a lot of assumptions in a probability question; some are "obvious", most aren't. The answer to an underspecified question can vary a lot depending on what you choose to assume.


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## James Storey (Jun 29, 2018)

If you pick up a 3x3 and turn it randomly after 2.99800 × 10^19 turns it will have a 50% chance of having been solved. I assumed that each time you turn the cube it goes into a random state but this should still be close to the answer. Turning at 10 TPS, this would take 9.5*10^10 years or 6.9 times the age of the universe.


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## Thom S. (Jun 29, 2018)

xyzzy said:


> Wrong for the specific part of the question you quoted. (Hint: If you flip a coin twice, what're the chances of getting the same outcome twice in a row? For bonus points: what if you get to flip the coin more than twice?)
> 
> There are often a lot of assumptions in a probability question; some are "obvious", most aren't. The answer to an underspecified question can vary a lot depending on what you choose to assume.



I think you need to explain something to me. I quoted all the information. The specific case is not important as it's a different case from all angles.
Gonna embarras myself but I try.
You have a 50% chance because the first outcome doesn't matter. What matters is the 50/50 chance on the second flip. 

Ok, I get that hint. Needed some days for it. The real chance is 1/206.


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## mencarikebenaran (Aug 12, 2018)

do u ever get unique OLL skip ?
i ever get :

1. OLL skip and H perm once
2. OLL skip and Z perm once
3. OLL skip and PLL skip (LL skip) once


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## xyzzy (Aug 12, 2018)

… ?

Unless you do some mysterious LL influencing in the last slot, the PLL cases you get after an OLL skip are distributed the way you'd expect. Skip, H perm and N perms would be 1/72 each; Z perm and E perm would be 1/36 each; the rest would be 1/18 each. If it feels like every time you get an OLL skip the resulting PLL case is different, that's merely because you haven't gotten enough OLL skips for a repeat PLL case to be statistically probable.


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## cuber314159 (Aug 12, 2018)

xyzzy said:


> … ?
> 
> Unless you do some mysterious LL influencing in the last slot, the PLL cases you get after an OLL skip are distributed the way you'd expect. Skip, H perm and N perms would be 1/72 each; Z perm and E perm would be 1/36 each; the rest would be 1/18 each. If it feels like every time you get an OLL skip the resulting PLL case is different, that's merely because you haven't gotten enough OLL skips for a repeat PLL case to be statistically probable.


Just to be pedantic z is 1 in 72 and h is 1 in 36. Maybe I should do some more intuitive forced OLL skip ( it's not viable for speedsolving but still fun) but it's pretty obvious that each of the PLLs will show up as normal.


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## xyzzy (Aug 12, 2018)

cuber314159 said:


> Just to be pedantic z is 1 in 72 and h is 1 in 36.


something something muphry's law 

Point stands either way, yeah.


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## Dancing Jules (Aug 31, 2018)

What is the probability of a PBL skip in square-1? My guess would be 1 in 20.736 (144²). Is that correct?


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## CyanSandwich (Feb 28, 2019)

I was wondering if anyone was able to generate a large number of skeletons and find average optimal insertions. More details if someone is actually up for this.


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## Underwatercuber (Mar 1, 2019)

CyanSandwich said:


> I was wondering if anyone was able to generate a large number of skeletons and find average optimal insertions. More details if someone is actually up for this.


If anyone can write up the program for it then i can run it in my computer for you


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## Max_Hult (Mar 27, 2019)

Hello

Today this CRAZY thing happened. I first got a green cross skip and the next solve was just a 1 move cross.
I really need the answer to what the odds for this to happen is.
So if your good at math please give me the answer.
Heres the scrambles. btw the scrambles were generated on cstimer so OMG!!!!!!!!!!!!!!!!!!!!!!!!!!!


4. 9.87 L2 D2 R2 D B2 F2 R2 D' F2 D L' R B L' U' F2 L B2 L 
5. (8.16) F2 B2 D2 F2 L' F2 R2 U' L' F2 U2 L2 F' U2 L2 B2 U2 B R2 D2


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## xyzzy (Mar 27, 2019)

Max_Hult said:


> Today this CRAZY thing happened. I first got a green cross skip and the next solve was just a 1 move cross.
> I really need the answer to what the odds for this to happen is.
> So if your good at math please give me the answer.


The probability of getting two ≤1-move crosses in two _specific_ solves is about (0.05%)^2 ~ 0.000025% ~ 1/4000000.

(Source: CubeZone cross study.)


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## WoowyBaby (Apr 1, 2019)

Hmm......
Thinking thinking thinking.....

If you get two of a _specific _VLS case in a row in two _specific _solves, chance is 1/46656

If you get _any _two VLS case in a row in two _specific _solves, chance is 1/216

If you get _any _two VLS case in a row within a session of 216 solves, chance is about (may not be exactly) 1/2

You get widely different numbers even though all of them are the chance of getting a VLS case in two consecutive solves.
1/46656 vs. 1/216 vs. ~1/2
(it doesn’t have to be VLS, I just picked a set where each human case is equally likely)


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## WillyTheWizard (Apr 3, 2019)

If a pll skip is 1/72 would 2 pll skips in a row mean 1/144? And what are the odds of an entire ll skip because that is literally what happened on my first ever solve


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## Etotheipi (Apr 3, 2019)

WillyTheWizard said:


> If a pll skip is 1/72 would 2 pll skips in a row mean 1/144? And what are the odds of an entire ll skip because that is literally what happened on my first ever solve


Th e chance of 2 pll skips in a row is 1/72 x 1/72 = 1/5184 the odds of a ll skip are 1/15552 you got very lucky.


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## Etotheipi (Apr 11, 2019)

So i was doing a solve, and when i got to LSE it was all correctly permuted but the UL and BD edges were misoriented. I calculated that the chance of 2 edges being misoriented but permuted and the rest solved to be about 1/940 including aufs. Correct me if im wrong.


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## whatshisbucket (Apr 12, 2019)

Etotheipi said:


> So i was doing a solve, and when i got to LSE it was all correctly permuted but the UL and BD edges were misoriented. I calculated that the chance of 2 edges being misoriented but permuted and the rest solved to be about 1/940 including aufs. Correct me if im wrong.


1/360 chance of correct permutation
15/32 chance of exactly 2 flipped edges
1/768 total.


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## Etotheipi (Apr 12, 2019)

whatshisbucket said:


> 1/360 chance of correct permutation
> 15/32 chance of exactly 2 flipped edges
> 1/768 total.


Ok lol i screwed up.


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## cubeshepherd (Apr 13, 2019)

What are the odds of a single, Ao5 and Ao12 all being the same time? I was just doing a little 3x3 and trying to practice it again, and I got a 15.56 single which helped lead to the Ao5 and Ao12 being the same.



Spoiler: Single, ao5 and 12 times/scrambles



Generated By csTimer on 2019-4-12
avg of 5: 15.56

Time List:
1. 15.18 F' L2 R2 U R2 U' L2 U' F2 D' B2 D' L B2 D2 B F' U2 L' R2 D' 
2. (14.62) R' F L2 F U' L' U2 F D' F2 U' R2 B2 L2 U2 L2 D' L 
3. 15.93 D2 F2 L2 R U2 L2 F2 L B2 R' U2 D' R' D F' L B L B D2 
4. (17.28) L F2 L2 U B2 U' F2 U' R2 B2 L2 D R F L D' R B D F' U' 
5. 15.56 D F2 L' B' U F2 D2 R2 L B D L2 D' F2 U F2 L2 F2 R2 D' F2


Generated By csTimer on 2019-4-12
avg of 12: 15.56

Time List:
1. 15.52 F2 L2 F2 R2 D2 F2 R2 D' L2 F2 U2 R' B2 F' L F' D F2 D2 U' 
2. (19.56) U' L2 D2 U' L2 U F2 U' F2 L2 U' R' U' B2 L F2 U B' D' U2 
3. 17.51 U2 R2 D2 R2 B D2 U2 L' U' B U R B D' L' U' F' 
4. 16.58 L2 D2 R' D2 U2 L' R D2 R' D B2 F2 D B2 R F D' B' U' 
5. (13.28) L2 F' L2 D2 B2 F' D2 F L2 D2 L' B D F2 R U B' F' L 
6. 13.62 U' L2 D' F2 D' B2 R D F' B2 U2 R2 D' F2 U B2 R2 D2 B2 U 
7. 13.78 F2 B R' D L' B2 R B' U L' R2 U2 R2 U' R2 B2 D R2 F2 D L2 
8. 15.18 F' L2 R2 U R2 U' L2 U' F2 D' B2 D' L B2 D2 B F' U2 L' R2 D' 
9. 14.62 R' F L2 F U' L' U2 F D' F2 U' R2 B2 L2 U2 L2 D' L 
10. 15.93 D2 F2 L2 R U2 L2 F2 L B2 R' U2 D' R' D F' L B L B D2 
11. 17.28 L F2 L2 U B2 U' F2 U' R2 B2 L2 D R F L D' R B D F' U' 
12. 15.56 D F2 L' B' U F2 D2 R2 L B D L2 D' F2 U F2 L2 F2 R2 D' F2


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## aerocube (Apr 13, 2019)

cubeshepherd said:


> What are the odds of a single, Ao5 and Ao12 all being the same time? I was just doing a little 3x3 and trying to practice it again, and I got a 15.56 single which helped lead to the Ao5 and Ao12 being the same.
> 
> 
> 
> ...


is it even possible to get odds on that? that seems incalculable as someone could just solve before the end and wait till the timer hit a certain number


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## cubeshepherd (Apr 14, 2019)

aerocube said:


> is it even possible to get odds on that? that seems incalculable as someone could just solve before the end and wait till the timer hit a certain number


That is true that someone could do that (although I am not sure why someone would purposely do that, unless they were bored), but that is not what happened from me and I was just checking to see if there is a way to calculate my questions above. Does anyone have any idea what the probability is (if there is a way to find out)?


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## whatshisbucket (Apr 14, 2019)

cubeshepherd said:


> That is true that someone could do that (although I am not sure why someone would purposely do that, unless they were bored), but that is not what happened from me and I was just checking to see if there is a way to calculate my questions above. Does anyone have any idea what the probability is (if there is a way to find out)?


Obviously information about the probability distribution of your times is necessary. If each time is 12.00 with probability 1, then obviously all your averages will also be 12.00. If you somehow know your distribution, a Monte Carlo simulation would probably be the best way to compute the probability.


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## iLikeCheese (Apr 26, 2019)

Chance of 2 sune PLL skips in a row? Just got it.


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## WoowyBaby (Apr 26, 2019)

1/72*4/216*1/72*4/216=0.00003%

If you did more than 2 solves or influenced LL or count antisune/fat sune as sune, any of these three things would make that number go way up.
If you did 200 solves then the chance of that happening is ~1% (not sure tho)

Just wondering, were those two solves faster than average?


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## iLikeCheese (Apr 26, 2019)

WoowyBaby said:


> 1/72*4/216*1/72*4/216=0.00003%
> 
> If you did more than 2 solves or influenced LL or count antisune/fat sune as sune, any of these three things would make that number go way up.
> If you did 200 solves then the chance of that happening is ~1% (not sure tho)
> ...


Idk I was just doing some solves (around 30) and it happened. I have gotten many low probability occurrences before, such as 3 Pll skips in a row, or 2 Oll skips in a row, those are easier to calculate. Also it wasn’t fat sunes or antisunes, so pretty rare  The two solves were both sub 7 btw, I think one was ~6.5 seconds.


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## 2x2 is fun (Jun 24, 2019)

chances of having a four mover 2x2 at a comp?


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## rubik2005 (Jun 24, 2019)

Oh! I forgot to post this a couple weeks ago. I just wanted to know what's the probability of getting a LL skip on a kibiminx because I actually got one.


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## whatshisbucket (Jun 24, 2019)

2x2 is fun said:


> chances of having a four mover 2x2 at a comp?


1847/3673775 for each scramble, which is about 1 in 1989.


rubik2005 said:


> Oh! I forgot to post this a couple weeks ago. I just wanted to know what's the probability of getting a LL skip on a kibiminx because I actually got one.


There is a 1 in 81 chance of skipping OLL and a 1 in 12 chance of skipping PLL for a 1 in 972 chance of a full LL skip.


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## Miklopy (Aug 4, 2019)

The chances of a cross skip is 1/15840
The calculation is:
1/24 x 1/22 x 1/20 x 1/18 x 6


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## Miklopy (Aug 4, 2019)

chinesed00d said:


> What are the chances of having a cross already done during a scramble?


The chances of a cross skip is 1/15840
The calculation is 1/24 x 1/22 x 1/20 x 1/18 x 6


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## xyzzy (Aug 4, 2019)

Miklopy said:


> The chances of a cross skip is 1/15840
> The calculation is 1/24 x 1/22 x 1/20 x 1/18 x 6


(i) Please don't reply to nine-year-old posts unless it's absolutely relevant and necessary.

(ii) This is almost correct, but it doesn't account for scrambles with multiple crosses already solved… or it would be, if you actually calculated your approximation correctly. Your calculation is off by a factor of two and you should've gotten 1/31680.

(iii) The exact probability is around 1/31704.


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## whatshisbucket (Sep 12, 2019)

Just got 6 center skips in 7 consecutive Master pyraminx solves. Each happens with 1/12 probability, so one should expect to see 6 or more out of 7 consecutive solves 78 times out of 12^7 solves, or roughly once every 459382 solves. Not quite as extraordinary as my accidental XXcross on megaminx, but quite rare.


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## ottozing (Mar 7, 2020)

http://www.cubezone.be/crossstudy.html

I'd like to know what the numbers are for "Quad Cross" aka Dual on two sets of colours (Half turn only because who cares about Quarter turn)

I suspect that the numbers would be basically the same as CN

I'm also interested because I know Max Park was Quad Cross for a very long time (I believe he's CN now)


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## Tranman64 (Mar 8, 2020)

What are the chances of having F2L skip?


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## Kit Clement (Mar 8, 2020)

ottozing said:


> http://www.cubezone.be/crossstudy.html
> 
> I'd like to know what the numbers are for "Quad Cross" aka Dual on two sets of colours (Half turn only because who cares about Quarter turn)
> 
> ...



Quoting this request, as it's a fairly interesting question and has implications for the relative gains of levels of color neutrality.


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## xyzzy (Mar 8, 2020)

My "first-order" estimate for quad CN cross is that it should be around 5.01 moves average, interpolating between the dual CN and full CN values on Lars's site.


```
dualcn = [53759, 806253, 8484602, 74437062, 506855983, 2031420585, 2311536662, 175751822, 3672]
fullcn = [30942374, 462820266, 4839379314, 41131207644, 239671237081, 543580917185, 151019930400, 258842496, 40]
cdf = [sum(dualcn[:k+1])/sum(dualcn) for k in range(9)] # this is exact
cdf2 = [1-(1-x)**2 for x in cdf] # computing quad CN from dual CN, assuming approximate independence
cdf3 = [1-(1-x)**3 for x in cdf] # computing full CN from dual CN, assuming approximate independence

extrapolated_quadcn_mean = sum(1-x for x in cdf2) # ~ 4.984
extrapolated_fullcn_mean = sum(1-x for x in cdf3) # ~ 4.753

# since we know the ground truth for the full CN mean, we can use that to roughly
# quantify how much the crosses on different axes fail to be independent
correction = sum(i*x for i,x in enumerate(fullcn))/sum(fullcn) - extrapolated_fullcn_mean

corrected_quadcn_mean = extrapolated_quadcn_mean + correction/2
print(corrected_quadcn_mean) # prints 5.012148701428693
```

Probably won't be too hard to compute a much better approximation using random sampling, or with enough CPU time, the true exact value.

---

Wrote a bit of code, spent a bit of CPU time. Quad cross with 2^27 ~ 134 million total samples:

cross colours: white, yellow, red, orange

depthsamplesproportion0​2826​0.000021​1​42502​0.000317​2​443883​0.003307​3​3827135​0.028514​4​24023111​0.178986​5​70399118​0.524514​6​35158823​0.261954​7​320330​0.002387​8​0​0.000000​
Mean: 5.01942(7) moves

My estimate above was pretty close! Also, nobody asked for this, but here's the same table for triple CN (@OreKehStrah may be interested?):

cross colours: white, red, green (three mutually adjacent choices)

depthsamplesproportion0​2162​0.000016​1​31567​0.000235​2​331994​0.002474​3​2887422​0.021513​4​18711235​0.139410​5​63047910​0.469744​6​47742095​0.355706​7​1463343​0.010903​8​0​0.000000​
Mean: 5.18663(7) moves

cross colours: white, yellow, red (two opposite choices + a third one)

depthsamplesproportion0​2068​0.000015​1​31820​0.000237​2​333504​0.002485​3​2897351​0.021587​4​18868616​0.140582​5​63934436​0.476349​6​46916485​0.349555​7​1233448​0.009190​8​0​0.000000​
Mean: 5.17570(7) moves

NOTE: 8-move crosses are vanishingly rare when you can do at least three cross colours and my sampling just happened to miss them. They're obviously not impossible.

To put all the information together:

crossesmean%*1*5.81200%2 (adjacent)5.408140.3%*2 (opposite)*5.387242.4%3 (adjacent)5.186662.4%3 (opp + another)5.175763.5%4 (all except adj)5.025978.4%*4 (all except opp)*5.019479.1%54.905490.4%*6*4.8095100%
The last column is the "percentage benefit relative to going from fixed cross to full CN", so e.g. going from fixed cross to quad CN is about 79% as good as going to full CN.


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## mencarikebenaran (Mar 8, 2020)

Tranman64 said:


> What are the chances of having F2L skip?


According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400). 
Skipping F2L and OLL, and corner CP has the odds of 1 in 37 trillion (1/37924385587200).


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## ProStar (Mar 8, 2020)

Chances for an Xcross to already be finished?


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## brododragon (Mar 8, 2020)

ProStar said:


> Chances for an Xcross to already be finished?


Dunno, but all you need to find is the chance of a pair skip and the chance of a cross skip, then multiply them.


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## Kit Clement (Mar 8, 2020)

brododragon said:


> Dunno, but all you need to find is the chance of a pair skip and the chance of a cross skip, then multiply them.



That assumes Independence, which is not true in this case.


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## brododragon (Mar 8, 2020)

Kit Clement said:


> That assumes Independence, which is not true in this case.


You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.


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## Kit Clement (Mar 8, 2020)

brododragon said:


> You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.



If you already have a cross, then there are fewer possibilities for the edge piece in the first pair, making it more likely to get that pair solved than if a cross already was there. If Y is the event of a solved pair, you're describing the event Y | X, and then independence is not necessary for multiplying.


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## brododragon (Mar 8, 2020)

Kit Clement said:


> If you already have a cross, then there are fewer possibilities for the edge piece in the first pair, making it more likely to get that pair solved than if a cross already was there. If Y is the event of a solved pair, you're describing the event Y | X, and then independence is not necessary for multiplying.


I think there was a misunderstanding. I'm talking about the chance of getting an F2L pair pre-solved *with cross accounted for. *I probably should've been clearer.


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## ProStar (Mar 8, 2020)

brododragon said:


> I think there was a misunderstanding. I'm talking about the chance of getting an F2L pair pre-solved *with cross accounted for. *I probably should've been clearer.



I'm talking about scrambling the cube, and there being a solved cross and a solved(not just paired up) F2L pair


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## Cuberstache (Mar 8, 2020)

brododragon said:


> You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.


Dude, Kit's a literal statistics professor. I don't think you should argue with him.


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## brododragon (Mar 8, 2020)

ProStar said:


> I'm talking about scrambling the cube, and there being a solved cross and a solved(not just paired up) F2L pair


That's what I'm thinking. Solved and placed.


CuberStache said:


> Dude, Kit's a literal statistics professor. I don't think you should argue with him.


Ya I think there was a misunderstanding between us.


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## Kit Clement (Mar 9, 2020)

First, to answer the X-cross probability question:

fixing 5 edges and 1 corner, permuting and orienting and permuting the remaining 7 edges and 7 corners can be done in this many ways: 7!*7!*2^6*3^6/2 = 592568524800.

That assumes a specific cross color and a specific pair though. To account for one cross color but any of the four pairs, multiply these states by 4: 2370274099200

Dividing by the full number of cube states results in the probability of an XCross on a specific color: 2370274099200/(12!*8!*2^11*3^7/2) = 5.48*10^-8

Multiply it by 6 for an x-cross on any color: 3.29*10^-7

And there you go, probabilities that are so small that I wonder what purpose you had for even asking the question in the first place.


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## ProStar (Mar 9, 2020)

Kit Clement said:


> First, to answer the X-cross probability question:
> 
> fixing 5 edges and 1 corner, permuting and orienting and permuting the remaining 7 edges and 7 corners can be done in this many ways: 7!*7!*2^6*3^6/2 = 592568524800.
> 
> ...



I think I'd better just hope for the 2 mover


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## mencarikebenaran (Mar 28, 2020)

got second OLL skip and H perm 

15.70 B2 U' B2 R2 D F2 L2 U R2 F2 D2 R' U' B D2 U' F' U' B L2 U'

x2 // inspection
F D F2 D L2 U F2 // cross
y' U' R' U2 R y U' L' U L // 1st pair
U' R' U R2 U' R' // 2nd pair
y2 U L' U2 L U L' U' L // 3rd pair
R U R' U2 R U2 R' y' U R' U' R // 4th pair
M2 U' M2 U2 M2 U' M2 // PLL
U // AUF

what's chance of getting OLL skip and H perm ?
and is it higher than OLL skip and Z perm ?


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## Ronxu (Mar 28, 2020)

weruoaszxcvnm said:


> what's chance of getting OLL skip and H perm ?
> and is it higher than OLL skip and Z perm ?


H: 1/216*1/72 (4-way symmetry)
Z: 1/216*2/72 (2-way symmetry)


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## mencarikebenaran (Mar 29, 2020)

ProStar said:


> Chances for an Xcross to already be finished?



did u see my post ? 

"According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400)"

im not expert in probability

maybe probability for X-cross is 1/914457600 

914457600 is 3657830400/4


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## Etotheipi (Mar 29, 2020)

weruoaszxcvnm said:


> did u see my post ?
> 
> "According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400)"
> 
> ...


Nope, thats not quite how probabilities work. =P


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## ProStar (Apr 11, 2020)

Chances of getting an EOLine skip?



weruoaszxcvnm said:


> did u see my post ?
> 
> "According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400)"
> 
> ...



Finishing F2L isn't an independent probability, but a dependent one. There's a different chance of having the 1st pair done and the 2nd one done.



Bump @ProStar 





Wait....


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## mencarikebenaran (Apr 12, 2020)

is OLL skip chance < PLL skip chance?


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## Username: Username: (Apr 12, 2020)

Yep. because the number of OLL cases are bigger than PLL cases. true.


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## ProStar (Apr 12, 2020)

weruoaszxcvnm said:


> is OLL skip chance < PLL skip chance?



Yes, OLL skip chance is 1/216 while PLL is 1/72


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## RedstoneTim (Apr 12, 2020)

ProStar said:


> Chances of getting an EOLine skip?


The probability of an EOLine skip is 0.00037% for random state.

Source: The EOLine section on Michal Hordecki's ZZ page


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## 1cubealot (Apr 26, 2020)

Hi

So a couple of days ago I was doing some 3x3 solves and for some reason I got 3 Y-perms in a row.

Ever since I've been getting Y-perms very commonly. Am I just lucky or unlucky?


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## Username: Username: (Apr 26, 2020)

1cubealot said:


> Hi
> 
> So a couple of days ago I was doing some 3x3 solves and for some reason I got 3 Y-perms in a row.
> 
> Ever since I've been getting Y-perms very commonly. Am I just lucky or unlucky?



I got 5 G Perms in 17 solves, haha


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## 1cubealot (Apr 26, 2020)

Username: Username: said:


> I got 5 G Perms in 17 solves, hah


Like in a row?


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## Cody_Caston (Apr 26, 2020)

1cubealot said:


> Am I just lucky or unlucky?


I dont know maybe look at pll probability


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## Username: Username: (Apr 26, 2020)

Cody_Caston said:


> I dont know maybe look at pll probability



Not possible, I got Gd Perm 5 times in a row.


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## 1cubealot (Apr 26, 2020)

Username: Username: said:


> Not possible, I got Gd Perm 5 times in a row.



Wha- how???


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## alexiscubing (Apr 26, 2020)

I have gotten the same LL 3 solves in a row lol. That is like 1 in 45000 or something


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## Owen Morrison (Apr 26, 2020)

1cubealot said:


> Hi
> 
> So a couple of days ago I was doing some 3x3 solves and for some reason I got 3 Y-perms in a row.
> 
> Ever since I've been getting Y-perms very commonly. Am I just lucky or unlucky?


If I were you I wouldn't look at the PLL just spam Y perm after OLL because you always get it.


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## TheRouxGuy (Apr 26, 2020)

I got this rare oll three in a row. It probability is 1 in 108.

Another time I got the Na perm twice in a row and then Nb perm thrice in a row.


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## PizzaCuber (Apr 26, 2020)

It just depends on what coll you get.


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## Ayce (Apr 26, 2020)

Believe it or not, my first ever solve in my life had a LL skip


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## 1cubealot (Apr 27, 2020)

Ayce said:


> Believe it or not, my first ever solve in my life had a LL skip


How did you react


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## DerpBoiMoon (Apr 27, 2020)

Probability of a 3 move solution from an official 7x7 scramble? It's in a reg or something


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## alexiscubing (Apr 27, 2020)

Ayce said:


> Believe it or not, my first ever solve in my life had a LL skip


My first 2 times finishing the cube had LL skips (not knowing LL and only knowing how to do F2L boosted this significantly)


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## xyzzy (Apr 27, 2020)

DerpBoiMoon said:


> Probability of a 3 move solution from an official 7x7 scramble? It's in a reg or something


Oh boy, this is a fun question.

Let's say our official scrambles are random-state. How many 3-move scrambles are there? There are 54 choices for the first move and 51 choices for the second move (always), and the number of choices for the third move depends on the first two moves but it's always between 48 and 51. So there are at least 54×51×48 = 132192 and at most 54×51×51 = 140454 3-movers. Divide this by the total number of possible 7^3 states (about 2×10^160), and you get an answer of about 7×10^−156.

But that's assuming that our official scrambles are random-state, which they are not! It is known (to me, at least) that random-move big cube scrambles require way more moves than what's currently used in TNoodle (60/80/100 moves for 5/6/7 respectively) to reach an approximately uniform distribution. One could naïvely estimate the probability of a 7^3 WCA scramble sequence not using three-layer moves at all as (2/3)^100 ~ 2.5×10^−18 (the true probability is actually somewhat lower because adjacent moves are not independent), and in this case the scramble will be functionally identical to just a 5^3 scramble. Now, if we assume that those hundred moves do reach an approximately uniform distribution on a 5^3, the probability of getting a 3-mover is now _at least_ (2/3)^100 × (36×33×30/(2.6×10^90)) ~ 3.4×10^−104. This is *fifty-two orders of magnitude* greater than the probability with a true random-state scramble.

---

*edit* The above has some minor mistakes and simplifications (which don't affect the qualitative conclusion that the scrambles being random-move rather than random-state changes the result of the calculation by many orders of magnitude), so here's the _real_ working.

For odd-order cubes, the number of 3-move _states_ is the same as the number of _canonical_ 3-move sequences, since the shortest "nontrivially" equivalent sequences are at least four moves long (U2 D2 L2 R2 = L2 R2 U2 D2). (Side note: On even-order cubes, the shortest nontrivially equivalent sequences are three moves long (U2 R2 U2 = R2 U2 R2 on 222).)

The number of canonical 3-move sequences is 28836 for 555 and 100764 for 777. This immediately gives:
\( P(\text{random-state 777 scramble can be solved in 3 moves})=\frac{100764}{1.95006\cdot10^{160}}=5.16724\cdot10^{-156} \)

TNoodle's random-move scramble generation looks very complicated, but most of that complexity seems to exist to deal with weird puzzles like megaminx (… except it no longer supports 12-gen mega scrambles anyway) and the core algorithm can be described simply as:
1. Pick a random move uniformly among the possible moves that doesn't cancel with the current scramble sequence. (Note: this checks for cancellations that can be obtained by rearranging commuting moves at the end, like L R L = L2 R, but not more complicated ones like L2 R2 U2 D2 L = U2 D2 R2 L'.)
2. Append that move to the scramble sequence and repeat until it is of the desired length.

In the case of the n×n×n cubes, this is fairly easy to implement directly, with a lot less complexity than the fully general version in TNoodle. See e.g. csTimer's or qqTimer's source code. But we don't care about that; we just want to count scramble sequences. If we assume every generated scramble sequence is equally likely (an assumption that turns out to be false, but still fairly close to the truth), then we can divide the number of 100-move 555 scramble sequences by the number of 100-move 777 scramble sequences to get the probability of a 777 scramble sequence not using any 3Uw*, 3Dw*, 3Lw*, 3Rw*, 3Fw* or 3Bw* moves. This works out to be
\( P(\text{scramble does not use 3*w* moves})\approx\frac{5.21753\cdot10^{150}}{7.74251\cdot10^{169}}=6.73881\cdot10^{-20}, \)
a bit more than an order of magnitude smaller than the fully naïve estimate of \( (2/3)^{100}\approx2.45965\cdot10^{-18} \) in the first section.

To get a proper lower bound, observe that after the first move (which always has an exactly 2/3 probability of not being 3*w*), the probability of a move not being 3*w*, conditioned on knowing the earlier moves and also on that the earlier moves are all not 3*w*, must be 11/17, 10/16, 9/15, or 8/14 (depending on how many moves along the same axis have been made), with the last one being the smallest. This gives a lower bound of
\( P(\text{scramble does not use 3*w* moves})>(2/3)\cdot(8/14)^{99}=5.79618\cdot10^{-25}. \)

Now, assuming that the scramble does not use 3*w* moves and that 100 moves is enough to scramble whatever can be scrambled almost uniformly, we have:
\( P(\text{scramble can be solved in 3 moves})>P(\text{no 3*w* moves and can be solved in 3 moves})\\
=P(\text{no 3*w* moves})P(\text{can be solved in 3 moves}|\text{no 3*w* moves})\\
\gtrsim(5.79618\cdot10^{-25})\cdot\frac{28836}{2.58264\cdot10^{90}}\\
=6.47163\cdot10^{-111} \)

This somewhat more rigorous lower bound is "only" 45 orders of magnitude smaller than what we would expect from random-state scrambles. There's still one handwaving step in the derivation where we assume that 100 moves is enough to scramble a 555, but heuristically, we expect that the deviation of this assumption from the truth is in our favour, i.e. it would actually _increase_ the probability of getting a 3-mover.

Code to calculate the number of canonical sequences / possible scramble sequences:

```
def count_scramble_sequences(size, length):
    nmoves_per_axis = size-1
    last = [1] + [0]*nmoves_per_axis
    for i in range(length):
        same_axis = [0] + [last[j-1] * (nmoves_per_axis-(j-1)) for j in range(1, nmoves_per_axis+1)]
        diff_axis = [0] + [sum(last) * (2*nmoves_per_axis)] + [0] * (nmoves_per_axis-1)
        last = [same_axis[j]+diff_axis[j] for j in range(nmoves_per_axis+1)]
    return sum(last) * 3**length

def count_canonical_sequences(size, length):
    nmoves_per_axis = size-1
    last = [1] + [0]*nmoves_per_axis
    for i in range(length):
        same_axis = [sum(last[:j]) for j in range(nmoves_per_axis+1)]
        diff_axis = [0] + [2*sum(last)] * nmoves_per_axis
        last = [same_axis[j]+diff_axis[j] for j in range(nmoves_per_axis+1)]
    return sum(last) * 3**length
```


----------



## 1cubealot (Apr 27, 2020)

Wow that is low


----------



## 1cubealot (Apr 27, 2020)

alexiscubing said:


> My first 2 times finishing the cube had LL skips (not knowing LL and only knowing how to do F2L boosted this significantly)


I hav never gotten an LL skip before


----------



## Username: Username: (Apr 27, 2020)

1cubealot said:


> I hav never gotten an LL skip before



Since a LL is sooooooooooooooo rare, I have never gotten it. I have like 3000 solves this month and not even a single one.



DerpBoiMoon said:


> Woah. Tysm but oof the ^ is kinda hard to understand, but tysm



^ means to bring up to X power.


----------



## DerpBoiMoon (Apr 27, 2020)

xyzzy said:


> Oh boy, this is a fun question.
> 
> Let's say our official scrambles are random-state. How many 3-move scrambles are there? There are 54 choices for the first move and 51 choices for the second move (always), and the number of choices for the third move depends on the first two moves but it's always between 48 and 51. So there are at least 54×51×48 = 132192 and at most 54×51×51 = 140454 3-movers. Divide this by the total number of possible 7^3 states (about 2×10^160), and you get an answer of about 7×10^−156.
> 
> But that's assuming that our official scrambles are random-state, which they are not! It is known (to me, at least) that random-move big cube scrambles require way more moves than what's currently used in TNoodle (60/80/100 moves for 5/6/7 respectively) to reach an approximately uniform distribution. One could naïvely estimate the probability of a 7^3 WCA scramble sequence not using three-layer moves at all as (2/3)^100 ~ 2.5×10^−18 (the true probability is actually somewhat lower because adjacent moves are not independent), and in this case the scramble will be functionally identical to just a 5^3 scramble. Now, if we assume that those hundred moves do reach an approximately uniform distribution on a 5^3, the probability of getting a 3-mover is now _at least_ (2/3)^100 × (36×33×30/(2.6×10^90)) ~ 3.4×10^−104. This is *fifty-two orders of magnitude* greater than the probability with a true random-state scramble.


Woah. Tysm but oof the ^ is kinda hard to understand, but tysm


----------



## xyzzy (Apr 27, 2020)

DerpBoiMoon said:


> Woah. Tysm but oof the ^ is kinda hard to understand, but tysm


7×10^−156 = 0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 000007
3.4×10^−104 = 0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00034

Scientific notation makes it so that you're not literally staring at a long string of zeros when dealing with very small or very big numbers.


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## 1cubealot (Apr 27, 2020)

xyzzy said:


> 7×10^−156 = 0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 000007
> 3.4×10^−104 = 0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00034
> 
> Scientific notation makes it so that you're not literally staring at a long string of zeros when dealing with very small or very big numbers.


0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 00034 = 
0.0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 034%


----------



## 2018AMSB02 (Apr 27, 2020)

Probability of cubeshape skip in square-1? I’ve seen it happen a few times so it shouldnt be too low


----------



## brododragon (Apr 27, 2020)

xyzzy said:


> Oh boy, this is a fun question.
> 
> Let's say our official scrambles are random-state. How many 3-move scrambles are there? There are 54 choices for the first move and 51 choices for the second move (always), and the number of choices for the third move depends on the first two moves but it's always between 48 and 51. So there are at least 54×51×48 = 132192 and at most 54×51×51 = 140454 3-movers. Divide this by the total number of possible 7^3 states (about 2×10^160), and you get an answer of about 7×10^−156.
> 
> But that's assuming that our official scrambles are random-state, which they are not! It is known (to me, at least) that random-move big cube scrambles require way more moves than what's currently used in TNoodle (60/80/100 moves for 5/6/7 respectively) to reach an approximately uniform distribution. One could naïvely estimate the probability of a 7^3 WCA scramble sequence not using three-layer moves at all as (2/3)^100 ~ 2.5×10^−18 (the true probability is actually somewhat lower because adjacent moves are not independent), and in this case the scramble will be functionally identical to just a 5^3 scramble. Now, if we assume that those hundred moves do reach an approximately uniform distribution on a 5^3, the probability of getting a 3-mover is now _at least_ (2/3)^100 × (36×33×30/(2.6×10^90)) ~ 3.4×10^−104. This is *fifty-two orders of magnitude* greater than the probability with a true random-state scramble.


Here's an expansion to that question: What event is most likely to generate a scramble that's the minimum amount of moves for said event? Here's the regulation:


> 4b3) Specification for a scramble program: An official scramble sequence must produce a random state from all states that require at least 2 moves to solve (equal probability for each state). The following additions/exceptions apply:
> 4b3a) For blindfolded events, the scramble sequence must orient the puzzle randomly (equal probability for each orientation).
> 4b3b) 2x2x2 Cube: The (random) state must require at least 4 moves to solve.
> 4b3c) Skewb: The (random) state must require at least 7 moves to solve.
> ...


----------



## xyzzy (Apr 27, 2020)

brododragon said:


> Here's an expansion to that question: What event is most likely to generate a scramble that's the minimum amount of moves for said event? Here's the regulation:




```
>>> 1847 / (3674160-1-9-54-321) # 222
0.0005027526182196787
>>> 315198 / (3149280-1-8-48-288-1728-10248-59304) # skewb
0.10241498803472124
>>> (38893230+3091458)/(435891456000-1-64-1153-17050-235144-3091458) # upper bound for squan
9.631987413274141e-05
>>> (38893230)/(435891456000-1-64-1153-17050-235144) # lower bound for squan
8.922692123419956e-05
>>> pyra_notips = [1,8,48,288,1728,9896,51808,220111,480467,166276,2457,32,0,0,0,0]
>>> pyra_tips = [pyra_notips[i]+2*4*pyra_notips[i-1]+4*6*pyra_notips[i-2]+8*4*pyra_notips[i-3]+16*pyra_notips[i-4] for i in range(16)]
>>> pyra_tips[6]/sum(pyra_tips[6:]) # pyraminx
0.002414913105050851
```

The values for squan are only lower/upper bounds because the exact distribution is not known for the WCA metric. (Wouldn't be hard per se to just calculate the exact value without determining the full distribution, but it's not relevant to your question.) The answer is clearly skewb, with about 10% of skewb scrambles being 7-movers.

All raw values from Jaap's Puzzle Page.


----------



## Cuberstache (Apr 27, 2020)

xyzzy said:


> about 10% of skewb scrambles being 7-movers.


Wow, incredible! I guess most 7-movers just aren't obvious to humans.


----------



## potatojuiceultra (Apr 27, 2020)

alexiscubing said:


> My first 2 times finishing the cube had LL skips (not knowing LL and only knowing how to do F2L boosted this significantly)


I was able to intuitively figure out f2l but the only last layer algorithm I would come up with was a sune. So my first solve either had a full LL skip or a sune or anti-sune with a PLL skip.
If I didn't get either of those I would just rescramble the cube.


----------



## nuclearaven (May 11, 2020)

chance of a LL skip with EO and CP done (so a 2GLL skip)? my initial thought was that its just 1/6th of a ZBLL skip but im not sure.


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## xyzzy (May 11, 2020)

nuclearaven said:


> chance of a LL skip with EO and CP done (so a 2GLL skip)? my initial thought was that its just 1/6th of a ZBLL skip but im not sure.


That is correct.

You can also calculate it directly as 1/3^3 (CO skip) × 1/(4!/2) (EP skip) = 1/324.


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## 2018AMSB02 (May 11, 2020)

What is the probability of having one edge paired with the center on any face on clock? 2 paired up? 3?


----------



## Cody_Caston (May 12, 2020)

What’s the probability of getting a skewb scramble with all corners solved


----------



## Kit Clement (May 12, 2020)

PingPongCuber said:


> What is the probability of having one edge paired with the center on any face on clock? 2 paired up? 3?



All 14 clock wheels are independent, so it's as simple as a binomial calculation. Assume the centers on each face are fixed, then you have 8 chances to get any pairs. It's easiest to go about this by first doing the chance of nothing pairing up at all to get the probability of at least one edge paired:

1 - (11/12)^8 = 50.14%

Having at least 2 or 3 matching on the same face gets more complicated, as we'd need to calculate the probability of getting it on one face, then use that probability to find whether it could happen on at least one of the faces. For just one face, we have 4 potential clocks to match a fixed center.

(4 choose 4)(1/12)^4(11/12)^0 + (4 choose 3)(1/12)^3(11/12)^1 + (4 choose 2)(1/12)^2(11/12)^2 = 0.03718

Now, since each face's edges/centers are determined independently, we can take the same approach of calculating the probability of neither face having this outcome, and then take the opposite probability:

1 - (1 - [above answer])^2 = 7.3%

We can do the same for 3 by taking the same approach but eliminating the case above for exactly 2 matching on one face.

(4 choose 4)(1/12)^4(11/12)^0 + (4 choose 3)(1/12)^3(11/12)^1 = 0.00217

1 - (1 - [above answer])^2 = 0.43%


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## xyzzy (May 12, 2020)

Cody_Caston said:


> What’s the probability of getting a skewb scramble with all corners solved


(Assuming true random-state and no scramble filter; this probably changes the probability by a few percent if you're using WCA-legal scrambles, which do have a scramble filter.)

Fix one tetrad of corners as a reference (e.g. the ones that are fixed to the core). Call this the first tetrad, and call the other tetrad the second tetrad.

P(first tetrad is oriented) = 1/3^4

P(second tetrad is permuted | first tetrad is oriented) = 1/4
(orientation of the first tetrad is related to the Klein 4-group coset of the second tetrad's permutation, so if the first tetrad is oriented, the permutation of the second tetrad must also be in the Klein 4-group)

P(second tetrad is oriented | first tetrad is oriented and second tetrad is permuted) = 1/3^3
(orientation of the first three corners can be freely chosen; the orientation of the fourth corner is determined by the rest)

P(all corners are solved) = 1/3^4 × 1/4 × 1/3^3 = 1/8748


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## Username: Username: (May 25, 2020)

Probability of the first layer solved with two oriented edges?


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## Cuberstache (May 25, 2020)

Username: Username: said:


> Probability of the first layer solved with two oriented edges?


What do you mean "with two oriented edges"? Two pieces outside the first layer oriented?


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## Username: Username: (May 25, 2020)

Yes.


CuberStache said:


> What do you mean "with two oriented edges"? Two pieces outside the first layer oriented?


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## Cuberstache (May 25, 2020)

Username: Username: said:


> Yes.


Exactly two or at least two?


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## Username: Username: (May 25, 2020)

CuberStache said:


> Exactly two or at least two?


Two exactly


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## Cuberstache (May 25, 2020)

Username: Username: said:


> Two exactly


Ok, I'll let someone else answer now since I don't really know how to calculate that, hahaha.


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## Username: Username: (May 27, 2020)

Lol no one answered my question for days.


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## Kit Clement (May 27, 2020)

Username: Username: said:


> Lol no one answered my question for days.



It would help if you gave some motivation for why anyone would want to know that probability. I could calculate it but I'm 0% interested in the result.


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## Username: Username: (May 27, 2020)

Kit Clement said:


> It would help if you gave some motivation for why anyone would want to know that probability. I could calculate it but I'm 0% interested in the result.


I'm just curious.


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## Devagio (May 27, 2020)

Username: Username: said:


> I'm just curious.


A back of the envelope calculation gives me 1.1x10^-9; I expect it to be off by less than 5%
Basically 1 in 24x22x20x18x24x21x18x15 scrambles will have a solved white face.
Approximating that all scrambles with one solved face is disjoint means about 6 in that many scrambles have a solved face. Then simply (8 choose 2)/128 EO cases have exactly 2 oriented edges.


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## Username: Username: (May 27, 2020)

Devagio said:


> A back of the envelope calculation gives me 1.1x10^-9; I expect it to be off by less than 5%


Thanks! even if it is not accurate.


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## Devagio (May 27, 2020)

Username: Username: said:


> Thanks! even if it is not accurate.


Not precise*
It is accurate.


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## N's-cvt (Jun 13, 2020)

I was re-watching J-perm's all 77 F2L cases video and he claimed that there are 167 total cases but he wouldn't explain why so over the years I have tried to get that number and I can't get it so leave a comment below so I can see you math, here is mine so far:
F2L consists of the cross edges being done with 4 edges and 4 corners that need to be solved. Because placement is an addition problem and orientation is a multiplication problem and solved positions are a subtraction problem I think the math would look like this (8+7+6+5+8+7+6+5) * 2 * 3 - 4 = 308


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## kubesolver (Jun 13, 2020)

Assuming that the two cases that differ only by the U layer are the same case I counted 167 cases excluding the solved state.

3 * 2 * (4 + 4 + 4 + 4*4) - 1

3 corner orientations *
2 edge orientations * (
4 relative edge.corner positions when edge and corner are in U layer:
+ 4 cases of edge in U layer, corner in D layer:
+ 4 cases edge in E layer, corner in U layer
+ 4 * 4 cases corner in D layer and edge in E layer )
- 1 solved position


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## SenorJuan (Jun 14, 2020)

In practice, a competent cuber would know far more than 167 cases. He/she would know how to solve every case from the y2 orientation, so only y or y' rotations would be needed to solve the case, and never y2, and in some cases a seperate algorithm for those y , y' cases would be known. Then add in good use of empty slots, and a considerable number on new ways to solve the same basic case become possible. On the other hand, not many people solve those cases known as 'Jeff2L' , where neither edge or corner are on the U layer. I did work out algs for a lot of these cases many years ago, mainly for adjacent slots, but only a few of them made it into my speedsolving repertoire. I have no idea how many 'algs' I actually use in the F2L stage, it could be over 500, far too many to try and total up.
There's also all the LL edge control variants. Even if you just use basic sledgehammer insert as an alternative to your normal insert, you're adding a lot of cases to your repertoire.


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## AMKuber (Jun 19, 2020)

The probability of getting cross solved is 1/190,080

This is because there are 12 edges on the cube and each has 2 possible orientations. this means each edge has a 1/24 chance of being solved, but because the first edge is already there the second edge has a 1/22 chance of being solved as one edge place is taken up already the third edge has a 1/20 chance of being solved and the fourth one has a 1/18 chance.


So the math would be 
1/24 x 1/22 x 1/20 x 1/18 = 190,080



To put this in perspective a LL skip on 3x3 is 1/15,552 and think about how many LL skips you have gotten, not many I assume. and you would need to get 12 (really 12.22222....) LL skips before you should have gotten a cross solved skip.


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## vidcapper (Jun 23, 2020)

Probability of x number of randomly paired edges on a 4x4?

I've rarely seen more than one...


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## xyzzy (Jun 23, 2020)

vidcapper said:


> Probability of x number of randomly paired edges on a 4x4?
> 
> I've rarely seen more than one...


*tl;dr* See the table at the bottom of this post.

First-order approximation: There are 12 places you can get a free edge pair, and the probability of each one actually having a free edge pair is 1/23 (whatever the first piece is, there are 23 possibilities for its partner and only one of those 23 will give you a free edge pair). The probabilities are _not_ independent, but since we're approximating, let's pretend they are, so we can use a binomial distribution.


# pairsprobability00.58659510.31996120.07999030.01212040.00124050.00009060.000005≥7<0.0000002

How good/bad is this approximation? Let's do a simulation.


```
import random

def test(num_samples, N=12):
    a = list(range(N))*2
    counts = [0]*(N+1)
    for i in range(num_samples):
        random.shuffle(a)
        counts[sum(a[j] == a[j+N] for j in range(N))] += 1
    return counts

>>> test(1000000)
[593992, 308682, 80931, 14315, 1845, 213, 22, 0, 0, 0, 0, 0, 0]
```

Unsurprisingly, the approximation breaks down near the tails (that's where the independence assumption happens to be least applicable), but at least for small number of pairs, it's not too far off. (We can further approximate the binomial distribution with a Poisson distribution with mean 12/23, which very counterintuitively actually improves (?!) the approximation a lot, as if most of the approximation errors cancel each other out. I don't understand this. There is probably some fancy group theory lurking behind this.)

It is in fact possible to compute the exact distribution via the inclusion-exclusion principle as well.

Let \( n \) be the number of edge pair locations in total. (12 for reduction, 9 for Yau after centres and 3/4 cross, 8 for Yau after centres and cross.) The number of ways of pairing up \( 2n \) items, labelled with two copies of \( 0 \) through \( n-1 \), such that no pair has items of the same label, is:
\( \displaystyle a_n:=\sum_{k=0}^n(-1)^k\binom nk(2n-1-2k)!! \)

(Note: !! means double factorial. The number of ways of pairing up \( 2n \) items, ignoring the order of the pairs and the order within the pairs, is \( \frac{(2n)!}{n!2^n}=(2n-1)!! \).)

If we want to count the number of ways to have \( m \) free pairs out of \( N \) total, we first choose which pairs are free (\( \binom Nm \)), then treat the remaining \( N-m \) pairs as an instance of the above problem (\( a_{N-m} \)). To wit:
\( \displaystyle(\text{# of ways to have }m\text{ free pairs out of }N)=\binom Nma_{N-m} \)

Now we can tabulate the values with \( N=12 \) and varying values of \( m \):

# pairs​# cases​probability​0​187778717632​0.59379647​1​97779555840​0.30919987​2​25548595776​0.08079012​3​4469273600​0.01413280​4​589382640​0.00186375​5​62574336​0.00019787​6​5580960​0.00001765​7​430848​0.00000136​8​29700​0.00000009​9​1760​0.00000001​10​132​0.00000000​11​0​0.00000000​12​1​0.00000000​

As a sanity check, these values are in good agreement with the distribution obtained via sampling.

edit: Fixed some incoherent nonsense. The calculations are still correct, of course.


----------



## vidcapper (Jun 24, 2020)

Thank you very much for this - very comprehensive!

They do seem to correspond with my observational experience. Offhand I don't recall getting more than 2, but I daresay others here have stories of getting rather more than that.


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## ProStar (Jul 1, 2020)

What are the chances of having F2L-1 solved?


----------



## DemonicCuberad (Jul 1, 2020)

Whats The Chance of getting an OLL+PLL skip normally after completing F2L.(do you need VCS?)


----------



## Nmile7300 (Jul 1, 2020)

DemonicCuberad said:


> Whats The Chance of getting an OLL+PLL skip normally after completing F2L.(do you need VCS?)


You mean a last layer skip?


----------



## DemonicCuberad (Jul 1, 2020)

I mean yeah. OLL+PLL Skip


----------



## ProStar (Jul 2, 2020)

DemonicCuberad said:


> I mean yeah. OLL+PLL Skip



That's called an LL skip, and the probability is 1 / 15552. Also this is a really widely-posted probability, make sure you do your research before asking a question


----------



## vidcapper (Jul 2, 2020)

I dare say this has been asked many times here before, but I can't seem to ask the right search questions...

What are the probabilities of the dot, 'L', bar & and cross after finishing F2L? The 'L' seems most common...


----------



## Cubingcubecuber (Jul 2, 2020)

I might be wrong but

Cross=5/30
L=12/30
Bar=8/30
Dot=5/30


----------



## xyzzy (Jul 2, 2020)

vidcapper said:


> I dare say this has been asked many times here before, but I can't seem to ask the right search questions...
> 
> What are the probabilities of the dot, 'L', bar & and cross after finishing F2L? The 'L' seems most common...


There are 2^4 = 16 possible ways of assembling the last layer edges, if we look at only the edge orientation. Only 8 of these 16 ways will be solvable (since you can't have an odd number of bad edges), and if you don't do any LL influencing when solving F2L (but you should), the 8 solvable ways will show up equally frequently. These 8 solvable ways are (imagine "x" is the last layer colour):


```
. x .
x x x
. x .
(cross)

. x .
x x .
. . .
(L)

. x .
. x x
. . .
(L)

. x .
. x .
. x .
(bar)

. . .
x x x
. . .
(bar)

. . .
x x .
. x .
(L)

. . .
. x x
. x .
(L)

. . .
. x .
. . .
(dot)
```

Thus the probabilities are:
Cross: 1/8
Dot: 1/8
Bar: 2/8 = 1/4
L: 4/8 = 1/2



Cubingcubecuber said:


> I might be wrong but
> 
> Cross=5/30
> L=12/30
> ...


Legit curious about how in the world you got those numbers.


----------



## vidcapper (Jul 2, 2020)

xyzzy said:


> There are 2^4 = 16 possible ways of assembling the last layer edges, if we look at only the edge orientation. Only 8 of these 16 ways will be solvable (since you can't have an odd number of bad edges), and if you don't do any LL influencing when solving F2L (but you should), the 8 solvable ways will show up equally frequently. These 8 solvable ways are (imagine "x" is the last layer colour):
> 
> 
> ```
> ...



Thanks for that - now I feel foolish that the calculation was so simple.


----------



## ProStar (Jul 2, 2020)

vidcapper said:


> Thanks for that - now I feel foolish that the calculation was so simple.



He makes everyone feel stupid, don't worry


----------



## ToastyTwurtle (Jul 2, 2020)

What is the chance of getting a last layer skip? Also, has anybody gotten one? Just wondering.


----------



## Jam88 (Jul 2, 2020)

I got LL skip once and also 


ProStar said:


> That's called an LL skip, and the probability is 1 / 15552. Also this is a really widely-posted probability, make sure you do your research before asking a question


----------



## Cubingcubecuber (Jul 2, 2020)

xyzzy said:


> Legit curious about how in the world you got those numbers.


30 ELL cases, 5 cross, 5 dot, 12 L and 8 Bar


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## ProStar (Jul 2, 2020)

ToastyTwurtle said:


> What is the chance of getting a last layer skip? Also, has anybody gotten one? Just wondering.



Search before asking a question. In this case, it was literally 3 posts ago


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## ToastyTwurtle (Jul 3, 2020)

ProStar said:


> Search before asking a question. In this case, it was literally 3 posts ago


okayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy that might be a little hard to depend on during solves.
I have done 1300 recorded solves and i haven't ever gotten one. I thought i was unlucky but i guess im "normally lucky"


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## Nmile7300 (Jul 3, 2020)

ToastyTwurtle said:


> okayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy that might be a little hard to depend on during solves.
> I have done 1300 recorded solves and i haven't ever gotten one. I thought i was unlucky but i guess im "normally lucky"


That is completely unrelated to what @ProStar said and makes no sense.


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## CodingCuber (Jul 3, 2020)

xyzzy said:


> There are 2^4 = 16 possible ways of assembling the last layer edges, if we look at only the edge orientation. Only 8 of these 16 ways will be solvable (since you can't have an odd number of bad edges), and if you don't do any LL influencing when solving F2L (but you should), the 8 solvable ways will show up equally frequently. These 8 solvable ways are (imagine "x" is the last layer colour):
> 
> 
> ```
> ...


But dot case can be avoided pretty much all the time as explained in one of @Owen Morrison 's videos.


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## Spacey10 (Jul 3, 2020)

ToastyTwurtle said:


> okayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy that might be a little hard to depend on during solves.
> I have done 1300 recorded solves and i haven't ever gotten one. I thought i was unlucky but i guess im "normally lucky"


That chances of you already getting one is about 1/15


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## Jam88 (Jul 3, 2020)

What is the chance of getting an EPLL skip with petrus?


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## ProStar (Jul 3, 2020)

Jam88 said:


> What is the chance of getting an EPLL skip with petrus?



The same chance as with any other method. Petrus doesn't influence EP at all, so it's still 1/12. Also this is another well-asked and answered question, please do your research


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## Owen Morrison (Jul 10, 2020)

This has probably already been asked before, but what are the chances of a PLL skip on megaminx?


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## Q-- (Jul 10, 2020)

Owen Morrison said:


> This has probably already been asked before, but what are the chances of a PLL skip on megaminx?


1/720


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## Owen Morrison (Jul 10, 2020)

Q-- said:


> 1/720


Thanks!


What about an OLL skip for megaminx?


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## Q-- (Jul 10, 2020)

Owen Morrison said:


> Thanks!
> 
> 
> What about an OLL skip for megaminx?


1/1296. I tried to link the post where this was answered but I can't put in a link without the message needing moderator approval. The last layer skip chance is 1/933120 if you wanted to know that too


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## Owen Morrison (Jul 10, 2020)

Q-- said:


> 1/1296. I tried to link the post where this was answered but I can't put in a link without the message needing moderator approval. The last layer skip chance is 1/933120 if you wanted to know that too


Thanks again!


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## NeoBridgeburn (Jul 15, 2020)

Sorry if this has already been asked, what are the chances that I consecutively get an OLL skip, then a Pll skip, and then an LL skip?


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## Q-- (Jul 15, 2020)

nairismic said:


> Sorry if this has already been asked, what are the chances that I consecutively get an OLL skip, then a Pll skip, and then an LL skip?


1/216 * 1/72 * 1/15552, or 1/241,864,704. Those are the same odds of getting 2 LL skips back to back.


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## Q-- (Jul 31, 2020)

For square-1 random state scrambles, does every cubeshape have an equal probability? I don’t know if some show up more than others, or if something like square and star cases show up less because they’re symmetrical.


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## xyzzy (Jul 31, 2020)

Q-- said:


> For square-1 random state scrambles, does every cubeshape have an equal probability? I don’t know if some show up more than others, or if something like square and star cases show up less because they’re symmetrical.


The shapes have equal probability (1/3678) if you treat different AUF/ADFs as different shapes.

But if you treat different AUF/ADFs as still being the same shape, then you need to adjust the probability by how many of the AUF/ADFs let you do a slice move. For example, with square-square, there are four combinations of AUF/ADF, so it has a probability 4/3678. (Since square is rotationally symmetric, doing a (3,0) does not produce a different shape, so we don't count that.) Shield-square has 3 choices of AUF and 2 choices of ADF, so the probability is 6/3678. (Note that this is treating shield-square and square-shield as different shapes; if you want to treat _those_ as the same too, the combined probability is 12/3678.) And so on.

This page has a list of the shapes and probabilities:





Avslutad hemsida


Avslutad hemsida




hem.bredband.net


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## vidcapper (Sep 13, 2020)

I know this has probably been answered elsewhere, but how is decided how long a scramble should be? 

3*3 is usually 20 moves, but how do we know that would be enough, rather than 30 moves, or 50, etc? Is it because 20 is God's Number for the 3x3? That number is unknown for bigger cubes though, so that can't be a factor in scramble length for them.


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## Q-- (Sep 13, 2020)

vidcapper said:


> I know this has probably been answered elsewhere, but how is decided how long a scramble should be?
> 
> 3*3 is usually 20 moves, but how do we know that would be enough, rather than 30 moves, or 50, etc? Is it because 20 is God's Number for the 3x3? That number is unknown for bigger cubes though, so that can't be a factor in scramble length for them.


Cube scrambles are always random state if feasible. For 3x3, since gods number is 20, scrambler programs can easily find a solution in 20 moves or less. This takes much less time than finding an optimal solution like 17 or 18 moves, so scrambles can be generated incredibly quickly.

For bigger cubes, random state is a bit harder. You don’t need God’s number to generate random state scrambles, because you can just set the searching range for an upper bound of gods number. This is why we have random states for 4x4 (and I think 5x5). 

For 6x6 and 7x7 random state is possible but it would be so long and so many moves more than the scrambles we use now that it would be impractical, since those take so long to scramble already.


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## PetraPine (Sep 13, 2020)

Lol I've only done like 40 mega solves and got a pll skip im so lucky


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## Jeff Dunham (Sep 23, 2020)

I tried calculating it but I’m not sure if it is correct.

I calculated # possible f2l-oll cases there were (last slot and last layer orientations):

(5*2) * (4*3) * 2^3 * 3^3 / 4 = 6480 combinations. This is why I did this. On the the permutation of the f2l edge counts: 
- 5 ways of permuting the f2l edge, and 2 ways of orienting.
- 4 ways of permuting the f2l corner, and 3 ways of orienting. (We only account for parity once).
- The final 4 edges can be oriented 2^3 ways as the final edge only has 1 way to be oriented.
- The final 4 corners likewise can be oriented 3^3 ways.
- Divide by 4 as the AUF doesn’t matter.

I just want to know if this is correct before calculating the number of WV cases


----------



## GenTheThief (Sep 23, 2020)

Well, if your end goal is to figure out the number of WV cases, you can just take the 8 OCLLs (T, U, L, Pi, H, S, As, O[solved]), account for the AUFs, and then mentally take out a pair. You have 32 cases, but H and O both have symmetry, reducing down to *27 cases* total (I'm pretty sure).


And WV is also only the LS cases where the pair is already formed, so you don't need to work out any permutations of the F2L edge or corner.

And WV also doesn't deal with _any_ permutation, so you don't have to calculate that for any of the LL pieces either.

And AUF _does_ matter because you can't AUF without changing the position of the F2L pair.

And then you also have to account for _symmetry_, because an H case has the same orientation from a y2 away, and you don't want to count that twice.


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## Jeff Dunham (Sep 23, 2020)

GenTheThief said:


> Well, if your end goal is to figure out the number of WV cases, you can just take the 8 OCLLs (T, U, L, Pi, H, S, As, O[solved]), account for the AUFs, and then mentally take out a pair. You have 32 cases, but H and O both have symmetry, reducing down to *27 cases* total (I'm pretty sure).
> 
> 
> And WV is also only the LS cases where the pair is already formed, so you don't need to work out any permutations of the F2L edge or corner.
> ...



Ah. I was trying to find out the number of total possible combinations of last f2l pair and the orientation of the last layer. Then find the number of ways to get a WV case to hence find the probability of getting a WV case for your last f2l case.


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## CodingCuber (Sep 23, 2020)

This is awesome work, Jeff but it should probably go in the probability thread.


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## Jeff Dunham (Sep 23, 2020)

CodingCuber said:


> This is awesome work, Jeff but it should probably go in the probability thread.


Oh I didn't notice there was such a thread. Thanks


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## Jeff Dunham (Sep 23, 2020)

For 3x3, what is the probability of getting a WV case out of all possible OLS cases (every single one of them). *By OLS I mean every case of solving the last f2l slot no matter the relative arrangements of the edge and corner and solving OLL at the same time.


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## Devagio (Sep 23, 2020)

Jeff Dunham said:


> For 3x3, what is the probability of getting a WV case out of all possible OLS cases (every single one of them). *By OLS I mean every case of solving the last f2l slot no matter the relative arrangements of the edge and corner and solving OLL at the same time.


So the two F2L pieces need to be paired in the top layer, which is 1 in (4/5)x(1/15).
Given the f2l edge is in top layer, the other 3 edges need to be oriented, which has probability 1/8
So overall, this is once in 150 solves.


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## Jeff Dunham (Sep 23, 2020)

Devagio said:


> So the two F2L pieces need to be paired in the top layer, which is 1 in (4/5)x(1/15).
> Given the f2l edge is in top layer, the other 3 edges need to be oriented, which has probability 1/8
> So overall, this is once in 150 solves.


Ah. I would've thought it would be slightly more frequent (maybe 1/60 or 1/70). Thanks.


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## Brayden_Speedcuber (Sep 23, 2020)

Here's a pi chart I made of the probability of each PLL.


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## Cuberstache (Sep 23, 2020)

Brayden_Speedcuber said:


> Here's a pi chart I made of the probability of each PLL.


How did you get these numbers? Most of them seem wrong.


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## Brayden_Speedcuber (Sep 23, 2020)

CuberStache said:


> How did you get these numbers? Most of them seem wrong.


I forget where I found them, but awhile back when I was learning PLL one PDF had the chances to get each PLL. For example I think H had 1/72 solves.


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## Cuberstache (Sep 23, 2020)

Brayden_Speedcuber said:


> I forget where I found them, but awhile back when I was learning PLL one PDF had the chances to get each PLL. For example I think H had 1/72 solves.


Well, you remembered some wrong. Most of them should be 4/72, not 5/72 or 5.5/72 (???)


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## xyzzy (Sep 23, 2020)

CuberStache said:


> Well, you remembered some wrong. Most of them should be 4/72, not 5/72 or 5.5/72 (???)


Gonna guess they're rounded to whole percents. 4/72 = 5.55555…% (Which is a weird thing to do, but hey.)


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## Cuberstache (Sep 23, 2020)

xyzzy said:


> Gonna guess they're rounded to whole percents. 4/72 = 5.55555…% (Which is a weird thing to do, but hey.)


Oh, that makes sense. They all added up to 93, so I wasn't really sure what's going on.


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## PetraPine (Sep 23, 2020)

Chance of Oll skip with 2 Gen Last side(ZZ right block or petrus last block)
((R U))
Also chance of Oll skip with 2 Gen Last side for megaminx?


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## WarriorCatCuber (Sep 23, 2020)

ObscureCuber said:


> Chance of Oll skip with 2 Gen Last side(ZZ right block or petrus last block)
> ((R U))
> Also chance of Oll skip with 2 Gen Last side for megaminx?


The chance of an OLL skip with 2-gen on 3x3 is 1/27


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## PetraPine (Sep 23, 2020)

WarriorCatCuber said:


> The chance of an OLL skip with 2-gen on 3x3 is 1/27


also if you get sune/antisune and recog it you could cancel oll while inserting last pair which is nice


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## YaleZ1023 (Sep 24, 2020)

riffz said:


> I just have to say this:
> 
> WOLFRAM ALPHA IS AMAZING
> 
> ...





DaijoCube said:


> After solving centers then edges, you most likely can't keep track of the scramble. Scramble was legit.


Hello, just thought somebody saying X said X said etc. looked cool.


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## Spacey10 (Sep 24, 2020)

YaleZ1023 said:


> Hello, just thought somebody saying X said X said etc. looked cool.


Woaj, near world record bump, 10 years!


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## Jeff Dunham (Sep 25, 2020)

I asked in the probability thread the probability of having a WV case at last slot , and the answer turned out to be about 1/150.
But in practice, since a lot of f2l algs involve an (R U' R') insert at the end, I'd imagine that WV would actually appear more than 1/150 of the time.

So when actually solving, does WV appear a decent amount in cases where it is justified to use it? Like at least 1/50 chance?


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## I'm A Cuber (Sep 25, 2020)

Jeff Dunham said:


> I asked in the probability thread the probability of having a WV case at last slot , and the answer turned out to be about 1/150.
> But in practice, since a lot of f2l algs involve an (R U' R') insert at the end, I'd imagine that WV would actually appear more than 1/150 of the time.
> 
> So when actually solving, does WV appear a decent amount in cases where it is justified to use it? Like at least 1/50 chance?


You will have a wv case 1/8 times
Edit: but you can influence eo to make it slightly more common


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## RadicalMacaroni (Oct 5, 2020)

Anybody know the probability of a 2x2 scramble with no bars already made?


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## SpeedyCube (Oct 6, 2020)

What are the chances of skipping PLL on a 2x2? I just got a 9.65 seconds for the first time, and part of it was no PLL except for turning the top layer one turn, which I don’t personally count as PLL. And, regardless of wether I mix the cube or the computer scrambles, I end up skipping the PLL quite often, I would think about 1/10 or more of the time. Is there a mathematical way to determine this?


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## Q-- (Oct 6, 2020)

SpeedyCube said:


> What are the chances of skipping PLL on a 2x2? I just got a 9.65 seconds for the first time, and part of it was no PLL except for turning the top layer one turn, which I don’t personally count as PLL. And, regardless of wether I mix the cube or the computer scrambles, I end up skipping the PLL quite often, I would think about 1/10 or more of the time. Is there a mathematical way to determine this?



Yep! It occurs 1/6 of the time. Here’s the math:

The corners when oriented can be arranged in 24 ways (4 places for the first corner times 3 places for the second times 2 places for the third and 1 for the last.) Considering all 4 possible AUFs brings it down to 6 (24/4). 
Out of six possible arrangements, 4 are Adjacent, 1 is diagonal, and 1 is solved. Therefore the odds of a PLL skip are 1/6.


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## SpeedyCube (Oct 6, 2020)

Ooohhhh, that makes sense, thanks! Yeah, that’s about what I’m seeing. Most of the time I have to do the adjacent perm, and the other two cases (diagonal and already solved) seem to be about equal. Now I know why!


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## RiceMan_ (Oct 6, 2020)

What are the chances of someone getting a world record average on 3x3 with the ZZ method? I think there is a 1/3 chances if people start using the method.


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## Nir1213 (Oct 6, 2020)

RiceMan_ said:


> What are the chances of someone getting a world record average on 3x3 with the ZZ method? I think there is a 1/3 chances if people start using the method.


yup but its gonna be hard


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## RiceMan_ (Oct 6, 2020)

Nir1213 said:


> yup but its gonna be hard


why?


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## Nir1213 (Oct 6, 2020)

RiceMan_ said:


> why?


cant just be 1/3 that would be too easy
maybe something like 1/200


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## RiceMan_ (Oct 6, 2020)

Nir1213 said:


> cant just be 1/3 that would be too easy
> maybe something like 1/200


I agree


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## qwr (Nov 3, 2020)

anyone have distributions for avg moves to make an ortega face (color neutral) and average cross moves (for white cross)


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## Tito (Nov 7, 2020)

what is the probability of getting a winter variation at a solve


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## Cubing Forever (Nov 7, 2020)

Tito said:


> what is the probability of getting a winter variation at a solve


1/8???


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## Cubing Forever (Nov 7, 2020)

Probabilities of all PLLs(Counting all cases including mirrors):
Ua: 1/18
Ub: 1/18
H: 1/72
Z: 1/36
Aa: 1/18
Ab: 1/18
E: 1/36
Ja: 1/18
Jb: 1/18
T: 1/18
F: 1/18
Ra: 1/18
Rb: 1/18
V: 1/18
Y: 1/18
Na: 1/72
Nb: 1/72
Ga: 1/18
Gb: 1/18
Gc: 1/18
Gd: 1/18

Either of the U Perms: 1/9
Either of the A perms: 1/9
Either of the J perms: 1/9
Either of the N perms: 1/36
Either of the 4 G perms: 2/9

(All symmetry based. Hope my math and grammar is correct. Coming from a total Math nerd here)

A question out of curiosity:
What is the probability of getting a Superflip as a scramble ? (either official or at home)

Mods please merge these. I double posted by accident.


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## Devagio (Nov 7, 2020)

Cubing Forever said:


> A question out of curiosity:
> What is the probability of getting a Superflip as a scramble ? (either official or at home)
> 
> Mods please merge these. I double posted by accident.


For a random state scrambler, 1 in 43 quintillion.


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## DNF_Cuber (Nov 21, 2020)

what is the likelihood of having a pre built center on 5x5?Also, what are the odds of a white or yellow center being pre built?(for yau)


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## 2018AMSB02 (Nov 21, 2020)

DNF_Cuber said:


> what is the likelihood of having a pre built center on 5x5?Also, what are the odds of a white or yellow center being pre built?(for yau)



Well, assuming it is in a random state, let’s do the math.

There are 9 center pieces on each center. 
Each one can be one of each center. The center one can be discounted because it can be the base center. The first corner one we do can be determined like this:
4 corner pieces per center.
6 colors. 6x4=24. This means that there are 24 possible pieces that could be there, and 4 of them are right, leaving a 1/6 chance it will be right. The next corner has 3/24 or 1/8 because there is one less correct center. Then 1/12 and finally 1/24. Multiply these and you get 1/13824 that all corners and the center are right. Then for edges, you can do the same thing because there are the same number as corners. So 1/13824 squared gets you 1/191102976 chance that a particular center will be solved, but for any of the 6 centers that goes up to:

1/31850496 chance that a center on the cube will be solved. 

For yellow/white it would have to be 3X as unlikely, 1/95551488, because yellow/white make up a third of the 6 centers.

These numbers seem too high, but I think I did everything right, so would someone check it?


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## DNF_Cuber (Nov 21, 2020)

maybe that is for just one center you calculated so it is 1/5308416
also I think it should be 3/23 for the second one,2/22 for the third and so on since you are eliminating one piece


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## RiceMan_ (Nov 22, 2020)

what is the probability of getting a sub-20 moves scrable


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## DNF_Cuber (Nov 22, 2020)

very likely, well over 99% I believe.


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## Nir1213 (Nov 22, 2020)

chances of getting a ll skip no auf, combined with cross being a one mover, combined with all the pairs except one inserted already in f2l?

ok that was a joke but seriously, im thinking is there a difference in probability with ll skips having auf, or no auf. I think that the no auf ones are more rare.


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## BenChristman1 (Nov 22, 2020)

RiceMan_ said:


> what is the probability of getting a sub-20 moves scrable


It depends on the scrambler you use.


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## DNF_Cuber (Nov 22, 2020)

read the gods number article on ruwix.com "This number may seem low, but theoretically it should be even lower. Only around 490,000,000 combinations require the full 20 moves to be solved. Although 490 million is a huge number, it is only a fraction of the 43 quintillion possible combinations (0.0000011328955% to be precise). " so .0000011 percent of positions take 20 moves to solve


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## Nir1213 (Nov 22, 2020)

DNF_Cuber said:


> read the gods number article on ruwix.com "This number may seem low, but theoretically it should be even lower. Only around 490,000,000 combinations require the full 20 moves to be solved. Although 490 million is a huge number, it is only a fraction of the 43 quintillion possible combinations (0.0000011328955% to be precise). " so .0000011 percent of positions take 20 moves to solve


that is a 1.1 in a million chance.

Not bad, if you think about it. The chances of being struck by lightning is 1 in a million, which means that the chance of the 20 move scramble would be less rarer.


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## Nir1213 (Nov 22, 2020)

also i have another one: What is the possibility of getting the super-flip on a 3x3?


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## xyzzy (Nov 22, 2020)

DNF_Cuber said:


> read the gods number article on ruwix.com "This number may seem low, but theoretically it should be even lower. Only around 490,000,000 combinations require the full 20 moves to be solved. Although 490 million is a huge number, it is only a fraction of the 43 quintillion possible combinations (0.0000011328955% to be precise). " so .0000011 percent of positions take 20 moves to solve


Please don't cite Ruwix as a source. The accuracy of the pages there is somewhat questionable.


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## 2018AMSB02 (Nov 22, 2020)

Nir1213 said:


> also i have another one: What is the possibility of getting the super-flip on a 3x3?



1. Someone asked that just a few post behind you...
2. 1/45 quintillion, as with any particular state


----------



## Kit Clement (Nov 22, 2020)

DNF_Cuber said:


> read the gods number article on ruwix.com "This number may seem low, but theoretically it should be even lower. Only around 490,000,000 combinations require the full 20 moves to be solved. Although 490 million is a huge number, it is only a fraction of the 43 quintillion possible combinations (0.0000011328955% to be precise). " so .0000011 percent of positions take 20 moves to solve



On top of Ruwix being a terrible source, the question was about the length of the scramble, not the optimal solution to that scramble's state.


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## Nir1213 (Nov 22, 2020)

PingPongCuber said:


> 1. Someone asked that just a few post behind you...
> 2. 1/45 quintillion, as with any particular state


hmm ok but what about a probability of a cube having 2 or more edges flipped while the rest of the cube is solved?


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## Nir1213 (Nov 22, 2020)

whats also the probability of getting a 2 mover in a 3x3?

Very rare, i know, but its still allowed in WCA regulations and rules.


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## 2018AMSB02 (Nov 22, 2020)

Nir1213 said:


> whats also the probability of getting a 2 mover in a 3x3?
> 
> Very rare, i know, but its still allowed in WCA regulations and rules.



This ones pretty simple. There are approximately 45 quintillion possible combinations on the cube. 1 is solved. From there there are 6 sides, and on each side you can do 3 types of moves. This means that there are 18 states where it is 1 move from solved. Then on each of those 18 states there are 18 more moves you could do. But, 3 of those moves (the ones on the same face) would result in it being still 1 move away, so in reality 15 more moves. 18 x 15 is 270, so the odds of there being a 2 move scramble (assuming it is a random state scrambler) is 270/45000000000000000000


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## BenChristman1 (Nov 22, 2020)

Nir1213 said:


> hmm ok but what about a probability of a cube having 2 or more edges flipped while the rest of the cube is solved?





Nir1213 said:


> whats also the probability of getting a 2 mover in a 3x3?
> 
> Very rare, i know, but its still allowed in WCA regulations and rules.


Both of these are dumb questions.


----------



## Nir1213 (Nov 22, 2020)

PingPongCuber said:


> This ones pretty simple. There are approximately 45 quintillion possible combinations on the cube. 1 is solved. From there there are 6 sides, and on each side you can do 3 types of moves. This means that there are 18 states where it is 1 move from solved. Then on each of those 18 states there are 18 more moves you could do. But, 3 of those moves (the ones on the same face) would result in it being still 1 move away, so in reality 15 more moves. 18 x 15 is 270, so the odds of there being a 2 move scramble (assuming it is a random state scrambler) is 270/45000000000000000000


thanks!



BenChristman1 said:


> Both of these are dumb questions.


if its dumb you dont have to post and say about it.

also, based on bens profile, since his cube is 9x9, can any odd numbered cube be able to have a super flip, or its the same for even numbered?


----------



## 2018AMSB02 (Nov 22, 2020)

Nir1213 said:


> also, based on bens profile, since his cube is 9x9, can any odd numbered cube be able to have a super flip, or its the same for even numbered?



I mean any google search or video would tell you the same thing but yeah. My 17x17 is super flipped


----------



## Seth1448 (Nov 23, 2020)

What’s the probability of a last layer skip?


----------



## BenChristman1 (Nov 23, 2020)

Seth1448 said:


> What’s the probability of a last layer skip?





ProStar said:


> The probability is 1 / 15552. Also this is a really widely-posted probability, make sure you do your research before asking a question


----------



## Deleted member 54663 (Nov 23, 2020)

What is the probability of having a 2x2x2 block in 3x3 solved?


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## Devagio (Nov 24, 2020)

JP cubing said:


> What is the probability of having a 2x2x2 block in 3x3 solved?


Roughly 8 x 1/24 x 1/24 x 1/22 x 1/20; or 1 in 31k


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## Athefre (Nov 24, 2020)

I mentioned an idea before on Discord. I think a well-developed probability page on the wiki would be very useful. We could have all of the ones that are commonly asked and any others. The page could be structured by categories like steps, methods, states, and others. Then the main post of this topic can be edited to include a link to the wiki page. If someone still asks a common question, then they can be linked to the page.

For anyone that enjoys the probability subject, this would be a fun project and you would be helping the community.


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## erdish (Dec 23, 2020)

Devagio said:


> Roughly 8 x 1/24 x 1/24 x 1/22 x 1/20; or 1 in 31k



By simulation I get about1 in 10,000, I guess due to overlaps?


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## Deleted member 55877 (Dec 23, 2020)

Devagio said:


> Roughly 8 x 1/24 x 1/24 x 1/22 x 1/20; or 1 in 31k


Wow I got a scramble where a 2x2 block was solved on my 3x3 a month ago...


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## xyzzy (Dec 23, 2020)

erdish said:


> By simulation I get about1 in 10,000, I guess due to overlaps?


How large is your simulation?


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## erdish (Dec 23, 2020)

xyzzy said:


> How large is your simulation?



30 million scrambles -- but turns out there was a bug (bad shuffle). Now it reports 1 in 31,380, consistent with Devagio's calc within the margin of error. Sorry!


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## erdish (Jan 6, 2021)

Of the 43,252,003,274,489,856,000 permutations of the 3x3 cube, how many have the maximal cycle length 1260?

I have an estimate, just generating a random scramble/permutation I get about a 1 in 800 chance of getting cycle length 1260. So maybe around 5.4*10^16?

But is there a way to calculate the exact number?


----------



## Devagio (Jan 6, 2021)

erdish said:


> Of the 43,252,003,274,489,856,000 permutations of the 3x3 cube, how many have the maximal cycle length 1260?
> 
> I have an estimate, just generating a random scramble/permutation I get about a 1 in 800 chance of getting cycle length 1260. So maybe around 5.4*10^16?
> 
> But is there a way to calculate the exact number?


I calculated by hand to get ~8.58*10^15; this is off by a factor of 6 from your estimate.

What I did basically was 1260=2x2x3x3x5x7
The only way to have a 1260 cycle on a cube is if:
You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation.
(To prove that this is the only way is pretty tedious but straightforward, so I’m not including that here)

So the answer is (1/2) * (8c5 * 4!) * (3c3 * 2!) * (12c7 * 6!) * (5c2) * (3c2) * (18 * 3^4) * (12 * 2^7)


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## abunickabhi (Jan 7, 2021)

The probability that a 5-style alg repeats in the next scramble is 3/126,720, which is 0.0023%. Not bad right, E' L U S2 R E2 R' E S2 U' L'.


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## erdish (Jan 8, 2021)

Devagio said:


> I calculated by hand to get ~8.58*10^15; this is off by a factor of 6 from your estimate.
> 
> What I did basically was 1260=2x2x3x3x5x7
> The only way to have a 1260 cycle on a cube is if:
> ...



Nice, thanks! I like the compact notation. Still I wonder why it's pretty far off the estimate based on throwing the dice. There is another possibility re the possible set of cycle lengths, no?--the case where 1 edge remains fixed? FWIW, here are some stats on 1000 random permutations (hex, so that E=14 and F=15):

1000 random permutations of cycle length 1260
{edge cycle lengths} {corner cycle lengths} N
*{1 2 2 4 4 E E E E E E E} {9 9 9 F F F F F} 343*
*{1 4 4 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 169
{2 2 2 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 326*
*{2 4 4 4 4 E E E E E E E} {9 9 9 F F F F F} 162*


----------



## Devagio (Jan 8, 2021)

erdish said:


> Nice, thanks! I like the compact notation. Still I wonder why it's pretty far off the estimate based on throwing the dice. There is another possibility re the possible set of cycle lengths, no?--the case where 1 edge remains fixed? FWIW, here are some stats on 1000 random permutations (hex, so that E=14 and F=15):
> 
> 1000 random permutations of cycle length 1260
> {edge cycle lengths} {corner cycle lengths} N
> ...


Yep, all of these permutations are of the kind that I mentioned. (At least One two-Cycle in edges of length 4, Exactly one three-cycle in corners of length 9; cycles of size 1,2,2,7 in edges, And of size 3,5 in corners); so that part agrees in both theory and simulation. The mutual ratios of these are supposed to be 1/3, 1/6, 1/3, 1/6; so that seems correct too.
Turns out I made an error in plugging in the expression into a calculator; I re-did it on my phone calculator just now to get 5.15e16, which is quite close to your estimate.
Just try calculating the numerical value of my expression; because the expression seems correct.


----------



## erdish (Jan 8, 2021)

Devagio said:


> Turns out I made an error in plugging in the expression into a calculator; I re-did it on my phone calculator just now to get 5.15e16, which is quite close to your estimate.
> Just try calculating the numerical value of my expression; because the expression seems correct.



Terrific!

Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE.

So the answer is 51,490,480,088,678,400.

A question I would have is are the four basic types seen in the simulation the only ones possible? They are certainly the vast majority.


----------



## Devagio (Jan 8, 2021)

erdish said:


> Terrific!
> 
> Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE.
> 
> ...


Yep, they’re the only ones, and as mentioned earlier they’re a part of the single family that I mentioned; it’s actually quite easy to prove that.
Try to get all the factors of 1260 as orientation and permutation cycles, while keeping the number of swaps even, and keeping CO and EO possible.


----------



## erdish (Jan 9, 2021)

Devagio said:


> Yep, they’re the only ones, and as mentioned earlier they’re a part of the single family that I mentioned;



For example, what about this set of cycles which happens:

*ORBITS {1 2 2 4 4 14 14 14 14 14 14 14} {9 9 9 15 15 15 15 15 15}
CYCLES {1 2 2 2 2 7 7 7 7 7 7 7} {3 3 3 5 5 5 5 5 5}*

That would be four 2-cycles of edges (two of which have a net flip) and not fit the prescription?:

"The only way to have a 1260 cycle on a cube is if: You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation."


----------



## Devagio (Jan 10, 2021)

erdish said:


> For example, what about this set of cycles which happens:
> 
> *ORBITS {1 2 2 4 4 14 14 14 14 14 14 14} {9 9 9 15 15 15 15 15 15}
> CYCLES {1 2 2 2 2 7 7 7 7 7 7 7} {3 3 3 5 5 5 5 5 5}*
> ...


This is two 2-cycles of edges (one of which as a net flip). Consequently there are 4 edges that are a part of 2-cycles. [Just like there aren't seven 7-cycles of edges, there is only one, but seven edges are a part of it]


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## erdish (Jan 12, 2021)

Devagio said:


> This is two 2-cycles of edges (one of which as a net flip). Consequently there are 4 edges that are a part of 2-cycles. [Just like there aren't seven 7-cycles of edges, there is only one, but seven edges are a part of it]



Ah, thanks. Taking it a step further, your result means that one out of 840 permutations has cycle length N=1260. Since each generates 1260 distinct permutations before repeating, it's possible that the set of all permutations generated by all N=1260 cases covers the entire range (~43 quintillion). Is that the case?


----------



## Devagio (Jan 12, 2021)

erdish said:


> Ah, thanks. Taking it a step further, your result means that one out of 840 permutations has cycle length N=1260. Since each generates 1260 distinct permutations before repeating, it's possible that the set of all permutations generated by all N=1260 cases covers the entire range (~43 quintillion). Is that the case?


No. A simple counter example is T-perm. Since it has a single 2cycle of corners (parity), it cannot be generated by a set of moves that doesn’t produce parity applied any number of times.


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## erdish (Jan 14, 2021)

Devagio said:


> No. A simple counter example is T-perm. Since it has a single 2cycle of corners (parity), it cannot be generated by a set of moves that doesn’t produce parity applied any number of times.


Come to think of it, the set of 1260 permutations generated from each will have a fixed distribution of cycle lengths that doesn't cover all possible cycle lengths. E.g., none will have cycle length 8 ,11, 16, 22, 24, 33... So any perm with such a cycle length would be another counterexample.


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## mencarikebenaran (Jan 14, 2021)

Probability getting "+ OLL"?


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## mencarikebenaran (Jan 14, 2021)

Got 2 LL skip today.
No WV, No VLS, No AUF.


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## Scollier (Jan 14, 2021)

weruoaszxcvnm said:


> Got 2 LL skip today.
> No WV, No VLS, No AUF.



According to my calculations, the probability of getting two LL skips is 1 in 31,104.

That makes me kind of skeptical.


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## mencarikebenaran (Jan 14, 2021)

Scollier said:


> According to my calculations, the probability of getting two LL skips is 1 in 31,104.
> 
> That makes me kind of skeptical.


I can make a reconstruction .
But u need to wait because i forget what i did on 1st solve


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## ProStar (Jan 14, 2021)

Scollier said:


> According to my calculations, the probability of getting two LL skips is 1 in 31,104.
> 
> That makes me kind of skeptical.



Is that the probability of getting 2 LL skips in a row? Or simply 2 LL skips in a day?


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## ProStar (Jan 14, 2021)

weruoaszxcvnm said:


> Probability getting "+ OLL"?



If you mean a cross OLL, then iirc the probability is 1/8 if you only use the standard R U R'/R U' R' inserts


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## Scollier (Jan 14, 2021)

ProStar said:


> Is that the probability of getting 2 LL skips in a row? Or simply 2 LL skips in a day?



IDK I'm terrible at math...


----------



## DNF_Cuber (Jan 14, 2021)

Scollier said:


> IDK I'm terrible at math...


that looks like you did 1 LL skipX2, which IDK what that calculates. 2 LL skips in a row would be 15552 ^2 (I think there are 5 digits actually)


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## ProStar (Jan 14, 2021)

I haven't taken any classes for probability calculation, so I may have this wrong, but I think that the chance of getting an LL skip without AUF would be 1/15552 * 4 = *1/62,208*. That would make getting 2 of them in a row without AUFs 1/62208^2 = *3,896,835,264*. If you discount the fact that there was no AUF on both of them then it should be 1/15552^2 = *1/241,864,704*.

This is getting 2 of them in a row though, not getting two of them in a single session, which would dramatically increase the chances for this to happen, although you'd still be extremely lucky


Again, there's a decent chance that this math is totally wrong


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## vidcapper (Jan 17, 2021)

Had an unusual occurrence today... on a 5x5 solve, the last 2 centres were each completely encircling the other colour's centre! Wonder what the odds of that are?


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## Deleted member 55877 (Jan 17, 2021)

vidcapper said:


> Had an unusual occurrence today... on a 5x5 solve, the last 2 centres were each completely encircling the other colour's centre! Wonder what the odds of that are?


i'm not sure about this answer, but i think it's the same probability that both centers end are solved.

so i think (1/2) ^ 8 = 1/256 chance


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## vidcapper (Jan 17, 2021)

Alex Davison said:


> i'm not sure about this answer, but i think it's the same probability that both centers end are solved.
> 
> so i think (1/2) ^ 8 = 1/256 chance



Thanks, I should have figured that out myself. 

On a 7x7, I guess it would be (1/2)^24 then, 1 in 16777216. On a 19x19 ~4.97x10^86.


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## xyzzy (Jan 17, 2021)

Alex Davison said:


> i'm not sure about this answer, but i think it's the same probability that both centers end are solved.
> 
> so i think (1/2) ^ 8 = 1/256 chance


Same probability yes, but your calculation is wrong.

To calculate it correctly, it's easiest to treat the different orbits of centre pieces individually. Just to make it concrete, let's say the last two centres' colours are white and black. Among the t-centres, there are 4 white pieces and 4 black pieces, so there are 8! / (4! × 4!) = 70 ways of arranging them, each equally likely, and hence there's a 1/70 chance that the white pieces are on the black face and the black pieces are on the white face. Likewise for the x-centres, so there's also a 1/70 chance. These probabilities are independent, so together there's a 1/70 × 1/70 = 1/4900 chance of getting all the centres swapped.

This extends to bigger cubes readily: on a 7×7×7, there are six orbits, so the probability is 1/70^6. On a 9×9×9, there are twelve orbits, so the probability is 1/70^12. And so on.


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## carcass (Jan 20, 2021)

i just got 3 v perms with U2 aufs in a row, what are the odds of that?


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## Scollier (Feb 2, 2021)

I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.

So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.

And for finding the number of permutations in x amount of edges and x amount of corners, I believe you just multiply both of the number of permutations together?

Thanks.

And Note: This is finding the number of solvable permutations of course, e.g., not finding permutations of flipped pieces as well.

And also, I should have clarified, *This is for 3x3.*


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## Jam88 (Feb 2, 2021)

Scollier said:


> I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.
> 
> So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.
> 
> ...


surely there would be less than multiplied together? bc parities etc. idk tho
xyzzy can prob provide an answer


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## Scollier (Feb 2, 2021)

Jam88 said:


> surely there would be less than multiplied together? bc parities etc. idk tho
> xyzzy can prob provide an answer



Sorry, I should have clarified, this is for 3x3 only.


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## Kit Clement (Feb 2, 2021)

Scollier said:


> I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.
> 
> So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.
> 
> ...



The total number of permuations of a 3x3x3 is:

12! - ways to permute edges
2^12 - ways to orient edges
8! - ways to permute corners
3^8 - ways to orient corners

However, we have to divide by:

2 - since the last edge's orientation is determined by the previous 11
3 - since the last corner's orientation is determined by the previous 7
2 - since the permutation parity of the puzzle is always even (i.e. can't swap only 2 corners)

This gives:

12! * 2^(12-1) * 8! * 3^(8-1) / 2

So if you want to do this for a subset of x edges and y corners, fixing the rest, this function could be used to find the number of permutations of those pieces:

f(x, y) = x! * 2^(x-1) * y! * 3^(y - 1) / 2

Note that this is not the same as cases for LL, as counting specific cases takes symmetries and AUF into account. So if you're looking for literal permutations this will work, but it gets trickier if your purpose is for counting specific cases for a given alg set.


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## Deleted member 55877 (Feb 11, 2021)

Sorry if this was asked before, but what's the chances of L2C skip on 5x5? I just got one lol


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## xyzzy (Feb 11, 2021)

Alex Davison said:


> Sorry if this was asked before, but what's the chances of L2C skip on 5x5? I just got one lol


Previous page!









Probability Thread


Terrific! Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE. So the answer is 51,490,480,088,678,400. A...




www.speedsolving.com


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## Deleted member 55877 (Feb 11, 2021)

turns out i actually had answered the very question i am asking (attempted to, at any rate) ... brain fart


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## rubik2005 (Feb 11, 2021)

I don't think there's really a way to figure this out, but for some reason a lot of the misscrambles in comps result in really fast times like Max's OH solve at Nats. If someone who isn't really world-class has a misscramble, technically nobody would know about it since no one would notice an average solve. So that chances that someone like Max or Felik's 1: gets a misscramble cubed, and 2 solves it in WR or extremely close to it would be astronomical right? But it has happened multiple times, so I maybe wrong scrambles happen more often than we think?


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## xyzzy (Feb 11, 2021)

rubik2005 said:


> So that chances that someone like Max or Felik's 1: gets a misscramble cubed, and 2 solves it in WR or extremely close to it would be astronomical right?


While 1 is moderately unlikely in a well-run competition, 2 isn't all that unlikely for those two people! There's also some reason to believe that misscrambles caused by misreading a single move in a scramble sequence (as opposed to misscrambles caused by accidental move done in transit, misscrambles with 2 or more misread moves, etc.) are slightly biased towards being easier than typical.

Also:


rubik2005 said:


> *someone like* Max or Felik's
> […]
> WR *or extremely close to it*


Introduce enough fudge factors and you can get something that _sounds like_ it should be unlikely, but in reality is very likely. This is something you have to be very wary of when doing statistics.


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## Kit Clement (Feb 11, 2021)

Yeah, this is the classic blade of grass paradox. In a pre-COVID world, world class solvers were each often doing upwards of 100 solves across all events every month. Additionally, there are hundreds of world class cubers whose solves are closely followed by the community. So it may seem like a one-in-a-million occurrence to happen at that very point in time, but with the frequency that official solves happen at a high level, it's almost certain for it to happen in a way that is noticed eventually.


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## thelargeman2048 (Apr 18, 2021)

does anyone know the odds of skipping eoll when using coll?


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## RyanSoh (Apr 18, 2021)

If I am not wrong it should be 1 out of 12 solves.


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## PetrusQuber (Apr 18, 2021)

I don’t see the point in learning COLL unless you’re doing EO beforehand though, unless you’re really dedicated


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## thelargeman2048 (Apr 18, 2021)

coll is 42algs so instead of learning full oll i can learn EOLL which is 3 algs, COLL at 42 and EPLL at 4 algs vs OLL with 57 alone. also coll has about an 8 move count average. im also using pure eoll which changes nothing except corners so recognition wont even be that bad.

theres probably somthing im missing but i think COLL is better than oll in every way.


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## OreKehStrah (Apr 18, 2021)

thelargeman2048 said:


> coll is 42algs so instead of learning full oll i can learn EOLL which is 3 algs, COLL at 42 and EPLL at 4 algs vs OLL with 57 alone. also coll has about an 8 move count average. im also using pure eoll which changes nothing except corners so recognition wont even be that bad.
> 
> theres probably somthing im missing but i think COLL is better than oll in every way.


A lot of COLL is slower than OLL on average. Why not just do CFCE? It’s what I use.


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## Pyjam (Apr 18, 2021)

You may complete your set of COLL with easy cases of ZBLL.
So, if you know 3 algs per COLL case, the odds for EPLL skip are 1/4.


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## Athefre (Apr 18, 2021)

Pyjam said:


> You may complete your set of COLL with easy cases of ZBLL.
> So, if you know 3 algs per COLL case, the odds for EPLL skip are 1/4.



If going this direction, it is only two algs per COLL case that is required to always have that skip chance. Ua, Ub, or a skip.


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## LukasCubes (Apr 18, 2021)

thelargeman2048 said:


> does anyone know the odds of skipping eoll when using coll?


if you get a COLL for every solve, you get a 1/12 chance of a PLL skip or 8.3333333333333333333333333333333333333333333333% chance (8 1/3%)

I know because I know full COLL and use ZB as my main method while also knowing almost a quarter of full ZBLL even though I am currently in a break right now.


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## xyzzy (Apr 18, 2021)

thelargeman2048 said:


> also coll has about an 8 move count average.


It's more like 9.8 moves with move-optimal algs, 11-ish with speed-optimal algs.



thelargeman2048 said:


> theres probably somthing im missing but i think COLL is better than oll in every way.


It's worse in terms of overall move count.

6 moves for EO + 10 moves for COLL + 9 moves for EPLL = 25 moves (add 2 moves if you're using corner-preserving EO algs)
9 moves for OLL + 11 moves for PLL = 20 moves

It's better to just do the standard OLL/PLL (78 algs, ~20.5 moves), 2-look OLL with full PLL (31 algs, ~23 moves), or CLL/ELL (71 algs, ~20.5 moves). The logic is simple: less steps means less moves (usually).

(Move counts reported above are for FTM-optimal, which is a good (but imperfect) proxy for execution time with speed-optimised algs. Execution time is highly subjective, but optimal move counts aren't.)

---



thelargeman2048 said:


> does anyone know the odds of skipping eoll when using coll?


Also two people upthread answered this for EPLL, but the probability of skipping EO when using COLL is the same as if you're _not_ using COLL, i.e. 1/8 assuming no edge control. The cube has no brain and can't possibly know that you're about to use a COLL alg, so EO can't possibly depend on that.


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## thelargeman2048 (Apr 19, 2021)

thanks guys


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## cuber314159 (Jun 18, 2021)

If we assume I have done 100000 3x3x3 solves with full CFOP and no last layer influencing during F2L, what are the chances I have seen all the last layer cases?

I think it's probably easiest to do an estimate where we assume that the probability of getting each case is constant but obviously that isn't really the case


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## brododragon (Jun 18, 2021)

You could use HARCS to do 100000 solves and see what the last layer is for them


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## Q-- (Jun 19, 2021)

cuber314159 said:


> If we assume I have done 100000 3x3x3 solves with full CFOP and no last layer influencing during F2L, what are the chances I have seen all the last layer cases?
> 
> I think it's probably easiest to do an estimate where we assume that the probability of getting each case is constant but obviously that isn't really the case


TLDR, 99.999997%
I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:

Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).

The probability of NOT getting a specific case after 100000 solves is the probability of not getting that case, 3887/3888, times itself 100000 times.

You take the complement (1-answer) of this to find the chance of actually getting that case at some point in the 100000 solves, and multiply that by itself 3888 times (the number of cases) to get the chance of finding all of them in those solves, which is virtually 100%.

It seems odd that the odds are so high, but 100000 is a lot of solves so it makes sense. If I knew how many cases had each kind of symmetry I could give a real answer, and the odds would be a bit lower. This was a fun probability problem, so thank you.


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## cuber314159 (Jun 19, 2021)

Q-- said:


> TLDR, 99.99999997%
> I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:
> 
> Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).
> ...


Interesting, I was expecting it to be a lot lower actually, like how the odds of having 365 people in a room all having different birthdays are astronomically low, but I suppose you are probably right.


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## xyzzy (Jun 19, 2021)

Q-- said:


> You take the reciprocal of this


Complement (\( x\mapsto1-x \)), not reciprocal (\( x\mapsto1/x \)).

Haven't thought too carefully about it but your calculation seems correct. The final answer has two too many "9"s in it, though.

```
>>> (1-(3887/3888)**100000)**3888
0.9999999738084874
```
(the last four digits printed are wrong due to accumulated rounding error)


----------



## kubesolver (Jun 19, 2021)

xyzzy said:


> (1-(3887/3888)**100000)**3888


That's not fully correct calculation because the events aren't independent. 
If you replace 100000 in the formula with 1 you will get positive result.


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## xyzzy (Jun 19, 2021)

kubesolver said:


> That's not fully correct calculation because the events aren't independent.
> If you replace 100000 in the formula with 1 you will get positive result.


Right, yes, that is just an approximation due to non-independence. That part of the approximation gets more accurate as the number increases. (I'm not sure if there's a good way of calculating the exact answer without resorting to a very messy inclusion-exclusion calculation.)


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## kubesolver (Jun 19, 2021)

one way to calculate is that expected number of unreached positions is (due to linearity of expectation)
3888 * ((1-(3887/3888)**100000)) = 0.00000067366
Which is the upper bound on probability of missing one.


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## Q-- (Jun 19, 2021)

xyzzy said:


> Complement (\( x\mapsto1-x \)), not reciprocal (\( x\mapsto1/x \)).
> 
> Haven't thought too carefully about it but your calculation seems correct. The final answer has two too many "9"s in it, though.
> 
> ...


Yeah, that’s what I meant and used in the calculation, good catch.


----------



## ProStar (Jun 19, 2021)

Q-- said:


> TLDR, 99.999997%
> I may have completely messed up my math for someone like xyzzy to catch me on, but this is a large simplification of the problem:
> 
> Assume every case has a probability of 4/15552 of occurring (4x a last layer skip, or 1/3888).
> ...



(I'm not that great at math, so sorry if this is wrong)

Doesn't each LL case have different chances of appearing? Like, for example, an N-perm is way less likely to happen than a T-Perm, so you can't just say that you can do 21 solves and see all PLL cases (bad example but you get the general idea)


----------



## Q-- (Jun 19, 2021)

ProStar said:


> (I'm not that great at math, so sorry if this is wrong)
> 
> Doesn't each LL case have different chances of appearing? Like, for example, an N-perm is way less likely to happen than a T-Perm, so you can't just say that you can do 21 solves and see all PLL cases (bad example but you get the general idea)


Yes, that’s what I mentioned in the last paragraph about the symmetry. An N perm has 90 degree symmetry so it’s 1/72 as opposed to the 4/72 of a T perm. However, I couldn’t find a list of last layer cases by their symmetry so I simplified the problem.


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## Romy4 (Jun 24, 2021)

Got my fist ever last layer skip today (almost a pb cos my cross was rubbish), I’ve probably done under 5000 timed solves. What are the chances of getting a last layer skip? also how many last layer skips have people had? And how long have you been cubing to get that many last layer skips?


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## kubesolver (Jun 24, 2021)

Romy4 said:


> Got my fist ever last layer skip today (almost a pb cos my cross was rubbish), I’ve probably done under 5000 timed solves. What are the chances of getting a last layer skip? also how many last layer skips have people had? And how long have you been cubing to get that many last layer skips?


Eo skip is one in 8
Co skip is one in 27
Pll skip is one in 72
Roughly one in 15 thousands.

I got one in a speedsolve in so far which seems about fair considering number of solves I did in total


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## Waffles (Jun 26, 2021)

I’ve had 3 Last Layer skips ever, have around 5000 3x3 solves on cstimer and hardly ever use it, and do like probably around 100 times solves every night. 1 LL skip was on 5x5 too (have around 150 solves in that session). I know I’m very lucky but I actually noticed the possible LL skip on the 2nd one and did a different pair solution to get it (also never gotten a no-AUF LL skip) and I believe I influenced the 5x5 LL skip by doing 2 adjacent pairs with a multi slot in between edge pairing stages.


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## Cubing Forever (Jun 26, 2021)

Q-- said:


> However, I couldn’t find a list of last layer cases by their symmetry so I simplified the problem.


PLL: (Case(s): symmetry, probability)
A perms: no symmetry, 1/18 each
E: 180° symmetry, 1/36
F: no symmetry, 1/18
G perms: no symmetry, 1/18 each
H: 90° symmetry, 1/72
J perms: no symmetry, 1/18 each
N perms: 90° symmetry, 1/72 each
R perms: no symmetry, 1/18 each
T: no symmetry, 1/18
U perms: no symmetry, 1/18 each
V: no symmetry, 1/18 
Y: no symmetry, 1/18
Z: 180° symmetry, 1/36

Too lazy to do it for OLL.


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## Q-- (Jun 26, 2021)

Cubing Forever said:


> PLL: (Case(s): symmetry, probability)
> A perms: no symmetry, 1/18 each
> E: 180° symmetry, 1/36
> F: no symmetry, 1/18
> ...


Thank you, but I meant all 3000+ last layer cases. Oll and Pll symmetry are fairly easy to find.


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## yCArp (Jul 10, 2021)

Out of curiosity, I was wondering how many possible cases there are for F2L in one look. That means the entire F2L is solved in one algorithm, or in other words, solving all 4 pairs at the same time.


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## xyzzy (Jul 10, 2021)

yCArp said:


> Out of curiosity, I was wondering how many possible cases there are for F2L in one look. That means the entire F2L is solved in one algorithm, or in other words, solving all 4 pairs at the same time.


Approximately \( \frac14\cdot(3^4\frac{8!}{4!})\cdot(2^4\frac{8!}{4!})=914457600 \) (914 million) up to rotation, roughly half that up to rotation and reflection (457 million), roughly a quarter up to rotation and AUF (229 million), roughly an eighth up to rotation, reflection and AUF (114 million).


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## yCArp (Jul 10, 2021)

Thanks a lot. How about 3 pairs at a time? Does changing some of the relevant numbers in your formula get me the correct result?


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## teh yoshi (Aug 22, 2021)

If you had all the stickers necessary to sticker a blank cube, but placed them randomly, what are the odds that it'll be solvable in the standard color scheme?

What are the odds that it'll be solvable in any color scheme?


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## stwert (Aug 22, 2021)

Edit: I think my question fits better in a separate thread. It's not exactly probability.


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## kubesolver (Aug 23, 2021)

teh yoshi said:


> If you had all the stickers necessary to sticker a blank cube, but placed them randomly, what are the odds that it'll be solvable in the standard color scheme?
> 
> What are the odds that it'll be solvable in any color scheme?


Are center, edge and corner stickers distinguishable?


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## Cubing Forever (Aug 23, 2021)

teh yoshi said:


> If you had all the stickers necessary to sticker a blank cube, but placed them randomly, what are the odds that it'll be solvable in the standard color scheme?
> 
> What are the odds that it'll be solvable in any color scheme?


1 in 12 if the centers are in the correct order. If the centers are in the wrong order, probably 1 in 72.


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## kubesolver (Aug 23, 2021)

Cubing Forever said:


> 1 in 12 if the centers are in the correct order. If the centers are in the wrong order, probably 1 in 72.


I think you misunderstood the question. It's about random stickering and not random assembly. My guess is more like one in a 1 000 000 000.


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## xyzzy (Aug 23, 2021)

teh yoshi said:


> If you had all the stickers necessary to sticker a blank cube, but placed them randomly, what are the odds that it'll be solvable in the standard color scheme?
> 
> What are the odds that it'll be solvable in any color scheme?


kubesolver brought up the observation that standard speedcube stickers have distinguishable centres/edges/corners, so this question has two reasonable answers: one where all 54 stickers are the exact same shape, and one where you use only corner stickers on the corners, etc.

It's helpful to imagine that the 54 stickers are all distinctly labelled, even the ones with the same colour.

*1. All stickers are the same shape*

The colour scheme can be uniquely defined exactly when every centre gets a different colour.
\( \displaystyle P(\text{centres get different colours})=\frac{9^6}{\binom{54}{6}}\approx0.0206 \)

There are 6!=720 ways to arrange the colours on the centre pieces, of which 24 are a rotation of the standard colour scheme, so there's a 24/720 chance of the colour scheme being the standard one.
\( \displaystyle P(\text{centres get standard colour scheme})=P(\text{standard colour scheme}|\text{centres get different colours})P(\text{centres get different colours})\\
=\frac1{30}P(\text{centres get different colours})\approx6.86\cdot10^{-4} \)

Once the colour scheme has been defined, we can compute the conditional probability of getting solvable corners and edges. There are \( \tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8! \) legal cube states once the centres are fixed, and for each of those, we can freely permute the eight remaining white stickers among themselves, etc., so there's an 8! term for each colour.
\( \displaystyle P(\text{solvable}|\text{centres get different colours})=\frac{\tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8!\cdot8!^6}{48!}\\
\approx1.497\cdot10^{-14} \)
This conditional probability calculation remains the same even if we restrict to any one (or any nonempty subset) of the 30 possible colour schemes.

Putting these together:
\( \displaystyle P(\text{solvable})=P(\text{any colour scheme})P(\text{solvable}|\text{centres get different colours})\\
=\frac{\tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8!\cdot9!^6}{54!/6!}\\
\approx3.08\cdot10^{-16}\\
P(\text{solvable into standard colour scheme})=P(\text{standard colour scheme})P(\text{solvable}|\text{centres get different colours})\\
=\frac{\tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8!\cdot9!^6}{54!/6!}\cdot\frac{24}{6!}\\
\approx1.027\cdot10^{-17} \)

(Actually, now that I've worked it out, it wasn't really necessary to treat the centres specially at all. Whatever.)

*2. The stickers differ by orbit*

Same logic.

\( \displaystyle P(\text{solvable into any colour scheme})=\frac{\tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8!\cdot4!^6\cdot4!^6}{24!24!}\\
\approx4.10\cdot10^{-12}\\
P(\text{solvable into standard colour scheme})=\frac{\tfrac12\cdot2^{11}\cdot12!\cdot3^7\cdot8!\cdot4!^6\cdot4!^6\cdot24}{6!24!24!}\\
\approx1.368\cdot10^{-13} \)

These numbers are approximately 13000 times larger than in the earlier case, which makes sense: if the stickers can only go on the same piece type, there're (intuitively speaking) less ways to mess up.


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## IsThatA4x4 (Sep 10, 2021)

How could you best find the number of algs required in a step e.g. if you had invented OLL, PLL, LSE, 6CO, etc. for the first time how would you figure out how many algs would be needed?


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## ruffleduck (Sep 10, 2021)

Math, or HARCS.


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## xyzzy (Sep 10, 2021)

Consider these questions:

1. What type(s) of pieces are you solving in this step?
2. How many pieces of each type are you solving?
3. For each type, are you fully solving the pieces, or only solving them partially (e.g. orientation only)?
4. Are there constraints between the piece types? Constraints within a piece type?
5. What do you consider to be "equivalent" cases?



Spoiler: Example with CMLL



CMLL is a step in the Roux method where, after finishing the first two blocks, you solve the four remaining corner pieces.

1. What type(s) of pieces are you solving in this step?
Only one type: the U-layer corners.

2. How many pieces of each type are you solving?
Four.

3. For each type, are you fully solving the pieces, or only solving them partially?
Fully solving.

4. Are there constraints between the piece types? Constraints within a piece type?
Corner orientation must add up to 0 mod 3.

5. What do you consider to be "equivalent" cases?
If two cases are the same up to recolouring (e.g. if you did the same setup on a cube starting in a different orientation), we consider them to be equivalent. We also consider the cases to be equivalent after applying any number of U moves. (To wit: we don't care about AUFs before/after the alg.)

The four corner pieces have 4! = 24 permutations and 3^4/3 = 27 orientations, so there are 24×27 = 648 cube states to consider.

What the "don't care about AUF" equivalence relation really means is that, if cube state \( X \) and cube state \( Y \) satisfy \( X=U^aYU^b \) for some \( a,b\in\{0,1,2,3\} \), then we consider them to be equivalent up to AUF. This can be seen as a (right) group action of \( (\mathbb Z/4)^2 \) on the set of states, acting by \( (a,b):X\mapsto U^{-b}XU^a \). At this point, you can use this thing called Burnside's lemma to count the number of equivalence classes – the number of states that are distinct up to AUF.

"okay but if I knew group theory I wouldn't be asking this question in the first place"

So here's another trick you can use: consider the permutation and orientation separately. You can do permutation first, or orientation first, and they're both very valid approaches.



Spoiler: permutation first



As you should know, _up to AUF_, the corner permutation can only be three things: (i) solved, (ii) adjacent swap, or (iii) diagonal swap.

In cases (i) and (iii), no matter what AUF you're at, doing any solved-CP alg (e.g. Sune) or any diag-CP alg (e.g. Y perm) (respectively) will always solve CP (up to AUF). In these cases, you can work out all 3^4/3=27 ways to orient the corners and see which of them are equivalents pretty easily – the eight equivalence classes are CO skip (aka "O"), 3 clockwise (aka "antisune"), 3 anticlockwise (aka "sune"), T, U, L, H, and pi. This gives 8 CMLL cases for (i) and (iii) each.

In case (ii), the adjacent swap is uniquely determined (e.g. if it's a URF-UBR swap, you can't do a ULB-UFL swap alg to solve CP), so you can AUF the case to have the corner swap on the right side. Now, all 27 corner orientations will lead to different cases: you can't convert one to another via AUFs, because then the corner permutation would change. So there are 27 cases within (ii).

Putting the numbers together, we have 8+27+8 = 43 CMLL cases. One of them is the solved case, where you don't need an alg, so there are 42 algs needed here.





Spoiler: orientation first



There are eight possible corner orientations up to AUF:
O, A, S, T, U, L, H, pi.
(Corner orientation is only affected by AUF on one side: the four cube states X, U X, U2 X, U' X will always have exactly the same corner orientation, while the four states U, X U, X U2, X U' might have different corner orientations (that are equivalent up to AUF anyway).)

In all of these other than O and H, the corner orientation has no nontrivial rotational symmetry. Once you know the CO, the corner permutation must be one of 4! = 24 permutations. These all have equivalence classes of size 4: X, U X, U2 X, U' X have different permutations, but they're equivalent up to AUF. Thus there are six equivalence classes each for A, S, T, U, L and pi.

O has 90° rotational symmetry: no matter how you AUF it, it's still the same. As in the permutation-first case, there are three CMLL cases here: solved, O-adjacent (solve with a J perm), and O-diag (solve with a Y perm).

H has 180° rotational symmetry: if you do a U2 AUF, the corner orientation remains unchanged. There are four CMLL cases here: H CP skip, H front swap, H right swap, and H diag swap. (H back swap is equivalent to H front swap, and H left swap is equivalent to H right swap.)

This gives 6×6 + 3 + 4 = 43 cases, one of which is solved, so there are 42 algs needed.



CMLL is just a type of CLL, and the calculations above will apply to any type of CLL. That's why COLL, 2×2×2 CLL, Waterman CLL, etc. also have 43 cases (42 algs).



(okay wow typing that out was a lot more effort than I expected; I was going to do two or three examples but maybe not)

If you have something in particular you want to see, I can work that out in detail. If not… condensing a few hours of combinatorics classes into a single forum post is maybe not the best use of my time. Or your time, for that matter. You'll learn some basic combinatorics in school in a couple of years (usually taught together with probability); pay attention in your maths classes.

*edit*: See also this other post I made (which says mostly the same things, but with less detail). tl;dr: Be systematic, so as to ensure you don't miss out on anything.


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## ottozing (Oct 11, 2021)

Probably a very annoying stat to generate correctly, but one that I'm very interested in since it has interesting implications for CN CFOP solvers 

How likely is it to get a 3x3 scramble where at least 1 edge is already lined up with its center? (so 0-1 HTM from solved on 2 colors, or 0-2 QTM from solved)

How likely is it to get a 3x3 scramble where at least 1 edge is already lined up with its opposite center? (so 2 HTM from solved on 2 colors, or 3-4 QTM from solved)

The reason I'm curious about this is that as a CN CFOP solver, the way you typically filter is by looking for one of these 2 types of edges since they stand out the most & tend to give crosses under 7 moves (though the opposite edge is a lot less reliable). I'd be curious to know how often CN CFOP solvers are given a scramble where you're forced to semi-brute force-check a bunch of crosses before you find something nice (nice being 6 or under because 7 move crosses in theory should only happen 0.03% of the time according to cubezone's cross study)


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## 2018AMSB02 (Oct 11, 2021)

ottozing said:


> Probably a very annoying stat to generate correctly, but one that I'm very interested in since it has interesting implications for CN CFOP solvers
> 
> How likely is it to get a 3x3 scramble where at least 1 edge is already lined up with its center? (so 0-1 HTM from solved on 2 colors, or 0-2 QTM from solved)
> 
> ...


I believe I could figure this out.

There are 12 positions and 2 orientations of each edge, meaning that there is a 4/24 or 1/6 chance of any edge being lined up with its center. Because there are 12 edges, just do the calculator function binomial dsitribution and use this: 1-binompdf(12,1/6,0), which gives your your answer to the first question of an* 88.78% chance. *I think this is right, but maybe someone better at his will have to correct me.

For lined up with opposite center, as long as you aren't looking for the probability of it being lined up with opposite center OR it's own center, then I believe it should be the same.


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## qwr (Oct 11, 2021)

2018AMSB02 said:


> I believe I could figure this out.
> 
> There are 12 positions and 2 orientations of each edge, meaning that there is a 4/24 or 1/6 chance of any edge being lined up with its center. Because there are 12 edges, just do the calculator function binomial dsitribution and use this: 1-binompdf(12,1/6,0), which gives your your answer to the first question of an* 88.78% chance. *I think this is right, but maybe someone better at his will have to correct me.
> 
> For lined up with opposite center, as long as you aren't looking for the probability of it being lined up with opposite center OR it's own center, then I believe it should be the same.


idk if this is correct because the edges aren't independent of each other. 
also this should go in the probability requests not the wca stats, unless we are asking specifically about solves that occurred in wca, but I see no reason not to calculate the exact number


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## tsmosher (Nov 4, 2021)

If I have DBM, DFM, DRS, and all of the U layer edges (7 edges total) oriented and permuted randomly-- with the other edges all being solved-- what would be the lowest case count required to make EOdM (one step solving EO, DFM, and DBM simultaneously) entirely algorithmic?

Let's assume we can influence the good edge (since we know that 1 of these 7 edges will be a good edge) to be in a known location.

Also, let's assume we don't care about DRF, DRB, or any U corners.

We do, however, want to preserve the solved edges (including the FRE and BRE belt edges).

How does this case count change if we are instead on LS with 7 edges remaining (replacing DRS above with FRE)? (Wanting to preserve dL and dbR, i.e. Roux FB and SS. And with our good edge in a known location. And not caring about DFR or U corners.)

And how does the case count change if we attempt EOdM just after SB in a Roux solve (with DBM, DFM, and U layer edges unsolved-- wanting to preserve FB and SB but not U corners)? e.g., ZBRoux

Thanks for anyone who helps answer this.


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## Jeff Dunham (Mar 1, 2022)

I'm interested in learning Winter Variation, and was curious as to what the probability of attaining a winter variation (no specific case, just getting _a _case) in any given solve is.
I suppose this is equivalent to asking "knowing Winter Variation, what is the probability of an 'OLL Skip'" (yes I know that often it is just a cancellation into OLL but the premise is there.)


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## Garf (Mar 1, 2022)

Jeff Dunham said:


> I'm interested in learning Winter Variation, and was curious as to what the probability of attaining a winter variation in any given solve is.
> I suppose this is equivalent to asking "knowing Winter Variation, what is the probability of an 'OLL Skip'" (yes I know that often it is just a cancellation into OLL but the premise is there.)


Winter variation only works when you have a f2l pair stuck together and all the edges are oriented up to the point of last layer. I would suggest trying to figure out the probability of just that, then whatever WV case you want.


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## Jaym-er (Mar 8, 2022)

I'm in the process of generating a 2x2 alg set where you build half the base as a bar of one colour, and half the base with a bar of the opposite colour (example a solid bar of white and solid bar of yellow on the bottom).

Could someone tell me the chance of getting a scramble that starts with this solved and compare to chance of a CLL or EG-1. I'm also wondering how to figure out the average number of moves to solve this base from any scramble (would this be done by sampling or is there convenient math?).


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## Burrito (Mar 8, 2022)

Do the bars have to be solve completely or just oriented like Ortega?


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## DuckubingCuber347 (Mar 8, 2022)

Correct me if I am wrong but this method already exists:





NMCLL - Speedsolving.com Wiki







www.speedsolving.com





Algs would be the same as CLL.


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## Jaym-er (Mar 8, 2022)

GenZ Cubing said:


> Do the bars have to be solve completely or just oriented like Ortega?


Bars have to be solved completely.



TheCubingCuber347 said:


> Correct me if I am wrong but this method already exists:
> 
> 
> 
> ...


Similar but instead of using CLL and up to 3 AUFs, the opposite faces are solved with a direct alg. This makes for only 1 AUF, and algs will be better than the CLL alg


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## Cubing Forever (Mar 9, 2022)

Jaym-er said:


> Similar but instead of using CLL and up to 3 AUFs, the opposite faces are solved with a direct alg. This makes for only 1 AUF, and algs will be better than the CLL alg


that would be a lot of algs


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## Jaym-er (Mar 10, 2022)

Cubing Forever said:


> that would be a lot of algs


True. Roughly 175. I ended up generating over 1 Billion solutions when generating algs. Currently sorting through them to find the best one (I have a program that helps with this though).


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## Jaym-er (Mar 21, 2022)

I came up with:

States on 2x2: 7! 3^6 = 3,674,160

Approximate positions:

Base: 4*4 Top: 4! 3^3 and multiply by 6 for 6 faces (need to remove intersection)

Multiply Base x Top x 6 = 62,208

Calculating intersection:

1 case on opposite face R2 away.

The other possibility lies on the left or right faces (presuming bars are on DL and DR):

2 sides * 3 orientations * 2 permutations = 12

Total cases = 62,195

Probability of starting with a face from this algset solved = 62,195/3,674,160 = 1.69%

Wish I knew the chance of being one move away from this base now.

My quick guess would be approximately 1.69%*6 = 10.2%


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## Jaym-er (Mar 21, 2022)

Jaym-er said:


> I came up with:
> 
> States on 2x2: 7! 3^6 = 3,674,160
> 
> ...


Also worth mentioning that I calculated positions of a premade EG layer is approximately 3! 4! 3^3 this excludes intersections.

Works out to be roughly 6/16 * the chance of this bars method which means ~0.63% vs 1.69%


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## Jackson Tarr (Mar 28, 2022)

I found this because I built the second block and then realized It was two moves from solved. I didn’t try to do this it just happened. So CMLl and LSE skips with and AUF. Also sadly didn't time or have a scramble because I was practicing slow solving. 

So 42*23040 = 967,680 then taking the 967,680 and multiplying by 1/4 to account for the AUF, which gives a 1/241,920 chance.


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## Halqrius (Apr 1, 2022)

Not really probability, but how many permutation are there when you exclude symmetries?
For example, a scramble done white top green front (usually) gives a different permutation of pieces than done red top blue front. Usually these are counted differently, but how many permutations are there if you count them as the same?


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## Thom S. (Apr 23, 2022)

Imsoosm said:


> I'm not a maths expert, but I would probably say that's the number of possible permutations divided by the number of orientations, because if you want the number of scrambles for that, all permutations for the same scramble on different sides would be the same.
> 
> Technically if you do a scramble on different orientations, the pieces would be in different places, but the solution is still the same, but hey.
> 
> ...


Not like I am qualified but 24 was the number used for symmetry at the God's Number Project.


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## qwr (May 1, 2022)

Halqrius said:


> Not really probability, but how many permutation are there when you exclude symmetries?
> For example, a scramble done white top green front (usually) gives a different permutation of pieces than done red top blue front. Usually these are counted differently, but how many permutations are there if you count them as the same?


If you mean symmetries from cube rotations (and not say mirroring along an axis or diagonally), then the cube orientation (including starting orientation) is uniquely defined by color on top and color on front, so 6 * 4 = 24.


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## NigelTheCuber (May 1, 2022)

Probability of all centers solved on 4x4


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## xyzzy (May 1, 2022)

nigelthecuber said:


> Probability of all centers solved on 4x4


1. count the number of possible centre arrangements (and convince yourself that they're all equally likely)
2. count the number of possible centre arrangements where the centres are all solved
3. divide

All fairly straightforward computations; you can try working it out yourself. (Slightly less straightforward if you take 4b3 into account, which forbids the solved state and 1-move states from being WCA-legal scrambles.)


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## Bruce MacKenzie (May 2, 2022)

qwr said:


> If you mean symmetries from cube rotations (and not say mirroring along an axis or diagonally), then the cube orientation (including starting orientation) is uniquely defined by color on top and color on front, so 6 * 4 = 24.


You are correct. There are 24 rotational symmetries and the number of cube positions may be reduced nearly 24 fold by these symmetries. It is "nearly" 24 fold since some positions have rotational symmetry. Superflip for example has full cubic rotational symmetry. It doesn't matter how one orients the cube when applying a superflip maneuver, you get the same permutation. Kociemba has a good discussion of symmetry on his site.


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## Bruce MacKenzie (May 3, 2022)

Halqrius said:


> Not really probability, but how many permutation are there when you exclude symmetries?
> For example, a scramble done white top green front (usually) gives a different permutation of pieces than done red top blue front. Usually these are counted differently, but how many permutations are there if you count them as the same?


On Kociemba's site there is a table giving the number of cube patterns with various symmetries. He gives the symmetry reduced number of cubes with each symmetry in the M*ore Information* link. The number he gives is reduced by the full 48 element cubic symmetry group which includes mirrored symmetries. I took his reduced numbers and back calculated the total number of cubes with each symmetry and the number reduced by only the rotational symmetries:


Symmetry​Mirrored (1 YES, 2 NO)​Symmetries​Count​Oh Sym Eq Classes​O Sym Eq Classes​Oh​1​48​4​4​4​Th​1​24​20​10​10​T​2​12​48​12​24​D3d​1​12​48​12​12​C3v​1​6​128​16​16​D3​2​6​1,664​208​416​S6​1​6​30,960​3,870​3,870​C3​2​3​15,083,456​942,716​1,885,432​D4h​1​16​372​124​124​D4​2​8​1,152​192​384​C4v​1​8​2,688​448​448​C4h​1​8​4,224​704​704​C4​2​4​433,920​36,160​72,320​S4​1​4​1,312,512​109,376​109,376​D2d (edge)​1​8​8,832​1,472​1,472​D2d (face)​1​8​1,152​192​192​D2h (edge)​1​8​5,760​960​960​D2h(face)​1​8​11,892​1,982​1,982​D2 (edge)​2​4​278,784​23,232​46,464​D2 (face)​2​4​560,544​23,356​46,712​C2v (a1)​1​4​186,624​15,552​15,552​C2v (a2)​1​4​3,490,560​290,880​290,880​C2v (b)​1​4​577,536​48,128​48,128​C2h (a)​1​4​1,722,624​143,552​143,552​C2h (b)​1​4​577,392​48,116​48,116​C2 (a)​2​2​45,856,382,976​1,910,682,624​3,821,365,248​C2 (b)​2​2​15,286,488,192​636,937,008​1,273,874,016​Cs (a)​1​2​55,028,305,920​2,292,846,080​2,292,846,080​Cs (b)​1​2​2,546,491,008​106,103,792​106,103,792​Ci​1​2​45,862,360,944​1,910,931,706​1,910,931,706​TOTALS​​​164,604,321,936​4,948,260,778​9,407,837,992​
 So the number of cubes with no symmetry is

43,252,003,274,489,856,000 - 164,604,321,936 = 43,252,003,109,885,534,064 This number may be reduced 24 fold by the rotation symmetries:

= 1,802,166,796,245,230,586

Add in the 9,407,837,992 rotational reduced symmetric cubes gives: 

= 1,802,166,805,653,068,578 

(Elsewhere, Kociemba says that the number of symmetric cubes is 164,604,041,664 not 164,604,321,936. So the numbers he gives don't jive. There is an error somewhere but I don't know where.)


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## Jaym-er (May 3, 2022)

nigelthecuber said:


> Probability of all centers solved on 4x4


(Taking only centre piece positions as other piece types are inconsquential)

Positions where centres are solved:
4!^6*24

Total centre positions:
24!

Probability:
4!^6*24/24! = 1.35*10^14

~1 in 135 Trillion.


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## xyzzy (May 3, 2022)

Jaym-er said:


> Probability:
> 4!^6*24/24! = 1.35*10^14


Correct, but the division is in the wrong direction here.


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## Jaym-er (May 4, 2022)

xyzzy said:


> Correct, but the division is in the wrong direction here.


Oh yeah lol, I inverted it to make it more digestible but didn't write in my comment.


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## Jaym-er (May 4, 2022)

Imsoosm said:


> A question not so related to probability:
> How many megaminx LL cases are there? And (probability question) what are the chances of a megaminx LL skip (with auf)?


So I was curious about how to mathematically answer the megaminx LL skip:

Using a mathematical derivation for probabilities by solving EO-CO-CP-EP:

Edge orientation:

2^4 - each edge has 2 possible orientations, when you know the orientation of 4, the last is predetermined, hence 2*2*2*2 = 2^4, so chance of EO skip = 1/16

Corner orientation:

3^4 - same logic as edge orientation, except 3 orientations possible for each corner. Chance of CO skip = 1/81

Chance of OLL skip: 1/81*1/16 = 1/1296

As permutation of corners and edges are dependent on each other, I will use a 4LLL approach calculating skip chance of first CP (ignoring EP) then EP (dependent on knowing CP)

Corner permutation:

4*3 - taking the first corner as fixed and then permuting the remaining 4 corners around this. The first corner has 4 to choose from, the next has 3 to choose from. Once you know these corners the remaining two are fixed (this also means you can determine megaminx cp from 3 corners) Chance of CP skip = 1/12

Edge permutation:

5*4*3 - once corner permutation is known, the 5 edges need to be permuted around this. First edge position has 5 to choose from, next has 4, third position has 3. Knowing all corner positions and first 3 edges means the position of remaining 2 edges is fixed. Chance of EP skip = 1/60

Chance of PLL skip = CPskip*EPskip = 1/12*1/60 = 1/720

Chance of LL skip = OLL skip* PLL skip = 1/1296*1/720 = 1/933120

A little wordy but this gives a few extra probabilities and info.


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## Timona (May 7, 2022)

What are the chances of getting the same EG-1 case?
I got this scramble, F R' U F2 R' F R U2 F', I saw it was an EG-1 case, I learnt it and continued solving. Then some 15 solves later, I get this scramble, U' F U' F' R F' U2 F' U' and it's the exact same case that I just learnt some minutes ago. What are the chances of this happening, I don't know if it's rare of quite common. If it helps, I only have like 1470 solves.


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## IsThatA4x4 (May 7, 2022)

Can someone explain _why_ certain PLLs (eg N, Z, H) are rarer than others? Just curious.


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## Thom S. (May 7, 2022)

Timona said:


> What are the chances of getting the same EG-1 case?
> I got this scramble, F R' U F2 R' F R U2 F', I saw it was an EG-1 case, I learnt it and continued solving. Then some 15 solves later, I get this scramble, U' F U' F' R F' U2 F' U' and it's the exact same case that I just learnt some minutes ago. What are the chances of this happening, I don't know if it's rare of quite common. If it helps, I only have like 1470 solves.


The first solve doesn't really matter, but since none(since we are talking only about EG-1) cases have symmetry, every one of the 42 cases is as likely to happen.
So, if you do a solve, remember the EG Case, then do another, the Case you just had has a 1 in 42 chance of appearing(assuming you can force EG-1, you certainly can, I din't think you did but this is the way you asked the question. If you ask it differently, we can talk about full EG)


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## baseballjello67 (May 7, 2022)

What are the chances of getting a LL skip when you know full WV?


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## ruffleduck (May 7, 2022)

baseballjello67 said:


> What are the chances of getting a LL skip when you know full WV?


How would you apply WV? If you are forcing WV every solve to skip OLL, the chance of LL skip is 1/72 (PLL skip probability)


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## baseballjello67 (May 7, 2022)

WV only works when you have a cross solved on top. VLS solves OLL from any case. I am talking about WV (and let's assume that I will use WV whenever I can)


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## ruffleduck (May 7, 2022)

baseballjello67 said:


> WV only works when you have a cross solved on top.


You can force EO solved.


baseballjello67 said:


> I am talking about WV (and let's assume that I will use WV whenever I can)


So more specifically, you use WV whenever EO is solved. In that case, you will have 1/72 chance of LL skip whenever EO is solved. Depending on your solutions for other LS cases, your probability for non-EO solved cases may vary.


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## Jaym-er (May 9, 2022)

IsThatA4x4 said:


> Can someone explain _why_ certain PLLs (eg N, Z, H) are rarer than others? Just curious.


An easy way is to imagine the number of ways you can draw the top layer - each way you draw it is a different possible position and can contribute to its probability of occurring.

(For below examples consider white base)

With an F perm, you can draw it with the "bar" on 4 different sides, and the bar can be 4 different colours (red,blue,green or orange). This is 16 different positions.

With the h perm, you can draw it only with 4 different corner colours facing the front. All other top down drawings will give the same image. This is only 4 different positions.

This method is basically just an easier way of visualising. I can't really give a simple math version other than saying that the pieces just can't permute (ie. be drawn) in any new way.


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## Jaym-er (May 9, 2022)

Timona said:


> What are the chances of getting the same EG-1 case?
> I got this scramble, F R' U F2 R' F R U2 F', I saw it was an EG-1 case, I learnt it and continued solving. Then some 15 solves later, I get this scramble, U' F U' F' R F' U2 F' U' and it's the exact same case that I just learnt some minutes ago. What are the chances of this happening, I don't know if it's rare of quite common. If it helps, I only have like 1470 solves.


It sounds like you're not exactly sure what the scenario you want a probability for is.

I'll dissect this into two interesting bits:

Chance of any solve having a premade EG-1:

Approx 6*4*4!*3^3/(7!*3^7) = 4/2835 (there are intersections where they can be 2 solved EG-1 faces on a scramble so the actual figure is slightly lower than this) This is approx 1 in 709 solves that you will have a premade EG-1 case.

Chance of not having any premade EG-1 scrambles in 1470 solves:

(2831/2835)^1470 = 12.5%

Expected premade EG-1 faces in 1470 solves:

4/2835*1470 = 2.07

Again those figures were approximate.

Now let's say you do get a premade EG-1 scramble, what are the chances of getting the same case within 15 solves?

Chance of not getting a specific EG-1 scramble in any solve (this is different for the solved and diag oriented cases):

(7!3^7-6*4*4)/(7!3^7) = 11022384/11022480 (not gonna bother as % here, trust me it's low)

Chance of getting a specific EG-1 at least once within 15 solves:

1- (11022384/11022480)^15 = 0.013%

So approx 1 in 7600 times that you get a premade EG-1 scramble you will get that same case within your next 15 solves.


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## Bruce MacKenzie (May 9, 2022)

IsThatA4x4 said:


> Can someone explain _why_ certain PLLs (eg N, Z, H) are rarer than others? Just curious.


There are 4! * 4! / 2 = 288 PLL cube arrangements. These are of 22 types when reduced by the orientation of the cube to which they are applied. Apply a pattern then rotate the cube 90˚ and apply it again and you get two different patterns which are deemed the same. However, many of the PLL patterns have symmetry. Rotating the cube gives the same exact pattern not a new similar one. A pattern may have two fold, four fold or no symmetry. Patterns with no symmetry will be twice as likely as patterns with two fold symmetry and four times as likely as patterns with four fold symmetry. Here are the 21 PLL patterns and their probability:


​Name​Symmetries​Probability​​​​​1​Aa​1​0.0556 (1:18)​​​​​2​Ab​1​0.0556 (1:18)​​​​​3​E​2​0.0278 (1:36)​​​​​4​F​1​0.0556 (1:18)​​​​​5​Ga​1​0.0556 (1:18)​​​​​6​Gb​1​0.0556 (1:18)​​​​​7​Gc​1​0.0556 (1:18)​​​​​8​Gd​1​0.0556 (1:18)​​​​​9​H​4​0.0139 (1:72)​​​​​10​Ja​1​0.0556 (1:18)​​​​​11​Jb​1​0.0556 (1:18)​​​​​12​Na​4​0.0139 (1:72)​​​​​13​Nb​4​0.0139 (1:72)​​​​​14​Ra​1​0.0556 (1:18)​​​​​15​Rb​1​0.0556 (1:18)​​​​​16​T​1​0.0556 (1:18)​​​​​17​Ua​1​0.0556 (1:18)​​​​​18​Ub​1​0.0556 (1:18)​​​​​19​V​1​0.0556 (1:18)​​​​​20​Y​1​0.0556 (1:18)​​​​​21​Z​2​0.0278 (1:36)​


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## Aluminum (May 31, 2022)

ProStar said:


> Chances for an Xcross to already be finished?


1 in 43 million


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## Imsoosm (May 31, 2022)

Aluminum said:


> 1 in 43 million


No, there are many different x-crosses that are possible. For each x-cross on white face, there are four different pairs that can be put in. And that's not to account for all the other pieces. You can also have 24 orientations for each x-cross, and you have 6 faces, so to start off you would get 576 different x-crosses NOT including the other pieces.


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## hyn (May 31, 2022)

Imsoosm said:


> No, there are many different x-crosses that are possible. For each x-cross on white face, there are four different pairs that can be put in. And that's not to account for all the other pieces. You can also have 24 orientations for each x-cross, and you have 6 faces, so to start off you would get 576 different x-crosses NOT including the other pieces.



24 orientations?


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## Anto (May 31, 2022)

hydynn said:


> 24 orientations?


6 possible faces at the bottom times 4 possible rotations
But do we count cube rotations actually ?


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## Thom S. (May 31, 2022)

Anto said:


> 6 possible faces at the bottom times 4 possible rotations
> But do we count cube rotations actually ?


You need to.
Let's say we take Cross + FR as an xCross on the YO orientation. Take that times(divided by) 24, then you have all possible ways an XCross can occur. And that is how we need to calculate chance in this case.


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## Bruce MacKenzie (Jun 2, 2022)

Aluminum said:


> 1 in 43 million





ProStar said:


> Chances for an Xcross to already be finished?


For any particular manifestation of the XCross, one has one corner cubie and five edge cubies. One may calculate the total number of arrangements these six cubies may be in:

24 * ( 24 * 22 * 20 * 18 * 16 ) = 24 * 3,041,280 = 72,990,720 

So the probability of getting a randomly generated cube with that pattern is 1 in 72,990,720

There are six faces and the pattern may be applied to a face four ways, giving a total of 24 XCross patterns. Thus the probability of randomly having an XCross cube is 1 in 3,041,280.


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## Aluminum (Jun 3, 2022)

chances of pll skip with no auf? I just got one


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## SpeedCubing RDJ (Jun 3, 2022)

Bruce MacKenzie said:


> For any particular manifestation of the XCross, one has one corner cubie and five edge cubies. One may calculate the total number of arrangements these six cubies may be in:
> 
> 24 * ( 24 * 22 * 20 * 18 * 16 ) = 24 * 3,041,280 = 72,990,720
> 
> ...


Wow.


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## IsThatA4x4 (Jun 3, 2022)

Aluminum said:


> chances of pll skip with no auf? I just got one


1/288


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## Imsoosm (Jun 4, 2022)

What are the chances of getting a PLL skip (including AUF) on a kilominx? Also LL skip?


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## IsThatA4x4 (Jun 4, 2022)

Imsoosm said:


> What are the chances of getting a PLL skip (including AUF) on a kilominx? Also LL skip?


I know nothing about kilo, but this should be simple. *Somebody please correct me if I'm wrong about this because I've probably made a silly mistake somewhere!*
There are 5 corners.
In PLL, we only need to worry about their permutation. Because the last corner's permutation is determined by the other 4, there are really only four corners, giving us
5 x 4 x 3 x 2 (x 1) aka 5! total PLLs (120).
Because a skip has a rotational symmetry of 5, the chance of a skip is 1/(120/5) = 1/24.

For LL, we also care about orientation too. Because the last corner's orientation is determined by the other 4, there are again really only 4 corners, so the amount of total OLLs is 3^4 = 81 (this does not account for symmetry ofc).
So the total number of LL cases (not reduced) is 81 x 120 = 9720.
The skip has a rotational symmetry of 5, so the chance of an LL skip is 1/1944.


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## Abram Grimsley (Jun 16, 2022)

What is the probability of getting a last-layer skip on megaminx after orienting the last-layer edges?


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## IsThatA4x4 (Jun 16, 2022)

Abram Grimsley said:


> What is the probability of getting a last-layer skip on megaminx after orienting the last-layer edges?


Giving it a go (mega may work differently so take this with... I don't know, a ball of salt?)

We need to consider:
CO
CP
EP

For CO:
Discounting AUFs, there are 3^4 cases, and I believe counting AUFs is incorrect in this scenario, so I'm leaving it there (we'll see why)

For CP:
There should be 5! (5x4x3x2x1) CPs

For EP:
Again, 5!

Because a swap of only two pieces is impossible, half of these cases are impossible.
This leaves us with (3^4 x (5!)^2) / 2 LLs (discounting AUFs), however this is not our answer, as a skip has a rotational symmetry of 5, so the chance of a skip is:
1 / (3^4 x (5!)^2) / 10
= 1/116640

This is what I came to, however it seems too small (?). Other people correct me if this is not the chance.


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## Abram Grimsley (Jun 16, 2022)

Thanks man. That means I got really lucky on that solve


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## Jaym-er (Jun 16, 2022)

IsThatA4x4 said:


> *Somebody please correct me if I'm wrong about this because I've probably made a silly mistake somewhere!*
> 5 x 4 x 3 x 2 (x 1) aka 5! total PLLs (120).
> Because a skip has a rotational symmetry of 5, the chance of a skip is 1/(120/5) = 1/24.





IsThatA4x4 said:


> For CP:
> There should be 5! (5x4x3x2x1) CPs
> 
> For EP:
> ...


You can't simply divide by 5 for a megaminx as some permutation cases can occur by a multiple other than 5. I've explained how to derive megaminx permutation in a post a month earlier in this thread.

See here:


Jaym-er said:


> So I was curious about how to mathematically answer the megaminx LL skip:
> 
> Using a mathematical derivation for probabilities by solving EO-CO-CP-EP:
> 
> ...


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## ruffleduck (Jun 16, 2022)

IsThatA4x4 said:


> I don't know, a ball of salt?


Here's my dodecahedron of salt : )

CO: 3^4
CP: 5! / 2
EP: 5! / 2
AUF: 5

AUF / (CO * CP * EP) = 5 / (3^4 * (5! / 2)^2) = 1 / 58320


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## fani (Jun 16, 2022)

so today I was just cubing in the bus (yeah in a bus) and a woman was TOO SURPRISED when she saw that I could solve the cube in under 25 seconds. She asked if she could scramble the cube and ofc I said yes. She was scrambling it for like 2 mins. It was the easiest solve in my life. It was a one move away cross, an H perm and F2L SKIP ( sadly I wasnt timing ) She scrambled it again and I got an OLL SKIP ( wasnt timing again tho )


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## Imsoosm (Jun 16, 2022)

fani said:


> so today I was just cubing in the bus (yeah in a bus) and a woman was TOO SURPRISED when she saw that I could solve the cube in under 25 seconds. She asked if she could scramble the cube and ofc I said yes. She was scrambling it for like 2 mins. It was the easiest solve in my life. It was a one move away cross, an H perm and F2L SKIP ( sadly I wasnt timing ) She scrambled it again and I got an OLL SKIP ( wasnt timing again tho )


Non-cubers always think that if they scramble the cube more it becomes harder to solve.


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## IsThatA4x4 (Jun 16, 2022)

zzoomer said:


> CP: 5! / 2
> EP: 5! / 2


_I hope you have more salt to take with this..._

I don't know if you need to divide both by 2. If you take CP and EP together, only one of them matters for odd swaps. E.g. if we focus on corners first, we can allow any possible CPs, however looking at edges, there is now a 50/50 chance that the last 2 edges are swapped, which would be impossible. The case count for CP isn't affected by this I think (?).

As for what @Jaym-er said, what makes it so that I can't divide by 5? Just curious.
Because the way I see it, if we first calculate every possible "ZBLL" disregarding symmetry, any giving case's probability should be 1 / total case count / rotational symmetry. Some cases will have different symmetries, so it won't always be 5 but in the case of a skip it is. Where am I going wrong here?


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## ruffleduck (Jun 16, 2022)

IsThatA4x4 said:


> _I hope you have more salt to take with this..._
> 
> I don't know if you need to divide both by 2. If you take CP and EP together, only one of them matters for odd swaps.


Unlike 3x3, EP and CP are independant on megaminx. Odd swaps are impossible


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## IsThatA4x4 (Jun 16, 2022)

zzoomer said:


> Unlike 3x3, EP and CP are independant on megaminx. Odd swaps are impossible


I see! Interesting...
Good thing you had salt


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## Imsoosm (Jun 20, 2022)

What are the chances of getting almost 2 almost exactly same solutions from a random state scramble in 4 solves??

Generated By csTimer on 2022-06-20
avg of 5: 2.237

Time List:
1. (3.176) F2 R2 F U' R2 U' F' U R' 
2. 1.884 *F U F' R F' R2 U' R' F' *
3. 2.094 U' F' R2 U2 F U2 F' U' R2 
4. 2.732 R U R2 F' U F' R U2 R2 U' 
5. (1.849) *U2 F' U R U R2 U' F2 U2*

Bold scrambles are the ones, they have the exact same EG-1 case and the solution is almost exactly same.


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## Zeke Mackay (Jun 21, 2022)

Not exactly probability but I've been stumped on calculating this.

How many cases are there for last bar of 7x7 last 2 centers? 

If you aren't able to visualize it here's an explanation: One center has (at least) a 4x5 bar of center pieces built on left side and the other center has at least 20 correct center pieces and no more than 5 incorrect center pieces on it. 

Ignoring AUF


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## xyzzy (Jun 22, 2022)

Zeke Mackay said:


> Not exactly probability but I've been stumped on calculating this.
> 
> How many cases are there for last bar of 7x7 last 2 centers?
> 
> ...


Who better to answer this question than the madlad who genned algs for all 6×6×6 last bar cases…

Most cases do not have any AUF symmetry; the only ones that do are last bar skip and the one with two x-centres unsolved diagonally opposite each other.

The total number of cases (without reducing by AUF) is \( \binom51\binom51\binom51\binom62=1875 \). (The \( \binom51 \) factors correspond to the t-centres and obliques, while the \( \binom62 \) factor corresponds to the x-centres.) Reducing by AUF, all but three of these cases have no AUF symmetry, one has 4-way symmetry (skip) and two have 2-way symmetry (diag x-centres), so the number of cases up to AUF is \( 1872/4 + 1 + 2/2 = 471 \).



IsThatA4x4 said:


> As for what @Jaym-er said, what makes it so that I can't divide by 5? Just curious.
> Because the way I see it, if we first calculate every possible "ZBLL" disregarding symmetry, any giving case's probability should be 1 / total case count / rotational symmetry. Some cases will have different symmetries, so it won't always be 5 but in the case of a skip it is. Where am I going wrong here?


Your reasoning here is correct.



IsThatA4x4 said:


> I know nothing about kilo, but this should be simple. *Somebody please correct me if I'm wrong about this because I've probably made a silly mistake somewhere!*
> There are 5 corners.
> In PLL, we only need to worry about their permutation. Because the last corner's permutation is determined by the other 4, there are really only four corners, giving us
> 5 x 4 x 3 x 2 (x 1) aka 5! total PLLs (120).
> Because a skip has a rotational symmetry of 5, the chance of a skip is 1/(120/5) = 1/24.


As pointed out by others (in the context of megaminx), this is off by a factor of 2 because only even permutations are possible.


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## Puzzlerr (Jun 25, 2022)

no clue what i'm doing in a thread that involves math, but what are the odds of getting 3 pll skips in one average 

sidenote, the last 2 plls skips were in a row


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## xyzzy (Jun 25, 2022)

Puzzlerr said:


> no clue what i'm doing in a thread that involves math, but what are the odds of getting 3 pll skips in one average
> 
> sidenote, the last 2 plls skips were in a row


Assuming no OLLCP/1LLL/etc., every individual PLL skip has a 1/72 chance of occurring; the chance of exactly three out of five solves having a PLL skip is
\( \binom53\left(\frac1{72}\right)^3\left(\frac{71}{72}\right)^2\approx0.000026\approx\frac1{38000} \)

(The probability of getting at least three PLL skips is about the same, since the chance of getting four or five PLL skips is much lower.)

But also the probability of getting three PLL skips in _some_ average-of-5 increases to 100% asymptotically as you do more solves, and if you count rolling averages (as most people do for unofficial solves), this probability also increases faster.


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## ottozing (Jun 30, 2022)

(Apologies if the statistics/probabilities for this request are already documented and I've somehow missed it)

For FMC solving I'm curious to know the % chance of the following EO cases occurring on a random scramble

1. The probability of getting a 4 bad edges case solvable in 3 moves or less
2. The probability of getting an 8 bad edges case solvable in 4 moves or less
3. The probability of getting a 6 bad edges case solvable in 4 moves or less

I basically want to know how often 2 moves to 4 bad & 1 move to 6 or 8 bad could yield a 5 move or less EO using NISS


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## GodCubing (Jun 30, 2022)

That one dude who exposed the WCA for not using random state did this I think it was pochman, but I will have to check. I know he analysed the frequency of 12 flips, but doing the other EO cases wouldnt' be that hard, but in the context of FMC it would be CN EO, aka EO on all 3 axis so I suppose this is useless (I doubt pochman analysed EO on all 3 axis). I could do fixed orientation by myself but that wouldn't be relevant


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## ottozing (Jun 30, 2022)

GodCubing said:


> That one dude who exposed the WCA for not using random state did this I think it was pochman, but I will have to check. I know he analysed the frequency of 12 flips, but doing the other EO cases wouldnt' be that hard, but in the context of FMC it would be CN EO, aka EO on all 3 axis so I suppose this is useless (I doubt pochman analysed EO on all 3 axis). I could do fixed orientation by myself but that wouldn't be relevant


Fixed orientation would be fine I think since for the purposes of what I'm looking for, I want to know how common these outcomes are on a single axis. I don't really care what's going on with all 3 axis at once

Knowing the data for these 3 scenarios would give me a good idea of how much better it is to check 2 moves to 4 bad followed by NISSing VS checking 1 move to 8 or 6 bad. I already suspect that 2 moves to 4 bad followed by NISSing is more likely to produce a 5 move EO than 1 move to 6 bad based on the current EO data I have, but I'm less sure of how it compares to 1 move to 8 bad given it has more cases in the 2-4 move range


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## mencarikebenaran (Jun 30, 2022)

xyzzy said:


> Assuming no OLLCP/1LLL/etc., every individual PLL skip has a 1/72 chance of occurring; the chance of exactly three out of five solves having a PLL skip is
> \( \binom53\left(\frac1{72}\right)^3\left(\frac{71}{72}\right)^2\approx0.000026\approx\frac1{38000} \)
> 
> (The probability of getting at least three PLL skips is about the same, since the chance of getting four or five PLL skips is much lower.)
> ...


are you a mathematician?


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## xyzzy (Jun 30, 2022)

mencarikebenaran said:


> are you a mathematician?


I mean I majored in mathematics but that calculation is something any high schooler (11th/12th grade?) should be able to do.


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## Peter Mc (Jul 1, 2022)

ottozing said:


> (Apologies if the statistics/probabilities for this request are already documented and I've somehow missed it)
> 
> For FMC solving I'm curious to know the % chance of the following EO cases occurring on a random scramble
> 
> ...


My computer tells me the answers are:
1. 32/99 = 32.3%
2. 13/45 = 28.9%
3. 52/231 = 22.5%

Full distribution:
2 bad edges: [0,0,40,22,4,0,0]
4 bad edges: [2,24,134,276,56,3,0]
6 bad edges: [0,0,8,200,556,160,0]
8 bad edges: [0,1,20,122,284,68,0]
10 bad edges: [0,0,0,0,0,54,12]
12 bad edges: [0,0,0,0,0,0,1]

Read this as: the i'th number in the row is the number of positions with that many bad edges whose minimal solution takes i moves.


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## ottozing (Jul 2, 2022)

Peter Mc said:


> My computer tells me the answers are:
> 1. 32/99 = 32.3%
> 2. 13/45 = 28.9%
> 3. 52/231 = 22.5%
> ...



Thanks for doing this, as well as posting the additional data! 

Knowing this now, I'll have to think about how I want to approach NISS EO. On the one hand, 2 moves to 4 bad certainly performs slightly better than 1 move to 8 bad & quite a bit better than 1 move to 6 bad. On the other hand, Doing NISS EO's with only one turn on one of the sides of the scramble makes it easier to check other EO's for DR triggers

This is also not factoring in 1-2 moves to 2 bad which is very likely to give 4-5 move EO's, though these tend to only work well when the other axis don't have very many bad edges when looking for short solutions to DR minus trigger

I have a lot to think about


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## Imsoosm (Jul 16, 2022)

How many cases are there when you have a belt and oriented yellow and white sides? I'm including parities (M2 U2 M2) and excluding AUFs and rotation symmetry (z2/x2). I'm currently working on a method and I'm genning some algs, I want to know how many exactly I have to gen. I know its going to be a lot though


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## IsThatA4x4 (Jul 16, 2022)

Imsoosm said:


> How many cases are there when you have a belt and oriented yellow and white sides? I'm including parities (M2 U2 M2) and excluding AUFs and rotation symmetry (z2/x2). I'm currently working on a method and I'm genning some algs, I want to know how many exactly I have to gen. I know its going to be a lot though


Using the batch solver gives...

It doesn't because it's taking to long to calculate
Fine I'll do it myself

Every piece is oriented, so we only care about permutation.
There are 8 corners and 8 edges to consider:

the corners it's 8!, for the edges it's 8! Again

Since we can't only swap two edges, we divide by 2

So it should be 8!^2 / 2 = 812851200 cases excluding AUF and symmetry

Good luck genning 800 million algs, I'll see you at the end of time

Edit: I see we're not counting AUFs, this would get pretty complicated because each case has different symmetries, but dividing by 4 would get pretty close, which is still around 200 million cases


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## ruffleduck (Jul 16, 2022)

Imsoosm said:


> How many cases are there when you have a belt and oriented yellow and white sides? I'm including parities (M2 U2 M2) and excluding AUFs and rotation symmetry (z2/x2). I'm currently working on a method and I'm genning some algs, I want to know how many exactly I have to gen. I know its going to be a lot though


Sounds like PBL but on 3x3. Should be the same number as sq1: 21^2 = 441

edit: you were pretty vague with your terminology ("oriented" relative to what?) and I may have misinterpreted what you are looking for.


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## IsThatA4x4 (Jul 16, 2022)

zzoomer said:


> Sounds like PBL but on 3x3. Should be the same number as sq1: 21^2 = 441
> 
> edit: you were pretty vague with your terminology ("oriented" relative to what?) and I may have misinterpreted what you are looking for.


This definitely matters whether it's PBL or PBL + separation but the number for PBL is actually higher.
The number you described is when both U and D layers have even parity, but both could also have odd parity (e.g. opp/adj PBL).
There are 2 different scenarios:
*Even/even *= 22^2 (because we have to include solved), -22 because both PLLs being the same would create duplicate cases = 462
*Odd/odd *= 22^2 again, but -22 again because of duplicates = 462

So we get a total of 462 x 2 = 924 PBL cases

924 is definitely more feasible to gen than 200 million, so I hope this is what @Imsoosm meant


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## ruffleduck (Jul 16, 2022)

IsThatA4x4 said:


> This definitely matters whether it's PBL or PBL + separation but the number for PBL is actually higher.
> The number you described is when both U and D layers have even parity, but both could also have odd parity (e.g. opp/adj PBL).
> There are 2 different scenarios:
> *Even/even *= 22^2 (because we have to include solved), -22 because both PLLs being the same would create duplicate cases = 462
> ...


Thanks for the correction 22^2 instead of 21^2. I didn't multiply by 2 because I was taking into consideration Imsoosm's request to exclude z2 rotational symmetry


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## SYUTEO (Jul 28, 2022)

Are the chances of a PLL skip higher if you learn full OLL?
I always wondered if that's true as I seem to get PLL skips more often the more OLLs I learn, or maybe I am just lucky.


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## OreKehStrah (Jul 28, 2022)

SYUTEO said:


> I always wondered if that's true as I seem to get PLL skips more often the more OLLs I learn, or maybe I am just lucky.


Ever so slightly yes. It increases the chances of you having 1LLL by using more OLLs.


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## DGCubes (Jul 28, 2022)

No. Assuming you're not using different OLL algorithms to influence what PLL you get (e.g. choosing between R U' L' U R' U' L or R U R' U R U2 R' for the sune OLL), you have a 1/72 chance of skipping PLL regardless of whether you complete OLL with 1 or 2 algs.

EDIT (reply):


OreKehStrah said:


> Ever so slightly yes. It increases the chances of you having 1LLL by using more OLLs.


While this is true, this does not increase the chances of a PLL skip, unless the alg you're doing is purposefully affecting PLL (like COLL or ZBLL algs).


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## ottozing (Aug 2, 2022)

FMC related request:

For DR minus 4C4E, does anyone know the chances of getting a case that is solvable in x moves, with x being anywhere from 1-7? I wouldn't mind knowing the numbers for 4C2E and 3C2E as well but they're much less important, at least to me

I'm mostly curious because I just missed optimal by 1 on an attempt where I decided to look for more DR's instead of looking closely at the DR's I had, which I think was a mistake given the quality of the DR's (they were very good) and the quality of the scramble/EO's (they were pretty bad)
It'd be nice to know more or less how likely you are to get these move counts from DR triggers to know whether or not it's likely to be a waste of your time or not (especially considering the corners could end up being awful)


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## ottozing (Aug 8, 2022)

ottozing said:


> FMC related request:
> 
> For DR minus 4C4E, does anyone know the chances of getting a case that is solvable in x moves, with x being anywhere from 1-7? I wouldn't mind knowing the numbers for 4C2E and 3C2E as well but they're much less important, at least to me
> 
> ...


Bumping for visibility. Would really appreciate if anyone could help me with this


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## xyzzy (Aug 8, 2022)

ottozing said:


> FMC related request:
> 
> For DR minus 4C4E, does anyone know the chances of getting a case that is solvable in x moves, with x being anywhere from 1-7? I wouldn't mind knowing the numbers for 4C2E and 3C2E as well but they're much less important, at least to me
> 
> ...


Optimal move count to complete DR (U-D) from EO already being solved (F-B). F and B quarter turns are allowed in solving DR, i.e. EO can be temporarily broken.


# c, # e0 move1 move2 moves3 moves4 moves5 moves6 moves7 moves8 moves9 moves10 movesavg moves0c0e​1​0​0​0​0​0​0​0​0​0​0​0.00​2c0e​0​0​0​0​0​24​16​16​0​0​0​5.86​3c0e​0​0​0​0​0​0​0​112​0​0​0​7.00​4c0e​0​0​0​0​0​0​64​220​128​8​0​7.19​5c0e​0​0​0​0​0​0​0​112​448​0​0​7.80​6c0e​0​0​0​0​0​0​0​0​136​480​0​8.78​7c0e​0​0​0​0​0​0​0​0​192​144​0​8.43​8c0e​0​0​0​0​0​0​2​2​28​36​18​8.77​0c1e​0​0​0​0​0​8​24​0​0​0​0​5.75​2c1e​0​0​0​0​0​48​376​1112​256​0​0​6.88​3c1e​0​0​0​16​120​552​1520​1328​48​0​0​6.16​4c1e​0​0​0​8​64​392​2120​5960​4712​184​0​7.15​5c1e​0​0​0​0​0​0​440​4344​11064​2072​0​7.82​6c1e​0​0​0​0​0​0​32​1664​11448​6552​16​8.25​7c1e​0​0​0​0​0​0​32​488​4144​5920​168​8.53​8c1e​0​0​0​0​0​0​0​96​776​1616​264​8.74​0c2e​0​0​0​0​2​16​62​80​8​0​0​6.45​2c2e​0​0​0​0​0​48​532​3540​4916​372​0​7.53​3c2e​0​0​0​0​0​184​1712​7272​9048​600​0​7.43​4c2e​0​2​16​84​476​2486​10520​28446​26718​1812​0​7.18​5c2e​0​0​0​0​64​704​6056​31392​48376​7488​0​7.59​6c2e​0​0​0​0​12​284​2832​19912​58196​21996​256​7.96​7c2e​0​0​0​0​16​144​1104​8200​31136​15696​152​8.09​8c2e​0​0​0​0​2​16​86​1222​6350​6440​332​8.39​0c3e​0​0​0​0​0​0​0​24​128​72​0​8.21​2c3e​0​0​0​0​0​0​16​1072​7024​4424​8​8.27​3c3e​0​0​0​0​16​144​1056​5576​14056​4208​32​7.84​4c3e​0​0​0​8​64​496​3776​21360​52480​15832​64​7.84​5c3e​0​0​0​0​64​768​6864​37088​66664​13984​8​7.69​6c3e​0​0​0​0​40​768​6904​37984​75024​17192​72​7.73​7c3e​0​0​0​0​0​240​2792​17920​43368​10936​8​7.82​8c3e​0​0​0​0​0​8​216​2936​11320​4752​32​8.07​0c4e​0​0​0​0​0​0​1​2​9​46​12​8.94​2c4e​0​0​0​0​0​0​0​0​648​2844​428​8.94​3c4e​0​0​0​0​0​0​32​368​2880​4336​224​8.56​4c4e​0​0​0​0​0​0​80​1582​12368​14944​426​8.48​5c4e​0​0​0​0​0​24​528​5976​22592​9984​96​8.08​6c4e​0​0​0​0​12​148​1240​9412​24336​7920​52​7.90​7c4e​0​0​0​16​104​624​3112​9648​9320​696​0​7.25​8c4e​0​0​1​2​9​88​772​2922​2082​144​0​7.23​

Interesting that 0c1e and 2c0e are better on average than 3c1e, but they're also harder to set up to (so setting up to 0c1e or 2c0e isn't necessarily better overall).

The actual thing you asked for, 4c4e, is:
0.3% solvable in at most 6 moves
5.7% solvable in at most 7 moves
47.7% solvable in at most 8 moves
98.6% solvable in at most 9 moves
100% solvable in at most 10 moves


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## ottozing (Aug 9, 2022)

xyzzy said:


> Optimal move count to complete DR (U-D) from EO already being solved (F-B). F and B quarter turns are allowed in solving DR, i.e. EO can be temporarily broken.
> 
> 
> # c, # e0 move1 move2 moves3 moves4 moves5 moves6 moves7 moves8 moves9 moves10 movesavg moves0c0e​1​0​0​0​0​0​0​0​0​0​0​0.00​2c0e​0​0​0​0​0​24​16​16​0​0​0​5.86​3c0e​0​0​0​0​0​0​0​112​0​0​0​7.00​4c0e​0​0​0​0​0​0​64​220​128​8​0​7.19​5c0e​0​0​0​0​0​0​0​112​448​0​0​7.80​6c0e​0​0​0​0​0​0​0​0​136​480​0​8.78​7c0e​0​0​0​0​0​0​0​0​192​144​0​8.43​8c0e​0​0​0​0​0​0​2​2​28​36​18​8.77​0c1e​0​0​0​0​0​8​24​0​0​0​0​5.75​2c1e​0​0​0​0​0​48​376​1112​256​0​0​6.88​3c1e​0​0​0​16​120​552​1520​1328​48​0​0​6.16​4c1e​0​0​0​8​64​392​2120​5960​4712​184​0​7.15​5c1e​0​0​0​0​0​0​440​4344​11064​2072​0​7.82​6c1e​0​0​0​0​0​0​32​1664​11448​6552​16​8.25​7c1e​0​0​0​0​0​0​32​488​4144​5920​168​8.53​8c1e​0​0​0​0​0​0​0​96​776​1616​264​8.74​0c2e​0​0​0​0​2​16​62​80​8​0​0​6.45​2c2e​0​0​0​0​0​48​532​3540​4916​372​0​7.53​3c2e​0​0​0​0​0​184​1712​7272​9048​600​0​7.43​4c2e​0​2​16​84​476​2486​10520​28446​26718​1812​0​7.18​5c2e​0​0​0​0​64​704​6056​31392​48376​7488​0​7.59​6c2e​0​0​0​0​12​284​2832​19912​58196​21996​256​7.96​7c2e​0​0​0​0​16​144​1104​8200​31136​15696​152​8.09​8c2e​0​0​0​0​2​16​86​1222​6350​6440​332​8.39​0c3e​0​0​0​0​0​0​0​24​128​72​0​8.21​2c3e​0​0​0​0​0​0​16​1072​7024​4424​8​8.27​3c3e​0​0​0​0​16​144​1056​5576​14056​4208​32​7.84​4c3e​0​0​0​8​64​496​3776​21360​52480​15832​64​7.84​5c3e​0​0​0​0​64​768​6864​37088​66664​13984​8​7.69​6c3e​0​0​0​0​40​768​6904​37984​75024​17192​72​7.73​7c3e​0​0​0​0​0​240​2792​17920​43368​10936​8​7.82​8c3e​0​0​0​0​0​8​216​2936​11320​4752​32​8.07​0c4e​0​0​0​0​0​0​1​2​9​46​12​8.94​2c4e​0​0​0​0​0​0​0​0​648​2844​428​8.94​3c4e​0​0​0​0​0​0​32​368​2880​4336​224​8.56​4c4e​0​0​0​0​0​0​80​1582​12368​14944​426​8.48​5c4e​0​0​0​0​0​24​528​5976​22592​9984​96​8.08​6c4e​0​0​0​0​12​148​1240​9412​24336​7920​52​7.90​7c4e​0​0​0​16​104​624​3112​9648​9320​696​0​7.25​8c4e​0​0​1​2​9​88​772​2922​2082​144​0​7.23​
> ...



Thanks a ton for doing this! I'll analyze this myself later today (btw, when I said 4C4E I was originally referring to what you've listed as 4C2E, but I should be able to get the info I need from this data table)

I already see some interesting things for me to look into outside the scope of normal DR triggers (4C4E/4C2E/3C2E) which might be worthwhile


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## xyzzy (Aug 9, 2022)

ottozing said:


> btw, when I said 4C4E I was originally referring to what you've listed as 4C2E


Ah, right, could've figured that out from context. I thought it was more common for XcYe to mean X corners twisted and Y _E-slice_ edges out of place, rather than Y edges unsolved altogether.

Thank goodness I made the full table instead of just computing the results for one row, huh.


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## Timona (Aug 17, 2022)

Idk if this is the right thread, but I want to know something. The average cross move count is 7 STM or maybe less. So I want to know the (Average or optimal, correct me if I'm wrong) movecount in STM for
- an xcross
- an xxcross
- EOLine
- EOCross
- EOxcross
- cross too, Cus I feel my estimations are wrong


For cross, xcross and xxcross, I'm regarding colour neutrality. For EOLine, EOCross and EOxcross, I'm ignoring colour neutrality and assuming a fixed orientation (Yellow top, Blue front )


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## Thom S. (Aug 17, 2022)

I think to have heard that EOCross can always be solved in 10 moves.


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## Bruce MacKenzie (Aug 18, 2022)

Thom S. said:


> I think to have heard that EOCross can always be solved in 10 moves.


I did a cosets at depth calculation on the EO_Cross subgroup (ie the cube subgroup with the Up face edges solved). The deepest coset in the quarter turn metric is 9, the deepest coset in the face turn metric is 8 and the deepest coset in the slice turn metric is 7.



Depth​QTM​FTM​STM​0​1​1​1​1​10​15​21​2​73​158​275​3​500​1,394​2,815​4​3,078​9,809​21,136​5​15,528​46,381​88,454​6​57,180​97,254​76,592​7​91,654​34,966​786​8​21,849​102​​9​207​​​


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## xyzzy (Aug 18, 2022)

Bruce MacKenzie said:


> I did a cosets at depth calculation on the EO_Cross subgroup (ie the cube subgroup with the Up face edges solved).


EOCross is cross _and_ edge orientation. So the intersection of (subgroup with U-face edges solved) and ⟨U, D, L, R, F2, B2⟩.


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## Bruce MacKenzie (Aug 18, 2022)

xyzzy said:


> EOCross is cross _and_ edge orientation. So the intersection of (subgroup with U-face edges solved) and ⟨U, D, L, R, F2, B2⟩.


According to the WIKI database here EOCross is:


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## xyzzy (Aug 18, 2022)

Bruce MacKenzie said:


> According to the WIKI database here EOCross is:
> 
> View attachment 20449


The light-grey-coloured edges are meant to be ones that are oriented (wrt ⟨U, D, L, R, F2, B2⟩). Compare with, say, the pEOLine image, where the top four edges are left as dark grey.

(But also dang, these images are indeed confusing. There isn't even a description of what the edge colouring means.)


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## Bruce MacKenzie (Aug 18, 2022)

Bruce MacKenzie said:


> I did a cosets at depth calculation on the EO_Cross subgroup (ie the cube subgroup with the Up face edges solved). The deepest coset in the quarter turn metric is 9, the deepest coset in the face turn metric is 8 and the deepest coset in the slice turn metric is 7.
> 
> 
> 
> Depth​QTM​FTM​STM​0​1​1​1​1​10​15​21​2​73​158​275​3​500​1,394​2,815​4​3,078​9,809​21,136​5​15,528​46,381​88,454​6​57,180​97,254​76,592​7​91,654​34,966​786​8​21,849​102​​9​207​​​


As xyzzy pointed out the above calculation is for the CROSS subgroup cosets not the EO_CROSS subgroup cosets. I added the orientation of the other eight edge cubies and redid the analysis:



Depth​QTM​FTM​STM​0​1​1​1​1​10​15​21​2​85​178​313​3​744​1,982​4,329​4​6,194​21,041​55,730​5​49,009​204,732​643,385​6​360,175​1,645,039​5,444,086​7​2,277,947​8,477,633​16,008,710​8​9,529,222​12,917,628​2,173,545​9​11,407,300​1,061,851​120​10​699,495​140​​11​58​​​
 I dumped out a representative member for ten of the depth 10 FTM cosets and solved them with a partial cube solver configured to solve to the EO_Cross subgroup and in each case got an optimal 10 move FTM solution so I am fairly confident in my numbers.



Generator: EO_CROSS FTM Antipodes​EO_CROSS Optimal FTM Solution​F' L' B' L' D' L2 R U F' U2 F' R U R' U2 R2 U2 F' U F​R F B D' U2 L D R B U​D F L' D R D2 L' U' B2 L2 B2 U L B L2 U' L' U2 B L B​R F' U' B D' L' F U' R B'​R U B F U2 L2 D B L U2 L' U' B' L' U2 B2 L B2 L2 B L'​R U B' U2 F2 R' D F' L2 B'​B' L2 F L D L B U B' R2 U R' U R B2 R' B' R2 B2 U2 R'​R U' L2 F' U2 L' B' D' R2 F​F U D2 B U R D2 L2 F U' F L' F U L' F' U L U2 L2​R U R L D' F L' B U2 L​R U F U2 L2 D U L' U2 B' L2 U B2 L' B2 L' B L U2 B L2 U L2​R U B U2 F2 L D F R2 B​U' D2 B L F L D L R' U F' R U F2 U2 R F2 R' F' U' F'​R U' L B' L D F' R U B​R D U2 B D2 L2 U D2 L' F L' F L' F2 L' D L2 D2 F' L' D​R U B D' L' F' U B' F2 L2​B F U' R' U' D F2 D2 B' R' B' U B2 U2 B R' U2 B2 R2 B' R' U2​R U B D' B' L2 B L U F​U L' F' L F2 U' L' U' F' U2 L U' F' B' R2 L2 D' U' R' D L'​R F' U B R D F U L' B2​


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## Imsoosm (Aug 19, 2022)

What's the average movecount just for EO? I'm starting to learn EO for FMC/ZZ so I want some statistics on how it would affect my FMC solves.


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## ruffleduck (Aug 20, 2022)

Imsoosm said:


> What's the average movecount just for EO? I'm starting to learn EO for FMC/ZZ so I want some statistics on how it would affect my FMC solves.


It depends on neutrality. IIRC, fixed (1 axis) averages low 4 HTM and fully neutral (3 axes) averages low 3


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## vidcapper (Aug 21, 2022)

This may be the wrong place or been posted before, but...








Rubik's Cube solution unlocked by memorising 3915 final move sequences


For the first time, a speedcuber has demonstrated a solution to the Rubik’s cube that combines the two final steps of the puzzle’s solution into one




www.newscientist.com





Wow, i can't even remember all 16 algs for 4LLL! 

Also...


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## ottozing (Aug 24, 2022)

Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction

1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)

2 - For 4C4E, every case has between 0 and 8 pairs formed (4 possible top pairs and 4 possible side pairs). For all of these cases (example: 3 top pairs + 1 side pair, or 2 top pairs 1 side pair etc.), what % of those cases are x moves or less with x being from 0-6 moves? For info on what a pair is in this context, see this document by Rodney Kinney > https://docs.google.com/document/d/...SJpMArtRrDT-p_CwM/edit#heading=h.o0b5b27es870

(I'm looking to basically confirm a suspicion of mine that top pairs matter much more than side pairs. As for interchangeable vs non interchangeable pairs, I'm already pretty convinced they don't matter, and even if they do they're quite a bit harder to NISS predict so I'm reluctant to implement that into my FMC attempts)


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## Puzzlerr (Aug 24, 2022)

whats the odds of me getting 2 last layer skips? Not in a row, just in the span of a year
because that happened to me
and my last one was timed


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## White KB (Aug 25, 2022)

@ottozing How do I get Platinum Member status? Or is it just a legacy feature?

But more on point:


Puzzlerr said:


> whats the odds of me getting 2 last layer skips? Not in a row, just in the span of a year
> because that happened to me
> and my last one was timed


Well, it really depends on how many solves you did. If you did 3 solves in a year, it would be fairly low, but if you did 3 million in a year it would be higher. The number of LL possibilities are 4!*4!*2^4*3^4/(2*3*2*4), which simplifies to 62,208.
Then one would have to divide by 4 to get the proper denominator since an LL skip usually means the last layer is solved minus parity. This yields a result of 15,552 possible cases, disregarding AUF.

Thus, the chances on any given solve of getting an LL skip are 1/15,552, or 0.0064300412%. This does not take "forcing" an LL skip into account, as often when a cuber is solving the cube they may run into a situation wherein the cube needs U2 R U2 R' to be solved, and the cuber (who would typically solve the case using U R U' R') realizes that they can get an LL skip if they use U2 R U2 R' to solve it. If we pretend that LL is not predictable at all whatsoever while solving the final F2L pair, then we can proceed. It is also good to note that ZZ / Petrus solvers are advantaged in this respect because at some point in the solve EO is done, meaning that the chances of getting an LL skip on any given solve multiply by 8 because all the edges are oriented. (1 in 1944; 0.0514403292%) DR solvers are further advantaged because the DR stage of a solve requires the corners to be oriented as well, thus multiplying the chances of an LL skip by an additional factor of 27. (1 in 72; 1.3888888888...%) HTR users enjoy a 1 in 4 chance of an LL skip, if you don't count AUF. (A 25% chance!)

Putting all of this aside, we can look at the chances of a CFOP solver [who is unable to see the last layer] getting a last layer skip.
To find the chances of this person getting at least 2 LL skips, we have to first determine the chances of them not getting any. We can put the number of solves in any given time period in as x, and then the chances of getting an LL skip based on this will be a function of x. The best equation I could come up with is:
1-\left(\left(\frac{15561}{15562}\right)^{x}\right)^{2}
which when put into the Desmos Graphing Calculator yields a probability chart.
If you can't access Desmos, here's an image of the equation:

I hope that was informative, but not too wordy. And I hope I was correct!


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## Imsoosm (Aug 25, 2022)

White KB said:


> @ottozing How do I get Platinum Member status? Or is it just a legacy feature?


Not completely sure but it's probably just an old member thing. (Kirjava is called Colourful)


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## Thom S. (Aug 25, 2022)

Imsoosm said:


> Kirjava is called Colourful


If you are a premium Member you can change what is says under your Username.
Kirjava used Colourful.


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## SollsIsCool (Aug 26, 2022)

how common is a accidental x cross


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## ruffleduck (Aug 26, 2022)

SollsIsCool said:


> how common is a accidental x cross


I assume you are asking what the probability is of having xcross solved provided cross is solved.
(4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944


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## turtwig (Aug 26, 2022)

ruffleduck said:


> I assume you are asking what the probability is of having xcross solved provided cross is solved.
> (4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944


I don't know where you got this number, but it's much higher. The chance that a given pair is solved is 1/8 * 1/2 * 1/8 * 1/3 = 1/384 (edge perm, edge orientation, corner perm, corner orientation) and there are 4 pairs (which does not mean the probability is 4x, although it would probably be close to that).

EDIT: I'm going to try to calculate this.

So there are 8! * 2^7 * 8! * 3^7 / 2 positions with a given cross solved, which is (edge permutation) * (edge orientation) * (corner permutation) * (corner orientation) / (parity).

For a given pair, there are 7! * 2^6 * 7! * 3^6 / 2 positions. Then, there are 4 * 7! * 2^6 * 7! * 3^6 / 2 positions where at least one pair is solved.

However, this overcounts positions where there are at least 2 pairs solved. Given 2 pairs, there are 6! * 2^5 * 6! * 3^5 / 2 positions. There are (4 choose 2) = 6 combinations of 2 pairs, so we subtract 6 * 6! * 2^5 * 6! * 3^5 / 2 to correct for overcounting positions with 2 pairs.

Now, we need to add back positions with 3 pairs (which have been counted (3 choose 1) - (3 choose 2) = 0 times). We add (4 choose 3) * 5! * 2^4 * 5! * 3^4 / 2.

Finally, positions with 4 pairs have been counted (4 choose 1) - (4 choose 2) + (4 choose 3) = 2 times, so we subtract 4! * 2^3 * 4! * 3^3 / 2.

Thus, there are 4 * 7! * 2^6 * 7! * 3^6 / 2 - 6 * 6! * 2^5 * 6! * 3^5 / 2 + 4 * 5! * 2^4 * 5! * 3^4 / 2 - 4! * 2^3 * 4! * 3^3 / 2 = 2358218126592 positions with at least one pair solved. Thus, our probability is 2358218126592 / (8! * 2^7 * 8! * 3^7 / 2) = 37908599/3657830400, which is about *1.04%*. I hope I did this right, I'm pretty rusty on this kind of math. This aligns with my intuition though, ending up with a pair solved accidentally is not too uncommon in my experience, although I sometimes try to preserve corners/edges during my cross solution without (consciously) solving the pair entirely, so the chance I get accidental pairs is probably higher because of that.

This doesn't include the minute chances of ending up with an xcross on another color, but that would be much harder to calculate and also is probably against the spirit of the question.


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## ruffleduck (Aug 26, 2022)

turtwig said:


> I don't know where you got this number, but it's much higher.


I multiplied the number of cases of a specific slot being solved by 4, thinking that would cover cases for any slot. Good catch


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## Bruce MacKenzie (Aug 26, 2022)

ruffleduck said:


> I assume you are asking what the probability is of having xcross solved provided cross is solved.
> (4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944


There are 24 ways to place the first corner and 16 ways to place the fifth edge. So:

24 * 16 / 4

since the xcross may be completed on any of the four corners of the face. 1 in 96 is the chance of a windfall xcross after forming the cross.


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## ottozing (Aug 27, 2022)

ottozing said:


> Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction
> 
> 1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)
> 
> ...


 Bumping this just to say that part one of question one has already been answered via some work Cale is currently doing - https://docs.google.com/spreadsheet...4TLW45M2ZkctQZAVxgQqEtlUs0/edit#gid=973034617


----------



## SollsIsCool (Aug 27, 2022)

ruffleduck said:


> I assume you are asking what the probability is of having xcross solved provided cross is solved.
> (4 * 7! * 7! * 2^6 * 3*6) / (8! * 8! * 2^7 * 3^7 / 2) = 1/1944


i got lucky then!!


----------



## ottozing (Aug 30, 2022)

ottozing said:


> Yet another set of FMC related requests. Looking to adjust my time management strategy and these would likely help me a bit with determining what I end up doing in terms of checking EO's & utilizing NISS prediction
> 
> 1 - What are the odds that after your EO is solved, the corner and edge state on at least one of the two axis' is 4C4E, 4C2E, or 3C2E collectively without adding any moves after the EO? These are the 3 triggers I sometimes NISS for since they can be quite easy (R for 4C4E, R U2 R for 4C2E, R U/U' R for 3C2E). Similarly, what are the odds that an EO case will be only one move away from these 3 pre mentioned states? (This may be trickier to calculate accurately since for an EO that ends with F there's 8 possible 1 move follow-ups, and for EO's ending in F B there are 16. It's also not as important as my first request)
> 
> ...


Bump again because I have a better 2nd question than the one I gave

Rather than knowing the % breakdowns, I think it would be better to know how often a 4C4E gives x number of pairs. My reason for this is that, hypothetically, if I know that 4 pairs is a 10% chance & 3 pairs is a 33% chance (random numbers for now since I don't actually know the odds), I could know that if I have 20 possible 5 moves to 4C4E to choose from that I'll likely have 2 which are 4 pairs & more likely to give a short finish when I NISS for those specifically


----------



## Noob's Cubes (Sep 2, 2022)

What are the chances of a complete x-cross skip?


----------



## Jorian Meeuse (Sep 2, 2022)

I dont have proof so if you dont believe me its fine, i probably also wouldnt belive it if someone said this but my first sub 6 (5.33) was with a premade x-cross, i think on yellow. As said in the easy/lucky/a lot of other things scrambles thread, i lost the scramble, but if i somehow retrieve it, ill make sure to post it here.


----------



## Bruce MacKenzie (Sep 3, 2022)

Noob's Cubes said:


> What are the chances of a complete x-cross skip?


The number of permutations of one corner cubie and five edge cubies may be calculated:

24 • ( 24 • 22 • 20 • 18 • 16) = 72,990,720

The X-Cross pattern may be applied to the cube in 24 orientations, so the chance of a random scramble giving an instance of X-Cross is 24 times that or 1 in 3,041,280.


----------



## ottozing (Sep 14, 2022)

ottozing said:


> Bump again because I have a better 2nd question than the one I gave
> 
> Rather than knowing the % breakdowns, I think it would be better to know how often a 4C4E gives x number of pairs. My reason for this is that, hypothetically, if I know that 4 pairs is a 10% chance & 3 pairs is a 33% chance (random numbers for now since I don't actually know the odds), I could know that if I have 20 possible 5 moves to 4C4E to choose from that I'll likely have 2 which are 4 pairs & more likely to give a short finish when I NISS for those specifically


bump

anyone?


----------



## Savi349 (Sep 26, 2022)

I have a quick question about the PLL sheet that was developed by Feliks Zemdegs. Does anyone know how the probabilities were calculated for the different algorithms? My solves do generally seem to follow the probabilities listed, but I was wondering how he came to these numbers – did he do his own solves and note down each PLL he encountered? Was it some sort of computer generation? Because I also added up the probabilities, however they do not add up to 1; where am I going wrong? 

Also, does a "skip" count as a PLL case, and therefore the probability of a "skip" is also part of these listed probabilities? Or does the PLL sheet somehow omit the probability of a skip? 
Any answers to any of my queries are greatly appreciated! My intense curiosity of this paired with my lack of confidence with maths is not a very pleasant combination. 
Thank you!


----------



## OreKehStrah (Sep 26, 2022)

Savi349 said:


> I have a quick question about the PLL sheet that was developed by Feliks Zemdegs. Does anyone know how the probabilities were calculated for the different algorithms? My solves do generally seem to follow the probabilities listed, but I was wondering how he came to these numbers – did he do his own solves and note down each PLL he encountered? Was it some sort of computer generation? Because I also added up the probabilities, however they do not add up to 1; where am I going wrong?
> 
> Also, does a "skip" count as a PLL case, and therefore the probability of a "skip" is also part of these listed probabilities? Or does the PLL sheet somehow omit the probability of a skip?
> Any answers to any of my queries are greatly appreciated! My intense curiosity of this paired with my lack of confidence with maths is not a very pleasant combination.
> Thank you!


Sure!

Basically you can think of full PLL as 72 cases. However, you also get some cases that are fully rotational symmetric, and can AUF the non-fully rotationally symmetric cases. So the 4 angles of T perm can be treated as 1 unique case by AUFing.

As such, any fully rotational case including a skip case is 1/72, while any other case has a probability of 4/72, which is equal to 1/18, because of the fact it has 4 AUFs.

Edit: Z perm and E perm have a probability of 1/36 because for solved cp and diag cp, the cp swap is rotationally symmetric, and those 2 PLLs have U2 angle symmetry.

Hope this helps!


----------



## Mastermind2368 (Sep 26, 2022)

(I only recently got interested in probability, so feel free to call me out if I got anything wrong)

To add on to the answer above, the original 72 cases is calculated from doing:

(4!*4!)/(2*4) 

Incase you haven't learnt about it yet, the "!" Means to make a number factorial. This means to multiple it by every smaller natural number. In other words, 4! = (4*3*2*1).

The two 4! Represent all the possibile corner and edge cases. Because each are independent, they're multiplied together, so that each corner case can be lined up with each edge case.

The total number produced by that is quite high though. But there are a couple of other things that need to be taken into account. Firstly, parity. You can never have an edge swapped with another edge, or a corner swapped with a corner. So this reduces the cases by half. Then secondly, remember that we can do any auf. Think of it like this, even though there's only one h perm alg, the aufs would turn it in to four different algs. If we were solving for 8 pieces of the cube that weren't on the same face, we couldn't do an auf like this, but since we can, it reduces the total amount of cases to an eighth of the original amount.

Therefore the total amount is (4!*4!)/(2*4) 
Which simplifies to: 576/8
Which is: 72

If you're interested in learning more about this stuff, I'd recommend watching Jperm's video about the total probability for the cube. He explains the concepts much better than I did


----------



## Savi349 (Sep 27, 2022)

OreKehStrah said:


> Sure!
> 
> Basically you can think of full PLL as 72 cases. However, you also get some cases that are fully rotational symmetric, and can AUF the non-fully rotationally symmetric cases. So the 4 angles of T perm can be treated as 1 unique case by AUFing.
> 
> ...





Mastermind2368 said:


> (I only recently got interested in probability, so feel free to call me out if I got anything wrong)
> 
> To add on to the answer above, the original 72 cases is calculated from doing:
> 
> ...


Thank you both so much! It's been really helpful having the maths explained like this, both explanations were beyond helpful! I love Jperm's videos — I didn't realise he had a video on probability, I'll make sure to check it out!


----------



## Hashbrown132 (Oct 17, 2022)

What are the odds of getting a solved layer on a 2x2 from a random scramble


----------



## Bruce MacKenzie (Oct 17, 2022)

Hashbrown132 said:


> What are the odds of getting a solved layer on a 2x2 from a random scramble


For any one face the odds are 1 in (21•18•15) or 1 in 5,670. Since there are six faces the odds are six times that or 1 in 945.


----------



## Herbert Kociemba (Nov 10, 2022)

Bruce MacKenzie said:


> For any one face the odds are 1 in (21•18•15) or 1 in 5,670. Since there are six faces the odds are six times that or 1 in 945.


I do not think that this is correct. The problem is that there is not always exactly one solved layer but there also are nontrivial cases where two layers are solved simultaneously. This makes the computation more complicated. The probability is less than your result because you count some situations more than once.


----------



## Osric (Nov 10, 2022)

Herbert Kociemba said:


> I do not think that this is correct. The problem is that there is not always exactly one solved layer but there also are nontrivial cases where two layers are solved simultaneously. This makes the computation more complicated. The probability is less than your result because you count some situations more than once.



If what you say is correct then the way I am calculating it is also wrong. Can we spot the problem?

There are 7! positions the corner cubies could be in, and 3^7 possible orientations. So the total number of permutations of the cubies of a 2x2 is 5040*2187. 

Let’s count all the cases where the top face is in a specific permutation and orientation that corresponds to START’s top face. If we allow symmetries of rotation only, there are 6*4 symmetries that are equivalent so 24 ways to rewrite the coordinate system of the cube so that we recognize any solved face. The remaining 3 corners can be in 3! positions and 3^3 orientations, for a total of 24*24*27 situations where one face is solved.

If by a ”random scramble” we mean a truly random choice among all the possible permutations (and not a “fairly scrambled”) cube, then the odds ought to be (24*24*27)/(5040*2187) = 0.14% or 1/709. 

This is higher than the other result and therefore also wrong but I can’t see how?

Osric


----------



## Osric (Nov 10, 2022)

Osric said:


> Let’s count all the cases where the top face is in a specific permutation and orientation that corresponds to START’s top face. If we allow symmetries of rotation only, there are 6*4 symmetries that are equivalent so 24 ways to rewrite the coordinate system of the cube so that we recognize any solved face. The remaining 3 corners can be in 3! positions and 3^3 orientations, for a total of 24*24*27 situations where one face is solved.



Ahem. Perhaps 3! != 24 :-D … and then the result becomes 0.035% or 1/2835. Oops


----------



## Herbert Kociemba (Nov 10, 2022)

Osric said:


> There are 7! positions the corner cubies could be in, and 3^7 possible orientations. So the total number of permutations of the cubies of a 2x2 is 5040*2187.


There are two ways to count. If you fix a corner, ther are 7! permutations and 3^6 orientations, so you get 3674160 positions. Or you do not fix one corner - I use this approach for my computation since you have more symmetries - you have 8! permuations and 3^7 orientations and you get 88179840, which is 24 times larger and corresponds to the 24 possible orientations o a cube in space. 

I tried a correct computation myself but I also see some possibility that I did something wrong. Here is my approach:

Let use the abbreviation L_U if *exactly* the layer U is correct and L_UF if *exactly* the layers U and F are correct for example. Then there are the following distinct cases:
L_URFDLB: all layers are solved, 1 case
L_UD, L_RL, L_FB: exactly two opposite layers solved, 3 cases
L_UR, L_UF, L_UL, L_UB, L_RF, L_RD, L_RB, L_FD, L_FL, L_DL, L_DB, L_LB: exactly two adjacent layers solved, 12 cases
L_U, L_R, L_F, L_D, L_L, L_B: exactly one layer solved, 6 cases

L_URFDLB: The positions are determined by the URF corner which are 24 different positions.

L_UD: The 24 positions of the URF corner define the solved U-layer. For some specific solved U layer there are 3 different possibilites for a solved D layer (not 4, one case results in L_URFDLB ). So L_UD has 24*3 positions.

L_UR: The 24 positions of the URF corner define the solved U-layer and the solved R layer. For some specific solved U and R layer there are 5 possible positions for the yet undetermined DLF and DLB corners (not 6, one case results in L_URFDLB). So L_UR has 24*5 positions.

L_U: The 24 positions of the URF corner define the solved U-layer. The 4 yet undetermined corners of the D layer have 4!*3^3 = 648 possibilities for each specific U layer. But since the 24*648 possibilites include also the cases L_UD and L_UR, L_UF, L_UL, L_UB and L_URFDLB we have to subtract 24*3 and 4 times 24*5 and 24*1 and we get 24*(648-3-4*5-1) = 24*624 different positions.

Now the total number of positions with at least one solved layer should be
N(L_URFDLB)+ 3*N(L_UD)+12*N(L_UR)+ 6*N(L_U)= 24+3*24*3+12*24*5+6*(24*624) = 91536

So the probability is 91536/88179840 = 1907/1837080 which is slightly less than 1/945

The probability for exactly one solved layer is btw.

6*N(L_U)/88179840=26/25515


----------



## Bruce MacKenzie (Nov 10, 2022)

Herbert Kociemba said:


> I do not think that this is correct. The problem is that there is not always exactly one solved layer but there also are nontrivial cases where two layers are solved simultaneously. This makes the computation more complicated. The probability is less than your result because you count some situations more than once.


You're right. The sets do overlap. The identity cube would be in all six sets for example. I don't know if the probability would be reduced significantly by this. This is a fairly small group. One could generate all the states and count how many have at least one solved face. I'll see what i can do.

I knocked out some code and found 2056 2x2 states with a least one face solved. That gives odds of 2056 in 3,674,160 or 1 in 1787.042847.
You are quite right. This is a very significant difference.

Since my last edit I have filled in some metrics:

There are 2056 states with at least one solved face. There are 2010 states with exactly one solved face. There are 45 states with exactly two solved faces and the identity cube with all six faces solved. There are 9 states with two opposing faces solved but not aligned with one another. I include these in the exactly one face solved count.

A clap on the back and a hearty "Well Done!" to anyone who can derive these numbers from first principles.


----------



## Jorian Meeuse (Nov 10, 2022)

Herbert Kociemba said:


> There are two ways to count. If you fix a corner, ther are 7! permutations and 3^6 orientations, so you get 3674160 positions. Or you do not fix one corner - I use this approach for my computation since you have more symmetries - you have 8! permuations and 3^7 orientations and you get 88179840, which is 24 times larger and corresponds to the 24 possible orientations o a cube in space.
> 
> I tried a correct computation myself but I also see some possibility that I did something wrong. Here is my approach:
> 
> ...


Is this for an actual first layer skip (as in half the cube being solved) or just a face skip where the permutation of that face isn't yet decided?

Edit: from the computation it actually seems quite obvious now that this is referring to a solved layer, not face.


----------



## Foreright (Nov 10, 2022)

Here’s one for Roux - I have 68 CMLL skips (ie. Top corners solved / oriented after block stage) out of my last 1250 or so solves so approx. 2% which is a fair bit better than expected (skip is 1/162 iirc).

I noted earlier when I was looking at smart cube stats that I have one LSE stage skip in my total 3500 or so solves recorded (ie. After CMLL the cube was completely solved). This was also with no AUF. What are the chances of that?


----------



## Herbert Kociemba (Nov 10, 2022)

Bruce MacKenzie said:


> You're right. The sets do overlap. The identity cube would be in all six sets for example. I don't know if the probability would be reduced significantly by this. This is a fairly small group. One could generate all the states and count how many have at least one solved face. I'll see what i can do.
> 
> I knocked out some code and found 2056 2x2 states with a least one face solved. That gives odds of 2056 in 3,674,160 or 1 in 1787.042847.
> You are quite right. This is a very significant difference.
> ...


----------



## Herbert Kociemba (Nov 10, 2022)

Are you sure these new values are correct? I mean now you get 2056/3673160 = 0.00056 fo at least one solved face which seems to be too far away from your original 1/945 = 0.0011 which is not correct but at least quite a good approximation because you just ignore the small overlaps. And my hopefully correct value is 1907/1837080 = 0.0010 for at least one solved layer. Why your new probability is only about half of this?

Edit: 1907/1837080 = 3814/3674160, so for your way of counting there should be 3814 and not 2056 states with at least one solved layer.
If you fix for example the DLB corner (what you essentially do if you take 3,674,160 positions in total ) the U, R and F layer counts should be different from your D, B and L layer counts.


----------



## Bruce MacKenzie (Nov 11, 2022)

Herbert Kociemba said:


> Are you sure these new values are correct? I mean now you get 2056/3673160 = 0.00056 fo at least one solved face which seems to be too far away from your original 1/945 = 0.0011 which is not correct but at least quite a good approximation because you just ignore the small overlaps. And my hopefully correct value is 1907/1837080 = 0.0010 for at least one solved layer. Why your new probability is only about half of this?
> 
> Edit: 1907/1837080 = 3814/3674160, so for your way of counting there should be 3814 and not 2056 states with at least one solved layer.
> If you fix for example the DLB corner (what you essentially do if you take 3,674,160 positions in total ) the U, R and F layer counts should be different from your D, B and L layer counts.


Well I'm fairly confident in my counts. Generating all members of the group is straightforward using radix encoding to compress the position permutation and the orientations to a number between 0 and 3,674,160. And testing for a solved face is trivial. I can not see how I could have under counted to that degree.
`-(void)test2: (NSString *)input
{
NSData *state,*item;
OSG_SYMTAG sym;
const OSG_SYMTAG *p;
NSUInteger index, atLeast1, exactly1, exactly2;
NSMutableSet *all,
*set1,
*set2;
BOOL flag1,
flag2;
all = [NSMutableSet set];
set1 = [NSMutableSet set];
set2 = [NSMutableSet set];
for(index = 0 ; index < 3674160 ; index++)
{
@autoreleasepool
{
state = [self t2_stateForIndex: index];
[all addObject: state];
flag1 = flag2 = YES;
for(sym = 0 ; sym < 24 ; sym++) // Test all orientations for UP face solved
{
item = [self conjugateOfState: state bySymtag: sym ];
p = [item bytes];
if(
flag1 == YES &&
p[UFR] == 0 &&
p[URB] == 0 &&
p[UBL] == 0 &&
p[ULF] == 0 )
{
[set1 addObject: state];
flag1 = NO; //only count once
}
if(
flag2 == YES &&
p[UFR] == 0 &&
p[URB] == 0 &&
p[UBL] == 0 &&
p[ULF] == 0 &&
p[DRF] == 0 &&
p[DBR] == 0)
{
[set2 addObject: state];
flag2 = NO; //only count once
}
}
}
}
atLeast1 = [set1 count];
[set1 minusSet: set2];
exactly1 = [set1 count];
exactly2 = [set2 count] - 1;
[self reportDone];
}
// Return a cube state with the edges and the UFR cubie solved as encoded by index
-(NSData *)t2_stateForIndex: (NSInteger)index
{
int i,n,p,
orient[7];
NSUInteger position, orientation,
perm[7];
OSG_SYMTAG state[20];
static NSUInteger last;
orientation = index % 729; //3^6 orientations
position = index / 729;
[self radixDecode: position
length: 7
toBuffer: perm];
orient[0] = 0;
for(n = 1 ; n < 7 ; n++)
{
orient[n] = orientation % 3;
orient[0] += orient[n];
orientation /= 3;
}
orient[0] = (21 - orient[0]) % 3;
for( n = 0 ; n < 13 ; n++)
state[n] = 0 ;
for( p = 0 , i = 0 ; i < 6 ; i++ )
for( n = i + 1 ; n < 7 ; n ++ )
if( perm[I] > perm[n] )
p ^= 1;
if( p == 1 )
state[UF] = state[UB] = [OhGroup symtagForKey: @"-x, y,-z"]; //balance parity to give a valid 3x3 state by swapping the UF and UB cubies
RBK_Cubie cubie, cubicle;
for( i = 0 , n = 13 ; n < 20 ; i++ , n++ )
{
cubicle = (RBK_Cubie)perm[I] + 13;
cubie = n;
state[n] = [self symtagToPutCubie: cubie
inCubicle: cubicle
withOrientation: orient[I] ];
}
return [NSData dataWithBytes: state length: sizeof(OSG_SYMTAG [20])];
}`[/I][/I][/I]


----------



## OreKehStrah (Nov 11, 2022)

Foreright said:


> Here’s one for Roux - I have 68 CMLL skips (ie. Top corners solved / oriented after block stage) out of my last 1250 or so solves so approx. 2% which is a fair bit better than expected (skip is 1/162 iirc).
> 
> I noted earlier when I was looking at smart cube stats that I have one LSE stage skip in my total 3500 or so solves recorded (ie. After CMLL the cube was completely solved). This was also with no AUF. What are the chances of that?






 there is a video of a person having an LSE skip in competition. Pretty neat!


----------



## Herbert Kociemba (Nov 11, 2022)

Bruce MacKenzie said:


> Well I'm fairly confident in my counts. Generating all members of the group is straightforward using radix encoding to compress the position permutation and the orientations to a number between 0 and 3,674,160. And testing for a solved face is trivial. I can not see how I could have under counted to that degree.


I do not understand this code. But at least we should reach an agreement what a "solved face" is. If you fix for example the DLB corner I still would count the U-face or better U-layer solved if we apply an U-move to the solved cube. Do you also count like this?


----------



## NigelTheCuber (Nov 11, 2022)

Jorian Meeuse said:


> I dont have proof so if you dont believe me its fine, i probably also wouldnt belive it if someone said this but my first sub 6 (5.33) was with a premade x-cross, i think on yellow. As said in the easy/lucky/a lot of other things scrambles thread, i lost the scramble, but if i somehow retrieve it, ill make sure to post it here.


you should participate in the lottery ig


----------



## NigelTheCuber (Nov 11, 2022)

what is the probability of at least 2 free pairs and pll skip on the same solve


----------



## Bruce MacKenzie (Nov 11, 2022)

Herbert Kociemba said:


> I do not understand this code. But at least we should reach an agreement what a "solved face" is. If you fix for example the DLB corner I still would count the U-face or better U-layer solved if we apply an U-move to the solved cube. Do you also count like this?


I did some testing since my last post and found there are indeed a lot of cases I was missing. The problem was with states with solved faces not color aligned with the UFR reference cubie. I thought that by looking at all the rotation conjugates I would catch these cases but that is not true. I changed the way I test for solved faces and now my numbers are:

At least 1 solved face: 3814
Exactly 1 solved face: 3753
Exactly 2 solved faces: 60
6 solved faces: 1 (the identity cube)

The number for at least 1 solved face agrees with your number I believe.


-(void)test2: (NSString *)input
{
NSData *state, *item;
NSArray *s6;
NSString *conj;
const OSG_SYMTAG *p;
NSUInteger index, 
atLeast1, 
exactly1, 
exactly2;
NSMutableSet *set1,
*set2;
BOOL flag1,
flag2;

s6 = [OhGroup subgroupNamed: @"S6xyz"];

set1 = [NSMutableSet set];
set2 = [NSMutableSet set];

for(index = 0 ; index < 3674160 ; index++)
{
@autoreleasepool
{
state = [self t2_stateForIndex: index];

flag1 = flag2 = YES;
for(conj in s6) //map each of the 6 faces onto the UP face
{
item = [self conjugateOfState: state 
bySymtag: [conj intValue]];
p = [item bytes];

if( flag1 == YES &&
p[UFR] == p[URB] &&
p[UFR] == p[UBL] &&
p[UFR] == p[ULF])
{
[set1 addObject: state];
flag1 = NO; //only count once

if( flag2 == YES &&
p[UFR] == p[DRF] &&
p[UFR] == p[DBR])
{
[set2 addObject: state];
flag2 = NO;
}

if( flag2 == YES &&
p[UFR] == p[DBR] &&
p[UFR] == p[DLB])
{
[set2 addObject: state];
flag2 = NO;
}

if( flag2 == YES &&
p[UFR] == p[DLB] &&
p[UFR] == p[DFL])
{
[set2 addObject: state];
flag2 = NO;
}

if( flag2 == YES &&
p[UFR] == p[DRF] &&
p[UFR] == p[DFL])
{
[set2 addObject: state];
flag2 = NO;
}

}
}
}
}

atLeast1 = [set1 count];
[set1 minusSet: set2];
exactly1 = [set1 count];
exactly2 = [set2 count] - 1;

[self report: @"\n\nExactly One"];
for( state in set1 )
[self report: [NSString stringWithFormat: @"\n%@",[self configurationForState: state]]];

[self report: @"\n\nTwo or more(i.e. Identity"];
for( state in set2 )
[self report: [NSString stringWithFormat: @"\n%@",[self configurationForState: state]]];


[self reportDone];
}


----------



## Herbert Kociemba (Nov 11, 2022)

Yes, my theoretical reasoning above also gave the 3814, so 1907/1837080 for the probability should be correct.

Sadly, the values for exactly one and two solved faces still differ. My 26/25515 for the probability corresponds to 3744 cases and not to 3753 for exactly 1 solved layer. And my 9 less cases here correspond to 9 additional cases with 2 solved layers. I have 69 here: 60 with two adjacent solved layers and 9 with two parallel solved layers. Is it possible that you did not recognize the second kind as two solved layers?


----------



## Bruce MacKenzie (Nov 11, 2022)

Herbert Kociemba said:


> Yes, my theoretical reasoning above also gave the 3814, so 1907/1837080 for the probability should be correct.
> 
> Sadly, the values for exactly one and two solved faces still differ. My 26/25515 for the probability corresponds to 3744 cases and not to 3753 for exactly 1 solved layer. And my 9 less cases here correspond to 9 additional cases with 2 solved layers. I have 69 here: 60 with two adjacent solved layers and 9 with two parallel solved layers. Is it possible that you did not recognize the second kind as two solved layers?


The nine cases with two opposing solved layers not color aligned with one another are counted as 1 solved layer in my count. If the layers aren't color aligned I deem one of the layers unsolved. If they are aligned then one has the identity cube. I was aware of those cases (I mentioned them in my first erroneous accounting).


----------



## SYUTEO (Nov 11, 2022)

Sorry if this is asked before but what are the odds of having a pyraminx scramble with all tips solved?


----------



## NigelTheCuber (Nov 11, 2022)

SYUTEO said:


> Sorry if this is asked before but what are the odds of having a pyraminx scramble with all tips solved?


1/81, one tip being solved has 1/3 chance and 3^4 is 81


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## Herbert Kociemba (Nov 11, 2022)

Bruce MacKenzie said:


> The nine cases with two opposing solved layers not color aligned with one another are counted as 1 solved layer in my count


Though I do not see the logic behind this we now have same numbers and I am glad that my combinatorical approach was correct. Thanks for confirming the numbers with your program.


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## Thom S. (Nov 11, 2022)

NigelTheCuber said:


> what is the probability of at least 2 free pairs and pll skip on the same solve


Do you even know yourself what at least 2 free pairs is?


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## NigelTheCuber (Nov 11, 2022)

Thom S. said:


> Do you even know yourself what at least 2 free pairs is?


R U R', R U' R' and mirrors


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## Bruce MacKenzie (Nov 11, 2022)

Herbert Kociemba said:


> Though I do not see the logic behind this we now have same numbers and I am glad that my combinatorical approach was correct. Thanks for confirming the numbers with your program.


I would argue that when a face (layer) is solved that defines the color scheme of the cube. For a second layer to be solved it must conform to that color scheme. If two parallel faces are solved wrt two different color schemes then they can't both be solved. One is solved and the other is one turn from being solved.

But this is a mere quibble. Our numbers agree. I congratulate you on performing a first principles analysis which I shied away from as "too complicated" and went with a brute force approach instead.


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## Imsoosm (Nov 12, 2022)

NigelTheCuber said:


> R U R', R U' R' and mirrors


There's actually more than that. You can use FreeFOP and have 3/4 of a cross and use F moves to insert pairs, or you finish some pairs first before completing the cross


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## NigelTheCuber (Nov 12, 2022)

Imsoosm said:


> There's actually more than that. You can use FreeFOP and have 3/4 of a cross and use F moves to insert pairs, or you finish some pairs first before completing the cross


Ok then ig my question is rly hard to answer, but like using standard cfop and doing the cross normally without forcing free pairs and stuff like that


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## yCArp (Nov 13, 2022)

May I know how many cases there would be for 4x4 Centres?


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## Herbert Kociemba (Nov 14, 2022)

xyzzy said:


> (i) Please don't reply to nine-year-old posts unless it's absolutely relevant and necessary.
> 
> (ii) This is almost correct, but it doesn't account for scrambles with multiple crosses already solved… or it would be, if you actually calculated your approximation correctly. Your calculation is off by a factor of two and you should've gotten 1/31680.
> 
> (iii) The exact probability is around 1/31704.


The recent post about the probability of a solved layer on a random 2x2x2 motivated me to do a paper and pencil computation for the probability for at least one cross on a random 3x3x3 which looks quite difficult on first sight if you want to include multiple crosses. I did not find anything better than the reply above in this probabilty thread.
I got 15471187/490497638400 which matches the depth 0 count for color neutral cross solving in http://www.cubezone.be/crossstudy.html (30,942,374 cases out of 980,995,276,800).
If anybody is interested I can give a relatively simple derivation here.


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