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Given $N$ bins and an unlimited number of balls, to be assigned one ball per time unit to a uniformly random bin. What is the expected time to achieve at least 2 balls in every bin?

From a previous question Probability that all bins contain strictly more than one ball? we know the probability of failure (of at least one bit fewer than two balls) at time interval $t$ is given by
$$ F(t) = 1 - \frac{N!}{N^t} \sum_i (-1)^i \binom{t}{t-i} \left\{ t-i \atop N - i \right\} $$ (where the braces are expressing a Stirling number of the second kind). And much easier reasoning gives the expected time to first success (all bins having two or more balls) as $$ \begin{array}{l} \sum_t t \left[ (1-F(t)) - (1-F(t-1)) \right] \\ = \left[ (1-F(1)) - (1-F(0)) \right] +2\left[ (1-F(2)) - (1-F(1)) \right] + \ldots \\ = F(0) - F(1) + 2F(1) -2F(2) + 3F(2) -3F(3) +\ldots \\ = \sum_t \left( 1 - \frac{N!}{N^t} \sum_i (-1)^i \binom{t}{t-i} \left\{ t-i \atop N - i \right\} \right) \end{array} $$ But it is sometimes easier to get a simpler expression for an expectation value than to express the underlying distribution cleanly, so I am hopeful that this answer can be simplified.

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Mark Fischler
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1 Answers1

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The expected number of tosses required to put at least 2 balls in each bin is $$\mathbb{E}(T)=\int_0^\infty 1-[1-(1+t/N)\exp(-t/N)]^N\,dt.$$ This can be proved by embedding your problem into a Poisson process.

Reference: Newman, Donald J.; Shepp, Lawrence (1960), "The double dixie cup problem", American Mathematical Monthly 67: 58–61.

  • Very good, this is the sort of improvement I was looking for. – Mark Fischler Jun 17 '14 at 05:51
  • The substitution $x=t/N$ yields the slightly simpler form

    $$ \mathbb E(T)=N\int_0^\infty\left[1-\left[1-(1+x)\mathrm e^{-x}\right]^N\right],\mathrm dt;. $$

     – joriki Feb 25 '24 at 13:21