Question

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7.$ Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?

$\dfrac{7}{36}$

$\dfrac{5}{24}$

$\dfrac{2}{9}$

$\dfrac{17}{72}$

$\dfrac{1}{4}$

Solution:

Rerolling one die, keeping two dice that sum to $s,$ wins with probability $\tfrac16$ when $s \le 6$ and $0$ otherwise. Rerolling two dice, keeping a die of value $v,$ wins with probability equal to the number of ways two dice sum to $7 - v,$ over $36;$ this is largest when $v$ is smallest. Rerolling all three wins with probability $\tfrac{15}{216}.$

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least $7$ (so rerolling one cannot reach $7$ ) while the smallest die is $1, 2,$ or $3$ (so keeping it beats rerolling all three).

Counting the ordered rolls with this property gives $42$ out of $216,$ a probability of $\dfrac{42}{216} = \dfrac{7}{36}.$

Thus, A is the correct answer.

2020 AMC 12A Problem 23

Solution:

Problem 23 in Other Years