TDM
Legend
Same. I was thinking about some supercube centres, but not all of them...
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Probability Thread
- Thread starter CubesOfTheWorld
- Start date
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- Tags
- cfop oll probability
goodatthis
Veteran
Well, you may think so, but an Examinx is not the next step up. By the regular orders of magnitude, it goes peta- exa- zetta-, but the one person who actually designed it named it wrong. Also, gigaminxes have 5 layers, not 6. I believe a 6 layered minx is called a super kilominx or star minx or something.
Examinx:
Based on Cmowla's formula, a super 13 layered minx would have 1.1*10^2934 combinations.
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what?
goodatthis
Veteran
Huh?
The person who designed the Examinx should have named it the zettaminx, since petaminx is 9 layers, and Examinx is 13.
TMOY
Veteran
If the one person who designed it has no clue about how that naming scheme works, we're not forced to stick to his error forever just because he chose the wrong name in the first place. His puzzle is a zettaminx, not an examinx, period.
goodatthis said:
Yes, U know, that was a typo, sorry.
IRNjuggle28
Veteran
What are the odds of getting a skip of orientation of the 8 edges other than the cross pieces? Because that's what just happened. I just got a solve that was rotationless, and was <RUL> except for one F move and one D move. This is a CFOP solve, not a ZZ solve. http://tinyurl.com/n5l3hy4
EDIT: actually, that's easy. The answer to getting a L8E orientation skip is just 1/2^7 or something.
EDIT2: ninja'd
What about the odds of having a scramble that's solvable with only two moves that aren't RUL? I guess another way to ask that would be what are the odds of having an EOline solvable in 2 non RUL moves?
EDIT: actually, that's easy. The answer to getting a L8E orientation skip is just 1/2^7 or something.
EDIT2: ninja'd
What about the odds of having a scramble that's solvable with only two moves that aren't RUL? I guess another way to ask that would be what are the odds of having an EOline solvable in 2 non RUL moves?
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uberCuber
Legend
I believe 1/2^7 = 1/128 = 0.78%. (Each edge has two possible orientations, so 1/2 for each of 7 edges, and the 8th edge's orientation is defined by the other 7).
Renslay
Legend
I was doing some skewb solves, when suddenly I had a last-5-centers skip. I solved 1 center + 4 corners first, then solved the last 4 corners with R' F R F', and then suddenly the skewb was solved. What is the probability for such a skip?
Cale S
Veteran
If you are left with 5 remaining centers, the U center can be in one of 5 places, so the chance of it being solved is 1/5. Given that the U center is solved, the F center can be in one of 4 places, so it has a 1/4 chance of being solved. Given the U and F centers are solved, the R center can be in one of 3 places, so it has a 1/3 chance of being solved. Once the U, F, and R centers are solved, L5C must be solved because there are two left and skewb can't have parity, so the chance of having a last five centers skip is 1/5*1/4*1/3 = 1/60. Another way to think of it is that there are 5 centers, so you could arrange them in 5! = 120 ways, but dividing by 2 (because skewb can't have parity) gives you 60 combinations, 1 of which is solved.
If you know Sarah's intermediate method, this is reduced to 1/12 because the U center is solved during the second step.
goodatthis
Veteran
So this is a tricky one.
What are the chances that 3/4 of the 4 f2l slots are simple 3 move inserts+AUF in between? Lets just assume we can optimally solve pair order to force 3 move inserts. But no multislotting algs, simply different pair order.
What are the chances that 3/4 of the 4 f2l slots are simple 3 move inserts+AUF in between? Lets just assume we can optimally solve pair order to force 3 move inserts. But no multislotting algs, simply different pair order.
Robert-Y
Legend
I'm guessing that the request is for at least 3 pairs, and that F2L pair skips are also taken into consideration?
Bindedsa
Veteran
Wouldn't any F2L pair skips after the first pair be considering accidental multislotting?
goodatthis
Veteran
Yep, they are.
guysensei1
Legend
If you had a set of 3x3 stickers (9 stickers of each sticker color, 6 colors), and you randomly stickered a cube, what are the odds that it will be solvable? ie, anytime there are 9 stickers of the same color on every face, it's 'solved', regardless of the color scheme.
What if we restricted the stickering such that centerpiece stickers must go on center pieces, corner stickers must go on corners, and edge stickers must go on corners (with a fitted sticker set or something in reality)?
What if we restricted 'solved' to only one specific color scheme?
What if we restricted the stickering such that centerpiece stickers must go on center pieces, corner stickers must go on corners, and edge stickers must go on corners (with a fitted sticker set or something in reality)?
What if we restricted 'solved' to only one specific color scheme?
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Stefan
Legend
guysensei1
Legend
Yikes. What about the other restrictions?
Stefan
Legend
Come on, my fish came with instructions.
goodatthis
Veteran
What is the probability that a 1x2x3 block will form in the LL during the solving of F2L? I realized that when I was at US Nationals, that happened 3 times.
Stefan
Legend
You mean just formed during LL and then perhaps broken again, or also still there at the end of F2L?
goodatthis
Veteran
Yeah, just formed at some point during F2L, even if it's in the middle of an F2L alg.
