Using the notation in the Wikipedia article on the hypergeometric distribution, I'm curious how one would obtain the maximum likelihood estimate for parameter $m$, the number of white marbles, given $T$ trials from the same urn. For convenience, I'll copy/paste the notation from the article:
Suppose you are to draw $n$ marbles without replacement from an urn containing $N$ marbles in total, $m$ of which are white. The hypergeometric distribution describes the distribution of the number of white marbles drawn from the urn, $k$.
Again, assuming I conduct $T$ trials, at each trial, I take $n$ balls from the urn, and $k_i$ is the number of white balls at trial $i$. Define $K = (k_1,\ldots,k_T)$. Then the likelihood function $L$: $L(m; K, N, n) = \prod_i^T \frac{\binom{m}{k_i}\binom{N-m}{n-k_i}}{\binom{N}{n}}$
Taking a hint from this post, I first tried to solve the inequality: $L(m;K,N,n) \geq L(m-1;K,N,n)$ when $T=1$. From this I obtained $m \leq \frac{Nk+k}{n}$ so the MLE should be $m = \left\lfloor \frac{Nk+k}{n} \right\rfloor$
Now, I'm stuck when I try to generalize to $T \geq 2$.
I first tried doing the same as above and I ended up with the following unwieldy inequality: $\prod_i^T \frac{m}{m-k_i} \geq \prod_i^T \frac{N-m+1}{N-m-n+k_i+1}$ which I'm not sure how to solve.
Then I tried to take the log of the likelihood and differentiate as if $m$ were defined over positive reals and I ended up with an equally unwieldy equation to solve: $\sum_i^T \left(\Psi(m+1) - \Psi(m-k_i+1) - \Psi(N-m+1) + \Psi(N-m-n+k_i+1)\right) = 0$ where $\Psi$ is the digamma function (i.e. the derivative of the log-gamma function).
My intuition tells me the solution to either of the above would look something like this: $m = \left\lfloor \frac{(N+1)\sum_i^T k_i}{Tn} \right\rfloor$ but I have no idea how to get here.
The motivation for this problem is pure curiosity, since I've never seen a MLE for the hypergeometric distribution in terms of $m$.