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Probability Thread

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?)
 
(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.
 
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)
 
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?
 
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.
 
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.
 
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!
 
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
 
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
 
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