jonlin
Veteran
I'm guessing that an LL skip is 1/933120 chance of happening. Now we have to wait for Simon to report back to us to see when he gets an LL skip on megaminx.
Probability Thread
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I'm guessing that an LL skip is 1/933120 chance of happening. Now we have to wait for Simon to report back to us to see when he gets an LL skip on megaminx.
Riley
Veteran
Are the chances of getting an LL skip with winter variation (as in you have winter variation in the current solve) 1/22? Because there are 21 PLL's you might get and 1 just solved last layer? (don't count AUF's) Thanks.
ben1996123
Banned
1/72 probly.
IanTheCuber
Regular
If you're just orienting the corners, then it would be 1/63.
Otherwise, no idea. You might also have to include half the F2L's, since mirrors make no extra contribution.
Otherwise, no idea. You might also have to include half the F2L's, since mirrors make no extra contribution.
Riley
Veteran
I'm trying to say, if you have a R U' R' insertion with 3 edges on the top layer that are already correct (like yellow facing up, if white was your cross color), what is the chance of an LL skip, using WV, so that the corners orient? No mirrors. And if it's still 63, can you please explain how you get that? Thanks!
Rpotts
Veteran
Same chance as a PLL skip - 1/72
IanTheCuber
Regular
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.
Godmil
Premium Member
Hmm, you know you could probably draw a probability distribution, like a gaussian or something with the 1/150 defining the width. Should be possible to get a rough estimate. But I don't know how to do it
MTGjumper
Veteran
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.
ben1996123
Banned
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%
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ben1996123
Banned
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?
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?
TMOY
Veteran
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.
CubeRoots
Veteran
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
Akash Rupela
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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?
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?
Rpotts
Veteran
1/27
1/162
6/27
1/96
1/8
1/162
6/27
1/96
1/8
cuBerBruce
Veteran
1/162, allowing AUF (1/648 if not allowing AUF)
1/48, allowing AUF (1/192 if not allowing AUF)
ben1996123
Banned
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.
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.
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cuBerBruce
Veteran
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.
SweetSolver
Veteran
What is the probability of one FACE being solved after a random scramble on a 2x2?
cubingawsumness
Veteran
Just curious, has anyone ever estimated the percentage of people in the US who can solve a 3x3?
