Fix $p \in (0, 1), N, M \in \mathbf{N}_{\geqslant 1}$. It is well-known that if $X \sim \textbf{Bin} (N, p)$ and $Z \sim \textbf{Bin} (M, p)$, then $Y = X + Z$ is marginally distributed as $Y \sim \textbf{Bin} (N + M, p)$.
Now, write $K_M (X \to Y)$ to denote the conditional distribution of $Y$ given $X$, i.e.
\begin{align} K_M (X \to Y) = {M \choose {Y - X}} p^{Y-X} (1 - p)^{M - Y + X}. \end{align}
One can then write out the joint distribution of $(X, Y)$ in two ways:
\begin{align} \Pi(X,Y) &= \mathbf{B}_{N, p} (X) \cdot K_M (X \to Y) \\ &= \mathbf{B}_{N + M, p} (Y) \cdot K_{M,N}^* (Y \to X), \end{align}
where by Bayes' rule, one can compute
\begin{align} K_{M,N}^* (Y \to X) = {Y \choose X} { {N + M} \choose N}^{-1} {{N + M - Y} \choose {N - X}}^{-1}, \end{align}
which is supported on $\max {(0, M - Y)} \leqslant X \leqslant \min { ( N, Y ) }$.
I remark quickly that it is not an accident that $K_{M,N}^*$ does not depend on $p$; all of the $p$ terms cancel out in the application of Bayes' rule. It is a priori not obvious to me why this should happen, but also not completely shocking; I suspect there's some intuitive explanation involving sufficient statistics or something to that effect. In any case, that's not important right now.
My question is then: does this conditional distribution $K_{M,N}^*$ have a name?
At a first glance it looks like a hypergeometric distribution of sorts, but on closer inspection, it doesn't quite match up. If it doesn't have a name, I'd still be happy with references which describe some of its properties in a bit of detail. It would also be nice to know if there is an efficient way to simulate this distribution.
A related question is what happens when one replaces every Binomial distribution above with a Negative Binomial distribution. In this case, the conditional distribution is given by
\begin{align} \tilde{K}_{M,N}^* (Y \to X) = {Y \choose X} { {N + M} \choose N}^{-1} {{N + M + Y} \choose {N + X}}^{-1} \cdot \frac{N + M + Y}{N + M} \cdot \frac{N}{N + X} \cdot \frac{M}{M + Y}. \end{align}
I would also be interested to learn about the properties of this distribution; I'll take what I can get at this stage.
Not an answer (i.e. not a reference / a name as requested) but an intuitive explanation why this is independent of $p$.
There are $M+N$ coins in a row, of which you know $Y=y$ are heads (and the rest are tails). Importantly, any of the ${M+N \choose y}$ ways of designating the subset of heads is equiprobable. Then $X =$ how many coins in the leftmost $N$ are heads?
Clearly $p$ does not appear anywhere in the above description. $p$ controls how likely it is to get $Y=y$ heads in the first place, and it disappears once you condition on $Y=y$.
Great question re: the name - I'd like to know too.