OtaMota
Member
1/2000Maybe someone already asked this, IDK
but what is the probability of getting a 4 mover on 2x2?
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1/2000Maybe someone already asked this, IDK
but what is the probability of getting a 4 mover on 2x2?
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?)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).
Thanks!1/62208 I think since LL skip is 1/15552 and each parity is 1/2.
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.Shame about the OLL parity (was it still a good time?)
Unless you solve OLL parity before last layer, that doesn't really count as an LL skip.(LL skip happened with OLL parity).
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.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?)
I've gotten that at least twice before, guess I'm just luckyNot 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).
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.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?
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.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.