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Paint it black II (Posted on 2007-04-17) Difficulty: 3 of 5
The eight corner cubes of the black cube formed in Paint it black (the cubes with half their faces already painted black) are assembled randomly into a smaller cube, with the outside of the new cube once again being painted black.

The eight cubes are then, again, disassembled and rebuilt randomly.

What is the probability that the outside of this cube is again completely black?

Bonus: What would the probability be if the smaller cube of eight cubes were a random selection of any eight cubes from the original black cube instead of only the eight corner cubes?

See The Solution Submitted by Dej Mar    
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Solution solution to main puzzle | Comment 1 of 2

Each small cube's history is independent of the other cubes' histories, as position doesn't matter in the 2x2x2 case, only the orientation of each cube. So we just need to find the probability that one cube will be showing all black faces at the end and raise it to the 8th power.

Initially, each of these 1x1x1 cubes will have three faces, adjacent to a common vertex, painted black.  At the middle stage, each of these little cubes can have 0, 1, 2 or 3 additional faces painted black. All the resulting 3-black-face cubes will be identical, as will all the 4-black-face cubes (they will have two adjacent white faces), and all the 5-black-face cubes (having one white face), and all the 6-black-face cubes (being all black).

The probability that a given cube will have only three black faces is 1/8. That's also the probability that it will become all black. The probability that it will have two white faces will be 3/8, the same as the probability that it'll have one white face.

If it has three black faces, the probability is 1/8 that it will show them all in the final step. If it has four black faces, then it will show all black faces outward if a certain two of its eight vertices is matched to the outside corner, for a probability of 1/4. If it has five black faces, the probability is 1/2 that its lone white face will not be one of the three outer faces. If it's all black then of course it will present all black faces to the outside with probability 1.

The probabilities of going into any given number of faces painted is then multiplied by the probability that it will show all black given that number, and the results added:

faces prob.  prob of showing all black   product
  3    1/8          1/8                    1/64
  4    3/8          1/4                    3/32
  5    3/8          1/2                    3/16
  6    1/8           1                     1/8

The total of the products is 27/64, and is the probability that any individual little cube will eventually be showing only black faces.

As these are independent, the final probability that all of them will be doing so is (27/64)^8 = 282429536481/281474976710656 or about 0.001003391277553333793548517860472 or 1 in 996.620184339649558935041985666275374.

The number is verified by a simulation program:

  REDIM cube(8, 6)
  FOR c = 1 TO 8
    FOR f = 1 TO 3
      cube(c, f) = 1
    GOSUB chooseFace
    cube(c, f1) = 1
    cube(c, f2) = 1
    cube(c, f3) = 1
  good = 1
  FOR c = 1 TO 8
    GOSUB chooseFace
    IF cube(c, f1) = 0 THEN good = 0: EXIT FOR
    IF cube(c, f2) = 0 THEN good = 0: EXIT FOR
    IF cube(c, f3) = 0 THEN good = 0: EXIT FOR
  totGood = totGood + good: ct = ct + 1
  PRINT totGood; ct; totGood / ct;
  IF totGood > 0 THEN PRINT ct / totGood:  ELSE PRINT


  f1 = INT(RND(1) * 6 + 1)
    f2 = INT(RND(1) * 6 + 1)
  LOOP WHILE f2 = f1 OR f2 + f1 = 7
    f3 = INT(RND(1) * 6 + 1)
  LOOP WHILE f3 = f1 OR f3 + f1 = 7 OR f3 = f2 OR f3 + f2 = 7

which when stopped after 197512 iterations, had 204 occasions of all-black resulting cubes, the fraction being  .001.032849 or 1 in 968.1961, statistically close enough to the calculated 1 in 996.62.

The bonus is harder, as the selection of small cubes takes place, rather than merely rotations of a given set of pieces, resulting in the selections' not being independent of each other. For example if a chosen piece was originally an edge piece on the 3x3x3 cube, then the other chosen pieces are less likely to be edge pieces than if the first had not been.

  Posted by Charlie on 2007-04-17 15:58:29
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