题目
甲乙轮流抛掷一枚质地均匀的硬币，若连续四次抛掷结果出现了”正正正反“则甲获胜，若出现了”正反反反“则乙获胜。求甲乙二人各自获胜的概率。
解析
设 S_{ A } 是甲获胜时的局面的构型之和，S_{ B } 是乙获胜时的局面的构型之和，N 为甲乙二人均不获胜的局面的构型之和。设正面记为 \mathtt{ Z }，反面记为 \mathtt{ F }，则有：
\begin{aligned}
S_{ A } & = \mathtt{ ZZZF } + \mathtt{ ZZZZF } + \mathtt{ FZZZF } + \mathtt{ ZZZZZF } + \mathtt{ ZFZZZF } + \ldots \\
S_{ B } & = \mathtt{ ZFFF } + \mathtt{ ZZFFF } + \mathtt{ FZFFF } + \mathtt{ ZFZFFF } + \mathtt{ FZZFFF } + \ldots \\
N & = 1 + \mathtt{ Z } + \mathtt{ F } + \mathtt{ ZZ } + \mathtt{ ZF } + \mathtt{ FZ } + \mathtt{ FF } + \mathtt{ ZZZ } + \mathtt{ ZZF } + \ldots
\end{aligned}
那么就有：
\begin{aligned}
1 + N ( \mathtt{ Z } + \mathtt{ F } ) & = N + S_{ A } + S_{ B } \\
N ( \mathtt{ ZZZF } ) & = S_{ A } \\
N ( \mathtt{ ZFFF } ) & = S_{ B } + S_{ A } ( \mathtt{ FF } )
\end{aligned}
令 \mathtt{ Z } = \mathtt{ F } = \dfrac{ 1 }{ 2 }，代入得：
\begin{aligned}
1 + N & = N + S_{ A } + S_{ B } \\
\dfrac{ 1 }{ 16 } N & = S_{ A } \\
\dfrac{ 1 }{ 16 } N & = S_{ B } + \dfrac{ 1 }{ 4 } S_{ A }
\end{aligned}
解得 S_{ A } = \dfrac{ 4 }{ 7 }, S_{ B } = \dfrac{ 3 }{ 7 }。
by CXY。