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Thread: Probability Thread

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    Quote Originally Posted by IanTheCuber View Post
    Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.
    I doubt that someone who averaged 37 would only get a sub-35 one in 150 solves. But a normal distribution should work, providing you chose the variance such that 1/150 solves was sub-35.

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    Quote Originally Posted by IanTheCuber View Post
    Here is a good one: What is the propability of getting a Sub-30 solve on 4x4, considering your average is about 37 seconds, and about 1/150 you get is Sub-35? I don't think there is enough info to make the answer out, but try. I am.
    Assuming the solve time, X is random and normally distributed:

    X ~ N(37, σ)
    P(X < 35) = 0.0067
    P(Z < -2/σ) = 0.0067
    P(Z < 2/σ) = 0.9933
    2/σ = 2.48
    σ = 2/2.48 = 0.806
    X ~ N(37, 0.806)
    P(X < 30) = P(Z < (30-37)/0.806) = P(Z < -8.685) = 0.0000000000000001894%
    Last edited by ben1996123; 03-29-2012 at 09:52 AM.

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    After solving the centres on a 4x4, how many possible states are there?

    Attempt:

    Edge permutations(24!) * Corner permutations(8!) * Corner orientations(3^7) = 54711040793092776444454502400000

    So the probability of a centres skip on 4x4 is 54711040793092776444454502400000/7401196841564901869874093974498574336000000000 = \frac{1}{135277939046250}

    Or do I need to divide something by 24 somewhere?

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    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.

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    Quote Originally Posted by TMOY View Post
    Well, it's pretty easy. Assume there are actually 73 different PLL cases, counting different orientations separately. Since 73 is prime, this is only possible if there are 73 pieces of the same kind on the LL; this is obviously not the case, hence your logic must be wrong

    Actually the correct result for the cube ls 1/72. But as Rpotts pointed out, we were talking about the megaminx, for which the correct result is 1/720. (There are 60 possible permutations of corners (5!/2, since only even permutations are possible), same for edges, hence 3600 possibilities disregarding AUF, hence the probability of a PLL skip is 1 out of 3600/5=720.)
    I know this was 2 weeks ago but i realised mistake, I thought there were 2 ways to get Hperm but in reality there was only one so i ended up with 73 instead of 72 possibilities

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    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

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    1/27
    1/162
    6/27
    1/96
    1/8
    3x3|CFOP| Single 7.98 | 8.80 | Avg5 10.47 | Avg12 11.82 | BLD | M2/OP | 1:51.71
    2x2|CLL| Avg12 3.73 |

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    Quote Originally Posted by Akash Rupela View Post
    Probability that corners are correctly solved after f2l?
    1/162, allowing AUF (1/648 if not allowing AUF)
    Quote Originally Posted by Akash Rupela View Post
    Probability that edges are correctly solved after f2l?
    1/48, allowing AUF (1/192 if not allowing AUF)

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    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
    centres: 6

    1/(24*16*24*81*6) = \frac{1}{4478976}

    not sure if its correct.
    Last edited by ben1996123; 04-19-2012 at 02:26 PM.

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    Quote Originally Posted by ben1996123 View Post
    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'
    Centers can be considered fixed. Corner and edge stickers of a given color must lie on the same face as the center of that color.

    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.

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