Let $X\sim NegBin(r,p)$ and $Y\sim NegBin(s,p)$ be independent and $p\in(0,1)$. The sum $Z:=X+Y\sim NegBin(r+s,p)$. Now, i did compute the conditional probability of $X$ under $Z$, that is $$\mathbb P(X=k|Z=n)=\frac{\binom{k-1}{r-1}\binom{n-k-1}{s-1}}{\binom{n-1}{r+s-1}}=\frac{\binom{n-1-(r+s-1)}{k-1-(r-1)}\binom{r+s-1}{r-1}}{\binom{n-1}{k-1}}.$$
In the special case that $r=s=1$ the r.v.s $X$ and $Y$ are geometrically distributed and the above conditional distribution is the uniform distribution on $\{1,2,\ldots,n-1\}$.
The second equality hints at a hypergeometric distribution where now for each $k$ with $n\geq k\geq r$ there is another distribution. I would have expected an beta-binomial distribution or something similar.
Is there any intuition behind the result?
$Z$ is the number of independent trials needed to obtain $r+s$ succeses, where each trial succeeds with probability $p$. The event $\{Z=n\}$ means that trial $n$ was a success, and we have to select $r+s-1$ of the first $n-1$ trials to also be successes. That explains the denominator ${n-1 \choose r+s-1}$. The numerator represents the event $\{X=k, \, Z=n\}$, since for this event we need to select $r-1$ successes among the first $k-1$ trials, and then find an additional $s-1$ successes among trials $k+1,\ldots, n-1$. In computing actual probabilities of these events there is an additional factor of $p^{r+s}(1-p)^{n-r-s}$ that appears both in the numerator and the denominator (so it cancels), since there are $r+s$ successes and $n-r-s$ failures in the first $n$ trials.