Begin with one cell, which can die, do nothing, transform to 2 or 3 cells, with probability 1/4 respectively. How's the probability of extinction? A colony begins with a cell, which can die, do nothing, transform to two or three cells, with probability 1/4 for each case at next time point. Children cells share the same property described above. What's the probability of this colony's extinction?
I got two solutions, $1$ and $\sqrt{2}-1$, from a simple recursive equation, $p=\frac{1}{4}(1+p+p^2+p^3)$. But I've no idea which one is correct.
Thanks for your help!
 A: You've done most of the work; what remains is to decide whether $1$ or $\sqrt2-1$ is the probability of eventual extinction. For that purpose, let $x_n$ be the probability of the event that the population goes extinct at or before the $n$-th time step.  Notice that these events form an increasing sequence with respect to $\subseteq$, so the probability of their union, the event of eventual extinction, is the supremum (and also the limit) of the increasing sequence of numbers $x_n$.  The issue is therefore whether this sequence ever gets above $\sqrt2-1$.
We have $x_0=0$ (since the population is initially a single cell, not extinct), and
$$
x_{n+1}=\frac14(1+x_n+{x_n}^2+{x_n}^3)
$$
(because of the rules for how the cells multiply or die or do nothing).  Notice that this equation says $x_{n+1}=f(x_n)$, where $f(x)=\frac14(1+x+x^2+x^3)$ is the function whose fixed points you already calculated. In particular, you know that $f(\sqrt2-1)=\sqrt2-1$.  But $f(x)$ is clearly an increasing function of $x$ as long as $x\geq0$.  So, since $x_0<\sqrt2-1$, we get, by induction on $n$, that $x_n<\sqrt2-1$ for all $n$.  Therefore, the probability of eventual extinction is not $1$ but $\sqrt2-1$.
A: I think I can shed some light on the recursion identity.
Let's call $p$ the probability that the population starting from one cell dies out. We have 4 scenarios:
1) The cell dies with probability $1/4$, hence extinction (term $1/4$).
2) It does nothing with the same probability, hence postponing the question (term $p/4$).
3) It spawns another cell, hence we study the extinction of two populations (term $p^2/4$)
4) It spawns two cells, hence we study the extinction of three populations (term $p^3/4$)
Thus, $$p=\frac 14 (1+p+p^2+p^3),$$
which gives us the roots $1$, $-1\pm \sqrt 2$.
Now we need to chose the correct root between $1$ and $\sqrt 2-1$. My intuition suggests that thanks to the expectation to generate half a cell on each step rules out the root $p=1$ (i.e. extinction almost surely), but I might be wrong. So my money is on $p=\sqrt 2-1$, and I would be glad if someone could give a formal proof of this result (or prove me wrong, of course).
A: I have translated this into a birt/death-process. Often one can set the master equations and derive the macroscopic relations from the microscopic. As there is no non-linearity (no liason of the cells) the rarction rates correspond to your probability of $1/4$ and we get 
$$C \xrightarrow{k} D$$
$$C \xrightarrow{k} 2C$$
$$C \xrightarrow{k} 3C$$
$$C \xrightarrow{k} C$$
$$\frac{d C}{d t}=-k\,C+kC+k\,C+k\,C$$
$$\frac{d C}{d t}=2k\,C$$
Not sure if there is any threat to the colony to ever die out. $C$ in the equations represents the Expectation value (first moment) of the living cells over the population.
The above calculation holds if you start with a large population. If you start with one cell you will need to set up the master equations with the transition probailities and follow the progress of how your probability distribution changes over each time step. At the start it is $1/4$ then less and less. Larger the population gets less the probability that it dies out. Once large enough (law of large numbers) it will follow the above equations.
