乒乓球概率

(1) $p_3$：3球后甲净胜分$\geq 2$。

3球后甲净胜只能是$\pm 1$或$\pm 3$，故净胜$\geq 2$等价于甲全胜。

$p_3 = p^3$。

同理，$q_3 = q^3$。

$p_4$：4球后甲净胜$\geq 2$，即净胜2或4。

净胜4：$p^4$；净胜2：甲3胜1负，$\mathrm{C}_4^3 p^3 q = 4p^3 q$。

故 $p_4 = p^4 + 4p^3 q$。

同理，$q_4 = q^4 + 4q^3 p$。

(2) 计算增量：

$p_4 - p_3 = p^4 + 4p^3 q - p^3 = p^3(p + 4q - 1) = p^3(3 - 3p) = 3p^3 q$。

$q_4 - q_3 = q^4 + 4q^3 p - q^3 = q^3(q + 4p - 1) = q^3(4p - p) = 3pq^3$。

由条件 $\dfrac{p_4 - p_3}{q_4 - q_3} = \dfrac{3p^3 q}{3pq^3} = \dfrac{p^2}{q^2} = 4$，

得 $\dfrac{p}{q} = 2$，即 $p = 2q = 2(1-p)$，解得 $p = \dfrac{2}{3}$。

(3) 设$X_k$为$k$球中甲胜的场数，服从二项分布$B(k,p)$。甲净胜分为$2X_k - k$。

$p_k = P(X_k \geq \frac{k+2}{2})$，$q_k = P(X_k \leq \frac{k-2}{2})$。

当$k = 2m$时： $p_{2m} = P(X_{2m} \geq m+1)$，$q_{2m} = P(X_{2m} \leq m-1)$。

当$k = 2m+1$时： $p_{2m+1} = P(X_{2m+1} \geq m+2)$，$q_{2m+1} = P(X_{2m+1} \leq m-1)$。

先证左边：$p_{2m+1} - q_{2m+1} < p_{2m} - q_{2m}$

由二项分布递推： $P(X_{2m+1} \geq m+2) = p \cdot P(X_{2m} \geq m+1) + q \cdot P(X_{2m} \geq m+2)$ $= P(X_{2m} \geq m+1) - q \cdot P(X_{2m} = m+1)$。

$P(X_{2m+1} \leq m-1) = q \cdot P(X_{2m} \leq m-1) + p \cdot P(X_{2m} \leq m-2)$ $= P(X_{2m} \leq m-1) - p \cdot P(X_{2m} = m-1)$。

两式相减： $p_{2m+1} - q_{2m+1} = (p_{2m} - q_{2m}) - q \cdot P(X_{2m} = m+1) + p \cdot P(X_{2m} = m-1)$。

注意 $P(X_{2m} = m+1) = \mathrm{C}_{2m}^{m+1} p^{m+1} q^{m-1}$，$P(X_{2m} = m-1) = \mathrm{C}_{2m}^{m-1} p^{m-1} q^{m+1}$，且$\mathrm{C}_{2m}^{m+1} = \mathrm{C}_{2m}^{m-1}$。

故 $q \cdot P(X_{2m} = m+1) - p \cdot P(X_{2m} = m-1) = \mathrm{C}_{2m}^{m+1} p^m q^m (p - q) > 0$。

因此 $p_{2m+1} - q_{2m+1} = (p_{2m} - q_{2m}) - \mathrm{C}_{2m}^{m+1} p^m q^m (p-q) < p_{2m} - q_{2m}$。

再证右边：$p_{2m} - q_{2m} < p_{2m+2} - q_{2m+2}$

同理，由两步递推： $P(X_{2m+2} \geq m+2) = p^2 P(X_{2m} \geq m) + 2pq P(X_{2m} \geq m+1) + q^2 P(X_{2m} \geq m+2)$ $= P(X_{2m} \geq m+1) + p^2 P(X_{2m} = m) - q^2 P(X_{2m} = m+1)$。

$P(X_{2m+2} \leq m) = q^2 P(X_{2m} \leq m) + 2pq P(X_{2m} \leq m-1) + p^2 P(X_{2m} \leq m-2)$ $= P(X_{2m} \leq m-1) + q^2 P(X_{2m} = m) - p^2 P(X_{2m} = m-1)$。

两式相减： $p_{2m+2} - q_{2m+2} = (p_{2m} - q_{2m}) + (p^2-q^2)P(X_{2m}=m) - q^2 P(X_{2m}=m+1) + p^2 P(X_{2m}=m-1)$。

代入 $P(X_{2m}=m) = \mathrm{C}_{2m}^m p^m q^m$，并利用 $\mathrm{C}_{2m}^{m+1} = \mathrm{C}_{2m}^{m-1}$，得：

$- q^2 \cdot \mathrm{C}_{2m}^{m+1} p^{m+1} q^{m-1} + p^2 \cdot \mathrm{C}_{2m}^{m+1} p^{m-1} q^{m+1} = -\mathrm{C}_{2m}^{m+1} p^{m+1} q^{m+1} + \mathrm{C}_{2m}^{m+1} p^{m+1} q^{m+1} = 0$。

故新增项为 $(p^2-q^2) \mathrm{C}_{2m}^m p^m q^m = (p-q) \mathrm{C}_{2m}^m p^m q^m > 0$。

因此 $p_{2m+2} - q_{2m+2} > p_{2m} - q_{2m}$。

综上，$p_{2m+1} - q_{2m+1} < p_{2m} - q_{2m} < p_{2m+2} - q_{2m+2}$。