Solution
The Arc Through the North Pole
Show the problem again
A circle has circumference 1. Three points are chosen independently and uniformly at random on it, dividing it into three arcs. A fixed point N (the "north pole") lies on one of these arcs (with probability 1, not at an endpoint). What is the expected length of the arc that contains N? Give a reduced fraction.
Worked solution
The answer is 1/2, not 1/3. Because the three random points are rotationally symmetric, the fixed point N behaves exactly like a fourth independent uniform point. The four points N, P₁, P₂, P₃ cut the circle into four arcs that are exchangeable, each with expected length 1/4. The arc of the original three-point partition that contains N is the union of the two four-point arcs adjacent to N (from the previous Pᵢ to N, and from N to the next Pᵢ). Its expected length is 1/4 + 1/4 = 1/2. This is the inspection paradox in geometric form: a fixed point is more likely to land in a long arc, so the arc you find it in is size-biased. With n random points the answer is 2/(n + 1) instead of 1/n.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.