MOOC第五单元作业¶
约 812 个字 预计阅读时间 3 分钟
1¶
第一题为学在浙大作业2第一题
（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
\[\begin{aligned}
P&=(1-p)q+[pq+(1-p)(1-q)]p(1-q)+[pq+(1-p)(1-q)]^2(1-p)q+\cdots \\
&=(1-p)q\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i}+p(1-q)\sum_{i=0}^{\infty}[pq+(1-p)(1-q)]^{2i+1} \\
&=\frac{(1-p)q}{1-[pq+(1-p)(1-q)]^2}+\frac{p(1-q)[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2} \\
&=\frac{(1-p)q+p(1-q)[pq+(1-p)(1-q)]}{1-[pq-(1-p)(1-q)]^2}
\end{aligned}\]
2¶
（1）¶
甲投注“A获胜”时，有
\[\begin{aligned}
E=pa+p(1-a)+2(1-p)a=p+2(1-p)a
\end{aligned}\]
甲投注“B获胜”时，有
\[\begin{aligned}
E=2p(1-a)+(1-p)(1-a)+(1-p)a=1-p+2p(1-a)
\end{aligned}\]
（2）¶
除甲外的其余\(N\)人中，有\(k\)人投注\(A\)的概率为\(\binom Nkp^k(1-p)^{N-k}\)。
若甲投注A，有甲的期望收益为：
\[\begin{aligned}E&=a\sum_{k=0}^N\binom Nkp^k(1-p)^{N-k}\cdot\frac{N+1}{k+1}+(1-a)p^N \\
&=a\sum_{k=0}^N\frac{(N+1)!}{(k+1)!(N-k)!}p^k(1-p)^{N-k}+(1-a)p^N \\
&=a\sum_{k=1}^{N+1}\frac{(N+1)!}{k!(N-k)!}p^{k-1}(1-p)^{N-k+1}+(1-a)p^N \\
&=\frac{a}p\left[\sum_{k=0}^{N+1}\binom{N+1}kp^k(1-p)^{N-k+1}-(1-p)^{N+1}\right]+(1-a)p^N \\
&=\frac{a}p(1-(1-p)^{N+1})+(1-a)p^N\end{aligned}\]
若甲投注B，有甲的期望收益为
\[\begin{aligned}E&=(1-a)\sum_{k=0}^N\binom Nkp^k(1-p)^{N-k}\frac{N+1}{N-k+1}+a(1-p)^N \\
&=(1-a)\sum_{k=0}^N\frac{(N+1)!}{k!(N-k+1)!}p^k(1-p)^{N-k}+a(1-p)^N \\
&=\frac{1-a}{1-p}(\sum_{k=0}^{N+1}\frac{(N+1)!}{k!(N-k+1)!}p^k(1-p)^{N+1-k}-p^{N+1})+a(1-p)^N \\
&=\frac{1-a}{1-p}(1-p^{N+1})+a(1-p)^N\end{aligned}\]
当
\[a\geq\frac{p-p^{N+1}}{1-(1-p)^{N+1}-p^{N+1}}\]
时，
\[\dfrac{a}{p}(1-(1-p)^{N+1})+(1-a)p^N\ge \dfrac{1-a}{1-p}(1-p^{N+1})+a(1-p)^N\]
甲的收益较大。
（3）¶
甲投注A获胜两场：
\[\frac14(N+1)+\frac12\cdot1+\frac14\cdot0\approx\frac14N\]
甲投注A一胜一负：
\[\frac14(N+1)+\frac14\cdot0+\frac14\cdot(N+1)+\frac14\cdot0\approx\frac12N\]
甲投注B获胜两场，收益为1
故甲投注A一胜一负的收益最大。