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1/1296. I tried to link the post where this was answered but I can't put in a link without the message needing moderator approval. The last layer skip chance is 1/933120 if you wanted to know that too

1/1296. I tried to link the post where this was answered but I can't put in a link without the message needing moderator approval. The last layer skip chance is 1/933120 if you wanted to know that too

For square-1 random state scrambles, does every cubeshape have an equal probability? I don’t know if some show up more than others, or if something like square and star cases show up less because they’re symmetrical.

For square-1 random state scrambles, does every cubeshape have an equal probability? I don’t know if some show up more than others, or if something like square and star cases show up less because they’re symmetrical.

The shapes have equal probability (1/3678) if you treat different AUF/ADFs as different shapes.

But if you treat different AUF/ADFs as still being the same shape, then you need to adjust the probability by how many of the AUF/ADFs let you do a slice move. For example, with square-square, there are four combinations of AUF/ADF, so it has a probability 4/3678. (Since square is rotationally symmetric, doing a (3,0) does not produce a different shape, so we don't count that.) Shield-square has 3 choices of AUF and 2 choices of ADF, so the probability is 6/3678. (Note that this is treating shield-square and square-shield as different shapes; if you want to treat those as the same too, the combined probability is 12/3678.) And so on.

This page has a list of the shapes and probabilities:

I know this has probably been answered elsewhere, but how is decided how long a scramble should be?

3*3 is usually 20 moves, but how do we know that would be enough, rather than 30 moves, or 50, etc? Is it because 20 is God's Number for the 3x3? That number is unknown for bigger cubes though, so that can't be a factor in scramble length for them.

I know this has probably been answered elsewhere, but how is decided how long a scramble should be?

3*3 is usually 20 moves, but how do we know that would be enough, rather than 30 moves, or 50, etc? Is it because 20 is God's Number for the 3x3? That number is unknown for bigger cubes though, so that can't be a factor in scramble length for them.

Cube scrambles are always random state if feasible. For 3x3, since gods number is 20, scrambler programs can easily find a solution in 20 moves or less. This takes much less time than finding an optimal solution like 17 or 18 moves, so scrambles can be generated incredibly quickly.

For bigger cubes, random state is a bit harder. You don’t need God’s number to generate random state scrambles, because you can just set the searching range for an upper bound of gods number. This is why we have random states for 4x4 (and I think 5x5).

For 6x6 and 7x7 random state is possible but it would be so long and so many moves more than the scrambles we use now that it would be impractical, since those take so long to scramble already.

I tried calculating it but I’m not sure if it is correct.

I calculated # possible f2l-oll cases there were (last slot and last layer orientations):

(5*2) * (4*3) * 2^3 * 3^3 / 4 = 6480 combinations. This is why I did this. On the the permutation of the f2l edge counts:
- 5 ways of permuting the f2l edge, and 2 ways of orienting.
- 4 ways of permuting the f2l corner, and 3 ways of orienting. (We only account for parity once).
- The final 4 edges can be oriented 2^3 ways as the final edge only has 1 way to be oriented.
- The final 4 corners likewise can be oriented 3^3 ways.
- Divide by 4 as the AUF doesn’t matter.

I just want to know if this is correct before calculating the number of WV cases

Well, if your end goal is to figure out the number of WV cases, you can just take the 8 OCLLs (T, U, L, Pi, H, S, As, O[solved]), account for the AUFs, and then mentally take out a pair. You have 32 cases, but H and O both have symmetry, reducing down to 27 cases total (I'm pretty sure).

And WV is also only the LS cases where the pair is already formed, so you don't need to work out any permutations of the F2L edge or corner.

And WV also doesn't deal with any permutation, so you don't have to calculate that for any of the LL pieces either.

And AUF does matter because you can't AUF without changing the position of the F2L pair.

And then you also have to account for symmetry, because an H case has the same orientation from a y2 away, and you don't want to count that twice.

Well, if your end goal is to figure out the number of WV cases, you can just take the 8 OCLLs (T, U, L, Pi, H, S, As, O[solved]), account for the AUFs, and then mentally take out a pair. You have 32 cases, but H and O both have symmetry, reducing down to 27 cases total (I'm pretty sure).

And WV is also only the LS cases where the pair is already formed, so you don't need to work out any permutations of the F2L edge or corner.

And WV also doesn't deal with any permutation, so you don't have to calculate that for any of the LL pieces either.

And AUF does matter because you can't AUF without changing the position of the F2L pair.

And then you also have to account for symmetry, because an H case has the same orientation from a y2 away, and you don't want to count that twice.

Ah. I was trying to find out the number of total possible combinations of last f2l pair and the orientation of the last layer. Then find the number of ways to get a WV case to hence find the probability of getting a WV case for your last f2l case.

For 3x3, what is the probability of getting a WV case out of all possible OLS cases (every single one of them). *By OLS I mean every case of solving the last f2l slot no matter the relative arrangements of the edge and corner and solving OLL at the same time.

For 3x3, what is the probability of getting a WV case out of all possible OLS cases (every single one of them). *By OLS I mean every case of solving the last f2l slot no matter the relative arrangements of the edge and corner and solving OLL at the same time.

So the two F2L pieces need to be paired in the top layer, which is 1 in (4/5)x(1/15).
Given the f2l edge is in top layer, the other 3 edges need to be oriented, which has probability 1/8
So overall, this is once in 150 solves.

So the two F2L pieces need to be paired in the top layer, which is 1 in (4/5)x(1/15).
Given the f2l edge is in top layer, the other 3 edges need to be oriented, which has probability 1/8
So overall, this is once in 150 solves.