$n$ people attend the same meeting, what is the chance that two people share the same birthday? Given the first $b$ birthdays, the probability the next person doesn't share a birthday with any that went before is $(365-b)/365$. The probability that none share the same birthday is the following: $\Pi_{0}^{n-1}\frac{365-b}{365}$. How many people would have to attend a meeting so that there is at least a $50$% chance that two people share a birthday?
So I set $\Pi_{0}^{n-1}\frac{365-b}{365}=.5$ and from there I manipulated some algebra to get
$\frac{364!}{(364-n)!365^{n}}=.5\iff (364-n)!365^{n}=364!/.5=…..$
There has to be an easier way of simplifying this.
Best Answer
Paul Halmos asked this question in his "automathography", I Want to Be a Mathematician, and solved it as follows: