Looking for a variation on negative binomial where one trial can give multiple successes I am trying to model a game, where individuals have a number of "hit points", and I know the probability distribution of how many hits they will take in a given round. I want to know (the distribution of) how many rounds it will take them to die.
This seems to me like it is a variation on a negative binomial distribution, where, rather than each trial having a probability of success, I have a number of successes. But I've not been able to find any discussion of this case based on the web searches I've done.
One option I did think of is to model this as an absorbing Markov chain, where the "number of hits remaining" is the state, and the probability of moving from state N to state N-i is given by the "number of hits" distribution for i (modified to make 0 an absorbing state, which is straightforward). I'm pretty sure that would give the answer I'm looking for, but it feels like a rather complex solution to the problem, and I don't know if I've missed a simpler solution using a standard distribution.
 A: Let $n$ be the initial number of hit points. Let $p_j$ be the probability of taking $j\leq n$ hits.
Let $\mathbf x^{(k)}=\begin{bmatrix} x_n{(k)};& \cdots & x_1{(k)};& x_0{(k)} \end{bmatrix}$ be your state variable at round $k$, with $x_l^{(k)}$ the probability of having $l$ hit points at the end of round $k$. $\mathbf x^{(0)}=\begin{bmatrix} 1;& 0 & \cdots & 0 \end{bmatrix}$.
The evolution of your state is given by:
$$ \mathbf x^{(k+1)}=\underbrace{\begin{bmatrix} p_0 & p_1 & p_2 & \cdots & p_n\\ 0 & p_0 & p_1 & \cdots & p_n+p_{n-1} \\ 0 & 0 & p_0 & \cdots & p_n+p_{n-1} +p_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}}_{P}\mathbf x^{(k)}$$
Call $f_k$ to the probability mass function for the number of rounds of the game. Play with $k=1,2,3,\dots$ and look for the pattern:
$$\begin{align}
f_1 &= p_n \\
f_2 &= p_0 p_n + p_1(p_{n-1}+p_n) + \dots + p_{n-1}(p_1+p_2+\dots+p_n)\\
\vdots
\end{align} $$
Writing $P$ as $\begin{bmatrix} T& \mathbf t_0 \\ \mathbf 0^T & 1 \end{bmatrix}$, one arrives at the expression from wikipedia:
$$f_k = \mathbf x^{(0)T} T^{k-1}\mathbf t_0 $$
They present the expected value as (I didn't check this last one):
$$ \mathbb E[K]= \mathbf x^{(0)T} (I-T)^{-1}\mathbf 1 $$

As a check, I computed $\mathbb E[K]$ for small $n$ to empirically find:
$$ \mathbb E[K] \stackrel{?}{=} \frac{\sum_{j=0}^{n-1} (-1)^j{n-1 \choose j}p_0^j}{\sum_{j=0}^n (-1)^j{n \choose j}p_0^j } $$
Which clearly indicates I took the long path, as you suspected in your question. Unfortunately, now I don't have the time to rethink the problem. So I leave you with what I have done so far. On my side, I had fun and learned some things on the way. Hope it helps.
