Ten points are placed independently and uniformly on a circle of circumference 1, cutting it into ten arcs. An eleventh point is then dropped uniformly and independently. What is the expected length of the arc containing it?

Ten points are placed independently and uniformly on a circle of circumference 1, cutting it into ten arcs. An eleventh point is then dropped uniformly and independently. What is the expected length of the arc containing it?

Approach: Compare the arc chosen by an independent point with an arc chosen uniformly from the ten. Long arcs are more likely to be hit, so weight each arc by its own length and compute the resulting mean.

2/11. Each arc has a Beta(1, 9) length distribution with mean 1/10 and variance 9/1100, so the second moment is E[L^2] = 9/1100 + 1/100 = 1/55. An independently dropped point lands in a given arc with probability equal to that arc's length, which is size-biased sampling, so the selected arc has mean E[L^2]/E[L] = (1/55)/(1/10) = 2/11 = 0.1818. That is 1.8 times the average arc of 1/10. This is the inspection paradox: sampling by hitting favours long intervals in proportion to their length, and the ratio E[L^2]/E[L]^2 is exactly the inflation factor.

Follow-up: What is the expected length of the arc that starts at one of the ten points instead?

Key concepts: size-biased sampling, inspection paradox, beta distribution, second moment.