[Combinatorics]
Problem: Each of the six faces of a cube is randomly colored red or blue with equal probability. Compute the probability that no three faces of the same color share a common vertex.

Solution: There are $2^{6}=64$ colorings of the cube. Let $r$ be the number of faces that are colored red. Define a monochromatic vertex to be a vertex of the cube for which the three faces meeting there have the same color. It is clear that a coloring without a monochromatic vertex is only possible in the cases $2 \leq r \leq 4$. If $r=2$ or $r=4$, the only colorings that do not have a monochromatic vertex occur when two opposing faces are colored with the minority color (red in the $r=2$ case, blue in the $r=4$ case). Because there are 3 pairs of opposite



faces of a cube, there are 3 colorings without a monochromatic vertex if $r=2$ and another 3 such colorings if $r=4$. For the $r=3$ colorings, of which there are 20, the only cases in which there are monochromatic vertices occur when opposing faces are monochromatic, but in different colors. There are $2^{3}=8$ such colorings, leaving $20-8=12$ colorings that do not have a monochromatic vertex. Therefore $3+3+12=18$ of the 64 colorings have no monochromatic vertex, and the answer is $\frac{\mathbf{9}}{\mathbf{3 2}}$.

Final answer: $\frac{9}{32}$