We have that $G(s) = \displaystyle(\frac { p } { 1-(1-p)s})^r $ . The extinction probability is the smallest non-negative $\alpha$ such that $\alpha = G(\alpha)$, which is when: $$ \alpha = \displaystyle(\frac { p } { 1-(1-p)\alpha})^r $$
I am struggling to go from here as I can't find a closed form for $\alpha$. Should I not be using the "$\alpha =G(\alpha)$" approach for these types of distributions because of this issue? Is there a better way including splitting the distribution up into geometric distributions?