adimare
Veteran
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:
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:
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.
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.
