# Peeling off the stickers to solve the cube



## adimare (Jan 1, 2010)

An old teacher from college visited me today to wish me a happy new year. When she saw my new un-stickered type C on the table she told me I should put the stickers on at random, and then try to solve it. I told her that would most likely result in an unsolvable cube, and proceeded to explain why, and then how to solve the cube, etc.

But that got me thinking in the following problem: if you put your 54 stickers completely at random on your cube, what's the probability that your cube will be solvable? I would consider a given state solvable if you can get a single color on each face by performing turns on the cube, the color scheme being irrelevant.

If you don't wanna try it out on your own this is what I came up with:


Spoiler



First, I calculated the different ways in which the cube could be stickered.
To figure this out I imagine placing a number in each cubie face that will be stickered, so I have 54 stickers and 54 positions where they can be applied. The number of ways this can be done is 54!. However, I must also consider the fact that I don't really have 54 different stickers, I have 6 sets of 9 stickers (stickers in each set are identical), which means that any random stickering can be made (9!)^6 different ways, so I only really have 54!/9!^6 ways to randomly sticker the cube.

What about cube rotations? Due to them, every single random stickering I produce will actually appear 24 times, so my final number of possible ways to sticker the cube is:

54! x 1/9!^6 x 1/24

Now I only need to figure out how many of those are solvable:
I consider a state solvable if you can get a single color in each face. The amount of different states a solvable cube can be in is:
8! x 3^7 x 12! x 2^10

But this is only for a particular color scheme, this number should be multiplied by the amount of color schemes that the cube can have with 6 sets of 9 stickers. If we were to choose a color for each face in the following order: U F D B L R. We'd have 6! ways to choose our scheme. However, many of those will be identical after a couple of cube rotations, so the actual number of color schemes is 6!/24, which is 30.

So my calculated probability for randomly stickering a cube into a solvable state would be:

[8! x 3^7 x 12! x 2^10 x 30] / [54! x 1/9!^6 x 1/24]

That's about 3.08x10^-16, or in percentage:

a 0.0000000000000308% probability

I'm fairly new to the cube, so any corrections would be appreciated.


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## xbrandationx (Jan 1, 2010)

my brain hurts


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## Muesli (Jan 1, 2010)

What is this I don't even...

I'm seriously impressed. I have no idea how to work out probabilities like this.


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## cincyaviation (Jan 1, 2010)

i usually leave this to the guys a wikipedia, ill probably have to learn how to do this in math class someday though


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## Lucas Garron (Jan 1, 2010)

I get exactly the same answer as you, although I just went for dividing by all permutations:



Spoiler



(6!*43252003274489856000*(9!)^6)/54!
= 40122452017152 / 130253249618151492335575683325
≈ 3.080341729 * 10^-16



Nice problem.


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## Hyprul 9-ty2 (Jan 1, 2010)

When I read the title at first, I was thinking "wrong forum?"

Very interesting


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## Swordsman Kirby (Jan 1, 2010)

Lucas Garron said:


> I get exactly the same answer as you, although I just went for dividing by all permutations:
> 
> 
> 
> ...



I did the method that Lucas did, though with a slightly different (and overall inconsequential) interpretation on the identical stickers.


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## nigtv (Jan 1, 2010)

That's an interesting thing to think about, I'll think through it some more before I spoiler it. Main thing I can think of that would make this a bad idea, unless you were paying attention, would be putting the same color on two centers, instantly ruining everything. I'll think this through though....

...it's 1 in 12 chance of being solvable if you sticker it properly then take apart and put back together again randomly, right?


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## bigbee99 (Jan 2, 2010)

nice post, I would have never been able to do this


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## adimare (Jan 2, 2010)

nigtv said:


> ...it's 1 in 12 chance of being solvable if you sticker it properly then take apart and put back together again randomly, right?



Right


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## miniGOINGS (Jan 6, 2010)

Found this while looking through the forums.


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## adimare (Jan 9, 2010)

miniGOINGS said:


> Found this while looking through the forums.



Goodbye false sense of originality


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## cmhardw (Jan 9, 2010)

Without looking at any other posts first I would say:



Spoiler



43252003274489856000 * 24 * (9!)^6 / (54!) ~ 1 * 10^-17

--edit--
Ooooh very neat, I forgot to include all possible color schemes. Yes I agree with the factor of 30 so readjusted:
43252003274489856000 * 24 * 30 * (9!)^6 / (54!) ~ 3 * 10^-16



Chris


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## Zane_C (Jan 9, 2010)

I'm very impressed with the some people and their mathematical knowledge on this forum. Well done to everyone who figured it out.


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## Tyrannous (Jan 9, 2010)

yeah i think your overall possible solvable states was very estimated, so a wide error was expected, too bad im only an applicable mathmatician and astrophysisist, not so good in statistical anylisis and solving lol


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## NanoCuber (Jan 9, 2010)

adimare said:


> An old teacher from college visited me today to wish me a happy new year. When she saw my new un-stickered type C on the table she told me I should put the stickers on at random, and then try to solve it. I told her that would most likely result in an unsolvable cube, and proceeded to explain why, and then how to solve the cube, etc.
> 
> But that got me thinking in the following problem: if you put your 54 stickers completely at random on your cube, what's the probability that your cube will be solvable? I would consider a given state solvable if you can get a single color on each face by performing turns on the cube, the color scheme being irrelevant.
> 
> ...



Wow, you are REALLY smart. I can't beleive you worked this out. It seems like a really interesting concept.


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## Tomk (Jan 9, 2010)

This was my method, I haven't read the other solutions yet so excuse me if someone else has used the same method



Spoiler



*Solvable Permutations*
We start with the normal 43 quintillion permutations but since it could have any colour scheme we need to find out how many different colours schemes there could be.

Let the six colours A,B,C,D,E and F. A will allways be on the front face on the solved cube. A and B could either be opposite (when we rotate the cube so C is on the up face *or* adjacent when we rotate the cube so that it is on the up face

For the first instance we have 3! ways of stickering the remaining faces and for the second 4! giving us 30 colour schemes

30 x 43 252 003 274 489 856 

*Number of Permutations*
I realised that there would be 54! ways of stickering the cube but not all would be different as there are only 6 different coulered stickers. However on a side the stickers can be arranged in 9! different ways yet still appear the same. Applying this to the whole cube we get 54!/9!^6 ways to sticker it.

Edit : however we can have 24 different roations of the cube so we divide by 24 giving 54!/9!^6/24

*The answer*

So we do *solvable permutations/permutations*giving us * 54!/9!^6/24/30/43 252 003 274 489 856 * or *1/3 246 393 055 999 402*



Edit: missed some cube rotations in total permutations


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## miniGOINGS (Jan 9, 2010)

adimare said:


> miniGOINGS said:
> 
> 
> > Found this while looking through the forums.
> ...



Haha, yea, sorry about that.


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## CubesOfTheWorld (Mar 10, 2010)

what do numbers with exclamation marks next to them mean? does it mean, if you have 9!, that that means 9x8x7x6x5x4x3x2x1?


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## PatrickJameson (Mar 10, 2010)

CubesOfTheWorld said:


> what do numbers with exclamation marks next to them mean? does it mean, if you have 9!, that that means 9x8x7x6x5x4x3x2x1?



Yes, they are factorials.


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## Olivér Perge (Mar 10, 2010)

And what is the number if you sticker the cube without using the same colour on one piece?


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