Let $(X_n)_{n\in\mathbb{N}}$ be independent random variables that are equal to $1$ with probability $p$ and to $0$ with probability $q = 1-p$. A run is a sequence $(X_{k}, \cdots, X_{k+l})$ where $k,l\in\mathbb{N}$ such that $X_{k} = \cdots = X_{k+l} \neq X_{k+l+1}$. Let $L_{j}$ be the length of the $j^{th}$ run ($l+1$ here).
Calculating the expected value of $L_{2j}$, for $j\in\mathbb{N}^{*}$, will always yield $2$, whereas the expected value of $L_{2j+1}$ is $\frac{p}{q} + \frac{q}{p}$. Is there an intuitive explanation as to why even runs have an expectation that is independent of p? It is a result I find quite suprising.
Outline of proof of my statement for $L_{1}$ and $L_{2}$:
1) $$P(S_1 = k) = p^kq + q^kp$$ so $$E(S_{1})=q\sum_{k=1}^{+\infty}{kp^k} + q\sum_{k=1}^{+\infty}{kp^k}=q\frac{p}{q^2} + p\frac{q}{p^2} = \frac{p}{q} + \frac{q}{p}$$
2)$$P((S_1, S_2) = (k,l)) = P(X_{1}=\cdots=X_k=0, X_{k+1}=\dots=X_{k+l}=1, X_{k+l+1}=0) + P(X_1=\cdots=X_k=1, X_{k+1}=\dots=X_{k+l}=0, X_{k+l+1}=1)$$ So $$P((S_1, S_2)=(k,l)) = p^lq^{k+1} + q^{l}p^{k+1}$$ and summing yields $$P(S_2 = k) = p^{k-1}q^2 + q^{k-1}p^2$$ and $E(S_2) = 2$ !!!
The runs alternate between 0s and 1s, so the probability that the 2jth run is in 1s is the probability that the first run is in 0s - that is 1-p. In this case, the expected length of the run is $\sum_{n=0}^\infty p^n=1/(1-p)$.
Likewise, the probability that the 2jth run is in 0s is p, with expected length $\sum_{n=0}^\infty (1-p)^n=1/p$, so the total expectation is $(1-p)/(1-p)+p/p=2$.
The intuition is that a low chance of the run containing 1s precisely cancels the high expected length if it does.