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.
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.