Modelling bacterial extinction

This week’s Riddler Classic poses a question about bacteria (paraphrased for brevity):

Each bacterium in a colony splits into two copies with probability $p$ and dies with probability $(1-p)$. If the colony starts with one bacterium, what is the probability that the colony survives forever?

We can model the colony’s size as a Galton-Watson process. Let $X_t$ be the colony’s size in generation $t\in \{1,2,\dots \}$ and let $Y_{it}$ be the number of offspring generated by bacterium $i\in \{1,2,\dots ,X_t\}$. The $Y_{it}$ are independently and identically distributed, with $$Pr(Y_{it}=y)=\{\begin{array}{cc}p& \text{if}y=2\\ 1-p& \text{if}y=0\\ 0& \text{otherwise}\end{array}$$ for each $i$ and $t$. The colony’s size grows according to $$X_{t+1}=\sum _{i=1}^{X_t}{Y}_{it}$$ with $X_1=1$. Our goal is to compute $$\underset{t\to \infty }{lim}Pr(X_t>0)=1-\underset{t\to \infty }{lim}{q}_t,$$ where $q_t\equiv Pr(X_t=0)$ is the probability that the colony is extinct by generation $t$.

We can compute $q_t$ by conditioning on $Y_{11}$, the number of offspring generated by the first bacterium. If $Y_{11}=0$ then the colony is extinct from the second generation onwards. However, if $Y_{11}=2$ then there are two sub-colonies in the second generation that must be extinct in $(t-1)$ generations if the whole colony is extinct by generation $t$. These sub-colonies grow independently, so the probability that both are extinct in $(t-1)$ generations is $q_{t-1}^2$. Thus, by the law of total probability, we have $$\begin{array}{cc}{q}_t& =Pr(X_t=0|Y_{11}=0)Pr(Y_{11}=0)+Pr(X_t=0|Y_{11}=2)Pr(Y_{11}=2)\\ & =1\times (1-p)+q_{t-1}^2\times p\\ & =1-p+p{q}_{t-1}^2\end{array}$$ for $t\ge 2$. Defining $q\equiv {lim}_{t\to \infty }{q}_t$ and taking limits as $t\to \infty$ gives¹ $$q=1-p+p{q}^2,$$ which has solutions $$\begin{array}{cc}q& =\frac{1\pm \sqrt{1-4p(1-p)}}{2p}\\ & =\frac{1\pm \sqrt{(2p-1{)}^2}}{2p}\\ & =\frac{1\pm |2p-1|}{2p}.\end{array}$$ The larger solution exceeds unity when $p<0.5$, which we cannot have because $q$ is a probability. Thus $$\underset{t\to \infty }{lim}Pr(X_t>0)=1-\frac{1-|2p-1|}{2p}.$$ For example, if $p=0.8$ then the colony survives forever with probability $$1-\frac{1-|2\times 0.8-1|}{2\times 0.8}=\mathrm{0.75.}$$ If $p<0.5$ then extinction is guaranteed because each bacterium generates fewer than one offspring on average.