Let $(X^{(i)})_{i \in [n]}$ be independent identically distributed 2-dimensional Bernoulli variables, with $\Pr[X^{(i)} = (a, b)] = p_{a, b}$ for every $a, b \in \{0, 1\}$. I then want to lower bound the probability that the sum is equal to the mean, that is, I want to lower bound $$ \Pr[\sum_{i \in [n]}X^{(i)} = E[\sum_{i \in [n]} X^{(i)}]] \; . $$ Clearly, this requires that $E[\sum_{i \in [n]} X^{(i)}_j]$ is an integer for $j \in \{1, 2\}$. If the two coordinates are independent then the probability splits into two products of 1-dimensional Bernoulli variables which is easy to handle. The problem is when the two coordinates are correlated.
For 1-dimensional Bernoulli variables, $(Y_i)_{i \in [n]}$, with $E[Y] = p$, we get that $$ \Pr[\sum_{i \in [n]} Y_i = pn] = \binom{n}{pn} p^{pn}(1-p)^{n - pn} \ge \frac{\sqrt{2\pi}}{e^2} \frac{1}{\sqrt{np(1-p)}} \ge \frac{\sqrt{2\pi}}{e^2} \frac{1}{\sqrt{n}} $$ where we have used that $\binom{n}{pn} \ge \frac{\sqrt{2\pi n}n^n e^{-n}}{(e \sqrt{pn} (pn)^{pn} e^{-pn}) \cdot (e \sqrt{n - pn} (n - pn)^{n - pn} e^{-n - pn})}$ which follows from the approximation $\sqrt{2\pi k} k^k e^{-k} \le k! \le e \sqrt{k} k^k e^{-k}$. A reference for this approximation is https://en.wikipedia.org/wiki/Stirling%27s_approximation.
My hope is that one can prove something similar in the 2-dimensional case, namely, that there exists a constants $\gamma$ and $C$, such that, $$ \Pr[\sum_{i \in [n]}X^{(i)} = E[\sum_{i \in [n]} X^{(i)}]] \ge C n^{-\gamma} \; . $$