Amoeba splitting probability 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.
 A: 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 $q={1\over3}$.
A: HINT: Let $p$ be the probability that the amoebas die. Then we have 
$$p = \frac{1}4 + \frac{3}4p^2.$$
A: 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.
