Suppose we throw $m_b$ black balls and $m_w$ white balls independently and uniformly at random in $n$ bins. Let $\left\{ B_i \right\}_{i \in \left\{ 1,\ldots,n \right\}}$, $\left\{ W_i \right\}_{i \in \left\{ 1,\ldots,n \right\}}$ be the number of black and white balls that end up in bin $i$, respectively.
Let
$\left\{ M_i \right\}_{i \in \left\{ 1,\ldots,n \right\}}$
be the binary random variables such that
$$
\Pr(M_{i} = 1 \,|\, B_i>W_i ) = 1,\\
\Pr(M_{i} = 1 \,|\, B_i=W_i ) = 1/2,\\
\Pr(M_{i} = 1 \,|\, B_i<W_i ) = 0.
$$
In other words, $M_{i}$ indicates whether bin $i$ contains more
black balls than white balls (breaking ties u.a.r.).
In
Randomized Distributed Edge Coloring via an Extension of the
Chernoff--Hoeffding Bounds by
Alessandro Panconesi and Aravind Srinivasan, it is shown that the
Chernoff bound holds for dependent binary random variables such that
for each $I \subseteq \left\{ 1, \cdots, n \right\}$
$$
\Pr \left( \bigwedge_{i \in I} X_{i} = 1 \right)
\leq \prod_{i \in I} \Pr \left( X_{i} = 1 \right)
$$
They call the latter property negative 1-correlation.
My question is: are the r.v. $M_{i}$ negatively one correlated?
This is very intuitive, but I couldn't figure out any proof. All my attempts end up dealing with the explicit joint probability distribution of the $B_i$ and $W_i$, at which point I'm stuck...