Amoeba Extinction: Die, Stay, Split in Two, or Split in Three

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:

Part (a): The extinction probability

*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.)

Part (b): The cubic, its three real roots, and which one to keep

*Step 1 -- Put it in standard form.* The fixed-point equation $s = g(s)$ with $g(s) = \tfrac{1}{4}(1 + s + s^2 + s^3)$ is

$$4s = 1 + s + s^2 + s^3 \quad\Longleftrightarrow\quad s^3 + s^2 - 3s + 1 = 0.$$

*Step 2 -- Factor it.* Every generating function satisfies $g(1) = 1$, so $s = 1$ is always a root and divides out:

$$s^3 + s^2 - 3s + 1 = (s - 1)(s^2 + 2s - 1).$$

The quadratic factor gives $s = -1 \pm \sqrt{2}$, so the three real roots are

$$s_1 = -1 - \sqrt{2} \approx -2.4142, \qquad s_2 = \sqrt{2} - 1 \approx 0.4142, \qquad s_3 = 1.$$

*Step 3 -- Confirm numerically in Python.*

```python import numpy as np

# s3 + s2 - 3s + 1 = 0, coefficients in descending powers roots = np.roots([1, 1, -3, 1]) print(np.sort(roots)) # [-2.41421356 0.41421356 1. ]

# keep the real roots that are valid probabilities, then take the smallest real = np.sort(roots[np.abs(roots.imag) < 1e-12].real) valid = [r for r in real if -1e-12 <= r <= 1 + 1e-12] print(valid) # the two roots in [0, 1]: ~0.41421356 and ~1.0 print(min(valid)) # ~0.41421356 = sqrt(2) - 1

# cross-check by iterating the generating function from p0 = 0 g = lambda s: (1 + s + s2 + s3) / 4 p = 0.0 for _ in range(200): p = g(p) print(p) # ~0.41421356, the same root, reached from below ```

A symbolic check agrees: sympy.factor(s3 + s2 - 3*s + 1) returns (s - 1)*(s**2 + 2*s - 1) and sympy.solve returns $\{1,\; -1 + \sqrt{2},\; -1 - \sqrt{2}\}$.

*Step 4 -- Which root is the answer.* Two of the three roots are ruled out immediately and the third needs an argument:

• $s_1 = -1 - \sqrt{2} \approx -2.414$ is negative, and a probability cannot be. It is an artifact of clearing denominators, not a candidate.
• $s_3 = 1$ is a root of *every* such equation, because $g$ is a probability generating function and $g(1) = \sum_k P(\text{offspring} = k) = 1$. So the presence of the root $1$ carries no information at all, and it must be tested, never assumed.
• $s_2 = \sqrt{2} - 1$ is the answer, because the extinction probability is by construction the smallest root in $[0, 1]$. Let $p_n = P(\text{extinct by minute } n)$. Then $p_0 = 0$ and $p_{n+1} = g(p_n)$, and since $g$ is increasing on $[0,1]$ with $g(0) > 0$, the sequence $p_n$ is increasing and bounded above by $1$, so it converges to some fixed point $p^* = g(p^*)$. Induction gives $p_n \le q$ for every fixed point $q \in [0,1]$: it holds at $n = 0$, and if $p_n \le q$ then $p_{n+1} = g(p_n) \le g(q) = q$. Hence the limit is the smallest fixed point in $[0,1]$, namely $\sqrt{2} - 1$. The iteration in the code block above shows exactly this: starting at $0$ it climbs to $0.41421$ and stops, never reaching $1$.
• *When would $1$ be the right answer?* $g$ is convex on $[0,1]$ with $g(1) = 1$, so the graph of $g$ crosses the diagonal at most twice there, and $1$ is the smallest fixed point exactly when $g'(1) \le 1$. Since $g'(1) = \mu$, the mean number of offspring, the root $1$ is the extinction probability precisely when $\mu \le 1$. Here $\mu = 1.5 > 1$, so the curve dips below the diagonal before $1$, a second fixed point exists strictly inside $(0,1)$, and $1$ is rejected.

Part (c): Criticality

The offspring distribution puts mass $1/4$ on each of $0, 1, 2, 3$ children, so the mean offspring per amoeba per minute is

$$\mu = g'(1) = \frac{0 + 1 + 2 + 3}{4} = \frac{3}{2} = 1.5.$$

Since $\mu > 1$ the process is supercritical: extinction is possible but not certain, $P(\text{extinction}) = \sqrt{2} - 1 < 1$, and conditional on survival the population grows like $\mu^n = 1.5^n$. (For reference, $\mu < 1$ is subcritical and $\mu = 1$ is critical; in both of those cases extinction is certain, the critical case being the delicate one where the population dies out with probability $1$ despite a mean of exactly one child.) The survival probability is $1 - (\sqrt{2} - 1) = 2 - \sqrt{2} \approx 0.5858$.

Answer: (a) $P(\text{extinction}) = \sqrt{2} - 1 \approx 0.4142$. (b) The fixed-point equation is the cubic $s^3 + s^2 - 3s + 1 = 0 = (s - 1)(s^2 + 2s - 1)$, with the three real roots $-1 - \sqrt{2} \approx -2.4142$, $\sqrt{2} - 1 \approx 0.4142$ and $1$; numpy.roots([1, 1, -3, 1]) returns exactly these. The extinction probability is $\sqrt{2} - 1$, the smallest root in $[0, 1]$, because $p_n = P(\text{extinct by minute } n)$ starts at $p_0 = 0$ and increases to the smallest fixed point of $g$; the negative root is not a probability, and the root $1$ is present for every branching process (since $g(1) = 1$) and is the answer only when the mean offspring satisfies $\mu \le 1$. (c) $\mu = 1.5 > 1$, so the process is supercritical and survives with probability $2 - \sqrt{2} \approx 0.5858$.