3
$\begingroup$

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 $k$.

  • 1
    Are you looking for something like this: http://math.stackexchange.com/questions/59738/probability-for-the-length-of-the-longest-run-in-n-bernoulli-trials/59749#59749 ?2012-09-04
  • 0
    @ByronSchmuland: I think my proposition is simpler because I am not looking for the probability of the **longest** run exceeding a value but the probability that there will k or more **consecutive** successes. I'm not sure if they are different propositions, though.2012-09-04
  • 0
    Out of how many trials?2012-09-04
  • 0
    @Sasha: I run the experiment, if there is failure. I restart a chain and record the previous chain (with 0 successes), if there are two consecutive successes followed by one failure in the next chain, I stop again and start a new (third) chain. Now I evaluate an infinite number of these chains and count what fraction of these chains have k or more successes.2012-09-04
  • 0
    @Sasha : Since he used the word "geometric", I'm assuming that the number of trials is precisely the thing that is random. You just keep going until you get the first failure.2012-09-04
  • 0
    @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

1 Answers 1

3

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.

  • 0
    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