I am considering a standard extinction problem, that is:
A bacterial colony consists of individual bacteria. One of the following happens with each bacterium each second:
- The bacterium dies.
- The bacterium remains without a change.
- The bacterium splits in two bacteria.
- The bacterium splits in three bacteria.
Each of the above happens with equal probability $\frac{1}{4}$. Estimate probability that the colony that initially consist of one bacterium will never die.
There is a well documented approach for solving this, e.g. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch14.pdf, http://math.uchicago.edu/~may/REU2015/REUPapers/Csernica.pdf. This approach is working with probability that the colony has died by timestep $n$.
However there is a slightly variant of this approach that has to be wrong for a reason I will explain below.
That is, let $p=P(\text{colony of 1 bacteria dies at any future timestep})$ (as opposed to "dies by timstep n"). Then we can write
\begin{equation} p = \frac{1}{4} + \frac{1}{4} p + \frac{1}{4} p^2+ \frac{1}{4} p^3 \end{equation}
as colony of one bacteria can either instantly die, then remain unchanged, then can split into two bacteria and then they would both have to die independently, etc.
Then we could solve that for $p$, but we would, as per usual solution, get multiple positive roots $p_1 = \sqrt{2} - 1$, $p_2 = 1$. The above doesn't explain which root to choose.
So clearly we actually we cannot write the above equation. Why?
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}x^2+\frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=\frac{1}{4}+\frac{1}{4}p+\frac{1}{4}p^2+\frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=\sqrt{2}-1$. Therefore $p=\sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.