On the issue of the birthday paradox,Let $p_{n}$ be the probability that in a class of $n$ at least $2$ have a their birthdays on the same day (exclude $29$ Feb). Use the inequality $1-x \leq e^{-x}$ to show that:

$p_{n} \geq 1- \exp{(-n(n-1)/730)}$ and then determine $n \in \mathbb N$ so that $p_{n} \geq \frac{1}{2}$

My ideas:

First $p_{n}=1-\frac{\frac{365!}{(365-n)!}}{365^{n}}$ using the inequality given to us.

$1-\exp{(-\frac{\frac{365!}{(365-n)!}}{365^{n}})}\geq p_{n}$, what am I supposed to do next? Use Stirling's Formula?


For $n\le 365$, $$ p_n=1-\prod_{i=1}^{n-1}\left(1-\frac{i}{365}\right)\ge 1-\prod_{i=1}^{n-1}e^{\frac{i}{365}}=1-e^{\sum_{i=1}^{n-1}\frac{i}{365}}=1-e^{-\frac{n(n-1)}{730}}. $$

  • $\begingroup$ I think a small comment saying $P(collision)=1-P(no collisions)$ will make this a perfect answer. Preemptive (+1) anyway ... $\endgroup$ – rtybase Feb 11 at 10:38
  • $\begingroup$ How did you get $\sum_{i=1}^{n-1}\frac{i}{365}=\frac{n(n-1)}{730}$? $\endgroup$ – MinaThuma Feb 11 at 11:17
  • 1
    $\begingroup$ @MinaThuma $1+2+3+\ldots + n=n(n+1)/2$. $\endgroup$ – d.k.o. Feb 11 at 16:48

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