Consider an iid. sequence of $p=1/2$ Bernoulli distributed random variables $(X_n)_{n\in\mathbb{N}}$. We let $X=1$ be called 'heads' and $X=0$ be called 'tails'. Furthermore, for $X_n=1$ let $l_n=\max\{m\ge 0\colon X_i=1, n-(m-1)\le i \le n\}$ be the maximum number of consecutive heads at time $n$. For $X_n=-1$ set $l_n=0$.
Now I want to prove that $P(l_n\ge k)=2^{-k}$ and also want to determine for $L_n:=\max_{1\le m\le n}l_m$ the probability $P(L_n\ge j)$.
The first one should not be too hard, but I have problems obtaining the right result.
For $k>n$ the first probability is obviously $0$ and for $k\le n$, $$ \mathsf{P}(l_n\ge k)=\mathsf{P}(X_n=1,X_{n-1}=1,\ldots, X_{n-k+1}=1)=2^{-k} $$ by independence. As for the second probability, $L_n$ is the length of the longest run of heads in $n$ trials. Its distribution can be calculated recursively, that is $$ \mathsf{P}(L_n\ge k)=1-2^{-n}S_n(k-1), $$ where $S_n(j)$ is the number of sequences of length $n$ in which the longest run of heads does not exceed $j$ and is given by $$ S_n(k)=\cases{\sum_{i=0}^k S_{n-i-1}(k), & $k<n$, \\ 2^n, & $k\ge n$.} $$