I think you're basically asking how the median compares to the mean for these two distributions.

For the binomial distribution the answer is negative: if $nr$ is an integer, then the mean, median, and mode coincide and equal $nr$.

Here's a positive answer for the binomial distribution in the case $p=1-\frac1{n}$. (I suppose you could check some other cases like $p=1-\frac2n$ in a similar way.)

We need to prove $$\binom{n}{n}p^n > \sum_{k=0}^{n-2}\binom{n}k p^k(1-p)^{n-k},$$ $$\left(1-\frac1n\right)^n > \sum_{k=0}^{n-2}\binom{n}k \left(1-\frac1n\right)^k\left(\frac1n\right)^{n-k},$$ $$\left(n-1\right)^n > \sum_{k=0}^{n-2}\binom{n}k \left(n-1\right)^k.$$ The last right-hand side is actually, per Wolfram Alpha $$ \sum_{k=0}^{n-2} (n-1)^k \binom{n}{k} = n^n-n (n-1)^{n-1}-(n-1)^n$$ so we need $$n^n<2(n-1)^n+n(n-1)^{n-1}$$ $$\left(1+\frac1{n-1}\right)^n<2+\frac{n}{n-1} = 3 + \frac1{n-1}$$ or, with $m=n-1$, $$\left(1+\frac1m\right)^m < \frac{3+\frac1m}{1+\frac1m} = \frac{3m+1}{m+1} = 3-\frac2{m+1}=3-\frac2n$$ This is true: the left-hand side is bounded above by $e$, and the right-hand side surpasses $e$ by $m=7$, and it can also be checked for $m\in\{1,\dots,6\}$.