probability calculation Given $m\cdot e$ balls, $b$ of which are black (suppose the rest are white balls). Randomly put the balls into $m$ baskets, with $e$ balls in each basket. What is the probability of the event that every basket has more white balls than black ones?
 A: let's consider a simpler question: for which values of the parameters does this probability tend to 0 or to 1?
Here are some basic estimates for the case where all the parameters tend to infinity and the ratio $r=\frac{b}{me}$ is fixed (or tends to some value). The case $r\ge \frac12$ is not interesting - clearly there will be some black majorities with high probability. If $r<\frac12$, then the probability of a black majority in a given basket is roughly
$$r^{e/2}(1-r)^{e/2} {e \choose e/2} \approx \frac{\big(2\sqrt{r(1-r)}\big)^e}{\sqrt{e}}$$
where the approximation is up to a multiplicative constant. For this approximation to hold we need to assume that $m$ grows quickly enough for the difference between sampling with and without replacement to be negligible. This is certainly the case when $m$ grows exponentially, which is the relevant regime.
In that regime we also have that the results for different baskets are asymptotically pairwise independent, hence we can get an upper and lower using union bound and the second moment method and conclude that when
$$m\ll  \frac{\sqrt{e}}{\big(2\sqrt{r(1-r)}\big)^e}$$
the probability of a black majority in some cells tends to 0, and when
$$m\gg  \frac{\sqrt{e}}{\big(2\sqrt{r(1-r)}\big)^e}$$
the probability tends to 1.
A: Since the first bin contains $k$ balls with probability
$$\frac{{e\choose k}{me-e\choose b-k}}{{me\choose b}},$$
we get the recursion relation
$$p_e(m,b)=\sum_{k=0}^{\lfloor e/2\rfloor}\frac{{e\choose k}{me-e\choose b-k}}{{me\choose b}}p_e(m-1,b-k)$$
with $p_e(m,b)$ denoting the probability that each of the $m$ bins filled with $e$ balls contains at least as many
white than black balls where the total number of black balls is $b$ and with the bins 
filled randomly in an obvious sense. (If we want strict inequality,
we have to replace the upper summation-bound $\lfloor e/2\rfloor$ by
$\lfloor (e-1)/2\rfloor$.
Using the obvious initial condition $p_e(1,b)=1$ if $b\leq e/2$ (respectively 
$b$ strictly smaller than $e/2$ if we wish strict inequality) and $p_e(1,b)=0$ otherwise,
we can compute $p_e(m,b)$ by an algorithm needing roughly the computation of $2mb+m$
binomial coefficients
and having a memory requirement $b$ (by computing $p_e(a+1,0),\dots,p_e(a+1,b)$
using the values $p_e(a,0),\dots,p_e(a,b)$.
