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Suppose that you have a game you can play that takes a variable amount of time to finish - suppose it takes between $A$ seconds and $B$ seconds to complete, and is uniformly distributed between the two.

Given $A$ and $B$, is there an easy way to compute the expected number of games you can play within some amount of time $T$?

Thanks!

templatetypedef
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  • This reminds me somewhat of how the exponential distribution is the waiting time between events which follow a Poisson distribution... I wouldn't know how to do it in this case, though. – Clarinetist Jun 14 '13 at 22:02

1 Answers1

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For small $T$ (less than a few multiples of $B$), you better do the integration explicitly. But if $T\gg B$, then we'd naively expect about $m:=\frac{2T}{A+B}$ games completed. That is, $m$ games take a time$E=m\cdot\frac{A+B}2$ with a variance of $V^2=\frac m{12}(B-A)^2$. Now if $T\gg V$, i.e. $T\gg \frac{(B-A)^2}{6(A+B)} $, the naive guess of $m$ should be quite close.

Hagen von Eitzen
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