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Probability Thread

Devagio

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Apr 21, 2020
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Terrific!

Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE.

So the answer is 51,490,480,088,678,400.

A question I would have is are the four basic types seen in the simulation the only ones possible? They are certainly the vast majority.
Yep, they’re the only ones, and as mentioned earlier they’re a part of the single family that I mentioned; it’s actually quite easy to prove that.
Try to get all the factors of 1260 as orientation and permutation cycles, while keeping the number of swaps even, and keeping CO and EO possible.
 

erdish

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Dec 15, 2020
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Yep, they’re the only ones, and as mentioned earlier they’re a part of the single family that I mentioned;
For example, what about this set of cycles which happens:

ORBITS {1 2 2 4 4 14 14 14 14 14 14 14} {9 9 9 15 15 15 15 15 15}
CYCLES {1 2 2 2 2 7 7 7 7 7 7 7} {3 3 3 5 5 5 5 5 5}


That would be four 2-cycles of edges (two of which have a net flip) and not fit the prescription?:

"The only way to have a 1260 cycle on a cube is if: You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation."
 
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Devagio

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For example, what about this set of cycles which happens:

ORBITS {1 2 2 4 4 14 14 14 14 14 14 14} {9 9 9 15 15 15 15 15 15}
CYCLES {1 2 2 2 2 7 7 7 7 7 7 7} {3 3 3 5 5 5 5 5 5}


That would be four 2-cycles of edges (two of which have a net flip) and not fit the prescription?:

"The only way to have a 1260 cycle on a cube is if: You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation."
This is two 2-cycles of edges (one of which as a net flip). Consequently there are 4 edges that are a part of 2-cycles. [Just like there aren't seven 7-cycles of edges, there is only one, but seven edges are a part of it]
 

erdish

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This is two 2-cycles of edges (one of which as a net flip). Consequently there are 4 edges that are a part of 2-cycles. [Just like there aren't seven 7-cycles of edges, there is only one, but seven edges are a part of it]
Ah, thanks. Taking it a step further, your result means that one out of 840 permutations has cycle length N=1260. Since each generates 1260 distinct permutations before repeating, it's possible that the set of all permutations generated by all N=1260 cases covers the entire range (~43 quintillion). Is that the case?
 

Devagio

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Ah, thanks. Taking it a step further, your result means that one out of 840 permutations has cycle length N=1260. Since each generates 1260 distinct permutations before repeating, it's possible that the set of all permutations generated by all N=1260 cases covers the entire range (~43 quintillion). Is that the case?
No. A simple counter example is T-perm. Since it has a single 2cycle of corners (parity), it cannot be generated by a set of moves that doesn’t produce parity applied any number of times.
 

erdish

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No. A simple counter example is T-perm. Since it has a single 2cycle of corners (parity), it cannot be generated by a set of moves that doesn’t produce parity applied any number of times.
Come to think of it, the set of 1260 permutations generated from each will have a fixed distribution of cycle lengths that doesn't cover all possible cycle lengths. E.g., none will have cycle length 8 ,11, 16, 22, 24, 33... So any perm with such a cycle length would be another counterexample.
 

ProStar

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I haven't taken any classes for probability calculation, so I may have this wrong, but I think that the chance of getting an LL skip without AUF would be 1/15552 * 4 = 1/62,208. That would make getting 2 of them in a row without AUFs 1/62208^2 = 3,896,835,264. If you discount the fact that there was no AUF on both of them then it should be 1/15552^2 = 1/241,864,704.

This is getting 2 of them in a row though, not getting two of them in a single session, which would dramatically increase the chances for this to happen, although you'd still be extremely lucky


Again, there's a decent chance that this math is totally wrong
 

vidcapper

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May 22, 2020
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Had an unusual occurrence today... on a 5x5 solve, the last 2 centres were each completely encircling the other colour's centre! Wonder what the odds of that are?
 

xyzzy

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Dec 24, 2015
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i'm not sure about this answer, but i think it's the same probability that both centers end are solved.

so i think (1/2) ^ 8 = 1/256 chance
Same probability yes, but your calculation is wrong.

To calculate it correctly, it's easiest to treat the different orbits of centre pieces individually. Just to make it concrete, let's say the last two centres' colours are white and black. Among the t-centres, there are 4 white pieces and 4 black pieces, so there are 8! / (4! × 4!) = 70 ways of arranging them, each equally likely, and hence there's a 1/70 chance that the white pieces are on the black face and the black pieces are on the white face. Likewise for the x-centres, so there's also a 1/70 chance. These probabilities are independent, so together there's a 1/70 × 1/70 = 1/4900 chance of getting all the centres swapped.

This extends to bigger cubes readily: on a 7×7×7, there are six orbits, so the probability is 1/70^6. On a 9×9×9, there are twelve orbits, so the probability is 1/70^12. And so on.
 
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