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After 1 success of a series of trials, the probability of success increases from 1/2 to 21/36. How do I calculate the expected number of trials until I reach 5 successes?

Naively, it seems to be a geometric distribution and a negative binomial distribution: I expect the first 2 trials to give 1 success, and I expect 7 trials to give just over 4 successes afterwards, so 9 trials leads to 5 successes.

To get a better answer, it seems I should calculate the probability of getting the first success on the n-th trial, then multiply by a binomial distribution of some sort. However, there is a (very small) chance of still not getting a single success in n+1 trials, and 4 more successes in arbitrarily many trials, and I don't see how to integrate across the two infinities.

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2 Answers 2

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I think you have the question slightly wrong. It's not "the number of trials in order to expect $5$ successes", it's "the expected number of trials until you have $5$ successes".

Hint: the number of trials until the $5$'th success is the number until the first success, plus the number after the first before the second, plus... Expected value of a sum is the sum of expected values.

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  • $\begingroup$ Ah. I think I understand. $\endgroup$
    – Jon Takagi
    Commented Jun 1, 2018 at 6:34
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The number of trials from 0 successes to 1 success, 1 success to 2 successes, etc. is geometrically distributed, with expected value 1/p.

Therefore, the expected number of trials for the 1st success is 1/(.5) = 2

The expected number of trials from the 1st to the 2nd success (and from 2nd to 3rd, etc.) is 1/(21/36) = 36/21 ~ 1.71

Therefore, the expected number of trials is 186/21 ~ 8.85

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