Question

Jason 掷三枚公平的标准六面骰。然后他查看结果，并选择其中一个骰子子集重新掷，子集可以为空，也可以是全部三枚。重新掷后，他获胜的条件是三枚骰子朝上的点数之和恰好为 $7$ 。Jason 总是采用使获胜概率最大的策略。他选择恰好重新掷两枚骰子的概率是多少？

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?

$\displaystyle \frac{7}{36}$

$\displaystyle \frac{5}{24}$

$\displaystyle \frac{2}{9}$

$\displaystyle \frac{17}{72}$

$\displaystyle \frac{1}{4}$

答案：A

知识点：骰子（概率）最优化分类讨论

难度评级：2380

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文字解答：

对任意初始结果，Jason 比较重新掷 $0,1,2$ 或 $3$ 枚骰子的最佳概率。重新掷三枚的概率为 $15/216=5/72$ 。若保留两枚、重新掷一枚，获胜概率可达 $1/6$ ，条件是某对保留骰子点数之和不超过 $6$ 。

若恰好重新掷两枚，则只保留一枚。保留点数 $1,2,3,4,5,6$ 时，获胜结果数分别为 $5,4,3,2,1,0$ ，总结果数为 $36$ 。这种选择只有在两个最小点数之和至少为 $7$ ，且最小点数为 $1,2$ 或 $3$ 时才可能最优。

满足条件的有序结果排序后为 $(1,6,6)$ 、 $(2,5,5),(2,5,6),(2,6,6)$ ，以及 $(3,4,4)$ $(3,4,5)$ $(3,4,6)$ $(3,5,5)$ $(3,5,6)$ $(3,6,6)$ 。按排列数计数，共有 $3+12+27=42$ 个结果，总结果数为 $216$ ，所以概率为 $42/216=7/36$ 。正确答案是 A。

For any initial roll, Jason compares the best probabilities from rerolling $0,1,2,$ or $3$ dice. Rerolling all three dice has probability $15/216=5/72$ . Rerolling one die has probability $1/6$ 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 $42/216=7/36$ . Thus, A is the correct answer.

2020 AMC 10A 第 25 题

答案：A

难度评级：2380

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