2020 AMC 10A Problem 25

First evaluate each option's chance of winning.

• Reroll none: wins only if the sum is already $7$.
• Reroll one die, keeping two dice with sum $s$: the rerolled die must show $7 - s$, possible exactly when $s \le 6$, with probability $\frac16 = \frac{36}{216}$.
• Reroll two dice, keeping one die showing $a$: the two new dice must sum to $7 - a$, which has $6 - a$ outcomes, probability $\frac{6-a}{36}$; at best $\frac{5}{36} = \frac{30}{216}$ when $a = 1$.
• Reroll all three: the number of ways three dice sum to $7$ is $\binom{6}{2} = 15$, probability $\frac{15}{216}$.

So if some pair of dice sums to at most $6$ (which includes every roll with total $7$), Jason keeps that pair and rerolls one die or none; he never rerolls two. If every pair sums to at least $7$, he compares keeping his smallest die $a$, worth $\frac{6(6-a)}{216}$, against $\frac{15}{216}$: keeping wins when $6 - a \ge 3$, i.e. $a \le 3$, and rerolling all wins when $a \ge 4$.

Hence he rerolls exactly two dice when the smallest die $a \le 3$ and the two smaller dice sum to at least $7$. With dice sorted $a \le b \le c$:

• $a = 1$: need $b \ge 6$: $(1,6,6)$, $3$ orderings.
• $a = 2$: need $b \ge 5$: $(2,5,5)$, $(2,5,6)$, $(2,6,6)$: $3 + 6 + 3 = 12$.
• $a = 3$: need $b \ge 4$: $(3,4,4)$, $(3,4,5)$, $(3,4,6)$, $(3,5,5)$, $(3,5,6)$, $(3,6,6)$: $3 + 6 + 6 + 3 + 6 + 3 = 27$.

Total $3 + 12 + 27 = 42$ of $216$ equally likely rolls: $\frac{42}{216} = \frac{7}{36}$.

The answer is $\boxed{\textbf{(A)}\ \frac{7}{36}}$.

Optimal-play problems have two layers: first find the best decision as a function of the observed state, then count the states leading to a given decision. The decision rule here collapses to a comparison of small fixed probabilities, and only two thresholds matter (does some pair sum to $6$ or less, and is the minimum at most $3$). Counting sorted triples and multiplying by orderings keeps the enumeration honest.

• Forgetting that rerolling all three dice (15/216) beats keeping a 4 or 5, or failing to count ordered outcomes when listing the favorable rolls.
• Overlooking that a pair summing to at most $6$ makes rerolling a single die ($\frac16$) strictly better than any two-die reroll, so those rolls must be excluded.