# Possibility that the cube is solvable



## celli (Dec 22, 2010)

I had a question and maybe you could tell me the answer: If you dissamble your cube, and you put the pieces back in random, what is the possibility that the cube is solvable? I've thought a lot on how to do that, but I just don't know. 

Please only post if you have something valuable to add. And no noob or :fp posts if the answer is very simple!


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## Stefan (Dec 22, 2010)

1/12


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## EricReese (Dec 22, 2010)

1/9001


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## flan (Dec 22, 2010)

1/3 possible orientations of a corner are correct

1/2 possible orientations of an edge are correct

1/2 possible permutations of a corner/edge are correct (T J F Y R N perms etc can exchange two corner swaps for two edge swaps)

all are independent of each other so

1/3x2x2 = 1/12


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## StachuK1992 (Dec 22, 2010)

Stefan said:


> 1/12


 
I'll expand.
There is a 1/3 chance that the corners will be oriented (twisted) in a fashion that they can all be 'correct.'
There is a 1/2 chance that the edges will be oriented (flipped) in a fashion that they can all be 'correct.'

There is a 1/2 chance that the corners and edges can move about each other in such a way that allows them all to be in the same place.
1/(3*2*2) = 1/12.

Edit:
FLAN NINJAS!!!ONE!!111!ELEVEN!!!


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## Stefan (Dec 22, 2010)

StachuK1992 said:


> There is a 1/2 chance that the corners and edges can *move about each other in such a way that allows them all to be in the same place*.


 
Huh?


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## vcuber13 (Dec 22, 2010)

http://en.wikipedia.org/wiki/Rubiks_cube#Mathematics


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## celli (Dec 22, 2010)

flan said:


> 1/2 possible permutations of a corner/edge are correct (T J F Y R N perms etc can exchange two corner swaps for two edge swaps)


 
I understand the 1/2 and 1/3 possible orientations, but why should you count this 1/2 again? And if you count it, why shouldn't you count the corners and edges apart? So 1/3x2x2x2 = 1/24


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## Sa967St (Dec 22, 2010)

celli said:


> I understand the 1/2 and 1/3 possible orientations, but why should you count this 1/2 again? And if you count it, why shouldn't you count the corners and edges apart? So 1/3x2x2x2 = 1/24


 
You only have P parity half of the time(2/4).

odd number of edge swaps + even number of corner swaps -> not possible
even number of edge swaps + odd number of corner swaps -> not possible
odd number of edge swaps + odd number of corner swaps -> possible
even number of edge swaps + even number of corner swaps -> possible


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## flan (Dec 22, 2010)

You count this 1/2 again because it's independent of orientation. If you have a cube that is unsolvable because two corners are switched you can never make it solvable by twisting corners or flipping edges. The only way you can make it solvable is by switching two corners or if you want switching two edges because edge permutation and corner permutation are not independent of each other.
Ie if you have 2 switched adjacent corners you can make the cube solvable by switching two edges leaving you with an F perm or similar which of course can be fixed with an alg. I hope this makes sense!


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## celli (Dec 22, 2010)

Hey thanks, that does make sense. I get it now! So you don''t coutn them apart because they're not independent of each other. If they were, then we should count them apart, is that what you're saying?


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## flan (Dec 22, 2010)

Yeah thats basically it


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## celli (Dec 22, 2010)

OK, thanks. I totally get it now! I was thinking of this a lot because of my pops, and I didn't know if I put the pieces back in where they belong. And I always put them back in wrong. But this chance is much lower I expected actually.


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## Rinfiyks (Dec 23, 2010)

celli said:


> But this chance is much lower I expected actually.


 Because you don't completely disassemble your cube every time you get a pop (might have just popped 1 edge), so you're more likely to reassemble it correctly.


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## Ranzha (Dec 23, 2010)

Rinfiyks said:


> Because you don't completely disassemble your cube every time you get a pop (might have just popped 1 edge), so you're more likely to reassemble it correctly.


 
Then again, if it pops out a 1x3x3 row, there is a chance of you getting all three parities.


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## oll+phase+sync (Dec 23, 2010)

Ranzha V. Emodrach said:


> Then again, if it pops out a 1x3x3 row, there is a chance of you getting all three parities.


that is true for "smaller" rows, too


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## Ranzha (Dec 23, 2010)

oll+phase+sync said:


> that is true for "smaller" rows, too


 
I was talking about 3x3x3.

Even so, to achieve all three parities from a solved cube, one must take out at least 3 pieces.


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## CharlesOBlack (Dec 23, 2010)

Ranzha V. Emodrach said:


> Then again, if it pops out a *1x1x3* row, there is a chance of you getting all three parities.


 

this statement is also true.


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## amostay2004 (Dec 23, 2010)

So I assume the probability for this on a 4x4 is 1/3?


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## Sa967St (Dec 23, 2010)

amostay2004 said:


> So I assume the probability for this on a 4x4 is 1/3?


 
The only kind of unsolvable parity you can get on even cubes is CO parity, so yes.


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## cuBerBruce (Dec 23, 2010)

Sa967St said:


> The only kind of unsolvable parity you can get on even cubes is CO parity, so yes.


 
Unless you're solving as a supercube. For a 4x4x4 supercube (where each center must must be solved to an exact location rather than just a particular face), the probability would be 1/6 (corners and centers must have the same parity).


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