Is a PLL skip lucky when using COLL?

part of the COLL "method" is to increase your odds of getting a PLL skip
The step after COLL is not PLL, so how could you skip it?

What? I don't see what's wrong with that. 'PLL' is a valid description of that step. It's like somebody saying 'I'm reading this book' and you saying 'No that's wrong, you're reading a novel'.
Unless I've the wrong end of the stick?
 
Here's how to make pseudo-COLL + EPLL skip non-lucky: Solve at least one edge along with the corners. I can already think of a couple of ways to achieve this right now, none of them too efficient, but it's feasible anyway.

If you haven't already figured out, with this approach, EPLL skip occurs 33.3% of the time (>20%).

part of the COLL "method" is to increase your odds of getting a PLL skip
The step after COLL is not PLL, so how could you skip it?

What? I don't see what's wrong with that. 'PLL' is a valid description of that step. It's like somebody saying 'I'm reading this book' and you saying 'No that's wrong, you're reading a novel'.
Unless I've the wrong end of the stick?
If you mean permutation of the last layer in the most literal sense, then yes, it is a valid description of that step. But when most, if not all cubers talk about PLL, they are referring to the step with 21 different cases, not 4.

Simply put, EPLL skips occur 1/12 of the time, PLL skips occur 1/72 of the time, hence EPLL != PLL when we're talking about skips here.
 
Last edited:
part of the COLL "method" is to increase your odds of getting a PLL skip
The step after COLL is not PLL, so how could you skip it?
I think he means EPLL as if that wasnt clear.
How does COLL increase your odds of getting an EPLL skip?
Odds of getting a PLL skip = a
Odds of getting an EPLL skip = b

COLL alters neither value. It merely increases the odds of getting a last step skip from a to b.
 
Back
Top