Discussion in 'Puzzle Theory' started by CubesOfTheWorld, Apr 8, 2010.

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1. ### FastCubeMasterMember

972
299
Jun 30, 2015
Brisbane, Australia
WCA:
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FastCubeMaster
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

2. ### Daniel LinMember

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

3. ### Cale SMember

2,238
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Jan 18, 2014
Iowa, USA
WCA:
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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

Daniel Lin likes this.
4. ### Daniel LinMember

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?

5. ### xyzzyMember

526
216
Dec 24, 2015
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: $\frac18(11088+3\cdot240+3\cdot48+144)=1512$

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6. ### Herbert KociembaMember

168
40
Nov 29, 2008
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.

Last edited: May 30, 2017

30
0
Jun 12, 2017
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))

Last edited: Jun 25, 2017
8. ### WACWCAMember

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Jun 2, 2015
Maryland
WCA:
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Will Callan
Probability of solved 2x2 face, any kind

9. ### LiquoriceMember

18
2
Feb 14, 2016
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?

10. ### RonxuMember

440
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Apr 19, 2013
Kirkkonummi, Finland
WCA:
2010RONK01
Rubikscuber96
0.5

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11. ### LiquoriceMember

18
2
Feb 14, 2016
Why? How do you know?

12. ### RonxuMember

440
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Apr 19, 2013
Kirkkonummi, Finland
WCA:
2010RONK01
Rubikscuber96
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.

13. ### UnderwatercuberMember

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

14. ### xyzzyMember

526
216
Dec 24, 2015
For 3bld, it's around 3.97%:

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.

15. ### greentgoatgalMember

136
23
Feb 6, 2017
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.

16. ### obelisk477Member

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.

17. ### applezfallMember

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

18. ### KAINOSMember

18
5
Jun 27, 2016
Somewhere in Earth
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)

19. ### RubixKidMember

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!

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Jan 18, 2014
Iowa, USA
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