Three points are chosen independently and uniformly on the circumference of a circle. What is the probability that the triangle they form contains the centre?

Three points are chosen independently and uniformly on the circumference of a circle. What is the probability that the triangle they form contains the centre?

Approach: Replace each point by the diameter through it. Conditional on the three diameters, only the choice of which end each point sits at matters, so count the eight sign patterns.

1/4. Condition on the three diameters through the points. Each point independently occupies one of the two ends of its own diameter with probability 1/2, giving 8 equally likely configurations. This symmetry argument makes the count immediate: exactly 2 of the 8 place the three points so that the triangle contains the centre, one for each of the two alternating arrangements around the circle, so the probability is 2/8 = 1/4. The complementary statement is the semicircle probability, that three uniform points all lie within some semicircle with probability 3/4, and for n points that probability is n/2^{n-1}.

Follow-up: What is the probability that four uniform points on the circle contain the centre in their convex hull?

Key concepts: symmetry argument, conditioning on the diameters, semicircle probability, equally likely configurations.