According to the table of conjugate distributions on Wikipedia, the hypergeometric distribution has as conjugate prior a beta-binomial distribution, where the parameter of interest is "$M$, the number of target members." I interpret "target members" to mean, I am modeling as hypergeometric the number of blue balls in a sample from an urn containing $N$ total balls of which $M$ are blue, $N-M$ non-blue.
But then I cannot make sense of the claim of conjugacy. After the data is observed, say $b$ blue balls in the sample, then it is known that $M>b$. But a beta-binomial distribution is supported on $0,…,F$ (for some parameter $F$). So how can the posterior for $M$ also be beta-binomial?
Best Answer
The problem with the Wikipedia article and the reference therein (Fink D., 1997) is that there is some key information missing.
Specifically, the given posterior is for $M-x$ (i.e. the number of target individuals in the population shifted by the number observed in the sample), not for $M$. Furthermore, the posterior parameter corresponding to the number of observations is missing and should be $N-n$ (i.e. the population size minus the sample size). These two corrections fixes the support problem that you correctly noticed, as shown below.
Suppose that $0 \leq X \leq n$ is the number of target individuals in a sample of size $n$ from a population of size $N$ with $0 \leq M \leq N$ total target individuals.
Then, $X \sim \text{HG}(n, M, N)$ with support in $[\max(0, n-N+M), \min(n, M)]$.
If $M \sim \text{BB}(N, \alpha, \beta)$ is the prior distribution of $M$, the posterior distribution for $M - x$ is also Beta-Binomial-distributed: $$M - x\,|\,x,\alpha,\beta \sim \text{BB}(N-n, \alpha + x, \beta + n - x)$$
If you write the probability mass function for $M$ you will find @Tim's answer above.
As an illustration, for $N = 20$ and $n = 10$, let's assume a non-informative prior distribution for $M$ with $M \sim \text{BB}(N, .5, .5)$. Suppose that we observe $x = 9$.
Created on 2018-10-10 by the reprex package (v0.2.1)
Note that the posterior support is correctly [x, N − n + x].