Amoeba's Survival

Amoeba survival

A population starts with a single amoeba. For this one and for the generations thereafter, there is a probability of 3/4 that an individual amoeba will split to create two amoebas, and a 1/4 probability that it will die out without producing offspring. What is the probability that the family tree of the original amoeba will go on forever?

The probability in question is 2/3. You can check the solution.

Solution 1

Good and general notations help solve the problem. Let p be the probability of a successful split for a single amoeba, and P the probability in question, the probability that an amoeba's family tree is infinite.

With the probability p we have a second generation of two amoebas. The probability that at least one of them will have an infinite family tree is 1 - (1 - P)², because (1 - P)² is the probability that both of them will perish undivided. Therefore,

P = p(1 - (1 - P)²)

because both sides of the equation represent the probability of the long term survival to the original amoeba.

Simipification yields

pP² + (1 - 2p)P = 0,

P·(pP + (1 - 2p)) = 0.

P may be 0 or, otherwise, satisfy the equation.

pP + (1 - 2p) = 0,

P = (2p - 1)/p.

By expermintal verification, if p > 1/2, there's a true population explosion that suggests that, in this case, P≠0. The formula we obtained also confirms that, if a generic amoeba divides with the probability not exceeding 1/2, it stands no chance to survive forever. However, for the specific case where p = 3/4, the probability of survival P = 2/3.

Remark

There's a nagging question whether relying on experimental results is a valid argument against $P=0.$ Let's modify the above a little. Let $P_n$ be the probability that the $n^{th}$ generation of amoebas is not empty. Then we have a recurrent relation:

$P_{n+1}=p\left(1-(1-P_n)^2\right).$

We shall prove by induction that $\displaystyle P_{n}\gt \frac{2p-1}{p}.$ First off, $P_0=1,$ as we a given that an amoeba is present at the outset. Let $\displaystyle P_{n}\gt \frac{2p-1}{p}.$ We proceed in several steps:

$\displaystyle \begin{align} &P_{n}\gt \frac{2p-1}{p}\\ &1-P_n\lt 1-\frac{2p-1}{p}=\frac{1-p}{p}\\ &(1-P_n)^2\lt \left(\frac{1-p}{p}\right)^2\\ &1-(1-P_n)^2\gt 1-\left(\frac{1-p}{p}\right)^2=\frac{2p-1}{p^2}\\ &P_{n+1}=p\left(1-(1-P_n)^2\right)\gt \frac{2p-1}{p}, \end{align}$

as promised.

Solution 2

(Chritopher D. Long)

This is a disguised biased random walk with an absorbing state at $0$ which starts at $1.$ If this random walk is at $N,$ considering $N$ moves corresponds to considering the fate of $N$ amoebas simultaneously. This yields $\displaystyle 1-\frac{1-p}{p} = 1-\frac{1/4}{3/4} = \frac{2}{3}$ by the usual arguments.

References

R. Blum et al., Mathemagic, Main Street, 2002

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