Note: cross-posted from https://stats.stackexchange.com/questions/566107/least-likely-sample-in-multivariate-hypergeometric-distributions
Let $U=(U_1,U_2,\dots,U_c)$ being an urn with $U_i$ balls of color $i \in [1,c]$ and $\Omega(U)$ be the set of possible draws from urn $U$. For $D \in \Omega(U)$ the probability of drawing $D$ is given by (multivariate hypergeometric pdf) $$p(D,U) = \dfrac{\prod_{i=1}^c \binom{U_i}{D_i}}{\binom{N_U}{N_D}}$$ with $N_U = \sum_{i=1}^c U_i$.
Main goal: Suppose I am now sampling from a given $D \in \Omega(U)$. I am looking for the sample $D^* \in \Omega(D)$ such that $p(D^*,U)$ is minimum. Alternatively, I would like to know if it exists $D^* \in \Omega(D)$ such that $p(D^*,U)<\alpha$ for some $\alpha \in [0,1]$.
In the case where $D=U$, it is easy to see that the problem can be reduced to the partition problem: we are looking for a subset of colors whose ball count in $U$ is as close as possible to $\frac{N_U}{2}$. I believe that, in the general case where $D\neq U$, greedy algorithms that only go through subsets of colors could be used to find $D^*$.
However, these observations rely on the following assumption: the least likely sample is achieved by drawing either all or none of the balls of each color i.e. $D^*$ always belongs to the subset of samples $\Omega'(D) = \{D' \in \Omega(D): \forall i \in [1,c], D'_i \in \{0,D_i\}\}$.
Question: How to prove (or disprove) the above assumption in the restricted case where $D=U$ or in the general case where $D\neq U$?
First approach: I tried to work with a more general but maybe simpler claim:
Let $D' \not\in \Omega'(D)$, by definition there is $i \in [1,c]$ such that $0<D'_i<D_i$. Let $D^0$ (resp $D^1$) be the sample we get from $D$ by setting $D^0_i=0$ (resp. $D^0_i=D_i)$ then either $p(D',U)>p(D^0,U)$ or $p(D',U)>p(D^1,U)$.
In other words: we can always get a sample that is least likely than $D'$ by either removing all the balls of a given color or taking the rest of the balls of a given color. The idea was to work with the ratio $\frac{p(D',U)}{p(D^0,U)}$ and $\frac{p(D',U)}{p(D^1,U)}$ and remove the dependencies to colors other than $i$.
I thought this claim would be easier to prove but there is still a lot of variables and I do not know how to simplify this further.
Second approach: In the case $D=U$. Reformulate the optimisation problem by developping $p(D',U)$ and using the $\log p(D',U)$ instead. After using Stirling's approximation I ended up with
$$\min_{X \in \Omega(D)} p(X,U) \Leftrightarrow \max_{X \in \Omega(D)} \sum_{i=1}^c \left[ X_i\log\left(\frac{X_i}{N_X}\right) + (U_i - X_i) \log\left(\frac{U_i- X_i}{N_U-N_X}\right) \right]$$
However, I am not sure it is even correct or where to go from here.
Any hint, suggestion or links to similar problems in the litterature are very welcome.