$1/2$ or $1$? probability that all bacteria will die

Suppose there is a bacterium in a bottle, it has $\frac{1}{3}$ chance to die and it has $\frac{2}{3}$ chance to split into 2 individuals, and the new individuals will follow this rule and so on. So here is the question, what is the probability that all bacteria are dead in the bottle?

Denote by p the probability that all the bacteria are dead. $$ p =\frac{1}{3}+\frac{2}{3}p^2$$ and it gives that $p = 0.5$ or $1$, so what is the next step? Which one is the answer? thanks.

Solution 1:

Another way to write this is as $$ p_{k}=\frac{1}{3}+\frac{2}{3}p_{k-1}^2, $$ or $$ p_{k}-p_{k-1}=\frac{2}{3}p_{k-1}^2-p_{k-1}+\frac{1}{3}=\frac{2}{3}\left(p_{k-1}-1\right)\left(p_{k}-\frac{1}{2}\right), $$ where $p_{k}$ is the probability that a bacteria and all its descendants are dead after $k$ generations. So if $p_{k}\in[0,1/2)$, $p_{k+1}>p_{k}$ (that is, the probability increases with each additional generation), while if $p_{k}\in(1/2,1)$, $p_{k+1}< p_{k}$ (the probability decreases). From this you can see that there are exactly two fixed points, at $p=1/2$ and $p=1$, and that the fixed point at $p=1/2$ is the attractive one.

Solution 2:

This is a supercritical Galton-Watson process (supercritical meaning each individual has an average of more than one offspring). It's a classical result that such a process survives with positive probability, so your p cannot be 1 and must be 1/2. You can probably find a proof in many places; I know it's in Durrett's PTE.