FastCubeMaster

Member
Since you can't reduce any cases to another cases by AUF, they are all equally likely.

So 1/216 given the pair is solved, or 1/27 given you have a UF case.
O, definitely makes sense but seems a bit weird for me cos a few cases keep occuring, more than others. I guess it's just chance

D

Daniel Lin

Guest
given a 2 random edges and 2 random corners being swapped on the cube

what's the probability of a least one corner and edge being adjacent?

R' F R U2 r2 F r U' r U2 (UFR and UR are next to each other)

u R' U' R' F R2 U' R' U' R U R' F' R E

Cale S

Member
given a 2 random edges and 2 random corners being swapped on the cube

what's the probability of a least one corner and edge being adjacent?

R' F R U2 r2 F r U' r U2 (UFR and UR are next to each other)

u R' U' R' F R2 U' R' U' R U R' F' R E
The corner pieces can be adjacent (3/7 chance), diagonal on a layer (3/7 chance), or on opposite corners of the cube (1/7 chance)

If the corners are adjacent, first edge has a 7/12 chance, and second has 6/11 chance

If the corners are diagonal, first edge has 6/12 = 1/2 chance, second has 5/11

If the corners are completely opposite, the numbers are the same as diagonal

1 - ((3/7)(7/12)(6/11) + (3/7)(1/2)(5/11) + (1/7)(1/2)(5/11))

73.38% chance of having adjacent corner and edge

D

Daniel Lin

Guest
ok 1 more question

how many total 2e2c cases are there if you count U/D, F/B, and L/R mirrors as the same?

so that
R U R' F' R U R' U' R' F R2 U' R' U'=
R' D' R F R' D' R D R F' R2 D R D=
L D L' F' L D L' D' L' F L2 D' L' D'

can someone do the burnsides lemma thingy?

xyzzy

Member
how many total 2e2c cases are there if you count U/D, F/B, and L/R mirrors as the same?

can someone do the burnsides lemma thingy?
Identity symmetry: $$(\binom82\cdot3)\cdot(\binom{12}2\cdot2)=11088$$
Mirror on one axis: $$(4\cdot3)\cdot(12+8)=240$$ (12 for edges in the slice preserved under mirroring, 8 for edges outside of that slice)
Mirror on two axes: $$(4)\cdot(12)=48$$
Mirror on all axes: $$(4\cdot3)\cdot(12/2\cdot2)=144$$
Burnside's lemma magic: $$\frac18(11088+3\cdot240+3\cdot48+144)=1512$$

Herbert Kociemba

Member
R U R' F' R U R' U' R' F R2 U' R' U'=
R' D' R F R' D' R D R F' R2 D R D=
L D L' F' L D L' D' L' F L2 D' L' D'
Is there a particular reason that e.g. F U F' L' F U F' U' F' L F2 U' F' U' should be counted as different? This belongs to a 90° rotation which cannot be generated by the three mirrors.

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Member
can someone find the probability of...

any 1x1x3 block formed after scramble
being 1 move away from a 1x1x3 block after scramble (in axis turn metric)
having at least one F2L pair anywhere on the cube after making a 2x2x3 block
having both F2L pairs anywhere on the cube after making a 2x2x3 block
2gr skip with phasing (and average movecount if able(htm))
2gr skip with 2 corners oriented (and average movecount if able(htm))

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Liquorice

Member
What is the probability of parity in 3x3 blindfolded?

What is the exact probability of getting an edges memo of length 11 (11 edges need to be solved, buffer does not count)? If buffer piece is solved, it adds 1 to the length. There may be a permuted flipped edge and a solved edge (11 +1 -1 = 11).
Length 12 (one edge is flipped and permuted or two edges are flipped/permuted and one is solved etc.)? Length 1, 2, 3 ... 20, 21, 22?
Corners memo length?

Ronxu

Member
Why? How do you know?
If the scramble has an odd number of quarter turns you have parity. This happens because a quarter turn does an odd number of both edge and corner 2-cycles. if you have an even number of quarter turns, then the total number of 2-cycles is even on both edges and corners and there's no parity. Both cases are equally likely.

Underwatercuber

Member
Probabilities of having a 3/4/5bld solve without parity or any cycle breaks.

xyzzy

Member
Probabilities of having a 3/4/5bld solve without parity or any cycle breaks.
For 3bld, it's around 3.97%:

One could "break cycles" in order to flip edges or twist corners, but (I think?) most people just use algs for that, so assuming that, we can ignore the orientation of the pieces entirely. In other words, we're asking for the probability that we have only one corner cycle and only one edge cycle. (Please correct me if this is wrong and not what you're asking for, because I have no idea how people do BLD.)

Consider the two cases based on permutation parity. Suppose the corner/edge permutation parities are even. (They must be either both even or both odd.)

P(one corner cycle | even parity) = (P(corner 7-cycle) + P(corner 5-cycle) + P(corner 3-cycle) + P(corners solved)) / P(even parity)
= [(C(8,7)6! + C(8,5)4! + C(8,3)2! + C(8,1)0!)/8!] ⋅ 2
= [1/(7⋅1!) + 1/(5⋅3!) + 1/(3⋅5!) + 1/(1⋅7!)] ⋅ 2
= 43/120

By the same reasoning:
P(one edge cycle | even parity) = [1/(11⋅1!) + 1/(9⋅3!) + 1/(7⋅5!) + 1/(5⋅7!) + 1/(3⋅9!) + 1/(1⋅11!)] ⋅ 2
= 48187/217728

As the corners and edges are conditionally independent once we know the parity, we can conclude that the probability of not having to break cycles when the parities are even is:
P(one corner cycle, one edge cycle | even parity) = P(one corner cycle | even parity) P(one edge cycle | even parity)
= 2072041/26127360
≈ 7.93%

Then we can do the same computation for the odd parity case:
P(one corner cycle, one edge cycle | odd parity) = 39784637/326592000
≈ 12.18%

And conclude:
P(one corner cycle, one edge cycle) = (1/2)(2072041/26127360) + (1/2)(39784637/326592000)
= 131370299/1306368000
≈ 10.06%
4bld is too hard to exactly calculate because it depends on how you orient the cube. Too lazy to calculate for 5bld now, but I think you can reuse the above calculation and replace 8 or 12 with 24 as appropriate for the wings, and for the centres you don't really need cycle breaks at all, I think.

greentgoatgal

Member
not cubing related, but interesting

if you buy a lottery ticket on a tuesday, you are more likely to die before the results are announced than you are to win on the wednesday
You are also more likely to get struck by lightning on your way to get the lottery ticket than you are to win the lottery.

obelisk477

Member
You are also more likely to get struck by lightning on your way to get the lottery ticket than you are to win the lottery.
If you drove to the gas station each time to buy a lottery ticket for the expected number of tickets you would need to win the lottery, you would be killed in a car wreck 6 or 7 times before actually getting a winning ticket.

applezfall

Member
What is the probability of an 4 move 2x2 solve?
And what's the rarest and least rare pll?

KAINOS

Member
What is the probability of an 4 move 2x2 solve?
And what's the rarest and least rare pll?
For the first question check out this website: https://www.jaapsch.net/puzzles/cube2.htm
The rarest PLLs are H and N perms, (probability=1/72) and other than E and Z (1/36) the rest of them have the same chance of happening(1/18). If you group variants as a single case the most common one would be G perm, though. (1/18*4=2/9)

RubixKid

Member
The chance of a last 4 centers skip on Skewb is 1/8

I got 4 in a row once which is a 1/4096 probability!