Last edited by ben1996123; 03-29-2012 at 10:52 AM.
After solving the centres on a 4x4, how many possible states are there?
Edge permutations(24!) * Corner permutations(8!) * Corner orientations(3^7) = 54711040793092776444454502400000
So the probability of a centres skip on 4x4 is 54711040793092776444454502400000/7401196841564901869874093974498574336000000000 =
Or do I need to divide something by 24 somewhere?
Seems correct to me. Once the centers are solved, the orientation of the cube is fixed, so you don't have to divide by 24.
probability that corners are correctly oriented after f2l=?
Probability that corners are correctly solved after f2l?
Probability that no corner is correctly oriented after f2l?
Probability that edges are correctly solved after f2l?
Probability that No edge is correctly oriented after f2l?
1/5/12 || 3x3=6.35/8.77/9.60 | 4x4=36.9x/42.xy/44.xy| 5x5=1:17.xy/1:26.xy/1:30.53
1/3 || 6x6 = 2:52.xy / 3:03.xy | 7x7 = 4:29.xy / 4:50.xy
3x3|CFOP| Single 7.98 | 8.80 | Avg5 10.47 | Avg12 11.82 | BLD | M2/OP | 1:51.71
2x2|CLL| Avg12 3.73 |
What is the probability that after a random 3x3 scramble, one side will be solved (not necessarily a layer, just a face)? Just curious because I got this scramble about a week ago: R2 B2 R2 D' L2 D F2 U F2 L2 D2 F U2 F D' F2 R' F D L2 F D'
7 greens on F. I think I had one other scramble like this some time last year too, but I've never had one with 8 or 9.
Edit: Might as well have a go at calculating it myself:
edge permutations: 4! = 24
edge orientations: 2^4 = 16
corner permutations: 4! = 24
corner orientations: 3^4 = 81
not sure if its correct.
Last edited by ben1996123; 04-19-2012 at 03:26 PM.
Consider a particular face such as the U face. Probability that the ULF sticker is on the U face is 4/24. The probability that the UFR sticker is on the U face, given that we know the ULF sticker already is, is 3/21. Similarly for the URB and UBL stickers, the probabilities can be given as 2/18 and 1/15. Similarly for the four U-color edge stickers, we get 4/24, 3/22, 2/20, and 1/18. Thus, the probability is (4/24)*(3/21)*(2/18)*(1/15)*(4/24)*(3/22)*(2/20)*(1/18) = 4!*4!/(24*21*18*15*24*22*20*18) = 1/(21*18*15*22*20*18) = 1/44906400.
The probability of at least one of the six faces being a solid color should be roughly (but not exactly) 6 times that or somewhere in the neighborhood of 1 in 7.5 million.