# I got 13 pll skips on my 2x2 in a row



## tfkscores (Jun 30, 2009)

i was using official scrambles and i dont mess up when i scramble them either. ended on 13 which is an unlucky number? best averages of my life all times were like under 7 seconds. really weird though i should share. sorry if this is the wrong forum.


----------



## JTW2007 (Jun 30, 2009)

13=unlucky.

Exception: PLL Skips.


----------



## tfkscores (Jun 30, 2009)

haha i know right? it was so weird though


----------



## cmhardw (Jun 30, 2009)

If you orient first and then permute when solving 2x2 then the probability of that is 1/6^13 = 1/13060694016

That is a one in thirteen billion chance approximately. As a comparison, that is like flipping heads on a coin 34 times in a row. That is also 54 times luckier than getting two full LL skips on 3x3x3 in a row.

Chris


----------



## StachuK1992 (Jun 30, 2009)

Which basically means that few of us believe you.


----------



## JTW2007 (Jun 30, 2009)

Stachuk1992 said:


> Which basically means that few of us believe you.



Which is also augmented by the fact that most people don't do avg. of 13s anymore.


----------



## cmhardw (Jun 30, 2009)

Stachuk1992 said:


> Which basically means that few of us believe you.



I won't say it's impossible, since it clearly is possible. I just think that getting such a string of PLL skips (also you did not mention anything about LL skips, so that is how I calculated it) is incredibly improbable. You basically just hit the super multi-million dollar jackpot on the lottery with those odds.

Chris


----------



## Stefan (Jun 30, 2009)

JTW2007 said:


> 13=unlucky


False.



cmhardw said:


> If you orient first and then permute when solving 2x2 then the probability of that is 1/6^13 = 1/13060694016


That's for getting 13 PLL skips in certain 13 solves. But what is the probability of getting 13 PLL skips in a row somewhere in let's say 100 solves?


----------



## tfkscores (Jun 30, 2009)

I just scrambled the cube the way it told me and after 3 pll skips in a row i started counting cause that was really weird. they are so much easier to get on the 2x2. it was really weird but i swear to you guys im not lieing. i guess iwas just really lucky.


----------



## blah (Jun 30, 2009)

StefanPochmann said:


> cmhardw said:
> 
> 
> > If you orient first and then permute when solving 2x2 then the probability of that is 1/6^13 = 1/13060694016
> ...



Exactly 13 or >=13?


----------



## tfkscores (Jun 30, 2009)

no i got exactly 13 skips. 14 was just a regular oll pll


----------



## Stefan (Jun 30, 2009)

blah said:


> StefanPochmann said:
> 
> 
> > cmhardw said:
> ...


I meant the second version, but now I'm interested in both (and btw I don't know the answers).


----------



## amostay2004 (Jun 30, 2009)

Do you still have the scrambles and solutions?


----------



## sooland (Jun 30, 2009)

I still find myself extremely skeptical. The odds of getting a pll skip 13 times in a row is less than the odds of getting a jackpot on a slot machine. (Really!)


----------



## tfkscores (Jun 30, 2009)

no i dont have the scrambles. idc if you believe me or not i just thought i should share. i did get it though and since im a slow 2x2 solver almost all of my times were under 7. for good people they wouldlve been like under 3 though lol.


----------



## Mike Hughey (Jun 30, 2009)

tfkscores said:


> no i dont have the scrambles. idc if you believe me or not i just thought i should share. i did get it though and since im a slow 2x2 solver almost all of my times were under 7. for good people they wouldlve been like under 3 though lol.



What method do you actually use?

I'm embarrassed because using Guimond, I'm generally barely under 7 seconds even when I get to skip the last step.


----------



## brunson (Jun 30, 2009)

Don't worry, Mike. The only reason I'm under 10 seconds is because I learned Guimond.


----------



## Mike Hughey (Jun 30, 2009)

brunson said:


> Don't worry, Mike. The only reason I'm under 10 seconds is because I learned Guimond.



Hey, I'm not the only one! That's nice. (I was averaging around 11 seconds with Ortega.)


----------



## tfkscores (Jun 30, 2009)

i just solve one side then oll and pll its really easy. i can average under 10 easily.


----------



## brunson (Jun 30, 2009)

tfkscores said:


> i just solve one side then oll and pll its really easy. i can average under 10 easily.


And here's how to do 3x3x3 in under five seconds: Cross, F2L, OLL, PLL. Just do it really fast.


----------



## Johannes91 (Jun 30, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > StefanPochmann said:
> ...


I don't know an easy way to calculate it on paper, so I asked my computer to do it for me and it gave 4.700277693455794e-9 for the first one and 5.627572299115469e-9 for the second one (the rationals have quite large numerators and denominators).

I guess you're more interested in the how than the what, though.


----------



## Rune (Jun 30, 2009)

StefanPochmann said:


> blah;200162
> Exactly 13 or >=13?[/QUOTE said:
> 
> 
> ...


----------



## dougbenham (Jun 30, 2009)

I would lol if we found out this guy was using CLL (or COLL whatever you call it) and didn't even realize it


----------



## panyan (Jun 30, 2009)

cmhardw said:


> the probability of that is 1/6^13 = 1/13060694016



if you did that thirteen times then it comes to 1/169000000000 chance, the chance of winning the lottery is 1/(1.4x10^51). That is an impressive pll skip scenario if it did happen


----------



## Rune (Jun 30, 2009)

"the chance of winning the lottery is 1/(1.4x10^51). "

Think it over!


----------



## Stefan (Jun 30, 2009)

Johannes91 said:


> I guess you're more interested in the how than the what, though.


LOL. Indeed.


----------



## tfkscores (Jun 30, 2009)

ok this isnt bs i actually did do it and i should've turned my camera on. it doesnt matter if you believe me cause i wouldnt even believe it. lol hope i didnt upset any1


----------



## Stefan (Jun 30, 2009)

panyan said:


> cmhardw said:
> 
> 
> > the probability of that is 1/6^13 = 1/13060694016
> ...


What???



uweren2000 said:


> "the chance of winning the lottery is 1/(1.4x10^51). "
> 
> Think it over!


What, you think someone ever won the lottery?


----------



## TomZ (Jun 30, 2009)

In an average of N, the streak of 13 can start at N-13 'points'. So the probability of getting a streak of 13 in an average of N is 1/6^13*(N-13). That means to have just a 1% chance you need to do 13 MILLION solves.

The streak of 13 or more is a bit trickier. That requires adding all the possible streaks from 13-N together, giving us the following sequence:

Sum(X from 13 to N) 1/6^X*(N-X)

I don't know of any algebraic way to calculate that (I'm not even sure it exists), but using some computer programming I generated this graph:






Pretty darn unlikely if you ask me.


----------



## Stefan (Jun 30, 2009)

TomZ said:


> So the probability of getting a streak of 13 in an average of N is 1/6^13*(N-13).


Wow is that wrong. And not only because you just said it's impossible to get 13 PLL skips in 13 solves.



TomZ said:


> The streak of 13 or more is a bit trickier. That requires adding all the possible streaks from 13-N together, giving us the following sequence:
> 
> Sum(X from 13 to N) 1/6^X*(N-X)


This would be wrong even if you plugged in the correct formula for the "=X" case. Btw, "13 to N" would've been clearer than "13-N".



TomZ said:


> Pretty darn unlikely if you ask me.


Just to clarify: I don't care whether he actually got that streak. I'm here solely for the probability riddle he caused.


----------



## brunson (Jun 30, 2009)

Hmm, I wonder if this approach would work:

Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.

Computation left as an exercise for the student. ;-)

Actually, I'll hack at it when I'm not at work.

Edit:
Someone critique this...

```
100(1/6)^13*100!
--------------------
13!(100-13)!(100-13)
```
Where:
100(1/6)^13 = P(13 skips in 100)
100! = permutations of 100 solves
13! = permutations of the 13 skips
(100-13)! = permutations of the non skips
(100-13) = the number of places in the 100 solves where your streak could occur

Actually, that result is greater than 1, so there's a problem in there somewhere.

Edit2: My denominator doesn't force the skips to be adjacent


----------



## tfkscores (Jun 30, 2009)

ummm....im repeating high school algebra 1 so i have no idea what you all are talking about.


----------



## Stefan (Jun 30, 2009)

brunson said:


> Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.


But what about all the PLL skips other than those 13 of the streak, somewhere else in the 100 solves?


----------



## brunson (Jun 30, 2009)

StefanPochmann said:


> brunson said:
> 
> 
> > Calculate the probability of 13 PLL skips in 100 solves (not too hard), then multiply by the number of permutations of 100 solves (100!), and divide by the number of permutations that have those 13 skips adjacent to one other.
> ...


Good point. Even if my calculation was correct it would only be the chance of *exactly* 13 skips and all in a row.


----------



## Stefan (Jun 30, 2009)

Yeah, there's even the possibility of more than just one streak of 13 skips.


----------



## panyan (Jun 30, 2009)

uweren2000 said:


> "the chance of winning the lottery is 1/(1.4x10^51). "
> 
> Think it over!



oh woops, it is 49! not 42! (i have never played the lottery )


----------



## Stefan (Jun 30, 2009)

panyan said:


> uweren2000 said:
> 
> 
> > "the chance of winning the lottery is 1/(1.4x10^51). "
> ...


You're... not exactly getting closer.

I suggest you follow Rune's advice and think it over. Independent of how you arrived at your number, do you not realize how tiny it is? Like I said earlier, it would pretty much mean that nobody ever won the lottery. If every person had lived since the big bang and filled out 1 lottery ticket every second, the probability that anyone ever won would still be almost zero. You gotta be kidding me.


----------



## Dene (Jun 30, 2009)

StefanPochmann said:


> LOL. Indeed.





StefanPochmann said:


> You gotta be kidding me.



I conclude that Stefan is deeply enjoying this thread


----------



## Carrot (Jun 30, 2009)

Dene said:


> I conclude that Stefan is deeply enjoying this thread



Then I conclude that you are right ^^


----------



## Herbert Kociemba (Jul 1, 2009)

The probabiltiy is about 5.627572299*10^-9 to get at least 13 PLL skips in a row within 100 which as almost exactly 73.5 times more than 1/6^13.

I do not think there is a closed expression for the result. You can describe the process by a Markov Chain with 14 states (for 0 to 13 skips in a row) and a 14x14 transition matrix. The 100 th power of this matrix gives the desired result.

View attachment 344

The graph shows the probability for at least 13 skips as a function of the number of trys. In good approximation it seems to be a linear function.


----------



## AvGalen (Jul 1, 2009)

Odder said:


> Dene said:
> 
> 
> > I conclude that Stefan is deeply enjoying this thread
> ...


And I conclude that although you don't have sufficient evidence, you are probably right.

But I also suspect that Stefan isn't convinced that he is deeply enjoying just because of that sample of 2. He is more interested in finding out how to calculate the probability that you are wrong about him deeply enjoying.

(also, he would like to have deeply and enjoying defined)


----------



## Stefan (Jul 1, 2009)

Thanks Herbert for the insight and confirmation of Johannes' result. I had suspected it would be something that complicated, I didn't see any easy way because of the heavy overlapping (e.g. solves 1-13 with solves 2-14 and 3-15 etc). And I wish my math teachers had been as knowledgeable. You make me wanna learn this stuff that I've heard about but never pursued.

Dene is right, Arnaud is mostly wrong, and needless to say, both increased the joy very much. Thanks guys.


----------



## AvGalen (Jul 1, 2009)

StefanPochmann said:


> Thanks Herbert for the insight and confirmation of Johannes' result. I had suspected it would be something that complicated, I didn't see any easy way because of the heavy overlapping (e.g. solves 1-13 with solves 2-14 and 3-15 etc). And I wish my math teachers had been as knowledgeable. You make me wanna learn this stuff that I've heard about but never pursued.
> 
> Dene is right, Arnaud is mostly wrong, and needless to say, both increased the joy very much. Thanks guys.


Well, what are the chances of having this many posts on a forum that has Pochmann-control activated and be always right?

I also share Stefans feeling that I want to know more about this stuff. Any good/fun sources for this, or are there just demotivating boring books ?

Found this source that might be useful for the topicstarter as well: http://www.webmath.com/lottery.html


----------



## rahulkadukar (Jul 1, 2009)

I got a OLL,PLL and then another OLL skip and I thought that was lucky


----------



## AvGalen (Jul 1, 2009)

rahulkadukar said:


> I got a OLL,PLL and then another OLL skip and I thought that was lucky


It really was lucky (http://www.speedcubing.com/records/records-rules.html rule 3) to get any of those three skips. Getting them all in a row .....


----------



## tfkscores (Jul 1, 2009)

how the hell did you find out what the chances are? im reapeating algebra 1 next year and am going to be a freshman. how did you figure this stuff out?


----------



## Johannes91 (Jul 1, 2009)

tfkscores said:


> how the hell did you find out what the chances are?


Who? Herbert explained quite clearly how he did it. If you aren't familiar with the terms, look them up.

http://en.wikipedia.org/wiki/Markov_chain

If you can't understand something, read the relevant articles first.

The way I did it wasn't as elegant. The state after any number of solves can be represented with two natural numbers: _n_, the length of the current streak, and _m_, the length of the longest streak so far. Doing one solve gets us from (n,m) to (0,m) with probability 5/6, and to (n+1,max(n+1,m)) with probability 1/6. At the beginning, we are obviously at state (0,0) with probability 1. Keeping track of all different states and the distribution is fairly easy. Here's the code I used, it takes a few seconds to run.



tfkscores said:


> im repeating high school algebra 1 so i have no idea what you all are talking about.
> im reapeating algebra 1 next year


So what? Are you not allowed to learn outside school?


----------



## tfkscores (Jul 1, 2009)

oh....my.....god....i have no idea what you just said lol. i suck at anything algebra related. apparently my brain isnt ready or something lol.


----------



## Stefan (Jul 1, 2009)

Johannes91 said:


> Here's the code I used, it takes a few seconds to run.


No perl?

Thanks for that recurrence relation, I probably would've ended up with the same approach had I put more thought into it. Now I just quickly did it in perl and confirm both of your numbers.



Herbert Kociemba said:


> View attachment 344
> 
> The graph shows the probability for at least 13 skips as a function of the number of trys. *In good approximation it seems to be a linear function.*


Small extra riddle: *Is it linear?* Explain your answer.

Herbert and Johannes are not allowed to answer, would be too easy for them.


----------



## Rune (Jul 1, 2009)

Johannes91 said:


> I asked my computer to do it for me and it gave 4.700277693455794e-9 for the first one and 5.627572299115469e-9 for the second one (the rationals have quite large numerators and denominators).



Does that mean that you got an exact expression for the probability?


----------



## moogra (Jul 1, 2009)

http://en.wikipedia.org/wiki/Binomial_probability

(100 C 13) * (probability of getting a PLL skip) ^13 * (probability of not getting a PLL skip) ^87 will be the probability of getting 13 PLL skips in 100.
If you want more, then you write a program
the general equation is
(totalnumber choose wantedNum) * (pllProbability)^wantedNum * (notPllProbability) ^ (totalNumber - wantedNum)
loop it from lower wantedNum bound to upper wantedNum

or just use binomialCdf on a TI-calculator (binomialpdf is 1 case)


----------



## Stefan (Jul 1, 2009)

moogra, you might want to mention that you're answering a completely different question, the math noobs here might not notice that otherwise.


----------



## Herbert Kociemba (Jul 1, 2009)

uweren2000 said:


> Johannes91 said:
> 
> 
> > I asked my computer to do it for me and it gave 4.700277693455794e-9 for the first one and 5.627572299115469e-9 for the second one (the rationals have quite large numerators and denominators).
> ...



With my Markovchain approach I am only able to find the expression for the second probability. The exact expression is

2836881009340468654612600338487002584885669527428222663495824930731/
504103876157462118901767181449118688686067677834070116931382690099920633856

but I do not think we get much insight from this.


----------



## Jacco (Jul 1, 2009)

Sorry, I haven't read the whole thread but I think that the chance is way higher to hear somebody talking nonsense instead of in fact having 13 2x2 PLL skips in a row. No offense, just experience


----------



## Tim Reynolds (Jul 2, 2009)

StefanPochmann said:


> Small extra riddle: *Is it linear?* Explain your answer.



The expected number of runs of 13 grows linearly--this is a result of the fact that expected value is additive. This way of counting counts a run of 14 as 2 runs of 13. But the probability that there is at least one grows slower than that (P=E-something representing probabilities of >1 run), and slower than linearly.

Besides, if it's linear, then P->infinity. But P approaches 1 asymptotically. So unless it's linear on some interval, which I assert it's not, no.


----------



## Swordsman Kirby (Jul 2, 2009)

http://i40.tinypic.com/rc1to6.gif

I wish I could get rid of the summation.


----------



## moogra (Jul 2, 2009)

StefanPochmann said:


> moogra, you might want to mention that you're answering a completely different question, the math noobs here might not notice that otherwise.



Oh I did not see the 13 skips *IN A ROW*. My answer is the probability of just getting 13 PLL skips in 100 solves.


----------



## Stefan (Jul 2, 2009)

Tim Reynolds said:


> Besides, if it's linear, then P->infinity.


Yep, that's what I was after. I found it neat that one could tell the function isn't linear without understanding it at all (Markov chains huh?), solely by keeping in mind that its result is a probability. Background information ftw! Just like in another thread where someone computed the probability to win the lottery as about 1/10^50. If he had just considered *what* he wanted to compute, he should've realized that *how* he computed it must be wrong.



Swordsman Kirby said:


> http://i40.tinypic.com/rc1to6.gif


False. You ignore cases like 13 skips followed by a nonskip followed by another skip.


----------



## JLarsen (Jul 2, 2009)

tfkscores said:


> ummm....im repeating high school algebra 1 so i have no idea what you all are talking about.



Dude, you're not the only one. Don't worry about it. Whenever I see probability, I see a long post of stuff, followed ussualy by Pochmann or Johannes correcting them, and whatever number I see at the end of their posts is the one I trust. Someone says how many cases, they do the math, I read the number at the end, I'm good.


----------

