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Consider an amoeba. At each iteration, it can split into 2, 3 amoebas, stay the same, or die. These 4 events occur with equal probability.

Let $E$ denote the event that all current amoeba dies, and $F_1, F_2, F_3, F_4$ denote the above 4 events, then we have

\begin{align} P(E) = P(E|F_1)P(F_1) + P(E|F_2)P(F_2) + P(E|F_3)P(F_3) + P(E|F_4)P(F_4) \\ P(F_i) = \frac{1}{4} \\ P(E|F_1) = 1 \\ P(E|F_2) = P(E)^2 \\ P(E|F_3) = P(E)^3 \\ P(E|F_4) = P(E) \\ P(E) = \frac{1}{4}[1 + P(E) + P(E)^2 + P(E)^3] \\ \end{align}

Solving this cubic, we find $P(E) = 1, 1 \pm \sqrt{2}$. We throw out the negative root, and we are left with $P(E) = 1, \sqrt{2} - 1$. In the book I am reading, it restricts the probabilities to $P(E) < 1$.

But intuitively, I do not understand why $P(E) = 1$ isn't possible. Can someone intuitively explain this?

In addition, using $P(E) = \sqrt{2} - 1$, does this mean that the expectation of the number of iterations that the amoeba population dies out is infinity?


One thought that occurred to me is that the expected number of amoeba after one iteration is $0.25(1 + 2 + 3 + 0) = 1.5$, and by induction we see that this continues to grow.

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On average one amoeba produces $1.5$ the next round, so if there are a large number of them the probability of extinction is very small. This shows you can't have extinction with probability $1$.

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  • $\begingroup$ I just edited into the OP what I think is what you stated. If on average one amoeba produces $< 1$ in the next round, would that make probability of 1 feasible and any probability $< 1$ not feasible? $\endgroup$
    – David
    Commented Sep 10, 2020 at 5:06
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    $\begingroup$ @David: that is correct. They would die out with probability $1$ even though it is theoretically possible they could go on forever. $\endgroup$ Commented Sep 10, 2020 at 5:08
  • $\begingroup$ So in the current case, what would be the expected number of turns until all the amoeba population dies out? It seems it's unbounded, so it becomes infinity? $\endgroup$
    – David
    Commented Sep 10, 2020 at 5:09
  • $\begingroup$ Yes, the expected number is undefined because you expect it (sometimes) never to go extinct and you can't put that into the average. $\endgroup$ Commented Sep 10, 2020 at 5:10

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