Survivor distribution following a zombie outbreak Suppose there is initially a population of $A$ humans and $B$ zombies. You are sitting on a nearby hill with a shotgun; however, you're not a very good shot, so each time you pull the trigger, one of the members of the population is hit uniformly at random.


*

*If you hit a zombie, the zombie dies.

*If you hit a human, the zombies pile onto the poor wounded human and turn them into a new zombie.
Undeterred by your lack of skill, you continue to take shots into the population until there are no more zombies. At this point, what can be said about the distribution of the number of human survivors?
For example, we can try to compute the probability $q(A,B)$ that there are no survivors at the end of the process. This quantity should satisfy the recurrence $$q(A,B) = \frac{A}{A+B} q(A-1, B+1) + \frac{B}{A+B} q(A, B-1)$$ with the initial conditions $q(0,k) = 1$ for $k\ge 0$ and $q(k,0) = 0$ for $k>0$. Is there a simple formula for $q(A,B)$ or reasonable asymptotics? What can be said about the probability $q(k,A,B)$ that there are $k$ survivors for $1\le k \le A$?
Also, looking at some data seems to suggest that we always have a "reverse unimodal" property $$q(0,A,B) > q(1,A,B) > \cdots < q(A-1, A, B) < q(A, A, B)$$ with $q(0,A, B) > q(A, A, B)$, i.e. it is likeliest to have zero survivors. Is it possible to show this in general?
 A: Possible approach / too long for a comment
Your recurrence for no survivors can be easily generalized to a recurrence for $k$ survivors.  Let $q(k,A,B)$ be the probability of ending with $k$ survivors starting with $A$ humans and $B$ zombies.  Then:
$$q(k,A,B) = \frac{A}{A+B}\, q(k,A-1, B+1) + \frac{B}{A+B}\, q(k,A, B-1)$$
In other words, the "form" of the equation is identical regardless of $k$, and that's because the recurrence is ultimately just based on the Law of Total Probability, applied to the event "$k$ survivors".
The boundary conditions however depend on $k$:
$$
q(k, A, 0) = \left. \begin{cases}
1, & k = A\\
0, & k \ne A
\end{cases}\right\|
 = \mathbb{I}_{\{k=A\}}.
$$
IMHO there is no real need to specify any boundary cases when $B>0$.  By definition you will keep shooting until $B=0$ anyway, and therefore reach some $q(k, A,0)$ state.  (Of course, for actual numerical calculations, it makes sense to specify $q(k, A, B) = 0$ whenever $A < k$ just to save computations.)
Wild guess: Since the "form" of the recurrence does not depend on $k$, and the boundary conditions contain only a single dot with non-zero value, I feel vaguely hopeful that there might be further simplifications available, e.g. by "back propagating" out from that single non-zero dot.  However, that's just a very vague gut feel, and even if it pans out, I would be surprised if the solution isn't a complicated summation and/or product form.
A: Too long for a comment. I think I can prove the following rather technical claim about the behavior of the zombie apocalypse started with large population.
Fix $\alpha, \beta > 0$. Then starting from the initial condition $(\lfloor \alpha n \rfloor, \lfloor \beta n \rfloor)$, denote by $(A_n, B_n)$ the pair of numbers of survivors/zombies after the $n$-th shot. If $T$ is the first time $n$ at which $B_n = 0$, then we consider the path $(Z_t)_{0 \leq t \leq T/n}$ defined by
$$ Z_t = \tfrac{1 - (nt - \lfloor nt \rfloor)}{n}(A_{\lfloor nt \rfloor}, B_{\lfloor nt \rfloor}) + \tfrac{nt - \lfloor nt \rfloor}{n}(A_{\lfloor nt \rfloor+1}, B_{\lfloor nt \rfloor+1}). $$
In other words, $(Z_t)$ is the piecewise linear path joining points $\frac{1}{n}(A_k, B_k)$'s and the speed is chosen so that each line segment from $\frac{1}{n}(A_k, B_k)$ to $\frac{1}{n}(A_{k+1}, B_{k+1})$ is traced at a uniform speed within the time interval $[k/n, (k+1)/n]$. Then

Technical Claim. As $n\to\infty$, $(Z_t)_{0 \leq t \leq T/n}$ converges in distribution to a deterministic curve $\gamma = (\gamma(t))_{0 \leq t \leq 2\alpha+\beta}$. Moreover, $\gamma(t) = (x(t), y(t))$ satisfies
  $$ x' = -\frac{x}{x+y}, \qquad y' = \frac{x-y}{x+y}, \qquad (x(0), y(0)) = (\alpha, \beta). $$
  Here, two paths on $\mathbb{R}^2$ are close if their graphs in $\mathbb{R}^3$ are close in Hausdorff distance.

Loosely speaking, the piecewise linear path joining $(A_n, B_n)$'s are close to the curve
$$ \frac{y}{x} + \log x = \text{constant}. $$
The following plot demonstrates this phenomenon in the case of $(a, b) = (1000, 500)$. The black line represents the curve $y/x + \log x = \text{const}$, where the constant is chosen so that it passes through $(a, b)$. Also, the colored zigzag curves represent 10 simulations of the history of zombie apocalypse.

Now, doing the same simulation with $(a, b) = (10000, 5000)$,

We clearly see concentration behavior emerges.
