Avoidance recurrence for two consecutive successes ($p_n=(1-q)p_{n-1}+q(1-q)p_{n-2},p_0=p_1=1$)

For independent and identically distributed trials with constant success probability $0<q<1$, let $p_n$ be the probability of avoiding two consecutive successes in the first $n$ trials. An acceptable sequence starts either with a failure, leaving $n-1$ unrestricted fresh trials to check, or with a success followed by a failure, leaving $n-2$. These disjoint cases prove the recurrence for $n\ge 2$.

Its characteristic roots areThey are distinct, and the initial conditions yieldIf $T$ is the completion time of the first consecutive pair, then $p_n=\mathbb{P}(T>n)$. The recurrence describes the whole tail distribution rather than only its mean.