Homework 5
Question 01
（1）在现规则下，对于每支球队来说，设先罚进球数为 \(X\)，其分布律如下：
\[
\begin{gather}
P\{X=0\}=(1-p)^2\\
P\{X=1\}=2p(1-p)\\
P\{X=2\}=p^2
\end{gather}
\]
设后罚进球数为 \(Y\)，其分布律如下：
\[
\begin{gather}
P\{Y=0\}=(1-q)^2\\
P\{Y=1\}=2q(1-q)\\
P\{Y=2\}=q^2
\end{gather}
\]
因此：
\[
\begin{aligned}
P\{\text{先罚获胜}\}&=(1-q)^2[2p(1-p)+p^2]+2q(1-q)p^2\\
&=p(1-q)(3pq-2q-p+2)\\
P\{\text{后罚获胜}\}&=(1-p)^2[2q(1-q)+q^2]+2p(1-p)q^2\\
&=q(1-p)(3pq-2p-q+2)\\
P\{\text{进球相同}\}&=(1-p)^2(1-q)^2+4pq(1-p)(1-q)+p^2q^2
\end{aligned}
\]
当 \(p=\frac{3}{4},q=\frac{2}{3}\) 时，有 \(P\{先罚获胜\}=\frac{17}{48},P\{后罚获胜\}=\frac{2}{9},P\{进球相同\}=\frac{61}{144}\)
在新规则下，对于每支球队来说，设先罚进球数为 \(X\)，其分布律如下：
\[
\begin{gather}
P\{X=0\}=(1-p)(1-q)=1-p-q+pq\\
P\{X=1\}=(1-p)q+p(1-q)=p+q-2pq\\
P\{X=2\}=pq
\end{gather}
\]
设后罚进球数为 \(Y\)，其分布律如下：
\[
\begin{gather}
P\{Y=0\}=(1-q)(1-p)=1-p-q+pq\\
P\{Y=1\}=(1-q)p+q(1-p)=p+q-2pq\\
P\{Y=2\}=pq
\end{gather}
\]
因此：
\[
\begin{aligned}
P\{\text{先罚获胜}\}&=(p+q-2pq+pq)(1-p-q+pq)+pq(p+q-2pq)\\
&=(1+3pq)(p+q-pq)-(p+q)^2\\
P\{\text{后罚获胜}\}&=(1-p-q+pq)(p+q-2pq+pq)+(p+q-2pq)pq\\
&=(1+3pq)(p+q-pq)-(p+q)^2\\
P\{\text{进球相同}\}&=(1-p-q+pq)^2+(p+q-2pq)^2+p^2q^2
\end{aligned}
\]
当 \(p=\frac{3}{4},q=\frac{2}{3}\) 时，有 \(P\{先罚获胜\}=P\{后罚获胜\}=\frac{41}{144},P\{进球相同\}=\frac{31}{72}\)
（2）在现规则下：
\[
\begin{aligned}
P\{先罚获胜\}&=p(1-q)+[pq+(1-p)(1-q)]·p(1-q)+[pq+(1-p)(1-q)]^2·p(1-q)+...\\
&=\frac{p-pq}{p+q-2pq}\\
P\{后罚获胜\}&=(1-p)q+[pq+(1-p)(1-q)]·(1-p)q+[pq+(1-p)(1-q)]^2·(1-p)q+...\\
&=\frac{q-pq}{p+q-2pq}
\end{aligned}
\]
在新规则下：
\[
\begin{aligned}
P\{先罚获胜\}&=p(1-q)+[pq+(1-p)(1-q)]·(1-p)q+[pq+(1-p)(1-q)]^2·p(1-q)+...\\
&=\frac{p-pq}{1-[pq+(1-p)(1-q)]^2}+[pq+(1-p)(1-q)]\frac{q-pq}{1-[pq+(1-p)(1-q)]^2}\\
&=\frac{p-pq+[pq+(1-p)(1-q)](q-pq)}{1-[pq+(1-p)(1-q)]^2}\\
P\{后罚获胜\}&=(1-p)q+[pq+(1-p)(1-q)]·p(1-q)+[pq+(1-p)(1-q)]^2·(1-p)q+...\\
&=\frac{q-pq}{1-[pq+(1-p)(1-q)]^2}+[pq+(1-p)(1-q)]\frac{p-pq}{1-[pq+(1-p)(1-q)]^2}\\
&=\frac{q-pq+[pq+(1-p)(1-q)](p-pq)}{1-[pq+(1-p)(1-q)]^2}
\end{aligned}
\]
Question 02
（1）\(P\{A\text{ 获得比赛胜利}\}=(\alpha+\beta)+\gamma^2(\alpha+\beta)+...=\frac{\alpha+\beta}{1-\gamma^2}\)
（2）\(P\{A\text{ 获得比赛胜利}\}=\alpha+\beta a+\gamma b\)
（3）\(a=\gamma + \beta\frac{\alpha+\beta}{1-\gamma^2},b=1-\frac{\alpha+\beta}{1-\gamma^2}\)
\(\therefore P\{A\text{ 获得比赛胜利}\}=\alpha+\beta\gamma+\beta^2\frac{\alpha+\beta}{1-\gamma^2}+\gamma-\gamma\frac{\alpha+\beta}{1-\gamma^2}=\frac{1-\beta+\beta\gamma^2+\beta^2}{1+\gamma}\)
在旧赛制下，\(P\{A\text{ 获得比赛胜利}\}=\frac{1}{1+\gamma}\)
从公平性角度来看，一般来说 \(-\beta+\beta\gamma^2+\beta^2<0\)，所以新赛制更为公平一些