AMC10 2020 A

Answer (A): There are 15 ways to roll a 7 with three dice (5-1-1, 4-2-1, 3-3-1, and 3-2-2 and their permutations) out of the $6^3$ possible rolls, so the probability of rolling 7 with three dice is $p_3=\frac{5}{72}$. Also, for $1\le a\le 7$, the probability of rolling $a$ with two dice is $p_2(a)=\frac{a-1}{36}$, and the probability of rolling any number on a single die is $p_1=\frac{1}{6}$. Because $p_2(a)<p_1$ for all $a<7$, Jason will always choose to reroll one die instead of two if possible; that is, if any two of the initial rolls sum to at most 6, then Jason is better off keeping those two rolls and rerolling the third, rather than keeping just one roll. Therefore Jason will choose to reroll two of the dice only if every pair of the initial rolls sums to at least 7. Furthermore, because $p_2(2)<p_3$ and $p_2(3)<p_3$, if all the rolls are at least 4, then Jason will choose to reroll all three dice. Therefore Jason will reroll two dice if and only if each pair of the initial rolls sums to at least 7, but at least one of the rolls is at most 3. These possibilities are given by 1-6-6, 2-$a$-$b$ where $a,b\in\{5,6\}$, and 3-$c$-$d$ where $c,d\in\{4,5,6\}$. Including permutations, this gives $3+12+27=42$ possibilities, so the requested probability is $\frac{42}{216}=\frac{7}{36}$.