If I uniformly distribute $n$ balls into $k$ bags, I am trying to work out the distribution of the number of bags which are empty.
Now I had thought that I could use that each bag has $\text{Binomial}(n, \frac{1}{k})$ balls and use this but these distributions are not independent so this doesn't work.
Any help will be greatly appreciated.
Define $P_k^n(m)$ as the probability, that there will be $m$ empty bags after $n$ balls were thrown into $k$ bags. Now, suppose, you have $k$ bags total and have already thrown $n - 1$ balls, and it resulted in $m$ bags remaining empty. Then, after the next ball is thrown, $m$ bags remain empty with probability $\frac{k - m}{k}$ and the number of empty bags will become $m - 1$ is $\frac{m}{k}$. So we have the following recurrence, that is sufficient to define all probabilities you search:
$$P_k^n(m) = \frac{k - m}{k}P_k^{n - 1}(m) + \frac{m}{k}P_k^{n-1}(m + 1)$$ $$P_k^0(k) = 1$$ $$P_k^0(m) = 0, \text{ if } m \neq k$$
It is also possible to get the following "closed form" for this distribution:
$$ \bbox[#EEE,6px,border:1.5px solid black]{\mathbb P\big(\text{# empty bags }=m\big)=\sum_{i=m}^k(-1)^{i-m}\binom{i}m\binom{k}{i}\left(1-{i\over k}\right)^n.} $$ This follows from the generalized inclusion exclusion principle. See Generalised inclusion-exclusion principle.
