[Combinatorics]
Problem: One face of a $2 \times 2 \times 2$ cube is painted (not the entire cube), and the cube is cut into eight $1 \times 1 \times 1$ cubes. The small cubes are reassembled randomly into a $2 \times 2 \times 2$ cube. Compute the probability that no paint is showing.

Solution: Call each $1 \times 1 \times 1$ cube a cubelet. Then four cubelets are each painted on one face, and the other four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted face can occur in six positions, of which three are hidden from the outside, so the probability that a particular painted cubelet has no paint showing is $3 / 6=1 / 2$. Thus the probability that all four painted cubelets have no paint showing is $(1 / 2)^{4}=\frac{1}{\mathbf{1 6}}$.

Final answer: $\frac{1}{16}$