You have to rethink your intuition about the binomial RV, s.t. the intuition for the hypergeometric RV makes sense. Let's do it step by step.
CURRENT INTUITION FOR BINOMIAL:
First of all let's question the intuition of the Binomial case. Let's say we have an urn with N balls, whereas R are red balls and N-R white. We want to draw red balls (successes). And we draw n times. The random variable X denotes the number of red balls we draw.
Currently your mental picture of drawing is:
- close your eyes
- draw (without looking)
- open your eyes; look at the ball ("ahaa it's red/white")
- put it back (since it's sampling with replacement)
- repeat n-1 times
For all draws i in {1, ... n} we have: E[Xi ] = R / N. With linearity of expectation we get E[X] = n * R / N.
NEW INTUITION FOR BINOMIAL:
So far so good. But now let's change your mental picture of the Binomial case. Just skip the "looking at the drawn ball":
- close your eyes
- draw (without looking)
- don't open your eyes, but take a picture of the ball (e.g. with your phone) to look at it later
- put it back (since it's sampling with replacement)
- repeat n-1 times
Here, during the drawing process you have no information about the drawn balls. Only when you look at the pictures later, you will know. Again, E[X] = n * R / N. In the binomial case, there is no difference, if you look at the drawn balls or not. BUT in the hypergeometrical case it does make a difference. So here is the mental picture:
NEW INTUITION FOR HYPERGEOMETRIC:
- close your eyes
- draw (without looking)
- keep the ball in a box (for you to look at it later; eyes still closed)
- repeat n-1 times
So you haven't looked at the balls yet. Now, what's the probability that the first ball is a red one? That's easy, it's R / N. You have R successes and draw them from N balls uniformly.
Now, what about the probability, that the second ball (let's say you have it in your hand, eyes closed) is red (You still haven't looked at the first ball yet): Ahaa, it's R/N again. Why? Because you have N unknowns (N-2 in the urne, 1 in the box and 1 in your hand) and still R successes (most of them in the urne, and maybe one in the box and/or in your hand - you don't know where these R balls are). The same is true for the third draw, etc. Since you don't know which ones you have drawn already, the probability of a red ball in your hand does not change.
So again E[Xi] = R/N. With linearity of expectation, we get E[X] = n * R / N (which is, what we were looking for).
This is, what @André Nicolas meant in his comment of the accepted answer. There's a difference between P["second ball red"] and P["second ball red" | "first ball red"] (respectively P["second ball red" | "first ball white"]).
Hope that helps.
Best Answer
A second order Taylor expansion around $\mathbb{E}(x)$ gives
$$log(x)\approx log(\mathbb{E}(x))+\frac{1}{\mathbb{E}(x)}\times(x-\mathbb{E}(x))-\frac{1}{2}\frac{1}{\mathbb{E}(x)^2}(x-\mathbb{E}(x))^2$$
Take expectations
$$\mathbb{E}(log(x))\approx \mathbb{E}(log(\mathbb{E}(x)))+\frac{1}{\mathbb{E}(x)}\times(\mathbb{E}(x)-\mathbb{E}(x))-\frac{1}{2}\frac{\mathbb{V}ar(x)}{\mathbb{E}(x)^2}$$
because the 2nd term vanishes and the expectation of a constant is the constant:
$$\mathbb{E}(log(x))\approx log(\mathbb{E}(x))-\frac{1}{2}\frac{\mathbb{V}ar(x)}{\mathbb{E}(x)^2}$$