Will they survive?

A single amoeba sits in a pond. Every minute, it does one of four things with equal probability:

1. Dies (probability 1/4)
2. Stays the same (probability 1/4)
3. Splits into 2 (probability 1/4)
4. Splits into 3 (probability 1/4)

Each offspring behaves the same way, independently. What's the probability the entire population eventually dies out?

One generation at a time

Before any equations, let's see what a single amoeba faces each minute. Click to watch the three possible fates:

Each outcome has the same probability — 1/3 in this simplified visualizer. The real model has four outcomes (die, stay, split into 2, split into 3) each with probability 1/4.

Tracking a full lineage

Now let's follow an entire amoeba family through multiple generations. Start with 1 amoeba and step forward:

Run several lineages. Some die quickly, some explode. The question isn't about any single trial — it's about the probability of eventual extinction across all possible futures.

Batch simulation

Now let's run hundreds of amoeba populations and see what fraction go extinct:

After enough trials, the extinction fraction converges to about 41%.

Watching the branching unfold

Step through a single branching process to see the tree structure:

Notice how extinction requires every branch to die off. Even one surviving lineage keeps the population alive.

Setting up the equation

Let $P(E)$ be the probability the population eventually dies out. Using the law of total probability, we condition on what happens in the first minute:

$P(E) = P(E \mid F_0) \cdot P(F_0) + P(E \mid F_1) \cdot P(F_1) + P(E \mid F_2) \cdot P(F_2) + P(E \mid F_3) \cdot P(F_3)$

where $F_k$ is the event that the amoeba produces $k$ offspring.

Solving the cubic

Rearranging $P(E) = \frac{1}{4}\bigl(1 + P(E) + P(E)^2 + P(E)^3\bigr)$:

$4P(E) = 1 + P(E) + P(E)^2 + P(E)^3$

$P(E)^3 + P(E)^2 - 3P(E) + 1 = 0$

We can factor this cubic. Notice $P(E) = 1$ is always a root (every branching process could go extinct with probability 1 if the mean offspring were $\leq 1$):

$(P - 1)(P^2 + 2P - 1) = 0$

Solving $P^2 + 2P - 1 = 0$ by the quadratic formula:

$P = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$

The three roots are:

Graphical interpretation

The extinction probability is where the curve $y = f(P) = \frac{1}{4}(1 + P + P^2 + P^3)$ intersects the line $y = P$. The smallest positive intersection gives the answer:

Why P = 1 isn't the answer

Since our cubic has $P = 1$ as a root, why don't we use it? In branching process theory, the extinction probability is the smallest nonnegative fixed point of the generating function $G(s) = \sum_k p_k s^k$. When the mean offspring $\mu > 1$, this smallest root is strictly less than 1. The root $P = 1$ corresponds to the "trivial" solution that always exists; the root $P = \sqrt{2} - 1$ is the actual extinction probability.

Competition appearances

• Quant finance interviews: Branching process extinction is a common Goldman Sachs / Two Sigma question
• IMO training: Recursive probability equations solved via polynomial roots
• Programming challenges: Simulating branching processes to verify theoretical extinction probabilities
• Actuarial exams: Branching processes model insurance claim cascades and epidemic spread

Takeaway

The amoeba problem is a beautiful application of conditional probability and self-similarity. By conditioning on the first generation, we convert the question into a polynomial equation. The extinction probability $\sqrt{2} - 1$ is the smallest nonnegative root. The technique generalizes to any branching process: write $P = G(P)$ where $G$ is the offspring generating function, and find the smallest root.