Suppose we are throwing $m$ balls into $n$ bins, where $m \gg n$. Let $X_i$ be the number of balls ending up in bin $i$, $X_\max$ be the heaviest bin, $X_\min$ be the lightest bin, and $X_{\mathrm{sec-max}}$ be the second heaviest bin. Roughly speaking, $X_i - X_j \sim N(0,2m/n)$, and so we expect $|X_i - X_j| = \Theta(\sqrt{m/n})$ for any two fixed $i,j$. Using a union bound, we expect $X_{\max} - X_{\min} = O(\sqrt{m\log n/n})$; presumably, we can get a matching lower bound by considering $n/2$ pairs of disjoint bins. This (not completely formal) argument leads us to expect that the gap between $X_{\max}$ and $X_{\min}$ is $\Theta(\sqrt{m\log n/n})$ with high probability.
I am interested in the gap between $X_\max$ and $X_{\mathrm{sec-max}}$. The argument outlined above shows that $X_\max - X_{\mathrm{sec-max}} = O(\sqrt{m\log n/n})$ with high probability, but the $\sqrt{\log n}$ factor seems extraneous. Is anything known about the distribution of $X_\max - X_{\mathrm{sec-max}}$?
More generally, suppose that each ball is associated with a non-negative score for each bin, and we are interested in the total score of each bin after throwing $m$ balls. The usual scenario corresponds to scores of the form $(0,\ldots,0,1,0,\ldots,0)$. Suppose that the probability distribution of the scores is invariant under permutation of the bins (in the usual scenario, this corresponds to the fact that all bins are equiprobable). Given the distribution of the scores, we can use the method of the first paragraph to get a good bound on $X_{\max} - X_{\min}$. The bound will contain a factor of $\sqrt{\log n}$ that comes from a union bound (via the tail probabilities of a normal variable). Can this factor be reduced if we're interested in bounding $X_{\max} - X_{\mathrm{sec-max}}$?