Probability Thread

[amostay2004]

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]

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

Four 2-cycles still yields 10. Numbering the corners not including the buffer from one to seven, you would shoot in a sequence such as 1, 2, 3, 2, 4, 5, 4, 6, 7, 6.

I don't understand your second question.

 

[cuBerBruce]

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

Four 2-cycles still yields 10. Numbering the corners not including the buffer from one to seven, you would shoot in a sequence such as 1, 2, 3, 2, 4, 5, 4, 6, 7, 6.

I don't understand your second question.

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.

 

[ilikecubing]

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

 

[riffz]

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.

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

 

[toastman]

Who the hell is that

Your little brother?

 

[toastman]

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

Four 2-cycles still yields 10. Numbering the corners not including the buffer from one to seven, you would shoot in a sequence such as 1, 2, 3, 2, 4, 5, 4, 6, 7, 6.

I don't understand your second question.

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

 

[riffz]

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

Good thing that's not possible.

 

[qqwref]

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

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

 

[timeless]

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

Four 2-cycles still yields 10. Numbering the corners not including the buffer from one to seven, you would shoot in a sequence such as 1, 2, 3, 2, 4, 5, 4, 6, 7, 6.

I don't understand your second question.

this has 12 shoots?

You don't need to figure out where to place that good edge when breaking in to a new cycle. You always place it at the beginning of the new cycle, which means that when ever you are finished with the new cycle your good edge will be in the right place again. If there are yet more cycles then you place it at the beginning of the new cycle and so forth...

The only thing you need to remember is to shoot to the beginning of the cycle again like this:

If your cycles are: 1-5-3-8-9 2-10-11 12-4-6
Then you would shoot like this: 1-5-3-8-9-2-10-11-2-12-4-6-12

I hope it helps...

 

[EnterPseudonym]

4 Z-perms in a row?

 

[miniGOINGS]

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

 

[EnterPseudonym]

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

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

 

[riffz]

this has 12 shoots?

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

 

[ilikecubing]

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

 

[Julian]

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.

 

[qqwref]

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.

 

[Julian]

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.

 

[qqwref]

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.

 

[Stefan]

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.