PLL skip? No problem, but...

Actually the title of the thread seems to imply that there's a problem with having 2 PLL skips in a row..

Anyway, I'm sure every cuber will happen to have 2 or more PLL skips in a row in their cubing lifetime...if they cube long enough...
 
Cride5 said:
Are you counting a solved LL requiring AUF as an LL-skip as well? Ie. would the probability of an LL-skip, including cases requiring U, U' and U2 still be 1/1944, or would it be 4/1944??
Click to expand...
I ignored AUF, i.e., counted all groups of 4 positions a U-move away from each other as 1 position. Counting them separately, there are 16 T-perms, the number is 3^3 * 4! * 4! / 2 = 7776 = 4 * 1944, and the probability is 4/7776 = 1/1944.

amostay2004 said:
Anyway, I'm sure every cuber will happen to have 2 or more PLL skips in a row in their cubing lifetime...if they cube long enough...
Click to expand...
... and because everybody uses a method that ends with PLL.

Edit: Also, I'm 100% sure I will have over 9000 F3L skips in a row, if I cube long enough.
 
Last edited:
Johannes91 said:
Cride5 said:
Are you counting a solved LL requiring AUF as an LL-skip as well? Ie. would the probability of an LL-skip, including cases requiring U, U' and U2 still be 1/1944, or would it be 4/1944??
Click to expand...
I ignored AUF, i.e., counted all groups of 4 positions a U-move away from each other as 1 position. Counting them separately, there are 16 T-perms, the number is 3^3 * 4! * 4! / 2 = 7776 = 4 * 1944, and the probability is 4/7776 = 1/1944.
Click to expand...

Aah, right. I get what you're saying! I like your example. There are four possible T perms, and for each one, there are four ways it can be rotated on the U face, giving 16 distinct cases for T-perm.

Just to make sure I'm getting the theory>
  • There are 4 corners, each with 3 possible orientations, combined that's 3*3*3*3 = 3^4 possibilities, but because only a third of corner orientations are reachable, that's (3^4)/3 = 3^3.
  • There are 4! ways to arrange (permute) the four corner pieces, and 4! ways to permute the edge pieces which gives 4!*4! ways to permute them independently. However, since only an even number swaps is possible only the even permutations are valid, giving (4!*4!)/2.
Thus the total number of cases is (3^3) * (4!*4!)/2 = 7776, since one of these cases involves the LL solved, and three more involve the LL solved, but rotated, the chances of a ZBLL skip = 4/7776 = 1/1944

... sorry for the repetition, just had to be sure :)
 
Last edited:
Yep, that's it. One nitpick: it's better to use "ZBLL" instead of "LL", the latter usually includes cases where EO isn't solved (adds a factor of 2^3).
 
Does it actually matter how many moves are made in a scramble? When I give it to somebody for a scramble and they don't know much about the cube, they always try to make it as scrambled as possible, they try to have it so that the same color is never 3 in a row. But does that actually make a difference between a quick 10 move hand scramble? More chance of a skip or easier F2L?
 
Dene said:
I got two in a row... in competition.
Click to expand...

Did you really?!
That's awesome.

I don't know if that was meant to be a joke, but if it wasn't.. Here's my guess as to which two solves it was.. 13.75, 13.92, your last two solves during the first round of Discovery Science Center. Only guessing those because they were much faster then your other solves at that competition and it helped you break your AuR. :p
 
You must log in or register to reply here.

Similar threads

Jteddy
Which Is Less Common?
Replies
6
Views
354
Wagnit
W
precompsubwr
Which do you think is better: PLL skip or OLL skip on 3x3
Replies
15
Views
1,999
CubeLite
CubeLite
DoAnKhang
What is the PLL skip chance?
Replies
2
Views
1,099
LBr
LBr
H
Should i skip beginner pll and skip to advanced pll but start learning with the ones that are in begnner?
Replies
2
Views
642
Caden Fisher
Caden Fisher
D
[Member Intro] I’m a pll skip
Replies
3
Views
919
AbsoRuud
AbsoRuud
Back
Top