I'm reading through some notes our Probability lecturer has uploaded, and he leaves a rather interesting variation on the classical Birthday Problem as an exercise for 'the adventurous Probabilist - with too much time on their hands'.
We let $B_k$ be the number of groups of $k$ individuals who all have the same birthday. From the phrasing of the first part of the problem, I feel he intends for the exercise to allow groups to overlap, i.e. if we have 3 people with birthday Jan. 1 and no other matches, then $B_3 = 1$, $B_2 = 3$, and $B_k = 0$ for $k > 3$.
The classical problem determines a value for $\mathbb{P} \{B_2>0\}$ and to find the least $n$ such that the probability of two people in a population of size $n$ sharing a birthday is greater that $\frac{1}{2}$. Through a standard combinatorial approach I've identified $n=23$.
Problem
We are asked now to determine $\mathbb{E}[B_2]$, $\text{Var}[B_2]$, $\mathbb{E}[B_3]$, and $\text{Var}[B_3]$. I think this is where a combinatorial approach breaks down, though I'm not sure.
I also wondered (beyond the realms of this exercise, but it seems an interesting problem - one which stumps me entirely), how we'd find the least $n$ such that the probability of three people in a population of size $n$ sharing a birthday is greater that $\frac{1}{2}$. I feel this could be approached using Chebyshev’s inequality though I'm yet to conclude how this could be applied successfully.
The lecturer's exercise also gives rise naturally to the question on how we could determine $\mathbb{E}[B_k]$, $\text{Var}[B_k]$ for general $k$. This seems like it could be quite a cumbersome exercise in combinatorics.
I would be very appreciative of anyone who can help with this exercise. Best, MM.