Probability Thread

[TDM]

Chances of not getting parity in 3 rounds of 4x4?
6x6?
(idk if the chances are different)

chance of no parity = 1/4
3 rounds = 15 solves
(1/4)^15 = 9x10^-10 (roughly one in a billion)

6x6 has the same for LL parity, I don't know about edge parity during pairing though

 

[cashis]

chance of no parity = 1/4
3 rounds = 15 solves
(1/4)^15 = 9x10^-10 (roughly one in a billion)

6x6 has the same for LL parity, I don't know about edge parity during pairing though

What about just OLL parity?
I guess it would be half?

 

[TDM]

What about just OLL parity?
I guess it would be half?

The probability of OLL parity on each solve is a half. The probability of no OLL parities (or all OLL parities) in 15 solves is (1/2)^15 = 0.00003 (about 1/33000).

 

[cuBerBruce]

Chances of not getting parity in 3 rounds of ... 6x6?

chance of PLL parity = 1/2 (per solve)
chance of OLL parity in at least 1 orbit = 3/4 (per solve)
3 rounds = 9 solves
no parities in 9 solves:
(1/8)^9 = 1/134217728

(What is the probability of a competition having 3 rounds of 6x6x6?)

 

[Myachii]

That's exactly my thoughts--it felt absolutely surreal. Without thinking about the exact math, it occurred to me that I may be the only person in the world this has ever happened to. It seemed so unbelievable that I wondered if I'd remembered the solve wrong, and done more of the solve than I thought before getting a skip, but I clearly remember the final moves of the solve being L' U L R' U2 R, which guarantees that I'm right about it being a LS+LL skip as it affects more than one F2L slot.

When it comes to massive skips, I've been quite lucky. I've been cubing for less than two years, and I have at least 5 LL skips on 3x3, not even counting this one on 4x4, and not counting the one or two I've gotten on big cubes. I don't get lucky in other ways too often; I don't get an abnormally large number of PLL or OLL skips, but I've had a few incredibly lucky solves. Some people go 5 years without getting a single one of what I must have close to 10 of by now.

Agreed. I've never had a LL skip and I've been cubing for 2 years and 2 months.

 

[Berd]

Agreed. I've never had a LL skip and I've been cubing for 2 years and 2 months.

Ive been cubing around a year and a quarter and I'm still LL skipless.

 

[CubeWizard23]

got my pb back in 09 on a AUF LL skip 25 sec

 

[AlexanderGZH]

1/190080

 

[Isaac Lai]

If you know full OLS, what are the chances of a OLL skip?

 

[Musicalboy2]

If you know full OLS, what are the chances of a OLL skip?

....100%?

 

[Isaac Lai]

....100%?

Oh wait ya nvm I was being dumb

 

[cashis]

What are the odds of having all but 4 edges solved after hoya stage?
e; WHAT ARE THE ODDS OF THIS HAPPENING TWICE IN A ROW

 

[PenguinsDontFly]

What is the chance of having 5 2x2 scrambles in a row with no bar? (by bar I mean a 1/2 face, not a 1/2 layer)

 

[guysensei1]

What is the probability of getting the 'perfect' BLD solve: no parity (this probability is 0.5 right?), no twisted pieces, no cycle breaks?

 

[not_kevin]

What is the probability of getting the 'perfect' BLD solve: no parity (this probability is 0.5 right?), no twisted pieces, no cycle breaks?

No cycle breaks means that there's only one cycle for the piece type in question, which also must include your buffer piece; since twisted pieces can be considered really silly one-piece cycles (eg, FU -> UF), both of these constraints end up being the same. Therefore, the only time you can have this happen is if you have a single cycle for each piece type that also involves your buffer piece, and the length of both cycles is odd (has an even number of targets, or has an odd number of pieces involved).

The number of times this can happen on corners is the sum of the number of n-cycles over the pieces - so 1 (for everything solved) + (7*3) * 6 [you can choose any non-buffer sticker for the first, and the second target has orientation pre-determined to prevent twisted corners] + (7*3) * (6*3) * (5*3) * 4 + (7*3) * (6*3) * (5*3) * (4*3) * (3*3) * 2 = 1247527. Similar logic for edges (basically, start counting down from 11 instead of 7, and there are 2 orientations, not 3) gives 21299668061, if my input to WolframAlpha was correct. Therefore, the number of 'perfect' BLD solves is 1247527 * 21299668061 = 2.657*10^16 (26571910997135147) - since the number of total positions is 43.252*10^19 (43252003274489856000), the probability of this occurring is about 0.061435% - about once every 1627 solves (0.061463%)!

 

[youSurname]

Do PLL occurrences change depending on what OLL algs you use. I'm seeing a lot of N perms.

 

[guysensei1]

Do PLL occurrences change depending on what OLL algs you use. I'm seeing a lot of N perms.

Negativity bias?

 

[obelisk477]

Do PLL occurrences change depending on what OLL algs you use. I'm seeing a lot of N perms.

It depends on how you're running in to them. If you're seeing them occur in full solves (from completely scrambled state to solved), then its just chance, and the OLL you use has no effect on the PLL. If you're using an LL scrambler, then sometimes they only scramble certain ways that affect the probabilities of certain OLLs and PLLs arising.

 

[guysensei1]

It depends on how you're running in to them. If you're seeing them occur in full solves (from completely scrambled state to solved), then its just chance, and the OLL you use has no effect on the PLL. If you're using an LL scrambler, then sometimes they only scramble certain ways that affect the probabilities of certain OLLs and PLLs arising.

Alternative he learnt OLLCP and uses it to get diagswaps every time

 

[IRNjuggle28]

Do PLL occurrences change depending on what OLL algs you use. I'm seeing a lot of N perms.

The simplest answer is no, it doesn't affect your PLL. It's possible to use OLL to influence PLL, but you have to do it on purpose. You aren't doing that, and if you were, you would be using OLL to force good cases, in which case you'd get less N perms, not more. It's just bad luck.