Probability Thread

[shadowslice e]

1) The skip chance us based on how many cases there are and how many cases you want. The study of it is combinatorics and there are some basic techniques which can get a rough estimate (and LL skip is not 1/72 it's more like 1/4000 off the top of my head).

2) The average movecount is done by finding the movecount for each case and then averaging no real quick way to do it.

 

[Rcuber123]

LL skip is 1 in 15552

 

[xyzzy]

LL skip is 1/72 and H PLL is 1/108. Why? How do you do this?

Step 1: Enumerate the cases in a way such that the cases are equiprobable.
Step 2: Form equivalence classes.
Step 3: Get the size of an equivalence class.
Step 4: There is no step 4.

For PLL, there are 4! 4! / 2 = 288 permutations in total. You consider some of them to be the same (e.g. up to rotation and AUF), so you form equivalence classes. The PLL skip case is an equivalence class with four elements (solved, U, U2, U'), so the probability of getting that is 4/288 = 1/72. Likewise, H perm is an equivalence class with four elements, so the probability of getting that is also 1/72. Z perm is an equivalence class with eight elements, so it has a probability 1/36, and so on.

You can argue about this in terms of symmetries, but it gets a bit confusing to think about why the two N perms each have probability 1/72 unless you have a good grasp of what's going on.

How do you know a step is N moves efficient on average? How do you say 1LLLL is 15 moves on avg and Cross is 4.9 moves efficient?

What shadowslice said, but specifically for anything that doesn't temporarily break up already existent progress (e.g. cross or EOline), we normally use a program to do a breadth first search starting from the solved case and applying moves from there. This avoids the need to explicitly solve every case (or indeed, to explicitly solve any case at all).

 

[LexCubing]

LL skip is 1 in 15552

Oops that was PLL skip when using COLL.

Step 1: Enumerate the cases in a way such that the cases are equiprobable.
Step 2: Form equivalence classes.
Step 3: Get the size of an equivalence class.
Step 4: There is no step 4.

You can argue about this in terms of symmetries, but it gets a bit confusing to think about why the two N perms each have probability 1/72 unless you have a good grasp of what's going on.

So how does symmetry affect probabilities and what step 4? Why did you even write it? What did you mean no step 4?

 

[xyzzy]

hello????

Anyway, cases that are more symmetric occur less frequently, with the ratio being determined by the size of the stabiliser. To understand what "symmetry" even means, of course you'll need some group theory for that, and I'm not going to compress a one-semester undergraduate course into a single forum post, because that's obviously impossible.

(Almost every "intro-level" video/text I've seen about group theory and Rubik's cubes is outright wrong in some subtle way, even if they get the correct result. This is clearly not easy to get right, so you might as well just use a different approach. That's why I gave a combinatorial approach rather than a group theoretic one.)

 

[LexCubing]

hello????

Anyway, cases that are more symmetric occur less frequently, with the ratio being determined by the size of the stabiliser.

I meant can you talk more about the confusing N perms. You say something about arguing in terms of symmetries?

Just to check you say Z PLL has 8 classes? So the Solved, U2, U' and another Solved, U, U2, U' because there's 2 Z PLLs? Is that what you meant by 8 since there's 2 Z PLLs, one going left and the other to the right?

What do you do when you combine both orientation and permutation?

Okay let's try this. 288 x 27 = 7776
A H pure flip has solved and U2 so 2/7776 so 1/3338?

 

[xyzzy]

Just to check you say Z PLL has 8 classes? So the Solved, U2, U' and another Solved, U, U2, U' because there's 2 Z PLLs? Is that what you meant by 8 since there's 2 Z PLLs, one going left and the other to the right?

The eight ways to have a Z perm: 1 2 3 4 5 6 7 8

"Z perm" is the equivalence class (singular "class", not plural "classes") consisting of these eight permutations of last layer pieces.

 

[LexCubing]

The eight ways to have a Z perm: 1 2 3 4 5 6 7 8

"Z perm" is the equivalence class (singular "class", not plural "classes") consisting of these eight permutations of last layer pieces.

What about when you start combing orientation and permutation?

Okay let's say a pure twist H ZBLL is

288 x 27 = 7776

2/7776 since we have solved and U2?

 

[xyzzy]

Okay let's say a pure twist H ZBLL is

288 x 27 = 7776

2/7776 since we have solved and U2?

The 7776 is correct; the 2 is wrong.

Pick any alg to generate the pure twist H ZBLL and call it X. Now look at these 16 possibilities: X, U X, U2 X, U' X, X U, U X U, U2 X U, U' X U, …, U' X U'. These give rise to exactly 8 distinct permutations/orientations of the last layer (try it out for yourself), so the probability is 8/7776 = 1/972.

 

[LexCubing]

The 7776 is correct; the 2 is wrong.

Pick any alg to generate the pure twist H ZBLL and call it X. Now look at these 16 possibilities: X, U X, U2 X, U' X, X U, U X U, U2 X U, U' X U, …, U' X U'. These give rise to exactly 8 distinct permutations/orientations of the last layer (try it out for yourself), so the probability is 8/7776 = 1/972.

Okay let me get this straight. First I calculate every possible valid cases. Look at the 16 possibilties and reduce it to the # of distinct cases.

Last question why that 16 possibilities? Any reason for that? I get the U X, ... X U since they're the AUFs but why do U X U, etc.? Won't U X U = U2 X?

 

[xyzzy]

Last question why that 16 possibilities? Any reason for that? I get the U X, ... X U since they're the AUFs but why do U X U, etc.? Won't U X U = U2 X?

The order of X and U cannot be switched. Compare: U X and X U.

 

[LexCubing]

The order of X and U cannot be switched. Compare: U X and X U.

I'm not asking U X vs X U.

I'm asking about U X U vs U2 X.
Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?

 

[xyzzy]

I'm not asking U X vs X U.

I'm asking about U X U vs U2 X.
Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?

No!!!!! They do not commute! Have you actually tried out both U X U and U2 X to see how they differ?

You get the same ZBLL case, but at a different angle. The angle matters.

 

[TDM]

I'm not asking U X vs X U.

I'm asking about U X U vs U2 X.

isn't that literally the same thing but with an extra U move added to the front of both...?

 

[AlphaSheep]

I'm not asking U X vs X U.

I'm asking about U X U vs U2 X.
Aren't they the same since you offset the U layer -> Alg -> Offset U layer so the total offset is U2. So it's the same w/ offset the U layer 2x -> Alg since the total offset is U2?

An AUF before depends on the angle of the case, for example, a U perm with the bar on the left vs the bar in front. An AUF afterwards depends on the colours, so for example an orange bar vs blue bar.

For a Ua perm, there are 4 possible colours for the bar, and 4 possible locations for the bar, which give the 16 possible Ua cases.

 

[LexCubing]

No!!!!! They do not commute! Have you actually tried out both U X U and U2 X to see how they differ?

You get the same ZBLL case, but at a different angle. The angle matters.

Okay thanks for the help.

 

[obelisk477]

What are the odds, on a cube with only 6 scrambled corners (they could be twisted as well), that the corners will be a set of two 3 cycles?

Asking for FMC

 

[xyzzy]

What are the odds, on a cube with only 6 scrambled corners (they could be twisted as well), that the corners will be a set of two 3 cycles?

Consider the possible cycles on the six unsolved corners.

• (5, 1):
  • Possible permutations with this cycle structure: \( \binom6{5,1}\cdot(5-1)!=144 \)
  • Orientations: \( 2\cdot3^4=162 \)
• (4, 2):
  • Permutations: \( \binom6{4,2}\cdot(4-1)!(2-1)!=90 \)
  • Orientations: \( 3^5=243 \)
• (3, 3):
  • Permutations: \( \binom6{3,3}\frac1{2!}\cdot(3-1)!^2=40 \)
  • Orientations: \( 3^5=243 \)
• (3, 1, 1, 1):
  • Permutations: \( \binom6{3,1,1,1}\frac1{3!}\cdot(3-1)!=40 \)
  • Orientations: \( 2^3\cdot3^2=72 \)
• (2, 2, 1, 1):
  • Permutations: \( \binom6{2,2,1,1}\frac1{2!2!}\cdot(2-1)!^2=45 \)
  • Orientations: \( 2^2\cdot3^3=108 \)
• (1, 1, 1, 1, 1, 1):
  • Permutations: \( 1 \)
  • Orientations: 1 for all twisted cw, 1 for all twisted ccw, \( \binom63=20 \) for three cw and three ccw; total 22.

Conditioned on the last six corners all being unsolved, the probability of getting a "good" (3, 3) case (i.e. both cycles aren't twisted) is \( 40\cdot3^4/(\text{sum of the stuff above})=3240/62680\approx5.17\% \).

 

[FastCubeMaster]

Does anyone know where I can find or know themselves the chances for each of the UF VLS cases occurring?
Just finished learning and curious to know

 

[TDM]

Does anyone know where I can find or know themselves the chances for each of the UF VLS cases occurring?
Just finished learning and curious to know

Since you can't reduce any cases to another cases by AUF, they are all equally likely.

So 1/216 given the pair is solved, or 1/27 given you have a UF case.