经销商的甲板

用对称性解决。定义：

• $E_1$：你的点数大于庄家
• $E_2$：你的点数等于庄家
• $E_3$：你的点数小于庄家

由于牌堆对称，有 $P(E_1)=P(E_3)$。先算 $P(E_2)$：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

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

因此

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

---

英文解析

We can solve this question using symmetry. There are three distinct outcomes that define the sample space. Let $E_1$ be the event that your number is larger than the dealer's; $E_2$ be the event that your number is equal to the dealer's; $E_3$ be the event that your number is smaller than the dealer's. By symmetry, $P(E_1) = P(E_3)$, so we only need to find $P(E_2)$. 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 $P(E_2) = \frac{3}{51}$.

Since $\sum_{\omega \in \Omega} P(\omega) = 1$, we can solve for $P(E_1)$

$$\begin{equation} \sum_{\omega \in \Omega} P(\omega) = P(E_1) + P (E_2) + P(E_3) = 1 \end{equation}$$ $$\begin{equation} 2P(E_1) + P(E_2) = 1 \end{equation}$$ $$\begin{equation} 2P(E_1) + \frac{3}{51} = 1 \end{equation}$$ $$\begin{equation} P(E_1) = \frac{8}{17} \end{equation}$$