Waiting time for two consecutive successes ($\mathbb{E}T=(1+p)/p^2$)

For independent trials of success probability $p\in (0,1]$, let $T$ be the first index completing two consecutive successes. If $e_0,e_1$ are the mean remaining times with no trailing success and with one trailing success, conditioning gives $e_0=1+(1-p)e_0+p{e}_1$ and $e_1=1+(1-p)e_0$. Solving yields $e_0=(1+p)/p^2$, so fair trials have mean waiting time six. Disjoint pairs give a geometric tail bound and justify finiteness before solving the equations. Overlapping pairs are not independent trials.

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.