赌金分别为 $a$ 和 $b$（$a,b\in {{\mathbf{N}}^{*}}$）的两位赌徒，进行每次单位1的赌博，已知赌徒甲每局胜率为 $p$（$p\in \left( 0,1 \right)$），赌徒乙每局胜率为 $1-p$，其中一位赌徒赌金输光赌局结束，则赌徒甲最终获胜概率是多少？
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无
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【答案】
略
【解析】
设赌徒甲在赌金剩余为 $n$ 时获胜概率为 ${{P}_{n}}$，其中 $n=0,1,2,\cdots ,a+b$，
则 ${{p}_{0}}=0$，${{P}_{a+b}}=1$，
且 ${{P}_{n}}=\left( 1-p\right){{P}_{n-1}}+p{{P}_{n+1}}$，特征方程为 $x=1-p+p{{x}^{2}}$，其特征根为 ${{x}_{1}}=1$，${{x}_{2}}=\dfrac{1-p}{p}$，
（1）若 $p=\dfrac{1}{2}$，则 ${{P}_{n}}=\lambda n+\mu $，
待定系数可得 $\left\{ \begin{align}
& {{P}_{0}}=\mu =0 \\
& {{P}_{a+b}}=\lambda \left( a+b\right)+\mu =1 \\
\end{align}\right.$，
解得 $\lambda =\dfrac{1}{a+b}$，$\mu =0$，故 ${{P}_{n}}=\dfrac{n}{a+b}$，则 ${{P}_{a}}=\dfrac{a}{a+b}$，
（2）若 $p\ne \dfrac{1}{2}$
故其通项为 ${{P}_{n}}=\lambda +\mu{{\left( \dfrac{1-p}{p} \right)}^{n}}$，
待定系数可得 $\left\{ \begin{align}
& \lambda +\mu =0 \\
& \lambda +\mu {{\left( \dfrac{1-p}{p}\right)}^{a+b}}=1 \\
\end{align}\right.$，
解得 $\mu =\dfrac{1}{{{\left(\dfrac{1-p}{p} \right)}^{a+b}}-1}$，$\lambda=-\dfrac{1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$，
故 ${{P}_{n}}=\dfrac{{{\left(\dfrac{1-p}{p} \right)}^{n}}-1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$，进而可求得 ${{P}_{a}}=\dfrac{{{\left( \dfrac{1-p}{p}\right)}^{a}}-1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$．
则 ${{p}_{0}}=0$，${{P}_{a+b}}=1$，
且 ${{P}_{n}}=\left( 1-p\right){{P}_{n-1}}+p{{P}_{n+1}}$，特征方程为 $x=1-p+p{{x}^{2}}$，其特征根为 ${{x}_{1}}=1$，${{x}_{2}}=\dfrac{1-p}{p}$，
（1）若 $p=\dfrac{1}{2}$，则 ${{P}_{n}}=\lambda n+\mu $，
待定系数可得 $\left\{ \begin{align}
& {{P}_{0}}=\mu =0 \\
& {{P}_{a+b}}=\lambda \left( a+b\right)+\mu =1 \\
\end{align}\right.$，
解得 $\lambda =\dfrac{1}{a+b}$，$\mu =0$，故 ${{P}_{n}}=\dfrac{n}{a+b}$，则 ${{P}_{a}}=\dfrac{a}{a+b}$，
（2）若 $p\ne \dfrac{1}{2}$
故其通项为 ${{P}_{n}}=\lambda +\mu{{\left( \dfrac{1-p}{p} \right)}^{n}}$，
待定系数可得 $\left\{ \begin{align}
& \lambda +\mu =0 \\
& \lambda +\mu {{\left( \dfrac{1-p}{p}\right)}^{a+b}}=1 \\
\end{align}\right.$，
解得 $\mu =\dfrac{1}{{{\left(\dfrac{1-p}{p} \right)}^{a+b}}-1}$，$\lambda=-\dfrac{1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$，
故 ${{P}_{n}}=\dfrac{{{\left(\dfrac{1-p}{p} \right)}^{n}}-1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$，进而可求得 ${{P}_{a}}=\dfrac{{{\left( \dfrac{1-p}{p}\right)}^{a}}-1}{{{\left( \dfrac{1-p}{p} \right)}^{a+b}}-1}$．
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