We have $X_1, ..., X_{100}$ i.i.d.r.v. having a Poisson distribution of parameter $\lambda$.
I'd like to compute the expected value $E(N)$ for the number of times two consecutive $X_i$ are zero:
$$N = \sum_{\qquad i \geq 1\\ X_i = 0\ \rm{and}\ X_{i+1} = 0} 1$$
I'm not sure if this is equivalent or not to calculate the expected value of the number of two consecutive wins in a binomial setting with $p=e^{-\lambda}$, and $n=100$. Is there a law for the k consecutive wins of a binomial setting?
As requested in comments:
It would be "the same" as each case is independent. Precisely what this means slightly depends on how you count.
If $100$ consecutive $0$s count as $99$ cases of two consecutive $0$s, then by linearity of expectation the expected number would be $$99p^2=99e^{-2\lambda}$$ and more generally $(n+1-k)\ p^k$