Probability Thread

[qbtronix]

What is the probability of a Roux CMLL + L6E Skip? I.e. a solved cube after just the 2 left and right blocks.

 

[Daniel Lin]

What is the probability of a Roux CMLL + L6E Skip? I.e. a solved cube after just the 2 left and right blocks.

if you mean COMPLETELY solved
then
1/(3^3*4!*2^5*6!/2)=1.33959191e-7

 

[shadowslice e]

if you mean COMPLETELY solved
then
1/(3^3*4!*2^5*6!/2)=1.33959191e-7

You need to factor in the fact you can have centres in any orientation (or yellow top, white bottom) so the number needs to be divided by 4 which is 3.34897977e-8

 

[qbtronix]

You need to factor in the fact you can have centres in any orientation (or yellow top, white bottom) so the number needs to be divided by 4 which is 3.34897977e-8

Dang. I'm not a Roux solver, (or even a fast CFOP solver) and, believe it or not, had this while fooling around on a hand-scramble. I've had a few CFOP LL skips, most with AUF, but being orders of magnitude less likely my only conclusion is that I must have scrambled it exceedingly poorly. I had had two beers in me so perhaps my mind was playing tricks on me.

 

[Daniel Lin]

how many total double 2cycles of edges are there on a 3x3?

like S' U2 M' U' M S U'

cases from different angles count as the same

EDIT: this is more of a combinatorics question

 

[xyzzy]

how many total double 2cycles of edges are there on a 3x3?

like S' U2 M' U' M S U'

cases from different angles count as the same

I'm assuming you don't count stuff like (M' U M U)3 U2 as a double 2-cycle because that would be too annoying to count. There are 289 cases up to rotational symmetry, or only 170 up to rotational and mirror symmetry.

Spoiler: Burnside's lemma stuff

There are five conjugacy classes of rotational symmetries: identity (1), 90° rotation about a face (6), 180° rotation about a face (3), rotation about an edge (6) and rotation about a corner (8). The numbers of 2+2-cycles preserved by any representative of each of these conjugacy classes are 5940, 6, 120, 100 and 0, respectively, so we can just plug this into Burnside's lemma to get (5940×1+6×6+120×3+100×6+0×8)/(1+6+3+6+8) = 289.

If you want to exclude mirror-identical cases, then you also have to consider the conjugacy classes of reflection symmetries: point reflection (1), axis-aligned plane reflection (3), 90° rotoreflection (6), face-diagonal-aligned plane reflection (6) and 60° rotoreflection (8). The numbers of 2+2-cycles preserved for these classes are 120, 156, 6, 100 and 0, respectively. Plugging the data from all ten conjugacy classes into Burnside's lemma, we get (5940×1+6×6+120×3+100×6+0×8+120×1+156×3+6×6+100×6+0×8)/(1+6+3+6+8+1+3+6+6+8) = 170.

 

[Daniel Lin]

I'm assuming you don't count stuff like (M' U M U)3 U2 as a double 2-cycle because that would be too annoying to count. There are 289 cases up to rotational symmetry, or only 170 up to rotational and mirror symmetry.

Spoiler: Burnside's lemma stuff

There are five conjugacy classes of rotational symmetries: identity (1), 90° rotation about a face (6), 180° rotation about a face (3), rotation about an edge (6) and rotation about a corner (8). The numbers of 2+2-cycles preserved by any representative of each of these conjugacy classes are 5940, 6, 120, 100 and 0, respectively, so we can just plug this into Burnside's lemma to get (5940×1+6×6+120×3+100×6+0×8)/(1+6+3+6+8) = 289.

If you want to exclude mirror-identical cases, then you also have to consider the conjugacy classes of reflection symmetries: point reflection (1), axis-aligned plane reflection (3), 90° rotoreflection (6), face-diagonal-aligned plane reflection (6) and 60° rotoreflection (8). The numbers of 2+2-cycles preserved for these classes are 120, 156, 6, 100 and 0, respectively. Plugging the data from all ten conjugacy classes into Burnside's lemma, we get (5940×1+6×6+120×3+100×6+0×8+120×1+156×3+6×6+100×6+0×8)/(1+6+3+6+8+1+3+6+6+8) = 170.

thanks!
you are awesome

if you counted the annoying stuff, would it be approximately double?

 

[xyzzy]

if you counted the annoying stuff, would it be approximately double?

The annoying stuff turned out to actually not be so annoying to count once I changed how I was counting the number of cases. There are 265 order-4 cases up to rotational symmetry, or total 289+265 = 554 cases including normal 2+2-cycle cases, so it's a bit less than double.

 

[Daniel Lin]

how many double 2cycles of corners are there?

less than edges I'm guessing?

 

[GenTheSnail]

how many total double 2cycles of edges are there on a 3x3?

like S' U2 M' U' M S U'

cases from different angles count as the same

EDIT: this is more of a combinatorics question

how many double 2cycles of corners are there?

less than edges I'm guessing?

Deciding if they are worth learning for BLD? Or just curious?

 

[Daniel Lin]

Deciding if they are worth learning for BLD? Or just curious?

I'm planning on making a list for BLD

The problem is, i don't have a way of organizing all 289 cases or even listing them all.

idk if it's worth learning all of them. But it saves you from having to do 3 comms, so learning at least a few would definitely be worth it (I know like 10 cases).

 

[xyzzy]

how many double 2cycles of corners are there?

less than edges I'm guessing?

Probably around 80 to 90 not counting the twisted cycle cases (e.g. [R', U' D L' U D']), roughly triple that if you include the twisted cycle cases.

(Not accounting for symmetry, there are 1890 cases, and since there are 24 rotational symmetries, we just divide 1890 by 24 to get an estimate of 78.75.)

 

[YouCubing]

probability of a cross solved on clock? I remember there was a scramble on the cubingtime weekly comp that had that

 

[Ronxu]

probability of a cross solved on clock? I remember there was a scramble on the cubingtime weekly comp that had that

2*(1/12)^5

 

[Mastermind2368]

What are the chances of ls and LL skip on ZZ?

 

[shadowslice e]

What are the chances of ls and LL skip on ZZ?

(2*4)/((3^3)(4!)(4!))=1/1944 (this is including ones which are an AUF from solved)
Not sure what you mean by ls and LL skip though but if you mean build F2L-1 and the rest is solved,
2/((3^4)(5!)(5!))= 1/583200

 

[xyzzy]

What are the chances of ls and LL skip on ZZ?

Yeah, this is too vague to answer without further information. Obviously, once you have three "pairs" done, you're not going to count whatever's left as an LS skip unless the F2L happens to be solved too, so for a real LS skip, the fourth pair has to be coincidentally solved together with the third. This happens much more rarely than doing three slots in fixed order (e.g. always doing LB-LF-RB) and finding that the fourth is also solved (which has probability 1/75).

Most common F2L algs for solving a single slot don't affect other slots, unless a piece you need for the pair is stuck in said other slot. Not only this, you'd normally prefer to solve the other pair first, unless it also has stuck pieces. This is a fairly rare situation, and even rarer is when you have something like this and you can't use keyhole-style multislotting to fix it.

Another way of thinking about it is that you're asking about the probability of finding an F2L solution as short as an F2L-1 solution. You have four choices of F2L-1, and each of them requires at most as many moves as solving the full F2L, so clearly, the chances of full F2L being as easy as all four of these choices has to be pretty tiny.

Of course human solvers exhibit a bias in the order in which the slots are solved, which complicates matters even further.

tl;dr: No easy closed form solution; do an experiment over a few thousand solves and count how many of them have "LS skips". That would be your answer for the probability of an LS skip, which you can divide by 1944 to get the approximate probability of an LSLL skip.

 

[Mastermind2368]

Yeah, this is too vague to answer without further information.[/QUOTE

I ment if you had your EO-Line, 2/4 pairs and when you put in the second last pair, you solved the last pair and the LL

 

[xyzzy]

The said "other information" wasn't pertaining to what you meant by "LS skip" (which I think is fairly universal), but to what F2L algs you use and how you solve F2L in general, because the answer to your question of the probability of an LS skip depends on that.

Take a look at this F2L case. If you use R2 U2 R' U2 R2 to insert the last two F2L edges at the same time, does that count as an LS skip? Even though there's no distinct "last slot"? Also note that if you solve the "third slot" differently, for example with R2 U' R' U R2, then there might no longer be an LS skip.

Multislotting and freestyle blockbuilding tricks severely complicate the notion of a "last slot", but even without these, the key consideration here really is whether you consider a certain fixed slot to be the "last slot", or if the "last slot" is always the last unsolved slot. In the former case, then skipping LSLL is as shadowslice answered: 1/75 multiplied by the probability of a ZBLL skip.

In the latter case, that's what I was trying to explain before, and there's no easy answer that can be obtained just from multiplying a bunch of small numbers together.

 

[CubingGenius]

What is the chance of having zero misoriented edges on two axis while the third one has at least two misoriented edges on a cube that is in a random, solvable state?