两名射手轮流向同一目标射击，射手甲和射手乙命中目标的概率都是 $\dfrac{1}{2}$．若射手甲先射，谁先命中目标谁就获胜，试求甲、乙两射手获胜的概率．
【难度】
【出处】
2000年上海交通大学保送生测试题
【标注】
【答案】
甲获胜的概率为 $\dfrac{2}{3}$，乙获胜的概率为 $\dfrac{1}{3}$
【解析】
甲只可能在第奇数枪获胜，乙只可能在第偶数枪获胜．
甲获胜的概率为$$\dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^4} \cdot \dfrac{1}{2} + \cdots = \dfrac{{\dfrac{1}{2}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}} = \dfrac{2}{3}.$$乙获胜的概率为$$\dfrac{1}{2} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^3} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^5} \cdot \dfrac{1}{2} + \cdots = \dfrac{{\dfrac{1}{4}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}} = \dfrac{1}{3}.$$
甲获胜的概率为$$\dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^4} \cdot \dfrac{1}{2} + \cdots = \dfrac{{\dfrac{1}{2}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}} = \dfrac{2}{3}.$$乙获胜的概率为$$\dfrac{1}{2} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^3} \cdot \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^5} \cdot \dfrac{1}{2} + \cdots = \dfrac{{\dfrac{1}{4}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}} = \dfrac{1}{3}.$$
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