# Hard combinatorics and probability question.

A large white cube is painted red, and then cut into $27$ identical smaller cubes. These smaller cubes are shuffled randomly.

A blind man (who also cannot feel the paint) reassembles the small cubes into a large one. What is the probability that the outside of this large cube is completely red?

Hint: Try to find first the probability that the corner cubes are put into the corners, the face cubes into the faces and the edge cubes into the edges. Then the probability that they have the correct orientation.

First, number the small cubes to make them distinguishable.

Without taking into account the orientation of the smaller cubes, there are 27! possible ways to reassemble them into a big cube.

• There is 1 way to place the center cube correctly
• There are 6! ways to place the face cubes correctly
• There are 8! ways to place the corner cubes correctly
• There are 12! ways to place the edge cubes correctly

So there are 6!8!12! correct cubes (without taking into account the orientation).

Now take such a cube.

• The center cube has probability 1 of being in the correct orientation.
• Each face cube has probability 1/6 of being in the correct orientation.
• Each corner cube has probability 1/8 of being in the correct orientation.
• Each edge cube has probability 1/12 of being in the correct orientation.

This means that the probability of getting a red cube must be

$$\frac{6!8!12!}{27!}\left(\frac{1}{6}\right)^6\left(\frac{1}{8}\right)^8\left(\frac{1}{12}\right)^{12}$$

• That's a little less than one chance in $2\times 10^{35}$, if my calculator was behaving. – Brian M. Scott Feb 22 '15 at 10:21
• @barakmanos: No, you are wrong and this answer is right. To begin with, there are only $24$ orientation-preserving symmetries of a cube, not $144$. And for a corner cube there are $3$ of them that will produce the correct colouring (rotation about the axis passing through the corner is OK). And so forth. – Marc van Leeuwen Feb 22 '15 at 11:19
• @MarcvanLeeuwen: I used a simpler way to compute the probability to get the small cube in the correct orientation : for the face cube, you can only interest you in face, hence 1/6. The rotations doesn't change the facts that only one face will be outside, and that each face is equiprobable. Same with the edge : there is 12 edges, and they are all equiprobable to be on the edge of the cube. And for the corner, there is 8 corners and all are equiprobable to be on the corner off the cube. Hence the result – Tryss Feb 22 '15 at 11:22
• @barakmanos: I don't follow. I don't see what it means to have more than one identical (not just indistinguishable) configuration, nor what reason there could be to count the same configuration multiple times (and if so, how many times?), nor what kind of counting of symmetries of a cube could have $144$ as answer. Maybe you could start explaining that last point. – Marc van Leeuwen Feb 22 '15 at 11:29
• @Tryss: Yes, I know, that's what I would do too. But I was trying to explain to barakmanos that even using the full set of symmetries leads to the same answer. – Marc van Leeuwen Feb 22 '15 at 11:31