Ball probability question Suppose I drop $k$ balls into $N$ urns, with the probability a ball falls into any one of the urns equal.
What is the probability that any $q$ urns contain at least $m$ balls?  Note: I'm not looking for the probability that a specific combination of $q$ urns contains $m$ balls, but rather the probability that any combination of $q$ urns contains at least $m$ balls.
 A: Without loss of generality, rank the urns $U_i$ from most to least full.
The probability that $U_1$ contains more than $m$ balls is (from the binomial distribution):
$$P(U_1\ge m)=\sum_{i=m}^k{n\choose i}\frac{(k-1)^{n-i}}{k^{n}}$$
For the next urn, at least $m$ balls were used up in the first urn so:
$$P(U_2\ge m|U_1\ge m)=\sum_{i=m}^{k-m}{n-m\choose i}\frac{(k-1)^{n-m-i}}{k^{n-m}}$$
And in general,
$$P(U_j\ge m|U_{j-1}\ge m)=\sum_{i=m}^{k-(j-1)m}{n-(j-1)m\choose i}\frac{(k-1)^{n-(j-1)m-i}}{k^{n-(j-1)m}}$$
And therefore
$$P(U_q\ge m)=\prod_{j=1}^q\sum_{i=m}^{k-(j-1)m}{n-(j-1)m\choose i}\frac{(k-1)^{n-(j-1)m-i}}{k^{n-(j-1)m}}$$
This is the probability that at least $q$ urns contain $m$ balls. To get the probability that exactly $q$ urns contain $m$ balls you need to subtract $P(U_{m+1}\ge m|U_m\ge m)$.
This may or may not simplify - have fun.
A: I was intrigued by the question, so here is my own version.
Suppose that the first $q$ urns contain $i$ balls, so the remaining $N-q$ urns contain $k-i$ balls. So, we want to find the probability the first $q$ urns contain at least $m\cdot q$ balls.
Then, we have the following:
$$ \displaystyle p=\sum_{i=m\cdot q}^k \dfrac {\dbinom{i+q-1}{i}\cdot \dbinom{ (k-i)+(N-q)-1}{k-i}}{\dbinom{k+N-1}{k}}$$ 
Some explanation:
$\dbinom{i+q-1}{i}:$ we shuffle $i$ identical elements (balls) in $q$ urns
$\dbinom{(k-i)+(N-q)-1}{k-i}$: we shuffle the remaining $k-i$ identical elements in the $N-q$ urns
$\dbinom{k+N-1}{k}:$ we shuffle $k$ identical elements in $N$ urns
It resembles hypergeometric distribution in a way.
