1. Standard memberMathurine
    sorozatgyilkos
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    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?
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    18 Apr '07 17:001 edit
    My first attempt comes out at about 1.83x10^-37. Which seems pretty small! Anywhere close?

    I'll omit the working for now to see what others come up with.
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    18 Apr '07 19:39
    Originally posted by mtthw
    My first attempt comes out at about 1.83x10^-37. Which seems pretty small! Anywhere close?

    I'll omit the working for now to see what others come up with.
    I get the same answer!
    Nice question.
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    19 Apr '07 12:19
    How about this for the general case (n >= 2)

    8! x (1/8)^8 x [12(n-2)]! x (1/12)^[12(n-2)] x [6(n-2)^2]! x (1/6)^[6(n-2)^2] x [(n-2)^3]! / (n^3)!
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    19 Apr '07 17:521 edit
    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.
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    19 Apr '07 20:14
    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..)
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    20 Apr '07 09:121 edit
    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.
    That's exactly how I did it. Once you've worked out the method, generalising it for an n x n x n cube isn't too hard.
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    20 Apr '07 13:522 edits
    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 🙂
  9. Standard memberDeepThought
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    20 Apr '07 17:51
    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.
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