# 444 avg WR - 41.29 Erik Akkersdijk



## David0794 (Sep 5, 2010)

> Erik 4x4 WR average - 41.29
> Erik 4x4 ER single - 36.77


@ Lemgo Open 2010, Germany.

Amazing


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## Andrew Ricci (Sep 5, 2010)

Whattt????


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## joey (Sep 5, 2010)

VIDEOS


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## Lucas Garron (Sep 5, 2010)

That's not faz!


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## krnballerzzz (Sep 5, 2010)

I demands a video.


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## nccube (Sep 5, 2010)

Where did you get that information from?


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## Joker (Sep 5, 2010)

VIDEO????
Damn, I thought Felik's record would last a lil longer.


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## David0794 (Sep 5, 2010)

nccube said:


> Where did you get that information from?



Speedcubers-forum from Germany.


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## StefanR (Sep 5, 2010)

Because some German Cubers was on the competition in Lemgo. For more Informations look here: http://www.speedcubers.de/forum/showthread.php?tid=3917


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## nccube (Sep 5, 2010)

Could you give that information in English. Not everyone in this forum speaks/read German


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## Shortey (Sep 5, 2010)

nccube said:


> Could you give that information in English. Not everyone in this forum speaks/read German



That is what he is doing. >_>


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## StefanR (Sep 5, 2010)

He's right.


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## StachuK1992 (Sep 5, 2010)

Woahhh, congratz!
Records-breaking is haywire!


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## Edward (Sep 5, 2010)

*Sees another world record*


Spoiler











*Sees that it's not Faz this time*


Spoiler











Congrats Erik!


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## flan (Sep 5, 2010)

congrasts erik! I was starting to think your WR setting days were over.


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## Robert-Y (Sep 5, 2010)

Congratulation Erik! I'll just stick to breaking my NR for now 


Also:



StefanR said:


> Because some German Cubers was on the competition in Lemgo. For more Informations look here: http://www.speedcubers.de/forum/showthread.php?tid=3917





nccube said:


> Could you give that information in English? Not everyone in this forum speaks/read German





Morten said:


> That is what he is doing. >_>



Did anyone else notice the funny misinterpretation of nccube's question due to its ambiguity?


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## Björn (Sep 5, 2010)

Final: 55.46, 41.68, 41.66, 38.68, 40.53 = 41.29


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## amostay2004 (Sep 5, 2010)

Are there any vids?


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## uberCuber (Sep 5, 2010)

I WANT VID!!!! D:

congrats Erik!!!


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## oprah62 (Sep 5, 2010)

inb4 Feliks sees this, practices hard, and owns it at AC.


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## nccube (Sep 5, 2010)

wow, when will there be a sub 40 4x4x4 avg?


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## Rpotts (Sep 5, 2010)

Lucas Garron said:


> That's not faz!



+1



Edward said:


> *Sees another world record*
> 
> 
> Spoiler
> ...



i lol'd



Robert-Y said:


> Did anyone else notice the funny misinterpretation of nccube's question due to its ambiguity?



yes


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## ariasamie (Sep 5, 2010)

Just wait for faz to get four 4x4 solves without parity error in a comp.
then, BAAAM! new avg WR !


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## Erik (Sep 5, 2010)

^^ I also thought my WR-days were over. This summer I practiced a bit and gotten a 41 avg of 12 once, but after that all the averages I did were kinda fail.

First round: 36 single, after cross I realized centers were wrong so I had to fix that .... :fp also 42.whatever ER avg then.

2nd round: 37 single, J perm locked it away, I think someone got that on video and will send it to me.

Final: 55 (thought I'd fail the rest too now) OLL par, 41, 41 PLL par, 38, 40 PLL par

Yaaaay  
I don't care how long it stands, at least the (almost) grandpa did a WR again


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## RyanO (Sep 5, 2010)

ariasamie said:


> Just wait for faz to get four 4x4 solves without parity error in a comp.
> then, BAAAM! new avg WR !



Getting 4 parity free solves in an average of 5 for 4x4 is pretty unlikely.


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## trying-to-speedcube... (Sep 5, 2010)

I thought this was about Erik getting the WR, not about Faz getting it back >_<

Congratulations Erik


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## FatBoyXPC (Sep 5, 2010)

Lol Erik you are so not even close to being a Grandpa!!! Good job though! Did you wear your hat?!?!


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## Anthony (Sep 5, 2010)

RyanO said:


> ariasamie said:
> 
> 
> > Just wait for faz to get four 4x4 solves without parity error in a comp.
> ...



lol.

At US Nats I had 9/10 possible parities in my solves.
My times were:
1:23.58 1:25.68 1:33.09 1:29.47 1:40.36 
1:29 avg.

The following weekend at TOS I had 1/10 possible parities.
My times were:
1:06.83 1:06.77 1:13.19 1:11.16 1:07.22
1:08 avg.


Anyway, congrats, Erik!


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## nccube (Sep 5, 2010)

And I thought that Faz's WR would stay for a long long time...


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## teller (Sep 5, 2010)

Woohoooo Erik!


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## flan (Sep 5, 2010)

nccube said:


> And I thought that Faz's WR would stay for a long long time...



lol none of faz's records stay for a long long time. They last about 2 months before he beats them again 

EDIT: This is ofcourse an exception


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## XXGeneration (Sep 5, 2010)

Congratulations!


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## guusrs (Sep 5, 2010)

Hi Erik 
Congratz , that's really great!
Grandpa? No you're just out of your diapers!
Gus


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## Lucas Garron (Sep 5, 2010)

RyanO said:


> ariasamie said:
> 
> 
> > Just wait for faz to get four 4x4 solves without parity error in a comp.
> ...


18.75% is unlikely?


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## JeffDelucia (Sep 5, 2010)

Lucas Garron said:


> RyanO said:
> 
> 
> > ariasamie said:
> ...



No parity solve is 1/4 not 1/2.


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## cubedude7 (Sep 5, 2010)

Wow, Faz got beaten. Congratz! Hopefully there will be a video soon.


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## waffle=ijm (Sep 5, 2010)

Why is everyone surprised that faz got beaten? Erik is a good cuber too =)

Congrats Erik!


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## Stefan (Sep 5, 2010)

Lucas Garron said:


> RyanO said:
> 
> 
> > ariasamie said:
> ...



On the other hand, shouldn't it be a four-way coin?
On the other other hand, there's usually more than one round in a competition.

I have no idea how to ask WA this...


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## Slash (Sep 5, 2010)

Faz just doesn't get 3 of the sum of avg ranks (officially).


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## Daniel Wu (Sep 5, 2010)

Wooo. Congrats to Erik.


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## Mr Cubism (Sep 5, 2010)

Really nice Erik! Congratz to ER and the amazing WR!:tu


You fullfilled my prediction, thanks! But I had already predicted that too!
http://www.speedsolving.com/forum/showthread.php?p=447807#post447807


I can also predict that I shall switch username tomorrow to Nostracubus 

Congratz again!


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## vcuber13 (Sep 5, 2010)

StefanPochmann said:


> On the other hand, shouldn't it be a four-way coin?
> 
> I have no idea how to ask WA this...



well,
for all 5 with out parity

\( \frac{1}{4}^5=1024^{-1} \)


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## hawkmp4 (Sep 5, 2010)

StefanPochmann said:


> Lucas Garron said:
> 
> 
> > RyanO said:
> ...


http://www.wolframalpha.com/input/?i=binomial+probability
That should do it.


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## JeffDelucia (Sep 5, 2010)

If I did it right the probability is 1/64..


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## Joker (Sep 5, 2010)

RyanO said:


> ariasamie said:
> 
> 
> > Just wait for faz to get four 4x4 solves without parity error in a comp.
> ...


"Getting lucky is not a crime"


waffle=ijm said:


> Why is everyone surprised that faz got beaten? Erik is a good cuber too =)
> 
> Congrats Erik!


Epic great cuber*


JeffDelucia said:


> If I did it right the probability is 1/64..


Is that supposed to be alot? The probability of a PLL skip is 1/72 . Might take a bunch of comps, but its definately possible.


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## vcuber13 (Sep 5, 2010)

JeffDelucia said:


> If I did it right the probability is 1/64..



i agree


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## Stefan (Sep 5, 2010)

hawkmp4 said:


> http://www.wolframalpha.com/input/?i=binomial+probability
> That should do it.



Ah, nice. Now if only it would give me an expression so I could further include it in asking for the probability of this occurring in at least one of two or three rounds...


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## Robert-Y (Sep 5, 2010)

I'm trying to think about this logically:

The chances of getting PLL parity is 1/2, right? And the chances of getting OLL parity is 1/2, right? So the chances of getting no parity is just (1/2)^2 which is 1/4.

And there are five ways having 4/5 solves without parity and 1/5 solves with:

NNNNP, NNNPN, NNPNN, NPNNN, PNNNN

Also there is one way of getting 5/5 without parity:

NNNNN

So the calculation is: (((1/4)^4)*(3/4)*5) + ((1/4)^5)

Which is: 1/64 (1.5625%)

Which is the same as Jeff's answer...


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## vcuber13 (Sep 5, 2010)

StefanPochmann said:


> hawkmp4 said:
> 
> 
> > http://www.wolframalpha.com/input/?i=binomial+probability
> ...



wouldnt it just be essentially 2/64 or 1/32, and 3/64?


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## Robert-Y (Sep 5, 2010)

The chances of it happening in no rounds out of 3 is 63/64 *3

Therefore the chances of it happening in at least one round would simply be 1 - (63/64 * 3) = 12097/262144 (~4.61%)


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## aronpm (Sep 5, 2010)

StefanPochmann said:


> hawkmp4 said:
> 
> 
> > http://www.wolframalpha.com/input/?i=binomial+probability
> ...



http://www.wolframalpha.com/input/?i=binomial(5,0.25)&a=*C.binomial-_*DistributionNoFluff-

The answer is at the bottom: 9.8%

For 2 or 3 rounds, then the probability can be found with binomial(2,0.09766) and binomial(3,0.09766), and checking where it says "at least one success", if I remember my statistics correctly.


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## Stefan (Sep 5, 2010)

Robert-Y said:


> Therefore the chances of it happening in at least one round would simply be 1 - (63/64 * 3) = 12097/262144 (~4.61%)



Yeah, that's how I had done it with my pocket calculator right away, but I'd still like to know whether I could do it more easily with wolfram alpha and have it as one nice link.


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## Faz (Sep 5, 2010)

Joker said:


> VIDEO????
> Damn, I thought Felik's record would last a lil longer.



You're kidding right? Congrats Erik!!!


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## Joker (Sep 5, 2010)

Robert-Y said:


> The chances of it happening in no rounds out of 3 is 63/64 *3
> 
> Therefore the chances of it happening in at least one round would simply be 1 - (63/64 * 3) = 12097/262144 (~4.61%)



The chance of one round is 1/64 isn't it...?
So once in 3 rounds would be 3/64.


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## Joker (Sep 5, 2010)

fazrulz said:


> Joker said:
> 
> 
> > VIDEO????
> ...



Yea, actually I was. Lol. (I was waiting for someone to ask)

EDIT:
Whoops typo on your name sry


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## scylla (Sep 5, 2010)

> Ah, nice. Now if only it would give me an expression so I could further include it in asking for the probability of this occurring in at least one of two or three rounds...



that would be 

1 - (1-1/64)^n (where n is the number of rounds)

which is 3,10% when n =2

and 4,61% when n = 3

And when you do 100 its almost 80% that you have at least one round with at least 4 no parities


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## vcuber13 (Sep 5, 2010)

shouldnt it be 100% after 64 rounds?


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## Erik (Sep 5, 2010)

fazrulz said:


> Joker said:
> 
> 
> > VIDEO????
> ...



Thanks man! Congrats to your crazy WR's too!


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## Joker (Sep 5, 2010)

Lol this thread looks like it should be in the "Puzzle Theory" thread with thee calculations lol.


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## Toad (Sep 5, 2010)

Wow Erik, well done!! You fully deserve a bit of limelight after that stupid 11 year old kid steals it all...


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## hawkmp4 (Sep 5, 2010)

vcuber13 said:


> shouldnt it be 100% after 64 rounds?


No.
If you flip a coin twice, are you guaranteed to get a heads?


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## scylla (Sep 5, 2010)

> shouldnt it be 100% after 64 rounds?



that's (I guess) the expected value not the probability

a probability of 100% can only be reach in the limit


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## aronpm (Sep 5, 2010)

vcuber13 said:


> shouldnt it be 100% after 64 rounds?



lmfao that's not how probability works


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## vcuber13 (Sep 5, 2010)

hawkmp4 said:


> vcuber13 said:
> 
> 
> > shouldnt it be 100% after 64 rounds?
> ...





scylla said:


> > shouldnt it be 100% after 64 rounds?
> 
> 
> 
> ...





aronpm said:


> vcuber13 said:
> 
> 
> > shouldnt it be 100% after 64 rounds?
> ...



lol i wasnt think prbability, but in theory it would right?


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## Toad (Sep 5, 2010)

vcuber13 said:


> shouldnt it be 100% after 64 rounds?




[10:47pm] <Toad> I always count my 3x3 solves in bunches of 72
[10:47pm] <Toad> if I have 71 in a row without a PLL skip I then wait til a comp before I solve a 3x3 again
[10:48pm] <Kirjava> I lol'd irl
[10:48pm] <Toad> :3


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## Stefan (Sep 5, 2010)

vcuber13 said:


> shouldnt it be 100% after 64 rounds?



Yeah, and *over* 100% after 65 rounds.


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## hawkmp4 (Sep 5, 2010)

vcuber13 said:


> lol i wasnt think prbability, but in theory it would right?


Probability IS theory. 1-(63/64)^64=.635


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## scylla (Sep 5, 2010)

> > Originally Posted by vcuber13
> > lol i wasnt think prbability, but in theory it would right?
> 
> 
> Probability IS theory. 1-(63/64)^64=.635


__________________

awesome, its 64% at 64 rounds !!! (feel like a nerd now)


anyway, Erik grats!


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## PhillipEspinoza (Sep 5, 2010)

Ya it has gotten off topic rather quickly. I say we drop that subject and carry it on elsewhere.

Congrats to Erik on an amazing 4x4 avg WR! Can't wait for the video (if there is one).


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## Joker (Sep 5, 2010)

Lol guys, you know what he means.
That's how it would work if all the laws of the universe where written in stone, but they aren't.
EDIT
I agree with the post above mine.


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## Aksel B (Sep 5, 2010)

Congratulations, Erik! 
Let's keep the off-topic discussion out of here


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## hawkmp4 (Sep 5, 2010)

Joker said:


> Lol guys, you know what he means.
> That's how it would work if all the laws of the universe where written in stone, but they aren't.
> EDIT
> I agree with the post above mine.



Yeah, we know what he meant. He was just wrong.

Congrats to Erik! I didn't think you were done, for what it's worth.


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## cmhardw (Sep 5, 2010)

Congrats to Erik for this great WR! It's exciting to see so many records being broken recently! 

Chris


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## Lucas Garron (Sep 5, 2010)

StefanPochmann said:


> Lucas Garron said:
> 
> 
> > RyanO said:
> ...


I have some ideas, but I'm too lazy. (Also, sorry, I was somehow in "OLL parity is the actual bad one" mode.)


```
CDF[BinomialDistribution[5, 0.25], 5] - CDF[BinomialDistribution[5, 0.25], 3]
```
is 1/64. Still decently likely to happen a few times at a large competition.


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## Zubon (Sep 6, 2010)

Just last night I was looking at the WCA rankings page and saw Erik's magnificent performance at the Alania Open in June which made the think that this "grandpa" is not quite finished yet.

Congratulations Erik!


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## bigbee99 (Sep 6, 2010)

Congrats, I can't wait to see the video!


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## Lucas (Sep 6, 2010)

scylla said:


> > > Originally Posted by vcuber13
> > > lol i wasnt think prbability, but in theory it would right?
> >
> >
> ...



That's because 0.64 is near a special number:

1 / (1 - 0.635) = 2.7397 (almost "e"; of course that's not coincidence)


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## That70sShowDude (Sep 6, 2010)

YEA ERIK! 
Congrats


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## eastamazonantidote (Sep 6, 2010)

Insane. Congrats Erik!!!


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## hawkmp4 (Sep 6, 2010)

Lucas said:


> \
> 
> That's because 0.64 is near a special number:
> 
> 1 / (1 - 0.635) = 2.7397 (almost "e"; of course *that's not coincidence)*



Oh really?


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## Dene (Sep 6, 2010)

The 4x4 King is back on his throne! Awesome job Erik


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## ReinForce (Sep 6, 2010)

LOLOL Erik's back to the show!


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## Andreaillest (Sep 6, 2010)

Another record! And to Erik nonetheless. Congrats, you're an epic cuber.


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## cmhardw (Sep 6, 2010)

hawkmp4 said:


> Lucas said:
> 
> 
> > \
> ...



Yes really. I don't get your question? Are you asking for the demonstration or the proof of this?


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## tx789 (Sep 6, 2010)

Feliks may not be happy


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## Zubon (Sep 6, 2010)

tx789 said:


> Feliks may not be happy



I don't think so. Faz is always a good sport. 

He is probably happier that there is some real competition for him now.


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## bigbee99 (Sep 6, 2010)

tx789 said:


> Feliks may not be happy


He congradulated him earlier in the thread...


fazrulz said:


> Joker said:
> 
> 
> > VIDEO????
> ...


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## RCTACameron (Sep 6, 2010)

Only 1 sub-40, yet it still beats Faz's avg which had 3. Just when we all thought he was slowing down, Erik is back.


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## Tim Major (Sep 6, 2010)

RCTACameron said:


> Only 1 sub-40, yet it still beats Faz's avg which had 3. Just when we all thought he was slowing down, Erik is back.



Erik didn't have a counting 48. That always helps 

Edit: And congrats Erik :tu forgot in my post


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## Zane_C (Sep 6, 2010)

Congrats, such a good avg.


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## hawkmp4 (Sep 6, 2010)

cmhardw said:


> Yes really. I don't get your question? Are you asking for the demonstration or the proof of this?


Yes, sorry, I realise that wasn't clear now. It's not apparent to me at all how e would be relevant at all to the calculation of that probability.


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## cmhardw (Sep 6, 2010)

hawkmp4 said:


> cmhardw said:
> 
> 
> > Yes really. I don't get your question? Are you asking for the demonstration or the proof of this?
> ...



Let \( N \) be a natural number. Now assume that the probability of some event occurring is \( \frac{1}{N} \)

We will run a trial of N attempts at this event, and we are interested in the probability that the event occurs at least once. This is done by finding 1 minus the probability that the event never occurs.

\( 1 - \left(1-\frac{1}{N}\right)^N \)

Now let's think about the number \( e \) for a moment. There are numerous definitions of \( e \), but one of which is the following:

Invest a sum of money for one year at an annual interest rate of 100%, compounded continuously. We are interested in the accumulation factor of that sum of money.

\( a(n) = \left(1+\frac{i}{n}\right)^n \) where n is the number of times you compound in 1 year, and i is the interest rate expressed as a decimal.

Then:
\( \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \) will express the accumulation factor for investing a sum of money for one year at an annual percentage of 100%, compounded continuously. This limit definition is also equal to \( e \). So \( e \) or approximately 2.71828 is the accumulation factor for this sum of money.

Now back to our probability question. We are interested in the probability:
\( 1 - \left(1-\frac{1}{N}\right)^N \)

Not only that, but let's examine what happens when we let N go to infinity. So we are really interested in:

\( \lim_{N \to \infty} 1 - \left(1-\frac{1}{N}\right)^N \)
\( \lim_{N \to \infty} 1 - \left(\frac{N-1}{N}\right)^N \)
\( \lim_{N \to \infty} 1 - \left[\left(\frac{N}{N-1}\right)^{-1}\right]^N \)


Now let's introduce a variable substitution. Let \( N=K+1 \)

Since we are examining the limit as N approaches infinity, this will be similar to investigating when K approaches infinity, as K is defined simply as the number before N.

\( \lim_{K \to \infty} 1 - \left[\left(\frac{K+1}{(K+1)-1}\right)^{-1}\right]^{K+1} \)

\( \lim_{K \to \infty} 1 - \left(\frac{K+1}{K}\right)^{-K-1} \)
\( \lim_{K \to \infty} 1 - \left[\left(1+\frac{1}{K}\right)^{K+1}\right]^{-1} \)

\( 1 - \left[\lim_{K \to \infty}\left(1+\frac{1}{K}\right)^{K+1}\right]^{-1} \)
\( 1 - \left[\lim_{K \to \infty}\left(\left(1+\frac{1}{K}\right)^{K}\times\left(1+\frac{1}{K}\right)\right)\right]^{-1} \)
\( 1 - \left[\left(\lim_{K \to \infty}\left(1+\frac{1}{K}\right)^{K}\right) \times \left(\lim_{K \to \infty} \left(1+\frac{1}{K}\right)\right)\right]^{-1} \)
\( 1 - \left(e \times 1 \right)^{-1} \)

\( 1 - e^{-1} \)
\( 1 - \frac{1}{e} \approx 0.632121 \)

Chris


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## FatBoyXPC (Sep 6, 2010)

Lol you guys got off topic so bad Erik never saw my question! Did you wear that hat at Lemgo?!


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## hawkmp4 (Sep 6, 2010)

Ah! Okay. Thank you.
I think I misunderstood Lucas- I thought that he was saying that the fact that 64 _trials_ gave a number related to e was coincidence. I understand now. Bad reading on my part.
Thank you again, for yet another wonderful explanation.


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## Erik (Sep 7, 2010)

fatboyxpc said:


> Lol you guys got off topic so bad Erik never saw my question! Did you wear that hat at Lemgo?!



In fact I did


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## tx789 (Sep 7, 2010)

Well done. How long will it last those?


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## KboyForeverB (Sep 7, 2010)

Edward said:


> *Sees another world record*
> 
> 
> Spoiler
> ...


LOL


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## Aksel B (Sep 7, 2010)

Is there a video?


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## FatBoyXPC (Sep 7, 2010)

Hey Erik, did anybody get anything on video? I'm guessing if they would have got your WR on video it would have been posted by now, but anything else?


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## Erik (Sep 7, 2010)

I haven't received anything, someone promised me to send the 37 (2nd round) to me but haven't gotten anything yet like I said. Also I don't think anyone got a solve of the finals on video at all


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## FatBoyXPC (Sep 7, 2010)

Lol looks like you need to make your girlfriend your permanent "Camera Guy"


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## Joker (Sep 7, 2010)

Erik said:


> I haven't received anything, someone promised me to send the 37 (2nd round) to me but haven't gotten anything yet like I said. Also I don't think anyone got a solve of the finals on video at all


Nobody filmed Erik? Fail...


fatboyxpc said:


> Lol looks like you need to make your girlfriend your permanent "Camera Guy"



Lol agreed. Don't wanna miss another WR


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## kutuan (Sep 14, 2010)

so there's really no video for this WR? 

congrats Erik, u are no near grandpa at all


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## Shortey (Sep 14, 2010)

Joker said:


> Erik said:
> 
> 
> > I haven't received anything, someone promised me to send the 37 (2nd round) to me but haven't gotten anything yet like I said. Also I don't think anyone got a solve of the finals on video at all
> ...



Well, his gf is probably competing aswell.


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## Erik (Sep 14, 2010)

Yes she mostly is, but even if she had time I don't think it'd be a good idea to ask her to film everything. Competing all day is exhausting enough as it is....


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