I needed a family of functions f(x; a) with a single parameter to transform probabilities based on a set of heuristics. The family should include both convex and concave functions. The family should also satisfy the following constraints:

f(0; a) = 0
f(1; a) = 1
f(x; 1) ≅ x
f(0; inf) ≅ step function at 0
f(1; 0) ≅ step function at 1

Practically speaking, f(x) = x^a functions well for these constraints, but I wanted something “symmetric” for a very specific meaning of the word:

f(1-f(x)) = 1 - x

In words, the graph should be symmetric around the line x + y = 1.

Despite not having any practical need, I spent a couple hours on it, chatted with a buddy who enjoys such puzzles, and even ended up emailing an old math professor of mine looking for an answer. My buddy, the brilliant Chris Poirel, ended up coming through: use a parameterized, transformed quadrant of the circle:

f(x; a) = ((1-(1-x)^a)^(1/a)

This is the “upper left” quadrant, moved over, and parameterized by how much curve it has. Large a looks like a box, while small a (0 < a < 1) starts to look like a star. And of course

f(x; 1) = x