2020 AMC 10A Problem 25

For any initial roll, Jason compares the best probabilities from rerolling $0,1,2,$ or $3$ dice. Rerolling all three dice has probability $\frac{15}{216}=\frac{5}{72}$. Rerolling one die has probability $\frac{1}{6}$ whenever some pair of kept dice has sum at most $6$. If he rerolls exactly two dice, he keeps one die. Keeping a die showing $1,2,3,4,5,6$ gives probabilities $5,4,3,2,1,0$ out of $36$, respectively. This can be optimal only when the two smallest dice sum at least $7$ and the smallest die is $1,2,$ or $3$. The sorted rolls satisfying this are $(1,6,6)$, $(2,5,5),(2,5,6),(2,6,6)$, and $(3,4,4),$ $(3,4,5),$ $(3,4,6),$ $(3,5,5),$ $(3,5,6),$ $(3,6,6)$. Counting permutations gives $3+12+27=42$ rolls out of $216$, so the probability is $\frac{42}{216}=\frac{7}{36}$. Thus, A is the correct answer.