Classic birthday problem turned on its head: With $N$ people, how many are likely to share most common birthday?

Solution 1:

The method to get to the most common shared birthday won't work as pointed out by Bram28. But it's easy to get to an estimate of the number, using the fact that on average on each day there are $\lambda = \dfrac{2000}{365}\approx 5.48$ birthdays. We then approximately have a Poisson distribution for the probability to have $n$ birthdays on some given day:

$$P(n) = \frac{\lambda^n}{n!}\exp(-\lambda)$$

This means that the expected number of days with $n$ birthdays will be $\mu(n) = 365 P(n)$. The number of days with $n$ birthdays will be approximately distributed according the the Poisson distribution with mean $\mu(n)$. This means that for $n$ such that $\mu(n) \geq \log(2)$, the probability of there being a day with $n$ birthdays will be larger or equal to 50%; this means that $n$ must chosen to be $13$ or smaller.

Solution 2:

I don't have any answer to your main question, but I do want to point out that your method may not find the date with the most shared birthdays. For example, suppose there are 10 people that all share April 10 as their birthday, and that no other date has that many shared birthdays. However, now also suppose that 900 people have their birthday in the first 6 months, and 1100 in the second six months ....