Consider the binomial distribution with parameters $n$ and $p$, $\text{Bin}(n,p)$. It has mean $m=np$, and the probability that a random variable $X\sim\text{Bin}(n,p)$, equals its expected value, is $$\Pr(X=m)=\binom{n}{m}p^{m}(1-p)^{n-m}$$
I want a lower bound on $\Pr(X=np)$, that is a function of $m$, and not dependent on $n$.
I have done this much: It can be easily showed that $\binom{n}{k}=\frac{n}{k}\cdot\frac{n-1}{k-1}\cdot\ldots\cdot\frac{n-k+1}{1}\ge\frac{n^k}{k^k}$ [as for any $0\le a\le k-1$, we have $\frac{n}{k}\le\frac{n-a}{k-a}$]. Thus I have $$\binom{n}{m}p^{m}(1-p)^{n-m}\ge (1-p)^{n-m}$$ This is still not a function of just $m$, it is still dependent on $n$. Any help in that direction will be great. Also it might be solved by using a different inequality that $\binom{n}{k}>\frac{n^k}{k^k}$. e.g., Stirling's approximation might help, I don't know.
The lower bound is zero.
The larger $n$ gets, the more numbers are possible, and the less likely any specific number becomes: even the mean, the most likely value of all. If you flip a fair coin $10$ times, there is a decent chance of getting exactly $5$ heads, but if you flip a coin a billion times, the probability of getting half a billion heads exactly is minuscule.