Intuition for hypergeometric variance? I'm trying to learn the major facts about a bunch of probability distributions, hypergeometric included. I can use the commonalities between it and a binomial to my advantage for thinking through some of the formulas; for example, the mean is basically exactly the same. But the variance has me a little bit more stumped. For reference, the variance of a hypergeometric can be written as
$n(\frac{r}{N})(1-\frac{r}{N})(\frac{N-n}{N-1})$
where $N$ is the number of events in the universe, $n$ is the number of trials, and $r$ is the number of possible successes.
Most of those terms are pretty straightforward to me. Again, comparing to a binomial, $n(\frac{r}{N})(1-\frac{r}{N})$ is basically $npq$. And even in that last term, the numerator has a pretty straightforward intuition: the closer the number of trials is to the possible number of trials, the lower the variance will be, down even to zero; in the classic example of drawing colored marbles from a jar, if you draw every single marble then of course you will have no variance.
But it's the denominator that I'm not so sure about. Now, I understand that the variance can be derived from the moment generating function and so on, but I'm trying to gain the intuition. And I can understand that one effect the denominator has is that the more marbles there are in the jar, the higher the variance will be. Although I'm struggling to understand (intuitively) why that would necessarily be the case.
So, in the end, there's two facts about the variance of a hypergeometric that I'm struggling to gain the intuition for: 1) the inverse relationship between variance and $N$ (although that would be the case even without the last term) and 2) what work the $N-1$ is doing. So even though the title refers to the variance of the hypergeometric in general, in particular I'm focused on the denominator of the last term of the variance.

Originally a comment:
In a sense when $n=1$ the $N−1$ in the denominator balances the $N−n$ in the numerator as you want then want the variance of a single sample to equal the Bernoulli variance of $p(1−p)$.
Similarly when $n=N−1$, the $N−1$ in the denominator balances the $n$ in the numerator as you want the all-but-one variance to equal the Bernoulli variance of the one not selected, given that they add up to a constant total.
So at each end $\frac{n(N−n)}{N−1}$ is intuitively right. The intuition is harder when $n$ is some intermediate value, and that is when you resort to analysis.
