I was given a really interesting problem at an interview. It seems like an elementary question but I just can’t think about it it properly. Apologies if the wording is a bit odd, I can’t think of better ways the phrase the question:
The six faces of a cube can be painted red, green or blue. What is the probability that each pair of opposite faces is coloured a distinct colour (i.e. one pair opposite faces is painted red, another pair painted green, the last pair painted blue) if:
a) the colour for each face is randomly chosen with probability $\frac13$?
b) the cube has exactly 2 red, 2 green, and 2 blue faces?
I am not too sure how to begin because of the geometry/symmetry involved. For example, for a), is the probability of a fully red cube $\frac1{3^6}$ or $\frac6{3^6}$? In fact, is the number of ways to have three pairs of identically coloured opposite faces of distinct colours $3!$, or is there more geometry involved? I feel like I’m overthinking somewhere.
Tldr; how do I account for overcounting due to the symmetry in the problem? Am I even overcounting in the first place?
Best Answer
The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $\frac{6}{729} = \frac{2}{243}$.
(b) Now there are only ${6 \choose 2} \cdot {4 \choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $\frac{6}{90} = \frac{1}{15}$.