2009 AIME II 第 8 题

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8.

戴夫掷一枚公平六面骰,直到第一次出现六点为止。琳达独立地掷一枚公平六面骰,也直到第一次出现六点为止。设互质正整数 mmnn 满足 mn\frac{m}{n} 等于戴夫与琳达的掷骰次数相差不超过一的概率。求 m+nm + n

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let mm and nn be relatively prime positive integers such that mn\frac{m}{n} is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find m+n.m + n.

答案:41
知识点:几何分布等比数列对称性
难度评级:2560
小提示:

第一次出现六点在第 kk 次的概率为 pk=(56)k116p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}

The chance the first six appears on roll kk is pk=(56)k116p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}

大提示:

平局概率计算 pk2\sum p_k^2,相差一的概率计算 pkpk+1\sum p_k p_{k+1};两者都是公比为 2536\frac{25}{36} 的等比级数

Compute pk2\sum p_k^2 for a tie and pkpk+1\sum p_k p_{k+1} for a difference of one; both are geometric series with ratio 2536\frac{25}{36}

解答:

某人的第一次六点出现在第 kk 次的概率为 pk=(56)k116p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}。两人次数相同的概率为 k=1pk2=136112536=111\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}\text{。}

琳达恰好比戴夫多掷一次的概率为 k=1pkpk+1\sum_{k=1}^{\infty} p_k p_{k+1} =56k=1pk2= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2 =566= \frac{5}{66},由对称性,戴夫恰好比琳达多掷一次也有相同概率。

总概率为 111+2566=6+1066=833\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33},所以 m+n=8+33=41m + n = 8 + 33 = 41

The probability that a player’s first six appears on roll kk is pk=(56)k116.p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}. The probability of a tie is k=1pk2=136112536=111.\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.

The probability that Linda needs exactly one more roll than Dave is k=1pkpk+1\sum_{k=1}^{\infty} p_k p_{k+1} =56k=1pk2= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2 =566,= \frac{5}{66}, and by symmetry the same holds with the players swapped.

The total probability is 111+2566=6+1066=833,\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33}, so m+n=8+33=41.m + n = 8 + 33 = 41.

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