Do OLLs effect which PLL you get?

Solved - Adjacent
Opposite - Adjacent
Adjacent - Solved
Adjacent - Opposite
Adjacent - Adjacent
Adjacent - Adjacent
Nice, that's actually a simpler way than I had in mind.

Basically your still going to have the same odds, it's just the OLL permutation is different than the PLL permutation, correct?
Not sure how you mean that, but I'd say PLL is simply independent of the previous steps. If you ignore the permutation of the LL pieces until you arrive at PLL, then you're not introducing any bias, and thus all 288 PLL cases have the same probability.

Btw, while that is obvious to me, I still liked the above explicit 2x2x2 experiment as it's more hands-on than the theoretic explanation. Hope you liked it, too. And thanks for asking that question which made me come up with it, that was a pretty good one.
 
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A certain OLL gives the probabilities for the following PLLs. Those probabilities are not neccesarily the same as the known probabilities for the different PLLs.
False. Before OLL, all 288 permutations of the LL pieces have the same probability, namely 1/288. The OLL algorithm then just translates cases to other cases, but all 288 cases end up with 1/288 probability again. Check Cyrus's illustrations for the 2x2x2 to see how the PLL cases are translated for example from adjacentSwap to diagonalSwap but their numbers stay the same.

I really do like such concrete questions or statements, making it easy to give concrete answers and giving me new perspectives on a topic that seemed simply clear to me already. I learned something in this thread.
 
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Solved - Adjacent
Opposite - Adjacent
Adjacent - Solved
Adjacent - Opposite
Adjacent - Adjacent
Adjacent - Adjacent
Nice, that's actually a simpler way than I had in mind.

Basically your still going to have the same odds, it's just the OLL permutation is different than the PLL permutation, correct?
Not sure how you mean that, but I'd say PLL is simply independent of the previous steps. If you ignore the permutation of the LL pieces until you arrive at PLL, then you're not introducing any bias, and thus all 288 PLL cases have the same probability.

Btw, while that is obvious to me, I still liked the above explicit 2x2x2 experiment as it's more hands-on than the theoretic explanation. Hope you liked it, too. And thanks for asking that question which made me come up with it, that was a pretty good one.

What I meant is (I may have this completely wrong): Basically, permutations come in pairs, & if you have one in a certain OLL you'll get the other in PLL, visa versa. These pairs are only pairs on certain OLL's with certain algorithms.

For example, lets say I have an OLL that switches adjacent corners, you can just look at the permutation because thats all that matters.

FRU/BRU = FLU/FRU = FRU/BRU = FLU/FRU etc.
FRU/BLU = Solved = FRU/BLU = Solved = FRU/BLU etc.
BRU/BLU = FLU/BLU = BRU/BLU = FLU/BLU etc.
 
FRU/BRU = FLU/FRU = FRU/BRU = FLU/FRU etc.
FRU/BLU = Solved = FRU/BLU = Solved = FRU/BLU etc.
BRU/BLU = FLU/BLU = BRU/BLU = FLU/BLU etc.
Ugh. Please don't use "=" for this, that's just wrong. An arrow like "=>" is ok and much clearer.

Basically, permutations come in pairs, & if you have one in a certain OLL you'll get the other in PLL, visa versa.
In a simplified analysis of the 2x2x2 this is correct, but not in a strict one and certainly not for the 3x3x3. The reason you just had case pairs was that you were only analyzing swaps. For the 2x2x2, already 3-cycles are a counter-example:

Solved => adjacentSwap => otherAdjacentSwap => Solved
 
FRU/BRU = FLU/FRU = FRU/BRU = FLU/FRU etc.
FRU/BLU = Solved = FRU/BLU = Solved = FRU/BLU etc.
BRU/BLU = FLU/BLU = BRU/BLU = FLU/BLU etc.
Ugh. Please don't use "=" for this, that's just wrong. An arrow like "=>" is ok and much clearer.

Basically, permutations come in pairs, & if you have one in a certain OLL you'll get the other in PLL, visa versa.
In a simplified analysis of the 2x2x2 this is correct, but not in a strict one and certainly not for the 3x3x3. The reason you just had case pairs was that you were only analyzing swaps. For the 2x2x2, already 3-cycles are a counter-example:

Solved => adjacentSwap => otherAdjacentSwap => Solved

Okay on a 3x3x3, if i have an algorithm for OLL that does an Acw, it's grouped, same with U's, & G's, so it's just a cycle.

Solved -> Accw -> Acw -> Solved

If I have a PLL, like the J perm, wouldn't 2x2x2 pairs work?
Solved-> J -> Solved -> J

Same with N, E, H, Z, F, R, V, Y, & T, or am I missing something still? It seems to me that this works some of the time on 3x3x3, although a better way to put it would've been: The OLL algorithm affects the PLL but not it's probability. Permutations come in Cycles of either 1, 2, or 3. One always leads to the next one, which depending on the number of Cycles, leads to the next one, or back to the first one.

So it's going to be one of these three:

A -> A -> A -> A etc.
A -> B -> A -> B etc.
A -> B -> C -> A etc.

There aren't any Permutations that have 4 cycles (correct me if I'm wrong).
 
U' R2 U R U R' U' R' U' R' U R'

And here's a 6-cycle:
U' R U R' U' R' F R2 U' R' U' R U R' F'

I only recently started learning full OLL, but I've noticed that almost every time I get this T shape OLL, it is followed by a T-perm.
O33.gif

Is this just some crazy coincidence, or are certain permutations more common depending on the OLL that preceded it?
Well I think, using 2 look OLL can improve your chances of getting an OLL skip, because;
When you take your first look and make the cross, there’s already a tiny chance (1/504) you skip the rest of OLL and PLL,

And you have the same probability as when you use normal OLL for the PLL skip (1/72) after your 2nd look
Adding up to 1/63
 
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