TITLE: Three points on a circle
QUESTION [7 upvotes]: If three points are randomly chosen on the boundary of a circle, what is the probability that there exists a diameter of the circle such that all three points lie on the same side of it?
I have a solution, but I'm very curious to see how others do it (and whether we get the same answer)!
Thanks.

REPLY [2 votes]: Here's my answer:
Suppose points $A$ and $B$ are placed first. The measure $\theta$ of the minor arc that has $A$ and $B$ as endpoints is uniformly distributed over the interval $(0,\pi)$. Given a value for $\theta$, though, we can know that the probability that point $C$ will be placed so that all three points are on the same side of a diameter is equal to $1-\frac{\theta}{2\pi}$. Averaging over the interval $(0,\pi)$ gives
$$
\frac{1}{\pi} \int_0^\pi (1-\frac{\theta}{2\pi})d\theta=\frac{1}{\pi}\left[\theta-\frac{1}{4\pi}\theta^2\right]_0^\pi=\frac{3}{4}
$$