Consider the probability distribution for a discrete random variable $X$ with support $\left\{0,1,\ldots,50\right\}$: $$ \Pr\left[X=k\right] = \frac{{600\choose k}{400\choose 50-k}}{1000\choose 50} $$ You might think of this as the probability mass function for the event that you draw $k$ black balls in a sample of size 50 from an urn containing 600 black balls and 400 blue balls. I need to show that this distribution is unimodal; I think I have a proof but it is inelegant and I'm wondering if anyone has a better method.
To begin with, I will discard the denominator and define $F(k) = {600\choose k}{400\choose 50-k}$. I have a theorem that says that a distribution that is log-concave is also unimodal so a natural approach is to consider $\log F$ and show that this is concave: $$ \log F(k) = \log {600 \choose k} + \log {400\choose 50-k} $$ Differentiating twice with respect to $k$ we obtain, $$ \frac{d^2}{dk^2}\log F(k)=-\psi^{(1)}(51-k) - \psi^{(1)}(601-k) - \psi^{(1)}(k+1) - \psi^{(1)}(k+351), $$ where $$ \psi^{(1)}(z) = \int_0^\infty \frac{t\exp\left(-zt\right)}{1-\exp\left(-t\right)} dt. $$ If we can show that $\psi^{(1)}(z) > 0$ then it will follow that $\frac{d^2}{dk^2}\log F(k) < 0$ and that therefore the distribution is log-concave. To see that $\psi^{(1)}(z) > 0$ consider the integrand: $\frac{t\exp\left(-zt\right)}{1-\exp\left(-t\right)}$; for $t>0$ the numerator and denominator are both positive and hence their ratio is also positive. Therefore, the integral is also positive.
This shows that $\log F(k)$ is concave and so $\Pr\left[X=k\right]$ is log-concave and therefore unimodal. I don't like my proof; does anyone have something better?
Set:
$$F(k):=\frac{P\left(X=k+1\right)}{P\left(X=k\right)}=\frac{\binom{600}{k+1}\binom{400}{49-k}}{\binom{600}{k}\binom{400}{50-k}}=\frac{600-k}{k+1}\frac{50-k}{351+k}$$
This function in $k$ can be shown to be strictly monotonically decreasing for $k\in\left\{ 0,1,2,\dots,49\right\} $.
Further $F(0)>1$ and $F(49)<1$ so by increasing $k$ probability $P(X=k)$ is first increasing, then reaches a peak, and then is decreasing.
Then it is enough to prove that no $k\in\left\{ 0,1,2,\dots,49\right\} $ exists with $F\left(k\right)=1$.
This all is also easy to check in e.g. Excel.