🦠 Amoeba Extinction Probability
A branching-process visualization
1. What can one amoeba do?
Every minute, each amoeba independently produces
\(0,1,2,\) or \(3\) amoebas, each with probability
\(\frac14\).
💀
0 amoebas
\(P=\frac14\)
● ●
2 amoebas
\(P=\frac14\)
● ● ●
3 amoebas
\(P=\frac14\)
\[
E[N]
=
\frac{0+1+2+3}{4}
=
\frac32
\]
Since the expected number of descendants is
greater than 1, extinction is not certain.
2. Define the extinction probability
Let
\[
q = P(\text{the entire population eventually dies out})
\]
Look only at what happens to the first amoeba.
0 →
The population is already extinct:
\(1\)
1 →
The one descendant must eventually die:
\(q\)
2 →
Both independent descendant populations must die:
\(q^2\)
3 →
All three independent descendant populations must die:
\(q^3\)
Since each case has probability \(\frac14\),
\[
\boxed{
q=\frac14(1+q+q^2+q^3)
}
\]
3. Solve the fixed-point equation
\[
4q=1+q+q^2+q^3
\]
\[
q^3+q^2-3q+1=0
\]
\[
(q-1)(q^2+2q-1)=0
\]
Therefore,
\[
q=1,\qquad q=-1+\sqrt2,\qquad q=-1-\sqrt2.
\]
A probability must lie in \([0,1]\), so the two possible
fixed points are
\[
q=1
\qquad\text{or}\qquad
q=\sqrt2-1.
\]
For a Galton–Watson branching process, the extinction
probability is the smallest fixed point
of the offspring generating function in \([0,1]\).
Probability of eventual extinction
\[
\boxed{q=\sqrt2-1}
\]
\[
q\approx0.41421356
\]
≈ 41.42%
Hence the probability that the population survives forever is
\[
1-q=2-\sqrt2\approx58.58\%.
\]
4. Watch the probability converge
Start with \(q_0=0\), then repeatedly apply
\[
q_{n+1}
=
\frac{1+q_n+q_n^2+q_n^3}{4}.
\]
This converges to the smallest fixed point,
\(\sqrt2-1\).