## ToMs Cube….like Rubik’s, but not…

Posted by Karl Sharman on May 27, 2014 – 1:41 am

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? The answer “Slim, very slim!” will not be accepted.

May 27th, 2014 at 6:47 am

Slender, very slender!

May 27th, 2014 at 8:45 am

If your next statement is “Anorexic, very….” there will be trouble….

May 27th, 2014 at 11:52 am

Not quite certain, but…….

((1! x 6! x 8! x 12!) / 27!) x (24/24) x (2/24)^12 x (3/24)^8 x (4/24)^6

maybe!

May 27th, 2014 at 2:25 pm

I agree with Curtis. Approx 1.83 * 10^-37.

Anorexia plus catabolic steroids.

May 27th, 2014 at 2:49 pm

Of the 27 cubes, 1 is the centre and is all white, 8 are corners and have 3 adjacent black faces, 6 are centre faces and only have 1 black face, 12 are middle edges and have 2 adjacent black faces.

There are 27!/(1! 6! 8! 12!) ways of placing the cubes into the big cube(that’s an extension of the n choose r function). However, that disregards the orientation of the cubes

For a corner cube there are 8 possible orientations (as only 1 of the 8 corners of the cube has 3 black faces), for a face centre cube there are 6 orientations (as the are 6 faces), for an edge cube there are 12 orientations (as there are 12 edges on a cube) and for the centre cube there is only 1 orientation.

So the probability is (1! 6! 8! 12! / 27!) /(1

^{1}6^{6}8^{8}12^{12})= 1/5465062811999459151238583897240371200

≈ 1.82981*10^-37

That is an incomprehensibly small probability. If the age of the universe was scaled to 1 second, then it would take 30 times the numerical age of the universe to try every unique combination assuming 1 combination per second.

May 27th, 2014 at 11:46 pm

Well done Curtis, and a good explanation from Chris.

I know I tagged this as a probabilities exercice, but I also see this problem more as a counting exercise. We will count the number of cube orientations and arrangements such that the outside of the larger cube is black, and then divide this by the total number of possible orientations and arrangements to obtain the required probability.

We will count without any consideration of symmetry. Other counting methods could be used, but as long as we are consistent we will obtain the correct probability.

Consider the four types of cubes upon disassembly:

a. 8 cubes with three faces painted black;

b. 12 cubes with two black faces;

c. 6 cubes with one black face;

d. 1 completely white cube.

Each cube of type (a) must be oriented in one of three ways, giving 38 possible orientations. Next, each corner cube must be placed in one of eight corners, giving 8! possible arrangements. Thus we have 38 x 8! possibilities in all.

Similarly, each cube of type (b) must be oriented in one of two ways, giving 212 possible orientations. Then, each edge cube must go to one of 12 edges, giving 12! possible arrangements. Thus we have 212 x 12! possibilities in all.

Also, each cube of type (c) has four possible orientations, and may be placed in one of six positions, yielding 46 x 6! possibilities.

Finally, the one white cube of type (d) may be oriented in 24 ways. (Four ways for each face.)

Thus the total number of correct re-assemblings is a = 38 x 8! x 212 x 12! x 46 x 6! x 24.

To find the total number of possible reassemblings, consider that each cube may be oriented in 24 ways, and there are 27! possible arrangements of the cubes, giving b = 2427 x 27! possibilities in all.

Therefore, the probability that the outside of the reassembled cube is completely black is

a/b = 1/(256 x 322 x 52 x 7 x 11 x 132 x 17 x 19 x 23) = 1/5465062811999459151238583897240371200 (ish) – not good in lay-man’s terms. Approx 1.83 × 10^−37

1/14,000,000 is the generally touted lottery winning probability.