Homework2¶
约 663 个字 预计阅读时间 2 分钟
问题一¶
1¶
现规则：
| 进球次数 | 概率 |
|---|---|
| 0 | \((1-p)^2\cdot (1-q)^2\) |
| 1 | \(2p(1-p)\cdot 2q(1-q)\) |
| 2 | \(p^2\cdot q^2\) |
可得
\[\begin{aligned}
P&=P_0+P_1+P_2=(1-p)^2\cdot (1-q)^2+2p(1-p)\cdot 2q(1-q)+p^2\cdot q^2 \\
&=6p^2q^2-6p^2q-6pq^2+8pq+p^2-2p+q^2-2q+1
\end{aligned}\]
当\(p=\frac{3}{4},q=\frac{2}{3}\)时，有
\[\begin{aligned}
P=\frac{61}{144}
\end{aligned}\]
新规则：
| 进球次数 | 概率 |
|---|---|
| 0 | \((1-p)(1-q)\cdot (1-q)(1-p)\) |
| 1 | \([p(1-q)+(1-p)q]\cdot [q(1-p)+(1-q)p]\) |
| 2 | \(pq\cdot qp\) |
可得
\[\begin{aligned}
P&=P_0+P_1+P_2 \\
&=(1-p)(1-q)\cdot (1-q)(1-p)+[p(1-q)+(1-p)q]\cdot [q(1-p)+(1-q)p]+pq\cdot qp \\
&=6p^2q^2-6p^2q-6pq^2+6pq+2p^2-2p+2q^2-2q+1
\end{aligned}\]
当\(p=\frac{3}{4},q=\frac{2}{3}\)时，有
\[\begin{aligned}
P=\frac{31}{72}
\end{aligned}\]
2¶
现规则：
加赛第一轮先罚球队获胜的情况为某一轮先罚球球进，后罚球不进，有
\[\begin{aligned}
P&=p(1-q)+[pq+(1-p)(1-q)]p(1-q)+[pq+(1-p)(1-q)]^2p(1-q)+\cdots \\
&=p(1-q)\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^i \\
&=\frac{p(1-q)}{1-pq-(1-p)(1-q)} \\
&=\frac{p(1-q)}{p+q-2pq}
\end{aligned}\]
加赛第一轮后罚球队获胜的情况为某一轮先罚球球不进，后罚球进，有
\[\begin{aligned}
P&=(1-p)q+[pq+(1-p)(1-q)](1-p)q+[pq+(1-p)(1-q)]^2(1-p)q+\cdots \\
&=(1-p)q\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^i \\
&=\frac{(1-p)q}{p+q-2pq}
\end{aligned}\]
新规则：
令加赛第一轮先罚球队为\(A\)，后罚球队为\(B\)
\(A\)获胜的情况为某一轮\(A\)罚球球进，\(B\)罚球不进，有
\[\begin{aligned}
P&=p(1-q)+[pq+(1-p)(1-q)](1-p)q+[pq+(1-p)(1-q)]^2p(1-q)+\cdots \\
&=p(1-q)\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i}+(1-p)q\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i+1} \\
&=\frac{p(1-q)}{1-[pq+(1-p)(1-q)]^2}+\frac{(1-p)q[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2} \\
&=\frac{p(1-q)+(1-p)q[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2}
\end{aligned}\]
\(B\)获胜的情况为某一轮\(A\)罚球球进，\(B\)罚球不进，有
\[\begin{aligned}
P&=(1-p)q+[pq+(1-p)(1-q)]p(1-q)+[pq+(1-p)(1-q)]^2(1-p)q+\cdots \\
&=(1-p)q\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i}+p(1-q)\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i+1} \\
&=\frac{(1-p)q}{1-[pq+(1-p)(1-q)]^2}+\frac{p(1-q)[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2} \\
&=\frac{(1-p)q+p(1-q)[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2}
\end{aligned}\]
问题2¶
1¶
\(A\)比赛胜利当且仅当\(A\)得分或\(A\)未得分时\(B\)未得分，有
\[\begin{aligned}
p_0&=(\alpha+\beta)+\gamma^2(\alpha+\beta)+\gamma^4(\alpha+\beta)+\cdots \\
&=(\alpha+\beta)\sum_{i=0}^{\infty}\gamma^{2i} \\
&=\frac{\alpha+\beta}{1-\gamma^2} \\
&=\frac{1}{1+\gamma}
\end{aligned}\]
2¶
第一回合\(A\)射门：
\[\begin{aligned}
a=\gamma+\frac{\beta}{1+\gamma}
\end{aligned}\]
第一回合\(A\)不得分：
\[\begin{aligned}
b=1-p=1-\frac{1}{1+\gamma}=\frac{\gamma}{1+\gamma}
\end{aligned}\]
3¶
\(A\)获胜的概率为
\[\begin{aligned}
p_1&=\alpha+a\beta+b\gamma \\
&=\alpha+(\gamma+\frac{\beta}{1+\gamma})\beta+\frac{\gamma^2}{1+\gamma} \\
&=\frac{\gamma^2\beta+\beta^2-\beta+1}{1+\gamma}
\end{aligned}\]
对于\(p_1\)，在\(\beta,\gamma\in (0,1)\)的情况下，有\(p_1>\frac{1}{2}\)，且\(p_0=\frac{1}{1+\gamma}>\frac{1}{2},p_0>p_1\)，故新赛制下\(A,B\)获得胜利的概率更接近，更加合理。