Which Is Less Common?

[Jteddy]

Which one of these is less common?
Getting the same OLL into PLL skip twice in a row, or getting the same OLL 4 times in a row? Both of these have happened to me before and both times it was a 1/54 OLL case.

 

[acrystal]

Which one of these is less common?
Getting the same OLL into PLL skip twice in a row, or getting the same OLL 4 times in a row? Both of these have happened to me before and both times it was a 1/54 OLL case.

depends on what oll you are talking about
some olls are rarer than others so you need to specify

 

[Jteddy]

depends on what oll you are talking about
some olls are rarer than others so you need to specify

I thought I did specify. I said it was a 1/54 OLL case.

 

[acrystal]

I thought I did specify. I said it was a 1/54 OLL case.

Getting the same OLL into PLL skip twice in a row = Getting the same 1LLL case twice in a row (if you know only 1 alg for that OLL case) = roughly 1/15522
Getting the same OLL 4 times in a row = roughly 1/157464

Getting the same OLL case 4 times in a row is significantly higher than getting the same LL case twice even at its rarest.

 

[Jteddy]

Getting the same OLL into PLL skip twice in a row = Getting the same 1LLL case twice in a row (if you know only 1 alg for that OLL case) = roughly 1/15522
Getting the same OLL 4 times in a row = roughly 1/157464

Getting the same OLL case 4 times in a row is significantly higher than getting the same LL case twice even at its rarest.

So how did you figure that out?

 

[acrystal]

So how did you figure that out?

1/15522 = (1/15522)^(2-1)
1/157464 = (1/54)^(4-1)

 

[Wagnit]

There are
- 51 OLL cases with 4/216 probability
- 5 OLL cases with 2/216 probability
- 1 OLL case with 1/216 probability

PLL skip probability is 1/72

Assuming you that there is no last layer manipulation (never considering a future last layer at any point in the solve),

For 2 solves the probability of same OLL into PLL skip twice is:
(51 * (4/216)^2 + 5 * (2/216)^2 + 1 * (1/216)^2) * (1/72)^2 = roughly 0.00000346

For 4 solves the probability of same OLL 4 times is:
51 * (4/216)^4 + 5 * (2/216)^4 + 1 * (1/216)^4 = roughly 0.00000603

If assuming OLL skip is an OLL case, change those 1's to 2's.

Becomes a quite a bit more difficult to calculate when there's more solves in sequence and now you have multiple cases with correlating probability (e.g. for 5 solves you could have solves 1-4 or solves 2-5 fit the 4 solves in a row requirement).