1. 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!

2. Originally Posted by Applejuice
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.

3. whats the probability of getting a 2x2 LL skip?
whats the probability of getting a pyraminx LL skip?

4. Originally Posted by TheAwesomeAlex
whats the probability of getting a 2x2 LL skip?
1/162

I'm really starting to enjoy this thread. ;D
What's the probability of having sune as OLL + A perm as pll?

6. Originally Posted by Applejuice
I'm really starting to enjoy this thread. ;D
What's the probability of having sune as OLL + A perm as pll?
P(Sune) = 1/54
P(A perm) = 2/18 = 1/9

P(Sune -> A) = 1/54 * 2/18 = 1/486.

Do you count Sune and Antisune the same? In that case, it would be twice as likely, 1/243.

7. Originally Posted by cuBerBruce
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.
holy crap

8. Originally Posted by TheAwesomeAlex
whats the probability of getting a pyraminx LL skip?
It's 1/12 (3 possible permutations, '4 possible orientations for each one of them).

9. what is the probability of getting an x-cross (unintentionally) after doing a regular cross?

10. Originally Posted by Smiles
what is the probability of getting an x-cross (unintentionally) after doing a regular cross?
About 1.0%. Determined by 10000 computer simulations and this:

Chance of specific F2L pair solved: 1/24 * 1/16.
Chance of specific F2L pair *not* solved: 1 - 1/24 * 1/16.
Chance of all four F2L pairs *not* solved: (1 - 1/24 * 1/16) ^ 4.
Chance of at least one F2L pair solved: 1 - (1 - 1/24 * 1/16) ^ 4 = 1.038%.
Note that the last two aren't exact, because the probabilities of the four pairs aren't independent, because where pieces of one pair are affects where pieces of the other pairs can be. But they seem to be fairly independent, as the experiment confirms.

For a description of the experiment and more results, see page 35 and the pages around it here:
http://www.stefan-pochmann.info/hume...oma_thesis.pdf

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•