抽牌选 Ace

Denote by $W_{k}$ the event that the $k$ - th card is an Ace and that one more card is an Ace among the first $k - 1$ cards. Note that $k \leq 50$ since there must be at least two cards (Aces) left after the top 50 cards are drawn. We need to find $1 \leq k \leq 50$ such that $\mathbb{P}\left(W_{k}\right)$ is maximized. Let $N_{k}$ be the number of ways in which the deck can be shuffled to have one Ace in the $k$ - th position, one Ace in a position from the set $\{1, 2, \ldots , k - 1\}$ , and two Aces in positions from the set $\{k + 1, \ldots , 52\}$ . Then,

$$\mathbb{P}\left(W_{k}\right) = \frac{N_{k}}{52!}$$

Denote by $M_{k}$ the number of ways in which we can choose the positions for four Aces. For each choice of positions for the four Aces, the Aces can be placed in those positions in 4! ways, while the other cards can be placed in the remaining positions in 48! ways. Therefore,

$$N_{k} = 4! \cdot 48! \cdot M_{k}$$

To calculate $M_{k}$ , note that: the $k$ - th position must be chosen; one position is in $\{1,2,\ldots ,k - 1\}$ and can be chosen in $k - 1$ ways; the other two positions are in $\{k + 1,k + 2,\ldots ,52\}$ and can be chosen in $\left( \begin{array}{c}52 - k \\ 2 \end{array} \right)$ ways. Thus,

$$\begin{array}{c}M_{k} = (k - 1)\cdot \left( \begin{array}{c}52 - k \\ 2 \end{array} \right) \\ = \frac{(k - 1)(52 - k)(51 - k)}{2} \end{array}$$

From (2.5- 2.7), it follows that the probability of winning, when choosing $k$ , is

$$\begin{array}{c}{{\mathbb{P}\left(W_{k}\right)=\frac{4!\cdot48!}{52!}\cdot\frac{\left(k-1\right)\left(52-k\right)\left(51-k\right)}{2}}}\\ {{=\frac{12(k-1)\left(52-k\right)\left(51-k\right)}{52\cdot51\cdot50\cdot49}}}\end{array}$$

The maximum value of $\mathbb{P}(W_{k})$ is obtained for the value of $k$ that maximizes the function

$$\begin{array}{c}f(k) = (k - 1)(52 - k)(51 - k) \\ = (k - 1)\cdot [51 - (k - 1)]\cdot [50 - (k - 1)]. \end{array}$$

Let $z = k - 1, M = 51$ , and

$$g(z) = z(51 - z)(50 - z).$$

Since $f(k) = g(k - 1)$ , the maximum of $f$ is attained at $k_{\star} = z_{\star} + 1$ , where $0\leq z_{\star}\leq 49$ is the integer for which the function $g$ is the largest; recall that $1\leq k\leq 50$ . The derivative of $g(z)$ is

$$g^{\prime}(z) = 3z^{2} - 2(2M - 1)z + M(M - 1)$$

and the solutions of the equation $g^{\prime}(z) = 0$ are

$$z_{1,2} = \frac{2M - 1\pm\sqrt{M^{2} - M + 1}}{3}.$$

Since $(M - 1)^{2}< M^{2} - M + 1< M^{2}$ , we find that

$$\begin{array}{l}{{z_{1}\in\left(\frac{M-1}{3},\frac{M}{3}\right)=(16.67,17),}}\\ {{z_{2}\in\left(M-\frac{2}{3},M-\frac{1}{3}\right)=(50.33,50.67).}}\end{array}$$

The quadratic function $g^{\prime}(z)$ is negative on the interval $(z_{1},z_{2})$ and positive everywhere else. This implies that $g$ is increasing on $(- \infty ,z_{1})$ , decreasing on $(z_{1},z_{2})$ , and increasing on $(z_{2}, + \infty)$ . Hence, the value $0\leq z_{\star}\leq 49$ where $g$ attains its maximum is either $\lfloor z_{1}\rfloor = 16$ or $\lceil z_{1}\rceil = 17$ . In other words, $z_{\star}\in \{16,17\}$ , which corresponds to $k_{\star}\in \{17,18\}$ . Since $f(17) = 19040$ and $f(18) = 19072$ , we conclude that it is optimal to choose is $k = 18$ .