This is answered in a paper by Raab and Steger (they switch your $n$ and $m$). The case $n=m$ is simpler and had been known before (see their introduction).
Intuitively, in order to find the "correct" answer, you calculate the expected number of bins $X_t$ which get at least $t$ balls; the value of $t$ such that $X_t \approx 1$ should be "close" to the expectation.
In order to make this rigorous, follow these steps:
- Find a critical value $t_0$ such that if $t \ll t_0$ then the probability that a given box gets at least $t$ balls is very close to $1$, and if $t \gg t_0$ then it is very close to $0$.
- When $t \gg t_0$, a union bound shows that with high probability no box gets at least $t$ balls.
- When $t \ll t_0$, the expected number of boxes with at least $t$ balls is very close to $m$, and so the probability that no box gets at least $t$ balls is very small.
- Deduce that most of the probability mass of the expectation is concentrated around $t_0$, and so the expectation itself is close to $t_0$.
When doing the calculations, we hit a problem in step 3: the number of expected boxes with at least $t$ balls is not $m$ but somewhat smaller. Raab and Steger show that the variable $X_t$ is concentrated around its expectation (using the "second moment method", i.e. Chebyshev's inequality), and so $\Pr[X_t = 0]$ is indeed small.
Most of the work is estimating the binomial distribution of the number of balls in each box, finding the correct $t_0$ for each case; the method fails when the number of balls is significantly smaller than the number of bins.
Here is some Sage code:
def analyze_partition(partition, bins):
last = partition
counts = 
for part in partition[1:]:
if part != last:
last = part
counts += 
counts[-1] += 1
counts.append(bins - sum(counts))
return multinomial(*partition) * multinomial(*counts)
def expected_max(balls, bins):
return sum([max(partition) * analyze_partition(partition, bins)
for partition in partitions(balls)
if len(partition) <= bins])/bins^balls
When plugging $10$ balls and $5$ bins, I get $$1467026/390625 \approx 3.76.$$