Two consecutive zeros in Poisson random variables, and consecutive wins in a binomial setting
Solution 1:
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 $(n+1-k)\ p^k$