A single amoeba sits in a petri dish. At each time step it independently either dies (probability 1/3), remains as one amoeba (probability 1/3), or splits into two amoebae (probability 1/3). What is the probability that the population eventually goes extinct?
The intuitive answer is "less than 1" — the amoeba has a 1/3 chance of splitting, which seems like it should guarantee survival with positive probability. The correct answer is 1: extinction is certain.
The Proof
Let p be the probability of eventual extinction starting from a single amoeba. Conditioning on the first event:
p = (1/3)(1) + (1/3)(p) + (1/3)(p²) The three terms: (1/3)(1) — amoeba dies immediately → extinct (1/3)(p) — amoeba stays → one amoeba, extinction probability p (1/3)(p²) — amoeba splits → two independent amoebae, both must go extinct
Solving 3p = 3 + 3p + p² → wait, let me be precise:
p = 1/3 + p/3 + p²/3 3p = 1 + p + p² p² − 2p + 1 = 0 (p − 1)² = 0 p = 1
The quadratic has a unique solution: p = 1. Extinction is certain. Despite having a positive probability of growth at every step, the population cannot sustain itself indefinitely because the expected offspring count (mean = 1) is not sufficient — what matters is the fixed-point structure of the generating function, which yields 1 as its only solution in [0,1] when the mean offspring count is ≤ 1.
The General Galton-Watson Result
This is a Galton-Watson branching process. The extinction probability is the smallest non-negative fixed point of the offspring generating function G(s). When E[offspring] ≤ 1, the fixed point at s=1 is the only solution in [0,1], so extinction is certain. When E[offspring] > 1, a second fixed point exists in (0,1), giving a positive survival probability.
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