Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 11F (Probability)

The number of heads has the binomial distribution. Choosing the $k$ head positions gives

Conditioned on exactly $k$ heads, every $k$-element set of head positions is equally likely. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such sets. A run containing all $k$ heads can begin at any of the positionsand each beginning determines one admissible set. Thus, for $k\ge 1$,

For the process to stop on toss $n$, the last toss must be the $k$th head, while the first $n-1$ tosses must contain exactly $k-1$ heads. Hence the negative binomial distribution gives

Let $E_j$ be the expected additional number of tosses when the current terminal run contains $j$ consecutive heads. Then $E_k=0$, and for $0\le j<k$,For $0<p<1$, iterating this recurrence from $j=k-1$ down to $0$ givesSolving,When $p=1$, the waiting time is deterministically $k$, which is also the continuous limit of this formula as $p\to 1$.