Suppose that $A, B, C$ are the results of 3 random dice rolls of a fair 6 sided dice. What is the probability that $4A+2B+C$ is a multiple of 8?
Let us work modulo 8. By using the fact that $4A$ is congruent to either $0$ or $4$ with equal probability and $2B$ is congruent to $2$ or $4$ with probability $1/3$ each and $6$ or $8$ with probability $1/6$ each, one can work out that $4A+2B$ is congruent to $0, 2, 4$, or $6$ each with probability $1/4$. Then, one can see the probability that $4A+2B+C$ is in any given congruence class will be $(3/4)(1/6)=1/8$.
Given that the distribution on congruence classes is uniform, there should be a better approach, either with a bijection or a symmetry. Can anybody suggest a solution that requires less brute force than the above calculation?
Work from right to left. $C$ has to be even or there's no chance. That happens with probability $\frac 12$. If $C$ is even, then $C \equiv 2B \pmod 4$ or there's no chance. That means for each even value of $C$, exactly half of the values of $B$ give you a chance.
Finally, if you have values of $B$ and $C$ that give you a chance (which happens with probability $\frac 14$), then you need $2B+C \equiv 4A \pmod 8$. Once again, for each acceptable combination of $B$ and $C$, exactly half of the possible choices of $A$ work. Thus, the overall probability of success is $\frac 18$.