In a room with n other people what is the expected value of n for k other people to share the same birthday as me?
I know from this very helpful wikipedia article that the probability for $k=1$ can be represented as:
$$q(n;d) = 1 - \left(\frac{d-1}{d}\right)^n$$
where d is some number of equally likely birthdays.
I found approximations for the canonical birthday problem extended to k collisions in this post. I'm looking for a similar generalization but applied to the same birthday as me variation.
I am planning on running some simulations so computability of the solution is a factor. A good approximation will be more than sufficient.
Using standard assumptions (birthdays i.i.d. uniformly across $d$ days), then
The expected number who share your birthday is $\mathbb E[K]=\frac{n}{d}$
If instead $k$ is fixed and you increase $n$ until $k$ matches occur, you have a version of the negative binomial distribution, the sum of $k$ geometric distributions each with expected value $d$, so $\mathbb E[N]=kd$
The Wikipedia article's same birthday as you variant in fact looks at the question of the minimum $n$ where $\mathbb P(K\ge 1)\ge \frac12$, and this would be $n = \Big\lceil\frac{\log(2)}{ \log(d)-\log(d-1)}\Big\rceil$. There is not such a simple calculation for the more general $\mathbb P(K\ge k)\ge \frac12$, but the number rises by close to $d$ each time $k$ increases by $1$. For example with $d=365$, you would get