Question

在某个游戏中， $4$ 名玩家各掷一个标准 $6$ 面骰。掷出最大点数的玩家获胜。如果最大点数出现并列，则这些并列者再次掷骰，如此继续直到一人获胜。Hugo 是其中一名玩家。已知 Hugo 赢得游戏，求 Hugo 第一次掷出 $5$ 的条件概率。

In a particular game, each of $4$ players rolls a standard $6$ -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo’s first roll was a $5,$ given that he won the game?

$\dfrac{61}{216}$

$\dfrac{367}{1296}$

$\dfrac{41}{144}$

$\dfrac{185}{648}$

$\dfrac{11}{36}$

答案：C

知识点：条件概率骰子（概率）分类讨论对称性

难度评级：2150

小提示：

使用条件概率 $P(Hugo\text{ 掷出 }5\mid Hugo\text{ 获胜})$ 。

Use $P(Hugo\text{ rolled }5\mid Hugo\text{ wins})$

大提示：

按其他玩家中有多少人也掷出 $5$ 分类。

Case on how many other players also rolled $5$

解答：

由对称性， $P(\text{Hugo 获胜})=\frac14$ 。因此所求条件概率等于 Hugo 第一次掷出 $5$ 并最终获胜的概率的四倍。

若 Hugo 第一次掷出 $5$ ，其他人不能掷出 $6$ 。按其他三人中也掷出 $5$ 的人数分类；若有 $t$ 人与 Hugo 并列，则 Hugo 后续获胜概率为 $\frac1{t+1}$ 。

按并列人数分类，得到 $\begin{gathered} P(\text{Hugo 掷出 }5\text{ 且获胜})\\ {}=\frac{1}{6^4}\sum_{t=0}^3 \binom3t\frac{4^{3-t}}{t+1}\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296} \end{gathered}\text{。}$

乘以 $4$ ，得到 $\frac{369}{1296}=\frac{41}{144}$ 。

所以正确答案是 C。

By symmetry, $P(\text{Hugo wins})=\frac14.$ Therefore, the requested conditional probability is four times the probability that Hugo first rolls $5$ and eventually wins.

If Hugo rolls $5,$ then no other player can roll $6.$ Case on how many of the other three players also roll $5.$ If $t$ other players tie Hugo, then Hugo wins the eventual tiebreaker with probability $\frac1{t+1}.$

Thus $\begin{gathered} P(\text{Hugo rolls }5\text{ and wins})\\ {}=\frac{1}{6^4}\sum_{t=0}^3 \binom3t\frac{4^{3-t}}{t+1}\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296}. \end{gathered}$

Multiplying by $4$ gives $\frac{369}{1296}=\frac{41}{144}.$

Thus, the answer is C .

2021 AMC 10B Fall 第 20 题

答案：C

难度评级：2150

小提示：

大提示：

解答：

其他年份的第 20 题