Probable direction of deviations from the expected value in binomial and hypergeometric cases Suppose I have an urn with N marbles, with frequencies p and q for red and black marbles, and with p > 0,5. I take a sample of r marbles.
It sounds intuitive to say that deviations from the mean should be expected to occur in the direction of the more frequent marble (say for example, if I have 90 red and 10 black, then with r = 10, 10 red should be more likely than 8 red, 2 black). That is, that $Prob(\hat{p} > p) > Prob(\hat{p} < p)$. Or equivalently, $Prob(\hat{p} \ge p) > Prob(\hat{p} \le p)$.
But that is not always the case. Take p = 0,51 and r = 2, for example. What's going on here is that the expected value is not a possible sample value, but it is close to it, so $\hat{p}=0,5$ will count as a case where $\hat{p} < p$. So let's get rid of this sort of cases by taking as an additional premise that $p .r \in \mathbb{Z}$. 
¿Will the result hold in this case?
What would be needed to answer positively is that


*

*For sampling with replacement:
$$\sum_{k=p.r}^{r} \binom{r}{k} p^k (1-p)^{r-k} > \sum_{k=0}^{p.r} \binom{r}{k} p^k (1-p)^{r-k}$$

*For sampling without replacement:
$$\frac{\binom{p.N}{p.r} \binom{N - p.r}{r - p.r}}{N \choose r} > \frac{\binom{(1-p).N}{(1-p).r} \binom{N - (1-p).r}{r - (1-p).r}}{N \choose r}$$
(could also use a cumulative hypergeometric distribution but this seems easier)
 A: 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 $m\in\{1,\dots,6\}$.
A: It's not true for sampling without replacement.  Consider e.g. $N=9$, $p = 2/3$, $r = 6$. Note that the possible numbers of red balls in the sample are  $3,4,5,6$, with $4$ corresponding to $\widehat{p} = p$.  Since $P(3) = P(5) + 2 P(6)$, we have $P(\widehat{p} > p) = P(5) + P(6) < P(3)$.
