444 avg WR - 41.29 Erik Akkersdijk

[Mr Cubism]

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!

 

[vcuber13]

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

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

 

[hawkmp4]

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.

18.75% is unlikely?

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...

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

 

[JeffDelucia]

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

 

[Joker]

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.

"Getting lucky is not a crime"

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

Congrats Erik!

Epic great cuber*

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.

 

[vcuber13]

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

i agree

 

[Stefan]

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...

 

[Robert-Y]

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...

 

[vcuber13]

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...

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

 

[Robert-Y]

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%)

 

[aronpm]

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...

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.

 

[Stefan]

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.

 

[Faz]

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

You're kidding right? Congrats Erik!!!

 

[Joker]

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.

 

[Joker]

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

You're kidding right? Congrats Erik!!!

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

EDIT:
Whoops typo on your name sry

 

[scylla]

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

 

[vcuber13]

shouldnt it be 100% after 64 rounds?

 

[Erik]

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

You're kidding right? Congrats Erik!!!

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

 

[Joker]

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

 

[Toad]

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