$|p- \dfrac xn|>|q- \dfrac xn|$ $\implies$ $p^x(1-p)^{n-x}<q^x(1-q)^{n-x}$?

Question

If $p,q \in (0,1)$ , and $ n \in \mathbb N$ be given and $x$ be given integer between $0$ and $n$ such that $|p- \dfrac xn|>|q- \dfrac xn|$ , then is it true that $p^x(1-p)^{n-x}<q^x(1-q)^{n-x}$ ? If this is not true then what additional property must be imposed on $p,q$ so as to make $|p- \dfrac xn|>|q- \dfrac xn|$ $\implies$ $p^x(1-p)^{n-x}<q^x(1-q)^{n-x}$ valid ? I am asking this becuase I know that for given $x$ and $n$ , $f(p)=p^x(1-p)^{n-x}$ increases for $0<p< \dfrac xn$ , attains maximum at $p=\dfrac xn$ and then decreases for $1>p> \dfrac xn$ . This function interested me while calculating MLE of $p$ from a Binomial distribution ...

Answer 1

Your condition does hold in the case where $x/n=1/2$, due to the symmetry in $f(p)$. That is, $$f(1/2 + h)=f(1/2 - h).$$

In the general case, I don't think there is a nice condition, since the growth/decay on one side of $x/n$ will be different than the decay/growth on the other side.