Amoeba Extinction: Die, Stay, Split in Two, or Split in Three
A single amoeba sits in a dish. Every minute, each amoeba alive does exactly one of four things, each with probability $1/4$ and independently of everything else:
- it dies,
- it stays as it is (one amoeba),
- it splits into two amoebas,
- it splits into three amoebas.
Every descendant behaves the same way in subsequent minutes. Starting from this one amoeba, what is the probability that the population eventually dies out completely?
Hints
- Let $p$ be the extinction probability starting from a single amoeba. Condition on what happens in the first minute.
- If the amoeba splits into $k$ independent amoebas, the whole population dies out iff each of the $k$ independent lines dies out, with probability $p^k$.
- $p = \tfrac{1}{4}(1 + p + p^2 + p^3)$. One root is $p = 1$; factor it out and take the root in $(0, 1)$.
Worked Solution
How to Think About It: Extinction of the whole family is a self-similar event: after the first minute you have $k$ independent amoebas, each of which starts an identical copy of the original problem. That gives a fixed-point equation for the extinction probability. The subtlety is that $p = 1$ always solves the equation, so you must argue which root is the true answer.
Quick Estimate: With probability $1/4$ the line dies immediately, so $p > 0.25$. With probability $1/4$ it stays put and you are back where you started, so effectively the relevant branches are die / split-2 / split-3 with weights $1/3$ each: $p \approx \tfrac{1}{3}(1 + p^2 + p^3)$, whose root near $0.41$ is already close to the exact $0.414$.
Formal Solution:
*Step 1 -- First-step analysis.* Let $p$ be the extinction probability starting from one amoeba. Because the descendants of different amoebas evolve independently, extinction from $k$ amoebas has probability $p^k$. Conditioning on the first minute:
$$p = \frac{1}{4}\cdot 1 + \frac{1}{4}\, p + \frac{1}{4}\, p^2 + \frac{1}{4}\, p^3.$$
*Step 2 -- Solve.* Multiply by 4 and rearrange: $p^3 + p^2 - 3p + 1 = 0$. Since $p = 1$ is a root, factor:
$$(p - 1)(p^2 + 2p - 1) = 0 \quad\Longrightarrow\quad p = 1 \ \text{ or } \ p = -1 \pm \sqrt{2}.$$
The candidates in $[0, 1]$ are $p = 1$ and $p = \sqrt{2} - 1 \approx 0.4142$.
*Step 3 -- Choose the right root.* Let $g(s) = \tfrac{1}{4}(1 + s + s^2 + s^3)$ be the offspring generating function and $p_n$ the probability of extinction by minute $n$. Then $p_0 = 0$ and $p_{n+1} = g(p_n)$, so $p_n$ increases to the smallest fixed point of $g$ in $[0, 1]$, which is $\sqrt{2} - 1$. (Equivalently: the mean number of offspring is $1.5 > 1$, so the process is supercritical and the extinction probability is strictly less than 1.)
Answer: $P(\text{extinction}) = \sqrt{2} - 1 \approx 0.414$.
Intuition
This is a branching (Galton-Watson) process with mean offspring $(0 + 1 + 2 + 3)/4 = 1.5 > 1$, so the population is supercritical and survives with positive probability, yet extinction is far from negligible because the first amoeba can simply die on minute one (probability $1/4$). The extinction probability is the smallest fixed point of the offspring generating function, $p = \sqrt{2} - 1 \approx 0.414$; the root $p = 1$ always exists but is the wrong one whenever the mean exceeds 1. Branching-process fixed points show up in epidemic modeling, order-book cascade models, and any "will this self-replicating thing fizzle out" question.