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个人生日匹配

Birthmate Problems

专题
Probability / 概率
难度
L4

题目详情

概率题：个人生日匹配：与生日悖论等概率的样本量。

英文原题

If rr persons compare birthdays in the pairing problem, the probability is PRP_{R} that at least 2 have the same birthday. What should nn be in the personal birthmate problem to make your probability of success approximately PRP_{R} ?

解析

“配对问题”里 rr 人至少一对同生日的概率近似

PR≈1−exp⁡(−r(r−1)2⋅365).P_R\approx 1-\exp\left(-\frac{r(r-1)}{2\cdot 365}\right).

“个人 birthmate”问题：你与 nn 个他人的生日比较，至少一人与你同生日概率为

1−(364365)n≈1−exp⁡(−n365).1-\left(\frac{364}{365}\right)^n\approx 1-\exp\left(-\frac{n}{365}\right).

令两者近似相等，得到

exp⁡(−n365)≈exp⁡(−r(r−1)2⋅365)⇒n≈r(r−1)2.\exp\left(-\frac{n}{365}\right)\approx \exp\left(-\frac{r(r-1)}{2\cdot 365}\right) \Rightarrow \boxed{n\approx \frac{r(r-1)}{2}}.

更精确可用 n≈−365ln⁡(1−PR)n\approx -365\ln(1-P_R)。

英文解析

PR≈1−exp⁡(−r(r−1)2⋅365).P_R\approx 1-\exp\left(-\frac{r(r-1)}{2\cdot 365}\right). 1−(364365)n≈1−exp⁡(−n365).1-\left(\frac{364}{365}\right)^n\approx 1-\exp\left(-\frac{n}{365}\right). exp⁡(−n365)≈exp⁡(−r(r−1)2⋅365)⇒n≈r(r−1)2.\exp\left(-\frac{n}{365}\right)\approx \exp\left(-\frac{r(r-1)}{2\cdot 365}\right) \Rightarrow \boxed{n\approx \frac{r(r-1)}{2}}.