抽两张牌比大小的胜率

Card Game

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

用一副标准 52 张牌（点数 2..10,J,Q,K,A）。洗牌后你先抽 1 张，庄家从剩下 51 张中抽 1 张。

若你的点数更大则你赢；若相等或更小则庄家赢。

问：你获胜的概率是多少？

A casino offers a simple card game using a standard deck of 52 cards. Each card’s value is one of $\{2,3,4,5,6,7,8,9,10,J,Q,K,A\}$ , and each suit is one of $\{\spades,\clubs,\hearts,\diamonds\}$ . The deck is thoroughly shuffled each time. You pick one card, then the dealer picks one card from the remaining 51. If your card’s value is higher, you win; if they are equal or yours is smaller, the house wins. What is your probability of winning?

解析

设 $E_2$ 为两张牌点数相等。

你抽到某点数后，剩下 51 张里同点数的牌还有 3 张，因此

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

由对称性，在非平局时你赢与庄家赢概率相同，所以

P(\text{你赢})=\frac{1-P(E_2)}{2}=\frac{1-3/51}{2}=\frac{8}{17}.

Original Explanation

Consider three outcomes:

$E_1$ : Your card is higher.
$E_2$ : Your card is equal.
$E_3$ : Your card is lower.

By symmetry, $P(E_1) = P(E_3)$ . We only need to find $P(E_2)$ .

Once you pick a card, there are 51 cards left. Only 3 of those have the same value as yours. Thus,

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

Hence,

P(E_1) \;=\; \frac{1 - P(E_2)}{2} \;=\; \frac{1 - \tfrac{3}{51}}{2} \;=\; \frac{8}{17}.