Calculating the extinction probability I am trying to solve the following problem. In a branching process the number offspring per individual has a binomial distribution with parameters 2, p. Starting with a single individual, calculate the extinction probability. 
I believe the solution to such a problem is evaluated using the equation $z=P(z)$
where of course $p(z)$ is pgf of the size of the nth generation.
$$z = (p(z-1) + 1)^n$$ 
Due to lack of enough appropriate examples i am unsure how to proceed from here. I believe due to the convolution of each generation depending on the last we have a recursive equation. 
So would it be correct to solve for $z_2$ using
$z_2 = (p(z_1-1) + 1)^2$ and $z_1 = (p(z_0-1) + 1)^1$.
 A: Here's a more direct solution: You have two attempts not to go extinct. Each succeeds if a) a descendant is produced with probability $p$ and b) that descendant's branch survives with probability $q$. So your survival probability $q$ must satisfy
$$q=(pq)^2+2pq(1-pq)=pq(2-pq)\;.$$
One solution is $q=0$, the other is $q=(2p-1)/p^2$. The crossover occurs when the two solutions coincide, i.e. at $p=1/2$. For $p\le1/2$, the survival probability is $0$ (which makes sense, since in that case the expected number of descendants is $\le1$), whereas for $p\gt1/2$ it is $(2p-1)p^2$, so the extinction probability is $1-(2p-1)/p^2=(p^2-2p+1)/p^2=((1-p)/p)^2$.
A: $$
\begin{align}
r &= \Pr(\text{survival}) \\[6pt]
& = \Pr((0\text{ offspring & survival})\text{ or }(1\text{ offspring & survival})\text{ or }(2\text{ offspring & survival})) \\[6pt]
& = \Pr(0\text{ offspring & survival}) + \Pr(1\text{ offspring & survival}) + \Pr(2\text{ offspring & survival}) \\[6pt]
& = 0 + \Pr(1\text{ offspring})\Pr(\text{survival}\mid 1\text{ offspring}) + \Pr(2\text{ offspring})\Pr(\text{survival}\mid 2\text{ offspring}) \\[6pt]
& = 0 + 2p(1-p)r + p^2 (1-\Pr(\text{extinction}\mid 2\text{ offspring})) \\[6pt]
& = 2p(1-p)r + p^2 (1 - \Pr(\text{both lines die out})) \\[6pt]
& = 2p(1-p)r + p^2 (1 - (1-r)^2).
\end{align}
$$
So we have a quadratic equation in $r$:
$$
r = 2p(1-p)r + p^2 (1 - (1-r)^2)
$$
The two solutions are $r=0$ and $r=\dfrac{2p-1}{p^2}$.
The second one is negative if $p<1/2$, so the probability in that case must be the first solution.  If $p=1/2$ then the two solutions are $0$.  If $p>1/2$, then must the solution be the second one?  It would be enough to show $r>0$ in those cases.  Obviously $r=1$ when $p=1$.
