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Let $X_n\sim \operatorname{Bern}(p)$.

How does one calculate that there will be $k$ more more successes in a row? I can only think of taking the complement of the Geometric c.d.f.. $1-\sum_{i=0}^{k-1} p^i(1-p)$

But this can get tedious for a large $k$.

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    @Wuschel The accepted answer does not solve (any plausible interpretation I can think of) the question you asked. This indicates you might want to rephrase your question.2012-09-06

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If the probability of success on each trial is $p$, then the probability that the first $k$ trials are successes is $p^k$. That's the probability that the length of the initial run of successes has length $k$ or more.

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    Or that any sequence of k consecutive trials result in a success. Which is the probability that a particular sequence has k or more consecutive successes.2012-09-04