This has been asked several times on this website but I am trying to come up with a formal approach to justify why the extinction probability cannot be 1.
The problem statement is as follows. You start off with one amoeba in a petri dish. Every minute every amoeba in the dish either dies, does nothing, splits into two or splits into three. All these events have equal probability of $1/4$. What happens to amoebas are independent of each other. What is the probability that the amoebas eventually die out?
Let $A_n$ be the number of amoebas at minute $n$. $A_0 = 1$. Define $\tau := \inf\{n \geq 1: A_n = 0\}$ and $p := P\{\tau < \infty\}$. By conditioning on what happens to the first amoeba, we can write
$$p = \frac{1}{4}(1 + p + p^2 + p^3)$$ This has two positive solutions, namely $\sqrt{2}-1$ and $1$. Based on some intuitive reasoning, we rule out $1$ as the solution. I want to make that intuition formal. I can write
$$E[A_{n+1} \mid \mathcal{F_n}] = \frac{3}{2}A_n$$ $$E[A_n] = \left(\frac{3}{2}\right)^n$$
I don't see how these contradict $\tau < \infty$ a.s. $A_n$ is a submartingale. I tried stopping it at $\tau$ but I don't see where to go from there.
Let $q_n=\mathbb P(A_n=0)$, so $p=\lim_{n\to\infty}q_n$. We can show that the PGF $G_n$ for the distribution of $X_n$ is the $n$-fold composition $G_n(t)=G^{(n)}(t)$, where $G(t)=\left(1+t+t^2+t^3\right)/4$ is the PGF for $A_1$.
Now $q_{n+1}=\mathbb P(X_{n+1}=0)=G_{n+1}(0)=G(G_n(0))=G(q_n)$, so taking $n\to\infty$ implies $p=G(p)$.
Claim: $p$ is the smallest non-negative root of $t=G(t)$.
Indeed, suppose the smallest root is $r$. Then $q_0=0\leq r$. Now if $q_n\leq r$, we have $$q_{n+1}=G(q_n)\leq G(r)=r.$$ So $q_n\leq r$ always, hence $p=r$.