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Suppose there are $n$ prisoners, and each day every prisoner independently has a probability $p$ of escaping. What is the expected length of time before all prisoners have escaped?

Someone asked me this question a while back, and I thought it would be relatively straightforward, but so far I've been struggling to find a solution.

My initial approaches have been with setting up recurrences, but I've been unable to arrive at something solvable (at least by methods I know). I was also told there were many ways to solve the problem, so seeing other approaches would be interesting.

user2034716
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    Poisson distribution... http://en.wikipedia.org/wiki/Poisson_distribution – the_candyman Mar 13 '14 at 16:15
  • For an individual prisoner, the time to escape is geometric. So aren't you looking to find the expected value of the maximum of $n$ geometric random variables? – soakley Mar 13 '14 at 16:51
  • Apparently there is no simple closed form. But search on this site to find formulas and approximations. – soakley Mar 13 '14 at 17:11
  • Related: http://math.stackexchange.com/questions/26167/expectation-of-the-maximum-of-iid-geometric-random-variables –  Mar 13 '14 at 17:25

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Let $Y$ be the time until they have all escaped. Then $$E(Y)=\sum_{i=1}^\infty \Pr(Y\ge i).$$ So let us calculate $\Pr(Y\ge i)$.

This is $1$ minus the probability that $Y\le i-1$.

The probability that a particular prisoner is gone at time $i-1$ or earlier is $1$ minus the probability she is not gone, that is, $1-(1-p)^{i-1}$. It follows that $$E(Y)=\sum_{i=1}^\infty \left[1-\left(1-(1-p)^{i-1}\right)^n\right].$$

We can expand using the Binomial Theorem, and sum the various related infinite geometric series. So we end up with a sum of $n$ terms. That sum may have a pleasant closed form.

André Nicolas
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