OLL skip:
• Each corner has 3 stickers so there's a probability of 1/3 that a corner is correctly oriented. There are 4 corners so the probability that all four of them is oriented would be (1/3)*(1/3)*(1/3)*(1/3).
BUT if you have 3 corners oriented, the 4th one HAS TO BE oriented as well. So the probability that all 4 corners are oriented is (1/3)*(1/3)*(1/3)=1/27
• You do the same thing for edges (only 2 possible orientations) and you find that the probability that all 4 edges are oriented is (1/2)*(1/2)*(1/2)=1/8
So the probability of an OLL skip is (1/27)*(1/8)=1/216
PLL skip:
• We're starting with a "reference" corner. What's the probability that its adjacent corner ends up being actually adjacent to it (headlights) ? There are 3 places this corner could go, so it has a probability of 1/3 of going to its correct location.
What about its other adjacent corner ? It only has 2 spots left, so it has a probability of 1/2 of going to its correct location.
In conclusion you have a probability of (1/3)*(1/2)=1/6 of having your corners solved.
• Since the corners are solved, it's a bit different for edges.
There are 4 spots our first edge can be, so it has a probability of 1/4 of being solved (ie being between its 2 corners, forming a bar like in U, J and F perms).
The second edges only has 3 spots left, so it has a probability of 1/3 of being solved.
We could do the same for our remaining 2 edges BUT since they are correctly oriented they HAVE TO BE solved (you can't have only two swapped edges on a cube).
So we're actually done: the probability of having all 4 edges solved is (1/4)*(1/3)=1/12.
So the probability of a PLL skip is (1/6)*(1/12)=1/72.
Corners solved after F2L:
They have to be both oriented and permuted, so that's a probability of (1/27)*(1/6)=1/162.
