# Solvable cube if stickers randomly applied? What's the chance?



## ruwix (Apr 1, 2016)

I know that I have 1/12 chance that my cube will be solvable if I take it apart and assemble it randomly (edge and corner parity combined).

The question is what is the probability of a solvable cube if I peel off the stickers and stick them back randomly. :confused:


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## 00 (Apr 1, 2016)

43252003274489856000*9!^6/54! = 278628139008/651266248090757461677878416625 or about 0.000000000000000043%


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## SFCuber (Apr 1, 2016)

Very low. (Depending on your color scheme). If you have an edge with 2 blue stickers then its already at zero. Same with corners.


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## Lucas Garron (Apr 2, 2016)

00 said:


> 43252003274489856000*9!^6/54!



Multiply by an extra factor of 6! if you don't care about the color scheme.


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## 00 (Apr 2, 2016)

Lucas Garron said:


> Multiply by an extra factor of 6! if you don't care about the color scheme.



Oh yes of course. The probablity is still pretty much zero, though.


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## Kit Clement (Apr 2, 2016)

Maybe a more difficult and related probability question: If an artist/advertiser/etc. puts an image of a cube in their work, only showing three faces, what is the probability that there exists a possible sticker arrangement for the other three faces where it is solvable? Would be interesting to see this for when forcing the standard color scheme and when not forcing it. 

The tricky part for me is that I'm not sure how to make a comprehensive list of the ways that stickering 3 sides can fail, gaining a list of restrictions on coloring the stickers, and then turning that into a counting problem.


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## JustinTimeCuber (Apr 2, 2016)

00 said:


> Oh yes of course. *The probablity is still pretty much zero*, though.



like your username...


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## Matt11111 (Apr 2, 2016)

Man, this looks like a good Mathcounts problem. I guess I should go solve it.


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## Dapianokid (Apr 4, 2016)

Kit Clement said:


> Maybe a more difficult and related probability question: If an artist/advertiser/etc. puts an image of a cube in their work, only showing three faces, what is the probability that there exists a possible sticker arrangement for the other three faces where it is solvable? Would be interesting to see this for when forcing the standard color scheme and when not forcing it.
> 
> The tricky part for me is that I'm not sure how to make a comprehensive list of the ways that stickering 3 sides can fail, gaining a list of restrictions on coloring the stickers, and then turning that into a counting problem.



I'm no expert and I haven't given this as much thought as you, but my main issue with solving this problem stems from figuring out which questions _not_ to answer. 
My take on the problem, correct me if I'm wrong, before asking what the probability is that there exists a possible sticker arrangement for the other three faces where it is solvable, ask: "How do we know that the visible 3 faces indicate an unsolvable position?"

Obviously, 27 stickers must be visible.
There may not be more than 9 of a color visible in total.
There may not be more than one of a color on any center.
There may not be more than 4 of the same color on any corner or edge.
On any one piece, there may not be more than one of a specific color.


That's all I've got so far. With three invisible pieces, it's impossible to tell if there is illegal orientation parity or permutation parity in the edges or corners.

Curious discussion!


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## 00 (Apr 17, 2016)

EminentCuber said:


> I absolutely love this part of the forum!
> 
> I've been hard at work making equations and quantifying all this data.
> 
> ...



This is not the correct answer... the 54 stickers aren't all unique, so there are fewer than 54! ways to arrange them.



> Once completely solved, this is your answer:
> 0.
> And I mean this legitimately.
> The computing power of the entire fleet of an average Airline would not be able to compute the answer to a quantifiable number other than 0.



No... the number is positive, which means it is not zero, and I can easily compute it to a lot of digits: 0.000000000000000000000000000000000000000000001422103989345743645216458545800499169872628252719370729936591517132547683030602326252059492655156448357673450533196070973736641912895463250288389528076468188229280281803031898896...


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## qqwref (Apr 17, 2016)

You don't take identical stickers into account, or different color schemes (the cube is still solvable with a weird color scheme - in the sense that you can turn the layers to get every face to have a solid color). Also, there's nothing "astonishing" about your result... and it is very easy to divide one integer by a larger one, you can do it by hand with paper and a pencil, or on any computer with the right software (even a programmable calculator or cheap smartphone).


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## EminentCuber (Apr 17, 2016)

I did make some mistakes in my mathematics, for which I apologize.


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## adimare (Apr 17, 2016)

EminentCuber said:


> We know the possibility of correct permutations is as follows:
> 43,252,003,274,489,856,000
> Because
> 12!/8! * 2^11 * 3^7


If you're not gonna explain an equation in any way, you shouldn't bother writing it down.
What if your teacher explained permutations to you by just saying:
"We know the that we can arrange 6 elements in 720 different ways because 6!"



EminentCuber said:


> If we take this number and divide it by the possible number of sticker arrangements: 54!. We can get the number the probability that if you randomly remove and put adhere stickers it will be a solvable permutation.


As others have pointed out: No.



EminentCuber said:


> Once completely solved, this is your answer:
> 0


It's not.



EminentCuber said:


> The computing power of the entire fleet of *an average Airline* would not be able to compute the answer to a quantifiable number other than 0.


What about an above average Airline?



EminentCuber said:


> ANALOGY
> [random facts about the weight of a Rubik's cube]


What?


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