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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?

Mathsguy123
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  • You can get properly sized parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. – joriki Feb 18 '20 at 22:40

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