Twenty-seven identical white cubes are assembled into a single cube, the outside of which is painted black. The cube is then disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man (who cannot feel the paint) reassembles the pieces into a cube. What is the probability that the outside of this cube is completely black?

As I understand it, there end up 4 different kinds of cubes in the bag.

8 corners (3 black), 12 edges (2 black), 6 faces (1 black), and one center.

Each must go to its assigned spot type, odds are 1!6!8!12!/27!.

Now on to orientations of each cube. Each cube has 24 potential orientations (if we assign a specific direction for front, top, etc). (Pick a front, then pick one of the 4 adjacent to be top)

Corners will be correct in 3 orientations for a 1-in-8 chance.
Edges will be correct for 2 orientations (1-in-12).
Faces only have to have the proper face pointed outward (1-in-6).
The Center cube isn't visible, rendering its orientation meaningless.

Combining all this, the odds become slightly less than one in 5.465 trillion trillion trillion. (5 followed by 36 digits)

EDIT: Actually calculating the chance, and not just the denominator, I get 1.8298 x 10^-37. The answer above is just a different way of stating the same value.

Here's one that might be worth looking at. Suppose all 27 cubes were painted with 3 black sides and 3 white sides, similar to a corner cube in the above. What are the odds of a monochromatic big cube now when the blind man constructs the big cube? (Not hard at all.)

For an even harder one I don't have an answer for yet, what are the odds in the original problem of an all white cube? (I'm not sure of any easy approach to this one..)

Originally posted by geepamoogle For an even harder one I don't have an answer for yet, what are the odds in the original problem of an all white cube? (I'm not sure of any easy approach to this one..)

1.046 x 10^-24, I think. I could easily have missed something.

No, I don't think there's an easy approach. Will post an explanation later, but it's not pretty 🙂

Originally posted by mtthw 1.046 x 10^-24, I think. I could easily have missed something.

No, I don't think there's an easy approach. Will post an explanation later, but it's not pretty 🙂

I looked at this and was wondered if there was some clever group theory approach which would save going through all the possibilities. I can't help thinking of the Rubik's cube looking at the problem and that was designed to illustrate group theory.