= Solution
The <Galton-Watson extinction fixed point> equation is
$$
(1+2q)^3=27q,\qquad
8q^3+12q^2-21q+1=(q-1)(8q^2+20q-1)=0.
$$
Its smallest root in $[0,1]$ is $q=(3\sqrt3-5)/4$, not the other allowed fixed point $1$. Consequently the large-depth escape probability is
$$
\boxed{1-q=\frac{9-3\sqrt3}{4}\approx0.95096.}
$$
For finite $N$, the exact escape probability is $1-q_N$ and is slightly larger. Indeed $\phi'(q)=2-\sqrt3<1$, and the <mean value theorem> on $[q_N,q]$ gives $0\leq q-q_N\leq q(2-\sqrt3)^N$. This also quantifies how quickly the finite-depth answer approaches the displayed limit.
Back to article page
