Standard error of exteremely biased coin OK, so I know that the typical standard error of a coin is estimated by $$\sigma_p=\sqrt{ \frac{p*(1-p)}n }$$
where $p$ is the estimated probability and and $n$ is the number of samples. This seems reasonable at high $n$ and $p \sim 0.5$; however, it seems unreasonable if I have $p = 1$ and $n = 20$, $\sigma_p = 0$.
Is there a better formula for standard error when $ p \sim 0$ or $p \sim 1$ and $n$ is low?
Note: this is a real-world problem and increasing $n$ is non-trivial.
Thanks!
 A: If $p = P(S) = 1,$ then $X \sim Binom(n, p),$ has $X \equiv n$ and $Var(X) = 0 = \sqrt{p(0)/n}$ so the formula for the variance works fine.
In public opinion polls, the margin of sampling error is often given as $\pm \sqrt{1/n},$ which comes from the largest possible variance at $p = 1/2.$
Then the margin of error for a 95% confidence interval (using the normal approximation for large $n$) is about 
$$\pm 1.96\sqrt{p(1=p)/n} = \pm 1.96\sqrt{1/4n} \approx \pm \sqrt{1/n}.$$
Thus, $n = 2500$ subjects give a margin of sampling error $\pm 2$%, and
$n = 1100$ subjects give a margin of sampling error of about $\pm 3$%.
In one sense, this vastly over-estimates the margin of error for cases
in which $p$ is near 0 or 1. But pollsters understand that non-sampling
errors (lack of response, unwillingness to give honest responses, sampled
population differing from target population, and so on) can be especially
serious for such extreme values of $p.$ So they use  $\pm \sqrt{1/n}$
anyway, hoping to cover all contingencies. 
Perhaps you have some intuition about
practical difficulties in sampling when $p$ is far from 1/2 that is responsible for your doubts about
the variance formula. But as an exact mathematical statement about
sampling error only, the formula is correct.
