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In a laboratory, there is one amoeba. Each second, an amoeba dies with probability $1/4$ and splits itself into two with probability $3/4$. What is the probability that at least one amoeba remains in the laboratory forever?

In a state with $n$ amoebas, the distribution of amoebas in the next second is according to the binomial distribution. There is $0$ amoeba with probability $(1/4)^n$, two ameobas with probability $n(1/4)^{n-1}(3/4)$, ..., and $2n$ amoebas with probability $(3/4)^n$. The state with $0$ amoeba ends there, but we have to recurse on the remaining states.

user336268
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    This is a classic conditional probability problem. Let $E$ be the event that amoeba will remain forever, $F$ be the event that it dies, and $G$ be the event that it will split in two. Applying the law of total probability and you can derive the rest from here. – harvey Aug 16 '16 at 21:16
  • when it splits , it exists in 3 versions or else only one of them remains as the original ? –  Aug 16 '16 at 21:33
  • @igael It's quite obvious (to me, at least) that 'amoeba splits into two' means 'it transforms into two amoebas', not 'it produces two new amoebas'. So a single split increases the numer of amoebas by 1, not by 2. – CiaPan Aug 17 '16 at 11:24
  • @CiaPan: if fact, my question was bad worded, sorry. Are there '2 versions' labelized 'old amoeba', one being the original and the other just a clone or both 2 are the original ? edit : and sorry twice, i was focuzed on the hope of a particular amoeba. One as a pointer and not as a number ... –  Aug 17 '16 at 12:12
  • @igael I suppose the two amoebas resulting from a split are not distingushable in this problem (in real life, probably, too; except by their position on the table). What's important here is each of them has the same probability of a forthcoming death or split in the next second, and that their future devlopment is independent. – CiaPan Aug 17 '16 at 12:35

3 Answers3

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Answering @Did 's comment to Sandeep Silval's post I propose the following:

Denote by $q_n$ the probability that after $n$ seconds there are no live amoebas left, given that there is one live amoeba right now. Then $$q_0=0;\qquad q_n={1\over4}+{3\over4}q_{n-1}^2\quad(n\geq1)\ .$$ It is obvious that the $q_n$ form an increasing sequence bounded above by $1$, so that we are sure that the limit $\lim_{n\to\infty} q_n=:q$ exists. In this case $q={1\over4}+{3\over4}q^2$, which implies $q\in\bigl\{{1\over3},1\bigr\}$. This suggests writing $q_n:={1\over3}-x_n$, which then leads to the recursion $$x_0={1\over3},\qquad x_n=\left({1\over2}-{3\over 4}x_{n-1}\right)x_{n-1}\ .$$ It is easy to see that the $x_n$ converge "linearly" to $0$, hence $q={1\over3}$.

Christian Blatter
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HINT: Let $p$ be the probability that the amoebas die. Then we have

$$p = \frac{1}4 + \frac{3}4p^2.$$

Sandeep Silwal
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    Right -- but this lacks a justification why the root p=1 is not the probability of extinction. – Did Aug 16 '16 at 22:24
  • As a hint for why $p=1$ is not correct, consider the mean number of amoeba in generation $n$ (as a function of $n$). – Math1000 Aug 16 '16 at 22:32
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    $p=1$ means that it is certain that the population will inevitably crash into extinction; but under the assumptions given, then it is plausible that the population will endure to grow large enough to make an eventual crash uncertain. – Graham Kemp Aug 17 '16 at 00:16
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    if the probability of eventual extinction at a population of 1 is p(1), then I agree that P(1) = 1/4 + (3/4) P(2) - but how do you justify that P(2) = P(1) ^2? If talking about eventual, not immediate extinction – Cato Aug 17 '16 at 10:04
  • right I see it, you've got two branches, both of which have to then become extinct – Cato Aug 17 '16 at 10:06
  • I'm not sure that infinite mean number of amoebae over all generations proves that p != 0. If the probability of survival was 1/n, but otherwise the expected population is n^2, I think you can see that expected future population tends to infinity with probability of survival tending to zero also – Cato Aug 17 '16 at 11:00
  • @AndrewDeighton A two-amoebas population are two 1-amoeba populations, each of them evolving separately (they never unite, mix, coagulate or mate). So the extinction of the 2-a.p. is a coincidence of extinction of two independent 1-a.p. which implies multiplication of probabilities: $P(2)=P(1)\cdot P(1)$. – CiaPan Aug 17 '16 at 11:18
  • @CiaPan - thanks for that, I understand that now. What I don't understand now is whether there is any rigorous explanation for the solution 1 being able to be rejected. I believe it can be shown that a non-zero expected population for all future times can co-exist with a probability of extinction of 1, and even an expected population tending to infinity is consistent with extinction=1 (yes really it is, the probability of survival just has to approach zero more slowly than infinity) - so why do we get p=1, and how do we rule it out as a solution. – Cato Aug 17 '16 at 14:01
  • As it is, the equation developed is valid, and the probability we need to find has to satisfy that equation, if someone tells you the answer, it is nthen implied that the value satisfies the equation. unfortunately satisfying the equation doesn't imply that it is the value we need to find. For that reason, we need to rule out P=1, which looking around the internet, seems to have been approached in more illustrative ways than rigorous ways. – Cato Aug 17 '16 at 14:10
  • @AndrewDeighton: your comments where hidden when I began my post ... –  Aug 17 '16 at 14:12
  • @igael, it's all very interesting, thanks for your input, it took a while for me to understand the relevance of the equation – Cato Aug 18 '16 at 09:06
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The short answer, the probability of an infinite hope of life for these amoeba is 0.

The reason of this claim is based on the world eternal and on the fact that an infinite product of positive reals inferior to 1 has the limit 0

Now, at each step there is a small probability that all the population P extincts , $p_i = 4^{-P_i}$ which is small but strictly positive. Then, the opposite probability is always inferior to 1 and its infinite product is 0.