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Probability Thread

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Dec 29, 2017
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Nice try FBI
WCA
2017PARE13
What is the probability of a cross skip?
TL;DR: I think 1/31,680. I'm not positive.

Many of us know that the chance you'll get an OLL skip in any given solve is 1/216; for PLL skip, 1/72; and for a full LL skip, 1/15,552. But what is the chance you will get a cross skip in a solve? I've never heard of anyone ever getting one. I tried to figure out the chance. Here's what I came up with.

The cube has 43,252,003,274,489,856,000 positions (12! x 8! x 2^11 x 3^7 / 2). How many of these positions have a cross on them? First of all, in a cross, all of the corners are completely irrelevant. We can start our calculation by using the number of positions of the corners, which is 88,179,840 positions (8! x 3^7). For there to be a cross, four edges matter and eight don't. For the four edges that matter, the cross edges, there are six options, because the cross could be on any of the six faces. The eight edges that don't matter could be permuted 8!/2 ways. This is because there are eight places to put the first irrelevant edge (because four spots are already occupied by the cross edges), seven places to put the second (because now there is a spot occupied by an irrelevant edge), six places for the third, and so on. The /2 at the end is because we have already chosen a position for the corners, and it's impossible to switch just two pieces on a cube. There are 2^7 ways that the eight irrelevant edges could be oriented, since the orientation of the eighth edge depends of the orientation of the other seven.

This means that there are, in total, 8! x 3^7 x 6 x 8! x 2^7 / 2 = 1,365,277,881,139,200 cube states that have a cross on them. Dividing this number by the total number of 3x3 positions gets us the chance that any random scramble will have a cross on it, thus, a cross skip. Ultimately this comes to 1/31,680, which is very close to half as likely as a last layer skip.

All of this is my own work (except figuring out 43 quintillion). I'm not positive that this is correct. I think my formula may not properly address cube states with multiple crosses on them. Any thoughts are welcome.
I have done around less than 12k solves on 3x3, and I have gotten both a LL skip, and a cross skip.
 
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Average # of cycles for 4x4 wings?
1/1 + 1/2 + 1/3 + … + 1/24 − 1 ≈ 2.776 if you don't count 1-cycles (pieces that are already solved).

Quick proof: If you start with an empty set, there'll obviously be 0 cycles. The kth element added will introduce a new cycle with probability 1/k, so this gives the harmonic number H_24 ≈ 3.776. To exclude the solved pieces, note that each element is a fixed point (i.e. the piece is solved) with probability 1/24, so the expected number of solved pieces is 1 and we subtract that from H_24.

A caveat is that this is assuming you don't change your orientation to influence the permutation, because that would make the problem intractable. Also, keep in mind stuff like the gambler's fallacy—just because you have three cycles already doesn't make it any more or less likely that you'll encounter a cycle break later. The probability of a cycle break only depends on how many pieces are left to trace, not on how many cycle breaks there already are.
 
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Chances of me solving a Skewb in four moves (It happened to me at a contest this morning.)

Basically when solving the white side, I accidentally solved the whole thing.
 
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Chances of me solving a Skewb in four moves (It happened to me at a contest this morning.)

Basically when solving the white side, I accidentally solved the whole thing.
0 with WCA-compliant scrambles:
  • 4b3c) Skewb: The (random) state must require at least 7 moves to solve.

But this is the probability thread, not the regulations thread, so check out Jaap's Puzzle Page for statistics on how many moves are needed. 2073 out of 3149280 states can be solved in four moves or less, so the probability is 2073/3149280 ~ 0.07%.
 
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