返回题库

经销商的甲板

The Dealer's Deck

专题
Probability / 概率
难度
L3

题目详情

你走进赌场并与庄家玩游戏。你和庄家各发一张公平、洗过的牌，共有 52 张牌。如果你的数字严格大于庄家的数字，你将获胜，否则你将失败。你获胜的概率有多大？

You walk into a casino and play a game with the dealer. You and the dealer are each dealt a card from a fair, shuffled deck of 52 cards. If you have a strictly larger number than the dealer you will win, otherwise you will lose. What is the probability that you win?

解析

用对称性解决。定义：

  • E1E_1：你的点数大于庄家
  • E2E_2：你的点数等于庄家
  • E3E_3：你的点数小于庄家

由于牌堆对称，有 P(E1)=P(E3)P(E_1)=P(E_3)。先算 P(E2)P(E_2)：在你拿到一张牌后，庄家从剩下 51 张牌中抽到同点数的只有 3 张，所以

P(E2)=351.P(E_2)=\frac{3}{51}.

因此

P(E1)=1−P(E2)2=1−3512=817.P(E_1)=\frac{1-P(E_2)}{2}=\frac{1-\frac{3}{51}}{2}=\boxed{\frac{8}{17}}.

Original Explanation

We can solve this question using symmetry. There are three distinct outcomes that define the sample space. Let E1E_1 be the event that your number is larger than the dealer's; E2E_2 be the event that your number is equal to the dealer's; E3E_3 be the event that your number is smaller than the dealer's. By symmetry, P(E1)=P(E3)P(E_1) = P(E_3), so we only need to find P(E2)P(E_2). We know that when we are dealt a card, there are 51 cards that the dealer could have, of which only 3 could be of the same number, so P(E2)=351P(E_2) = \frac{3}{51}.

Since ∑ω∈ΩP(ω)=1\sum_{\omega \in \Omega} P(\omega) = 1, we can solve for P(E1)P(E_1)

∑ω∈ΩP(ω)=P(E1)+P(E2)+P(E3)=1\begin{equation} \sum_{\omega \in \Omega} P(\omega) = P(E_1) + P (E_2) + P(E_3) = 1 \end{equation} 2P(E1)+P(E2)=1\begin{equation} 2P(E_1) + P(E_2) = 1 \end{equation} 2P(E1)+351=1\begin{equation} 2P(E_1) + \frac{3}{51} = 1 \end{equation} P(E1)=817\begin{equation} P(E_1) = \frac{8}{17} \end{equation}