Consider the preliminary question of getting a sequence of $$N$$ heads out of $$k$$ throws, with probability $$1-p(N,k)$$. This is given by the recurrence formula $$p(N,k) = \begin{cases} 1 &\text{if } k Indeed, my reasoning is that no consecutive $$N$$ heads out of $$k$$ draws can be decomposed according to the first occurrence of a tail out of the first $$N$$ throws. Conditioning on whether this first tail occurs at the first, second, ..., $$N$$th draw leads to this recurrence relation.
Next, the probability of getting the first consecutive N heads in $$m\ge N$$ throws is $$q(N,m) =\begin{cases} \dfrac{1}{2^N} &\text{if }m=N\\ \dfrac{1}{2^N}\dfrac{1}{2} &\text{if } N The first case is self-explanatory. the second case corresponds to a tail occuring at the $$m-N-1$$th draw, followed by $$N$$ heads, and the last case prohibits $$N$$ consecutive heads prior to the $$m-N-1$$th draw. (The two last cases could be condensed into one, granted!)
Now, the probability to get $$M$$ heads first and the first consecutive $$N$$ heads in exactly $$m\ge N$$ throws (and no less) is $$r(M,N,m) = \begin{cases} \dfrac{1}{2^N} &\text{if }m=N\\ 0 &\text{if } N Hence the conditional probability of waiting $$m$$ steps to get $$N$$ consecutive heads given the first $$M$$ consecutive heads is $$s(M,N,m) = \begin{cases} \dfrac{1}{2^{N-M}} &\text{if }m=N\\ 0 &\text{if } N The expected number of draws can then be derived by $$\mathfrak{E}(M,N)= \sum_{m=N}^\infty m\, s(M,N,m)$$ or $$\mathfrak{E}(M,N)-M$$ for the number of additional steps...