Given an all black 3x3 Rubik's cube. You are provided with 54 stickers (9 pieces per color) and then blindfolded. If you put the stickers on the Rubik's while being blindfolded. What is the probability that the cube that you have built can be efficiently solved as a standard 3x3 Rubik's cube.

  • $\begingroup$ One reason the probability for this is so low is that you could easily e.g. not use one of each color for the centers (and many similar effects for sides, corners, and interactions between all these). For comparison, if you were blindfolded with a Rubik's cube with all the sides & corners broken off and your task was to reassemble it, a reassembly would be 1 in 12 to be solvable. (See Remark 11.15 in these notes.) $\endgroup$ – aes May 29 '17 at 7:21

It is well known that the Rubik's Cube has $\frac{12!*8!*3^8*2^{12}}{12} = 43,252,003,274,489,856,000$ reachable states. For 6 colours, there are $6!$ possible colour schemes for the cube, i.e. $6!$ ways to apply one colour to each face. This means that there are $\frac{6!*12!*8!*3^8*2^{12}}{12}$ arrangements of the stickers that are valid. Note that some of these colour schemes differ only by a rotation of the cube as a whole, but that is exactly what we want because it does not matter which way up you create your cube.

There are $\frac{54!}{9!^6}$ ways in which you can apply 54 stickers, 9 of each colour, to the cube.

Therefore, the probability of making a valid solvable cube is $\frac{6!*12!*8!*3^8*2^{12}/12}{54!/9!^6}$, which is about 4.278e-19.

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    $\begingroup$ That probability is still probably higher than the probability of me solving a rubiks cube in my lifetime though $\endgroup$ – Beastly Gerbil May 28 '17 at 17:21
  • $\begingroup$ @BeastlyGerbil I provided this particular 5-step algorithm/guide to the guys at the local liquor store, and now all 6 of them can solve using the beginners method. Give it a shot, and best of luck! :) $\endgroup$ – DevNull May 28 '17 at 23:59
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    $\begingroup$ @BeastlyGerbil Just use the Devil's algorithm. It's ye good olde brute force method for dummies (read the description): youtube.com/watch?v=BWpdq-GXa0A $\endgroup$ – Simply Beautiful Art May 29 '17 at 1:21
  • $\begingroup$ If I'm not mistaken, both the 43... number and the 6! number account for whole-cube rotations, so you're double-counting the factor of 24 possible whole-cube rotations and you need to divide out a factor of 24. $\endgroup$ – user2357112 May 29 '17 at 3:18
  • $\begingroup$ @JaapScherphuis: Only side rotations, no center-slice rotations in that number, then? $\endgroup$ – user2357112 May 29 '17 at 4:57

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