[Unofficial] 15.98 Average of 5 w/Roux+L6E skip!

[RubiXer]

Descent average. Has anyone else gotten a L6E skip. What are the odds of L6E skip?

 

[Ross The Boss]

i once got an lse skip. it was actually on my, like, 5 roux solve ever... the solve was about 2 1/2 mins. good job.

 

[bgdgyfer]

I don`t do roux but, good solved anyway!

 

[RubiXer]

Thanks

 

[Sinapanis]

Hey Neil! Remember me from Caltech (I'm Everett). Nice average . I got my first sub-15 average today (14.837 avg). Watch out haha

 

[RubiXer]

Hey Neil! Remember me from Caltech (I'm Everett). Nice average . I got my first sub-15 average today (14.837 avg). Watch out haha

Hey what's up Everrett! I remember you. I got my first sub-15 avg the other day too. Add me on FB your privacy settings won't let me -.-

 

[nqwe]

chances are 1/11520 i guess.

 

[RubiXer]

chances are 1/11520 i guess.

Woah. I can't believe I got it on camera then!

 

[BlackStahli]

Not much of a Roux solver, but I'd say that you can count an LSE skip as all edges solved with either U face or M slice adjustments necessary, but not both.
oriented correctly is 1/2^6, which is 1/64
permuted correctly is 1/6C2 * 1/6C4 * 4, or 4/225

which should give 1/3600. Not sure though, I probably missed something >.<

 

[RubiXer]

Hmm. That seems to be more accurate. Thanks.

 

[JasonK]

I'm pretty sure Kir had a video of a solve that just went FB, SB, H-Perm. So CLL and DF+DB skip.
I wonder what the probability is of just being left with an EPLL after SB.

EDIT: Found the vid

 

[RubiXer]

I remember seeing that video a few weeks ago, and I've been left with an EPLL after CMLL but never after SB...

 

[phi]

Not much of a Roux solver, but I'd say that you can count an LSE skip as all edges solved with either U face or M slice adjustments necessary, but not both.
oriented correctly is 1/2^6, which is 1/64
permuted correctly is 1/6C2 * 1/6C4 * 4, or 4/225

which should give 1/3600. Not sure though, I probably missed something >.<

I have to disagree with this. The probability that all of the L6E are oriented correctly is 2/2^6 = 1/32, because the last edge doesn't have any freedom of orientation (unless you take apart and reassemble the cube). Frankly I don't quite understand the reasoning behind the expression 1/6C2 * 1/6C4 * 4, but I think we should look at the number mentioned in a previous post:

1/11520. (There are 6! ways to permute the 6 edges and 2^6 ways to orient them, but you have to divide that number of L6E cases by 2 because half of the time, after taking your cube apart and reassembling it randomly, two edges will have to be swapped in the end to make it solvable; and then divide by 2 again because half of the time, independently of the parity I just described, you will have to flip an edge to make it solvable.)

This number assumes that you already have all the centers and all the corners in exactly the right spot, which is not generally going to be the case. It may be the case, but equally likely (I think), you may have to do an M, or M', or M², or U, or U', or U². (This is based on the definition that either an M-slice or a U-turn, but not both, could be necessary and it still count as an L6E skip.) Here's the deal, though: In the first case (where all centers and corners are already right), your probability is exactly 1/11520. In the second case, where the corners are already exactly right, but you need M in order to get the centers right, the probability of all edges to be exactly right after that M is again exactly 1/11520, because they still all have to be in their respective correct positions and orientations. As a matter of fact, for each one of those 7 cases, the conditional probability is 1/11520. Therefore the probability of an L6E skip should be 1/11520 times the probability that your corners and centers will be in a state that allows you to solve them completely by doing just one of the following 7 moves: M, M', M², U, U', U², or nothing. This last probability is 7/16, isn't it? (There are 4 different ways for the corners to be positioned and 4 different ways for the centers to be positioned, corners and centers independently of each other.)

I therefore claim that the chance of an L6E skip is 1/11520 * 7/16 = 7/184320, or roughly 1 in 26331.

To summarize for those who don't like words, but math instead: 2/6! * 2/2^6 * 7/16 = 7/184320.

All this assumes that you don't do anything to influence the L6E until we actually get to that step, but realistically we all probably at the very least try to finish our CMLL in such a way that either the top or bottom center ends up on top, but personally I don't do it for all CMLL algs. Some people pick from different CMLL algs depending on how many bad edges they see when they start CMLL, so the probability will be a little bit better than the number I calculated, but depend on exactly what kind of look-ahead or L6E-preparation you do.

What does everyone else think?

 

[shadowslice e]

Well, if centres are done correctly the odds of an uninfluenced skip are 6!*2^5=1/23040. But if it is white/yellow facing up then it would be double. If no special orientation, then 1/92160

Also, nice 4 year bump.

 

[Thermex]

What are the chances of a CMLL skip? I was think of working on a Roux-like method and I thought it would be cool if you could put the last corner (not pair) into the second block and forced a CMLL skip.