Probability Thread

[OtaMota]

Maybe someone already asked this, IDK

but what is the probability of getting a 4 mover on 2x2?

1/2000

 

[SpeakCuba]

Not sure if it's been posted here before, but I just got my first LL skip; and it was on a 4x4 of all cubes. What's the probability of a 4x4 LL skip? (LL skip happened with OLL parity).

 

[bldcuber314]

Not sure if it's been posted here before, but I just got my first LL skip; and it was on a 4x4 of all cubes. What's the probability of a 4x4 LL skip? (LL skip happened with OLL parity).

1/62208 I think since LL skip is 1/15552 and each parity is 1/2. Shame about the OLL parity (was it still a good time?)

 

[FineGeoduck2211]

what is the probability of getting at least one solved bar on clock?

 

[SpeakCuba]

1/62208 I think since LL skip is 1/15552 and each parity is 1/2.

Thanks!

Shame about the OLL parity (was it still a good time?)

It was 'ok' for me (51.17), would have defiantly been PB or one of my best times without the parity. Kind of a normal solve, nothing special except for the LL skip.

 

[xyzzy]

(LL skip happened with OLL parity).

Unless you solve OLL parity before last layer, that doesn't really count as an LL skip.

1/62208 I think since LL skip is 1/15552 and each parity is 1/2. Shame about the OLL parity (was it still a good time?)

This isn't correct, because if you get OLL parity, you're not going to do the OLL parity alg from a random angle to get a random even-flipped-edge LL case afterwards; you're going to do it from an angle that'll solve EO if possible.

A full LL skip including AUF is 1 in (2^4 × 4!) × (3^3 × 4!) = 248832.
A full LL skip up to AUF is 4 in that, or 1 in 62208.
Getting OLL parity as 1LLL is 4 in that, or 1 in 15552, assuming you use only one OLL parity alg.

 

[BVCuber13]

Not sure if it's been posted here before, but I just got my first LL skip; and it was on a 4x4 of all cubes. What's the probability of a 4x4 LL skip? (LL skip happened with OLL parity).

I've gotten that at least twice before, guess I'm just lucky

I also got 3 PLL skips in a row today, my friend @Discomantis told me it's a 1/373,248 chance. Can someone verify this?

Times:
Generated By csTimer on 2024-10-11 (solving from 2024-10-11 11:32:14 to 2024-10-11 11:37:08)
mean of 3: 8.117

Time List:
1. 8.505 U R' L' U' F' U' B' L2 B2 R2 B2 U' F2 B2 U R2 U' R2 L' D' @2024-10-11 11:32:14
2. 7.307 B R2 U2 B2 D F2 D R2 D2 R2 B2 U2 R' B' F L' D2 B R2 U' F @2024-10-11 11:32:48
3. 8.539 R' D2 R' D2 B2 L' B2 R U2 R2 F2 R' U' L' F L2 D B2 U R D @2024-10-11 11:37:08

(Third one used niklas instead of sune so idk if that does anything to the probability)

 

[Celeritardum]

How do you account for symmetries and duplicate cases when calculating probabilities or positions on cubes? For example, if you account for AUF, and parity it seems like there are 72 pll cases, but there's actually only 22 distinct ones (including pll skip). Why is this? How do people find these numbers from scratch? Is it just pure probability theory, or casework with trial and error?

 

[bldcuber314]

How do you account for symmetries and duplicate cases when calculating probabilities or positions on cubes? For example, if you account for AUF, and parity it seems like there are 72 pll cases, but there's actually only 22 distinct ones (including pll skip). Why is this? How do people find these numbers from scratch? Is it just pure probability theory, or casework with trial and error?

This is the way I have always understood it, but I don't know if it is correct. I just did some checks and suddenly I'm not sure if it is the right explanation.

So don't rely on this, but I'll do my best and hopefully if it is correct you'll know.

Take the T-perm. There are 4 different ways to get a T-perm(white cross)- The case with a blue bar, green bar, red bar and orange bar. This means that the chance of a T-perm is 4/72 or 1/18.

Now take the Z-perm. There are 2 different ways to get a Z-perm- Orange/green swapped and red/blue swapped or red/green swapped and orange/blue swapped. This means that the chance of a Z perm is 2/72 or 1/36.

And I think you can do this in some form for all PLLs.

I'm not sure if this is the correct explanation as I'm no expert, but hopefully it helps.

 

[Garf]

This is the way I have always understood it, but I don't know if it is correct. I just did some checks and suddenly I'm not sure if it is the right explanation.

So don't rely on this, but I'll do my best and hopefully if it is correct you'll know.

Take the T-perm. There are 4 different ways to get a T-perm(white cross)- The case with a blue bar, green bar, red bar and orange bar. This means that the chance of a T-perm is 4/72 or 1/18.

Now take the Z-perm. There are 2 different ways to get a Z-perm- Orange/green swapped and red/blue swapped or red/green swapped and orange/blue swapped. This means that the chance of a Z perm is 2/72 or 1/36.

And I think you can do this in some form for all PLLs.

I'm not sure if this is the correct explanation as I'm no expert, but hopefully it helps.

We’ve taken the general pattern of each PLL and defined it as one pattern. Even though there’s like 4 Tperms you can get for LL on white cross (24 depending if you use full CN) we’ve learned to see each case as different bars, checkers, and headlights to recognize and execute each case.

 

[Celeritardum]

This is the way I have always understood it, but I don't know if it is correct. I just did some checks and suddenly I'm not sure if it is the right explanation.

So don't rely on this, but I'll do my best and hopefully if it is correct you'll know.

Take the T-perm. There are 4 different ways to get a T-perm(white cross)- The case with a blue bar, green bar, red bar and orange bar. This means that the chance of a T-perm is 4/72 or 1/18.

Now take the Z-perm. There are 2 different ways to get a Z-perm- Orange/green swapped and red/blue swapped or red/green swapped and orange/blue swapped. This means that the chance of a Z perm is 2/72 or 1/36.

And I think you can do this in some form for all PLLs.

I'm not sure if this is the correct explanation as I'm no expert, but hopefully it helps.

OHHH, I think I understand now. The specific color patterns of each case also matters! I was previously thinking only in terms of auf of a single case when actually (for example T-perm) it applies to multiple cases that just seem the same from a color-neutral perspective.
So it should be:
T - 4/72
F - 4/72
Ga - 4/72
Gb - 4/72
Gc - 4/72
Gd - 4/72
Aa - 4/72
Ab - 4/72
L - 4/72
J - 4/72
Ra - 4/72
Rb - 4/72
E - 2/72
Y - 4/72
V - 4/72
Na - 1/72
Nb - 1/72
H - 1/72
Z - 2/72
Ua - 4/72
Ub - 4/72
AUF - 1/72

Thanks!

 

[THE_V01DL0RD]

what are the odds of a kilo LL Skip no auf

 

[OtaMota]

what are the odds of a kilo LL Skip no auf

Hope I did my math right. I think it’s 1/972, but depending on if my assumption on the number of permutations is correct or not, it may be 1/1944. I know OLL skip is 1/81*, and then you would times by chance of PLL skip, which are either 1/12** (for the 1/972) or 1/24 (for the 1/1944)

* 3^4 for each of the corners orientation, one stays oriented so even tho there are 5 corners, you would only do 3^4

** 5! And then divide by 5 (for AUF) rereading OPs question, don’t do the crossed out part. After that, you possibly divide by 2, depending on whether or not a swap of only 2 corners is possible (I don’t think it’s possible, but I might be wrong)

Edit: just saw the no AUF part, so times 5 more rare. 1/4860 or 1/9720

 

[Jorian Meeuse]

Hope I did my math right. I think it’s 1/972, but depending on if my assumption on the number of permutations is correct or not, it may be 1/1944. I know OLL skip is 1/81*, and then you would times by chance of PLL skip, which are either 1/12** (for the 1/972) or 1/24 (for the 1/1944)

* 3^4 for each of the corners orientation, one stays oriented so even tho there are 5 corners, you would only do 3^4

** 5! And then divide by 5 (for AUF) rereading OPs question, don’t do the crossed out part. After that, you possibly divide by 2, depending on whether or not a swap of only 2 corners is possible (I don’t think it’s possible, but I might be wrong)

Edit: just saw the no AUF part, so times 5 more rare. 1/4860 or 1/9720

Correct, and it would be 1/4860 as a 2-swap is indeed not possible. See this post: https://www.speedsolving.com/threads/help-thread-hugebld-discussion.93969/post-1637388

 

[bcube]

How many cases are there for a DR state having 2 uniform faces? For example having U face white and D face yellow (DR being on U/D axis if that makes sense, I don´t want to include green/red/blue/orange uniform faces - I want to include only those faces on which is DR made - in this example U and D faces).

I am thinking of (((4!*4!)^2)/2)*4!

 

[CubeWizard23]

I wrote a program that calculates the shortest XCross solution for a scramble and ran it on ~100k scrambles. This is what i got:

Spoiler: Tables of Data and a Chart

For each of the six faces I cross tabulated the shortest Cross Solution with the Shortest XCross Solution, I was curious what the correlation was:

Spoiler: Heatmap of Cross vs XCrosses

Obviously this data is only a small subset, and not necessarily 100% representative. I am currently generating 1 million more scrambles, which will yield 2 million unique cube states (inverse scramble) and 48 million XCrosses to check.

I was looking for a thread or a webpage similar to the Cross Study by Lars Vandenbergh but my searching yielded surprisingly little.

The most interesting takeaway is that CN XCross (6.53 moves avg) is a half move better on average than DCN XCross (6.98 moves avg) which is only 0.37 moves better than single color (7.35).

That said, average isn't necessarily the best measure, almost 42% of the time Full CN has a 6 move or less XCross, compared to only ~20% for DCN and 10% for Single Color.

This tells me that as a DCN solver I should practice until i am at least able to find 7 move XCrosses. This should yield XCross in ~80% of my solves.

I'm sure other people have done this kind of thing before, don't shoot me. It was a fun coding project, and is general enough I can also do any cube substate, EO, EOCross, RouxFB, 2x2x3, etc.

 

[CubeWizard23]

I wrote a program that calculates the shortest XCross solution for a scramble and ran it on ~100k scrambles. This is what i got:

I analyzed the same dataset for Pseudo XCross. When looking at a Pseudo XCross you have to understand that it adds a move later in the solution, so its not 1:1 comparison to XCross, but I think we can agree a 6 move solution with an AUD later (assuming fluency in Pseudo F2L) is going to be easier than a 7 move XCross. I was expecting the Pseudo XCross to be shifted down by at least a full move, so these findings are rather interesting. It only saves slightly more than a half move on average. That said, it increases the number of "findable" cases for any depth, so it would probably be worthwhile being comfortable with Pseudo F2L and XCross and utilizing option select. (Means are shown on the bottom right of the tables)

Note: The way I defined Pseudo XCross actually includes the trivial case as well, in the event it is the shortest solution for its depth.

Spoiler: Tables and Chart

Update: ran the sim for an additional 100k solves (2.4 million states) and the distributions all hold within a 0.25%

 

[Constant Cuber]

I analyzed the same dataset for Pseudo XCross. When looking at a Pseudo XCross you have to understand that it adds a move later in the solution, so its not 1:1 comparison to XCross, but I think we can agree a 6 move solution with an AUD later (assuming fluency in Pseudo F2L) is going to be easier than a 7 move XCross. I was expecting the Pseudo XCross to be shifted down by at least a full move, so these findings are rather interesting. It only saves slightly more than a half move on average. That said, it increases the number of "findable" cases for any depth, so it would probably be worthwhile being comfortable with Pseudo F2L and XCross and utilizing option select. (Means are shown on the bottom right of the tables)

Note: The way I defined Pseudo XCross actually includes the trivial case as well, in the event it is the shortest solution for its depth.

Spoiler: Tables and Chart

View attachment 47389
View attachment 47390

Update: ran the sim for an additional 100k solves (2.4 million states) and the distributions all hold within a 0.25%

Great info ! I am curious about comparing movecount of EOCross to a standard CFOP cross + 1st pair. Would you be able to calculate?

 

[CubeWizard23]

Great info ! I am curious about comparing movecount of EOCross to a standard CFOP cross + 1st pair. Would you be able to calculate?

Here are the stats for EOCross over the same dataset.
It appears that CN EOCross has a comparable movecount to CN XCross and DCN Axis Neutral EOCross is slightly better (but not much) than DCN XCross based on the ~300k states analyzed here.

Spoiler: Tables and Charts for EOCross

I am going to make a thread for all this once I comb through all the data and make a more rigid study/report framework. Probably will do RouxFB, EOLine, 2x2x2, 2x2x3, and XXCross as well.

 

[Constant Cuber]

Here are the stats for EOCross over the same dataset.
It appears that CN EOCross has a comparable movecount to CN XCross and DCN Axis Neutral EOCross is slightly better (but not much) than DCN XCross based on the ~300k states analyzed here.

Spoiler: Tables and Charts for EOCross

View attachment 47468
View attachment 47469
View attachment 47470

I am going to make a thread for all this once I comb through all the data and make a more rigid study/report framework. Probably will do RouxFB, EOLine, 2x2x2, 2x2x3, and XXCross as well.

Thanks for running this analysis! Roughly speaking the EOcross is a bit less than 1 full move more than CFOP cross on a single axis. For a human solver I would expect a bit more than that, perhaps a good EOcross solver would be in the 8 to 9 move range.

Another way to interpret the results is that the most likely movecount for EOcross is 7 and 8 moves, while for the fully CN EOcross is most likely 6 and 7. This kind of suggests that the other axes will give you less bad edges, but perhaps we shouldn't expect a large improvement in bad edges, ie it seems to be pretty common that we would expect a reduction of 2 bad edges by changing axis, and we would experience 4 or 6 fewer bad edges only occaisionally.