Assume $p$ = probability a gene hit another gene. Such collision would create a new gene that could hit other genes. A collision could create either $3$ or $4$ new genes, each with probability $0.5$. Find maximum value of $p$ so that this branching process would be ensured to die out?
My attempt: Let $\pi_0 =$ probability that gene would eventually die out. Then we want $\pi_0=1$ is the solution to the equation: $\pi_0 = p + 0.5p(\pi_0^3+\pi_0^4)$, or $1 = p + p = 2p$. Thus $\fbox{$p=\frac{1}{2}$}$ is the max probability.
My question: Could someone please help verify if the above solution is correct?
Let $Z$ be the offspring distribution, then \begin{align} \mathbb P(Z=0) &= 1-p\\ \mathbb P(Z=3) &= \frac12 p\\ \mathbb P(Z=4) &= \frac12p.\\ \end{align} Hence $$\mu :=\mathbb E[Z] = \frac12p\cdot 3+\frac12p\cdot 4 = \frac72p. $$ The extinction probability is one precisely when $\mu\leqslant 1$, which is the case when $p\leqslant\frac27$.