Obtuse triangle from three random points on a circle

Three points are chosen independently and uniformly on the circumference of a circle. What is the probability that the triangle they form has an obtuse angle?

$\tfrac34$. The triangle is obtuse exactly when all three points fit in some semicircle (the centre lies outside it). For $n$ random points that probability is $\tfrac{n}{2^{n-1}}$, which is $\tfrac34$ for $n=3$; right triangles have probability zero.