if 4 BAD EDGES ARE ON THE E SLICE, AND ON THE F AND B SLICES THEN DO A y.
And that achieves what?
Consider the superflip. All 12 edges have wrong orientation. Do y rotation. Still all 12 edges have wrong orientation.
Exactly.
Keep this in mind:
After a y/y' rotation,
U/D edges on U/D are
not flipped
U/D edges on E are
flipped
E edges on U/D are
flipped
E edges on E are
not flipped
Yes I know that's extremely trivial, but I'm surprised by the number of people who don't realize this. Somehow people tend to think that doing a y/y' rotation is sort of a "black becomes white, white becomes black" move - it's actually a "gray" move, in which you have nonzero probabilities of getting more, fewer or the same number of bad edges.
I did a lot of calculations last night, probabilities and stuff. And this is what I found: If you have exactly 8 bad edges, the probability of you getting 6 or fewer bad edges after a y/y' rotation is 67.953%, which is somewhat considerably higher than 50%. The probability of you still getting 8 bad edges after that rotation would be 27.825%. Add those 2 figures up, and you can be *pretty* sure that after a y/y' rotation, you won't be worse off. But it's worth knowing this fact: P(10 or 12 bad edges after y/y' rotation | exactly 8 bad edges before y/y' rotation) = 4.222%; P(10 or 12 bad edges on a random scramble) = 3.271%.
I don't know about anyone else, but 68% just doesn't seem convincing enough to me. I think I'll stick to flipping those 8 edges when I get them. Should I get some numbers for cases when there are 10 bad edges as well? :rolleyes:
By the way, can someone do an independent calculation as well, just to confirm these figures? I may be completely wrong
P/S: Should I start this in a new thread? Anyone interested in discussing this?