Intuitive reason why the Fisher Information of Binomial is inversely proportional to $p(1-p)$ It confuses/blows my mind that the Binomial has variance proportional to $p(1-p)$. Equivalently, the Fisher information is proportional to $\frac{1}{p(1-p)}$. What is the reason for this? Why is the Fisher Information minimized at $p=0.5$? That is, why is inference most difficult at $p=0.5$?
Context:
I'm working on a sample size calculator, and the formula for $N$, the sample size needed, is an increasing factor of $p(1-p)$, the result of a variance estimation in the derivation.
 A: The Fisher information is the variance of the score function. And it is related to entropy. For a Bernoulli trial we are getting one bit for each trial. So this Fisher Information has similar properties as the Shannon Entropy, as we would expect. In particular the entropy has a maximum at 1/2 and the information has a minimum at 1/2. 
A: To see, in an intuitive way, that the variance is maximised at $p = 0.5$, take $p$ equals to $0.99$ (resp. $p = 0.01$). Then a sample from $X \sim \text{Bernoulli}(p)$ will likely contain many $1$'s (resp. $0$'s) and just a few $0$'s (resp. $1$'s). There is not much variation there.
A: The inference is "hard" for $p$ 'in the middle, because an sample with $\hat p$ near the middle is consistent with a wider range of $p$. Near the ends, it can't be so far away - because of the ends being "barriers" beyond which $p$ cannot go.
I think the intuition is easier when looked at in variance terms, though.
The intuition about the variance of a binomial being large in the middle and small at the ends is rather straightforward: near the endpoints there isn't room for the data to "spread out". Consider $p$ small  -- because the mean is close to 0, the variation can't be large -- for the data to average $p$ it can only get so far from the mean.
Let's consider the variance of a sample proportion in a series of Bernoulli trials. Here $\text{Var}(\hat p) = p(1-p)/n$. So holding $n$ fixed, and varying $p$, the variation is much smaller for $p$ near 0:
Sample proportion in binomial samples -- here $y$ is just random uniform; the blue case has mean 0.03, the black mean 0.5 (some jitter added so the points don't pile up too much and lose detail)

The corresponding probability functions:

In each case pay attention to the lines marking the mean. As the mean line becomes more 'jammed up' against the barrier, points below the mean can only get a small way below. 
As a result, points above the mean can't typically get too far above the mean (because otherwise the mean would shift!). Near $p = \frac{1}{2}$ the endpoints don't really "push it up" the same way it does when there's a barrier there. 

We see at the same time why the distribution must be skewed at the ends; for the random variable $\hat p$ to be even some of the time to be more than $p$ above the mean, there must be correspondingly more probability squished about as far below the mean as it can go. That looming barrier at 0 gives both a limit on the variability and leads to skewness.
[This form of intuition doesn't tell us why it takes that exact functional form, but it does make it clear why the variance must be small near the ends, and get smaller the closer to the ends you go.]
