Survivor distribution following a zombie outbreak

Solution 1:

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.

Solution 2:

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.