#### Devagio

##### Member
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.

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.

• Owen Morrison

#### erdish

##### Member
Yep, they’re the only ones, and as mentioned earlier they’re a part of the single family that I mentioned;

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

##### Member

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

#### erdish

##### Member
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

##### Member
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 and DNF_Cuber

#### erdish

##### Member
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.

• Devagio

#### weruoaszxcvnm

##### Member
Probability getting "+ OLL"?

#### weruoaszxcvnm

##### Member
No WV, No VLS, No AUF.

• CuberStache and BenChristman1

#### Scollier

##### Member
No WV, No VLS, No AUF.
According to my calculations, the probability of getting two LL skips is 1 in 31,104.

That makes me kind of skeptical.

• CuberStache

#### weruoaszxcvnm

##### Member
According to my calculations, the probability of getting two LL skips is 1 in 31,104.

That makes me kind of skeptical.
I can make a reconstruction .
But u need to wait because i forget what i did on 1st solve

• DNF_Cuber

#### ProStar

##### Member
According to my calculations, the probability of getting two LL skips is 1 in 31,104.

That makes me kind of skeptical.
Is that the probability of getting 2 LL skips in a row? Or simply 2 LL skips in a day?

• CuberStache

#### ProStar

##### Member
Probability getting "+ OLL"?
If you mean a cross OLL, then iirc the probability is 1/8 if you only use the standard R U R'/R U' R' inserts

• DNF_Cuber

#### Scollier

##### Member
Is that the probability of getting 2 LL skips in a row? Or simply 2 LL skips in a day?
IDK I'm terrible at math...

• DNF_Cuber

#### DNF_Cuber

##### Member
IDK I'm terrible at math...
that looks like you did 1 LL skipX2, which IDK what that calculates. 2 LL skips in a row would be 15552 ^2 (I think there are 5 digits actually)

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#### ProStar

##### Member
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

• Scollier and DNF_Cuber

#### vidcapper

##### Member
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?

• DNF_Cuber

#### Alex Davison

##### Member
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?

so i think (1/2) ^ 8 = 1/256 chance

#### vidcapper

##### Member

so i think (1/2) ^ 8 = 1/256 chance
Thanks, I should have figured that out myself. On a 7x7, I guess it would be (1/2)^24 then, 1 in 16777216. On a 19x19 ~4.97x10^86. • Alex Davison

#### xyzzy

##### Member
• 