Prove that when $\frac{{n-d\choose g}}{{n\choose g}}\geq\frac{1}{2}$ and $d=o(n)$, we have $g=o(n-d)$.

Question

Given that I have $\frac{{n-d\choose g}}{{n\choose g}}>\frac{1}{2}$ and $d=o(n)$, how do I prove (or explain) that $g=o(n-d)$? I can see it intuitively that for the numerator to be greater than half of the denominator, we must have $g\ll n-d$. Is there a more formal way to describe this?

Answer 1

\begin{align} \frac 12 &\leq\frac{\binom{n-d}g}{\binom ng}\\ &=\prod_{k=0}^{g-1}\frac{n-d-k}{n-k}\\ &=\prod_{k=0}^{g-1}\left(1-\frac{d}{n-k}\right)\\ &\leq\left(1-\frac dn\right)^g \end{align} from which \begin{align} g &\leq\frac{\log(2)}{\log(1+d/(n-d))}\\ &\sim\frac{\log(2)(n-d)}d\\ &\in o(n-d) \end{align} provided that $d\to\infty$.