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Probability Thread

CuberFles

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What are the odds for hitting edge parity on a 4 x 4? I'd say 40 %, since there are 5 possible situations after F3L (0, 1, 2, 3 or 4 dedges flipped) and only 2 of these are "wrong" (1 and 3). However, after several dozens of solves I can't seem to get any less than a 60 % chance to hit edge parity. I'm using the 3 x 3 reduction method (centers first, then pair the edges, then solve like a 3 x 3). Is my method invoking unfair amounts of parity cases? Is there anything I can do to hit parity less frequently?
 

obelisk477

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What are the odds for hitting edge parity on a 4 x 4? I'd say 40 %, since there are 5 possible situations after F3L (0, 1, 2, 3 or 4 dedges flipped) and only 2 of these are "wrong" (1 and 3). However, after several dozens of solves I can't seem to get any less than a 60 % chance to hit edge parity. I'm using the 3 x 3 reduction method (centers first, then pair the edges, then solve like a 3 x 3). Is my method invoking unfair amounts of parity cases? Is there anything I can do to hit parity less frequently?

Call the UB edge 1, the UR edge 2, the UF edge 3, and the UL edge 4, for the sake of this discussion. Now let us consider the ways that you can have parity.

0 Edges
1,2,3,4 are flipped - NO PARITY

1 Edge
1 is flipped - PARITY
2 is flipped - PARITY
3 is flipped - PARITY
4 is flipped - PARITY

2 Edges

1 and 2 are flipped - NO PARITY
2 and 3 are flipped - NO PARITY
3 and 4 are flipped - NO PARITY
4 and 1 are flipped - NO PARITY
1 and 3 are flipped - NO PARITY
2 and 4 are flipped - NO PARITY

3 Edges

1 is not flipped - PARITY
2 is not flipped - PARITY
3 is not flipped - PARITY
4 is not flipped - PARITY

4 ֵEdges

All four are not flipped - NO PARITY

If you count up each of the 'Parity' and 'No parity', you'll see that the split is 50%. The fact that you're seeing 60% is just unfortunate, and over a long number of solves, you'd see it settle to 50%.
 

xyzzy

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What are the odds of a last layer skip, oll skip, or pll skip on a 4x4 or even layered cube

OLL skip: 1 / (2^4 3^3) = 1/432
Full PLL skip: 1 / (4! 4!) = 1/576
PLL skip (allowing for AUF): 4 / (4! 4!) = 1/144
Full LL skip: (1/432) (1/576) = 1/248832
LL skip (allowing for AUF): (1/432) (1/144) = 1/62208
 

94matt

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HI guys just want to know that if you have all the corners and one side completed, how many possibilities do you have for the last 8 egdes?
 

guysensei1

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There are 8!/2 possible permutations and 2^7 possible orientations of the last eight edges. Multiplying those gives a total of 2,580,480 cases.
This only works if you are considering rotated versions of the same case as distinct. For example there are 8 'ways' for a clockwise U perm to appear that fits OP's description.
 
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Duncan Bannon

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Most of the people here probably know this, but the chance of a PLL skip is the same as getting a H perm. 1/72


Also, how many algs would there be for FULL 1LLL
 

Duncan Bannon

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Probability of CP skip: 1/6
Probability of AUF skip: 1/4
Total probability of AUFless PLL skip: 1/24.

Sweet, You the man, thanks.

Edit- that was really easy math Duncan, LOL
Edit2- Is the chance of PLL skip really 1/6? Because of the H cases wouldn’t it be 6/30?
Edit3- What are chances of full No AUF LL skip on 2x2? Thanks
 

Username

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Edit2- Is the chance of PLL skip really 1/6? Because of the H cases wouldn’t it be 6/30?

Yes, there are 4! ways to permute the four corners. Four of them are "solved" (disregarding AUF). 4!/4 = 3! = 6, hence 1/6

Similarly, there are 4! different ways to arrange four corners when they are oriented. Only one of them is completely solved, thus PLL skip with no AUF is 1/24
 
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