1. Probability of a PBL skip on 2x2?

probability that corners are solved after finishing F2L on 3x3?

2. 1/6^2
1/27*6

3. Originally Posted by toastman
I like this. Difficult, but a rough maximum to start the thinking: Assume a 20 move "naive" scramble. Probability that moves 11-20 are the inverse of 1-10 = 1 in 18^10 = 1 in 3.57 Trillion

The actual probability is likely way higher than this, as there's so many possibilities, e.g. where the scramble has junk like R L R' L' in it. I'm guessing by a factor of a hundred.

So, it *could* happen, given millions of cubers doing hundreds of thousands of solves in a lifetime. And I want to see the dude's face when it does. He'd be all like "Holy crap!"
This is wrong because scambles aren't just random strings of letters. First, a random combination is selected, then a scramble is generated that leads to that combination.
So the probability is one in ~43 quintillion.

4. Sorry for the double post, but I'm writing this from my iPod at school, and when I tried to edit my last post, It wouldn't let me scroll though it to get to the end.
Anyways...
What the probability of a relative cross solved on any colour?
How many combinatons are there when scambling using <R,U>?
After solving a 2x2x3 block in Petrus, how many combinations are there left?

5. Originally Posted by Julian
After solving a 2x2x3 block in Petrus, how many combinations are there left?
6 corners to permute (6!) * 3 orientations for all but one corner (3^5) * 7 edges to permute (7!) * 2 orientations for all but one edge (2^6) / Inability to get PLL parity (2)

=~ 28 billion - does this sound right?

6. Originally Posted by Antcuber
Is there any possibility that with a random scramble the cube will solve itself? how about being 1 move away? or a u2?
Yes.

To get a probability, you need to define "random scramble". For random state the answer is obvious. But for a random move sequence of 20 or more moves the exact answer seems practically impossible to determine (feel free to prove me wrong!).

Originally Posted by Julian
What the probability of a relative cross solved on any colour?
I'm not 100% sure what you mean by relative, but see http://www.cubezone.be/crossstudy.html.

Originally Posted by Julian
How many combinatons are there when scambling using <R,U>?
6*5*4 * 3^5 * 7! / 2 = 73,483,200

Originally Posted by Julian
After solving a 2x2x3 block in Petrus, how many combinations are there left?
6! * 3^5 * 7! * 2^6 / 2 = 28,217,548,800

7. After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??

8. 1/4: there are 4 ways to AUF, and 1 is right.

9. Originally Posted by Schmidt
After the OLL is finished, what is the probability that the PLL is in the right place (no U, U', U2, y, y', y2 needed to apply the alg) ??
Originally Posted by Kynit
1/4: there are 4 ways to AUF, and 1 is right.
I think the question perhaps needs a little clarification.

Generally a PLL has 4 angle cases and 4 AUF cases. For example, for the A-Perms, the algs I use require same color corner stickers facing me. If not, I have to rotate the cube, or at least the U layer so that same color corner stickers are facing me. I now have the correct angle case. After executing the A-Perm alg, I still might not have the U layer aligned properly with the bottom two layers. This can be handled by an AUF after the algorithm or by rotating the bottom two layers before the alg. If the Schmidt only cares about if the angle case is correct, the answer would be a little larger than 1/4. This is because the N-Perms and H-Perm only have 1 angle case, and Z -Perm and E-Perm have 2 angle cases. I view PLL skip as having 1 angle case and 4 AUF cases.

If we assume there is no attempt to influence the PLL case, and we only care about having the U layer correct for applying the alg, and don't care about if there will need to be an AUF or not, then the probabilty I get is 11/36 (or 85/288 if you want to consider AUF adjustments on PLL skips to matter).

If the question is for having to do no adjustment for either the angle case or for the AUF case, the probability becomes (11/36)*(1/4) = 11/144.

Of course, there are a number of assumptions here, such as you're not trying to influence the PLL case and you only use one particular alg for each PLL case (always applied from the same angle) regardless of which angle or top/bottom alignment situation you might get.

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