Random Points on a Circle and the Semicircle Problem

Random points on a circle land in a common semicircle with probability n over two to the power n minus one, three quarters when n equals three.

n points are placed independently and uniformly at random on a circle. The probability that all n points fall within some common semicircle is n over two to the power of n minus one. For three points that works out to three over four, or 0.75, and the same formula gives one for two points, one half for four points, and 5 over 16 for five points.

Each of the n points lands anywhere on the circle's circumference with a uniform distribution, independently of the others. A semicircle here means any arc spanning exactly half the circle's circumference, and the semicircle is not fixed in advance, it can be rotated to any position, so the question is whether some half of the circle, in any orientation, contains every one of the n points.

This distinction, a fixed semicircle versus a semicircle chosen after seeing where the points landed, is the entire difficulty of the problem, and getting it wrong is the single most common way candidates go astray.

For each point, define the event that all of the other n minus 1 points fall within the specific semicircle that starts exactly at that point and extends 180 degrees in a chosen direction, say clockwise. Since each of the other points independently has probability one half of landing in any fixed semicircle, this specific event has probability one half raised to the power n minus 1.

The key structural fact is that these n events, one for each point acting as the starting edge of its own clockwise semicircle, are mutually exclusive. If all n points genuinely fit within some common semicircle, exactly one point among them sits at the counterclockwise edge of the tightest arc containing all n points, and only that point's clockwise semicircle can contain the rest. Two different points cannot simultaneously be that edge point.