Dave投掷一颗均匀的骰子，直到第一次出现6点为止；Linda也独立投掷―颗均匀的骰子，直到第一次出现6点为止．设Dave投掷的次数等于或与Linda投掷的次数相差不超过1的概率为 $\frac{m}{n}$（其中 $m$，$n$ 是互素的正整数、求 $m+n$．
【难度】
【出处】
2009年第27届美国数学邀请赛Ⅱ（AIMEⅡ）
【标注】
【答案】
41
【解析】
Dave和Linda掷骰子直到第 $k$ 次才出现6点的概率均为前 $k-1$ 次都未出现6点而接下来一次时出现6点的概率．即 ${{P}_{k}}={{\left( \frac{5}{6} \right)}^{k-1}}\left(\frac{1}{6} \right)$．Dave需要1次就出现6点而Linda需要1或2次才出现6点的概率为 ${{P}_{1}}\left( {{P}_{1}}+{{P}_{2}} \right)$．Dave需要 $k1$ 次就出现6点而Linda需要 $k-1$，$k$ 或 $k+1$ 次才出现6点的概率为 ${{p}_{k}}\left({{p}_{k-1}}+{{p}_{k}}+{{p}_{k+1}} \right)$．于是，所求的概率应为 $\displaystyle {{p}_{1}}\left( {{p}_{1}}+{{p}_{2}}\right)+\sum\limits_{k=2}^{\infty }{{{p}_{k}}\left({{p}_{k-1}}+{{p}_{k}}+{{p}_{k+1}} \right)}$，也即
$\displaystyle \frac{1}{6}\cdot \left( \frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6} \right)+\sum\limits_{k=2}^{\infty }{{{\left( \frac{5}{6}\right)}^{k-1}}\left( \frac{1}{6} \right)\left( {{\left( \frac{5}{6}\right)}^{k-2}}\left( \frac{1}{6} \right)+{{\left( \frac{5}{6}\right)}^{k}}\left( \frac{1}{6} \right) \right)}$
$\displaystyle =\frac{1}{6}\cdot \left( \frac{6}{36}+\frac{5}{36}\right)+\sum\limits_{k=2}^{\infty }{{{\left( \frac{5}{6} \right)}^{k-1}}\left(\frac{1}{6} \right){{\left( \frac{5}{6} \right)}^{k-2}}\left( \frac{1}{6}\right)\left( 1+\frac{5}{6}+{{\left( \frac{5}{6} \right)}^{2}} \right)}$
$=\frac{11}{{{6}^{3}}}+\frac{91}{{{6}^{4}}}\cdot\frac{\frac{5}{6}}{1-{{\left( \frac{5}{6}\right)}^{2}}}=\frac{11}{{{6}^{3}}}+\frac{91}{{{6}^{3}}}\cdot\frac{5}{11}=\frac{576}{{{6}^{3}}\cdot 11}=\frac{8}{33}$．
故 $m+n=8+33=41$．
$\displaystyle \frac{1}{6}\cdot \left( \frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6} \right)+\sum\limits_{k=2}^{\infty }{{{\left( \frac{5}{6}\right)}^{k-1}}\left( \frac{1}{6} \right)\left( {{\left( \frac{5}{6}\right)}^{k-2}}\left( \frac{1}{6} \right)+{{\left( \frac{5}{6}\right)}^{k}}\left( \frac{1}{6} \right) \right)}$
$\displaystyle =\frac{1}{6}\cdot \left( \frac{6}{36}+\frac{5}{36}\right)+\sum\limits_{k=2}^{\infty }{{{\left( \frac{5}{6} \right)}^{k-1}}\left(\frac{1}{6} \right){{\left( \frac{5}{6} \right)}^{k-2}}\left( \frac{1}{6}\right)\left( 1+\frac{5}{6}+{{\left( \frac{5}{6} \right)}^{2}} \right)}$
$=\frac{11}{{{6}^{3}}}+\frac{91}{{{6}^{4}}}\cdot\frac{\frac{5}{6}}{1-{{\left( \frac{5}{6}\right)}^{2}}}=\frac{11}{{{6}^{3}}}+\frac{91}{{{6}^{3}}}\cdot\frac{5}{11}=\frac{576}{{{6}^{3}}\cdot 11}=\frac{8}{33}$．
故 $m+n=8+33=41$．
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