Probability Thread

[Julian]

Probability of a CP skip for SQ1 = 1/36?

 

[ben1996123]

Probability of a CP skip for SQ1 = 1/36?

4!^2 permutations, divide by 16 for AU/DF = 1/36 probability for skip.

 

[cuBerBruce]

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

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.

 

[Applejuice]

Whats the probability of having 2 centers solved on a 5x5 after scramble? Less or more than having 1 side done on a 3x3?

 

[Bob]

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.

 

[CubeRoots]

5/18818646 for 2 centers, 1/7484400 for one side, i think

 

[cuBerBruce]

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]

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.

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]

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]

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]

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]

Wearing black shoes on an odd numbered solve? what if that completely blows your 50% streak and you never solve again? Risky move.

 

[CubeRoots]

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

 

[Noahaha]

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.

 

[Applejuice]

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]

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...

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.

 

[applemobile]

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]

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.

 

[Rune]

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!

 

[cuBerBruce]

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.