# Clopper-Pearson for non mathematicians

I was wondering if anyone can explain to me the intuition beyond the Clopper-Pearson CI for proportions.

As far as I know, every CI includes a variance in it. However, for proportions, even if my proportion is 0 or 1 (0% or 100%), the Clopper-Pearson CI can be calculated. I tried looking at the formulas, and I understand it has something with percentiles of the Binomial distribution and I understand that finding the CI involves iterations, but I wondered if anyone can explain the logic and rational in "simple words", or with minimum math ?

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You can get a Clopper–Pearson 95% (say) confidence interval for the parameter $\pi$ working directly with the binomial probability mass function. Suppose you observe $x$ successes out of $n$ trials. The p.m.f. is
$$\Pr(X=x)= \binom{n}{x}\pi^x(1-\pi)^{n-x}$$
Increase $\pi$ until the probability of $x$ or fewer successes falls to 2.5%: that's your upper bound. Decrease $\pi$ until the probability of $x$ or more successes falls to to 2.5%: that's your lower bound. (I suggest you actually try doing this if it's not clear from reading about it.) In the long run, bounds calculated this way cover the true value of $\pi$, whatever it is, at least 95% of the time.