- 18 Apr '07 16:41
**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?** - 19 Apr '07 17:52 / 1 editAs 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. - 19 Apr '07 20:14Here'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..) - 20 Apr '07 09:12 / 1 edit

That's exactly how I did it. Once you've worked out the method, generalising it for an*Originally posted by geepamoogle***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.*n*x*n*x*n*cube isn't too hard. - 20 Apr '07 13:52 / 2 edits

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

No, I don't think there's an easy approach. Will post an explanation later, but it's not pretty - 20 Apr '07 17:51

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.*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