Will the amoebas die out?

Idea. Look one minute ahead. After the first minute there are 0, 1, 2 or 3 amoebas, each starting an independent copy of the same process. That gives an equation for the extinction probability $q$.

Step 1: the equation. A colony of $k$ amoebas dies out when each of the $k$ independent families dies out, which has probability $q^k$. Conditioning on the first minute, $$q=\frac{1}{4}\left(1+q+q^2+q^3\right).$$

Step 2: solve. This is $q^3+q^2-3q+1=0$. Since $q=1$ is a root, it factors as $$(q-1)(q^2+2q-1)=0,$$ with roots $1$ and $-1\pm \sqrt{2}$. The candidates in $[0,1]$ are $1$ and $\sqrt{2}-1\approx 0.414$.

Step 3: which root. Let $q_n$ be the probability that the colony is extinct by minute $n$, and $f(s)=\frac{1}{4}(1+s+s^2+s^3)$. Then $q_0=0$ and $q_{n+1}=f(q_n)$. Since $f$ is increasing, if $r\ge 0$ is any root of $f(r)=r$, then $q_n\le r$ implies $q_{n+1}=f(q_n)\le f(r)=r$. So every $q_n$ stays below the smallest root, and $q_n$ increases to it. The extinction probability is the smallest root: $$q=\sqrt{2}-1\approx \mathrm{0.414.}$$

A check. Each amoeba leaves on average $\frac{0+1+2+3}{4}=\frac{3}{2}$ amoebas. When this mean is above 1 the colony survives with positive probability, so the answer must be below 1. When the mean is at most 1 (except in the trivial case where every amoeba always stays as one), extinction is certain.