N Points on a Circle in a Semicircle

This probability question considers a set of independently and uniformly chosen points on the circumference of a circle and asks for the chance that they can all be covered by a single semicircle. The setup focuses on rotational symmetry, random placement on a continuous space, and how a global geometric condition ("all lie in some semicircle") can be translated into a more manageable event. Variants and generalizations of this style occasionally appear in quant interviews for trading and research roles when probing geometric intuition alongside probabilistic reasoning.

Solving it leans on conditioning, symmetry arguments, and viewing the configuration in terms of gaps or arcs between ordered points on the circle. Candidates are expected to recognize that one can fix a reference point without loss of generality and convert a geometric requirement into an inequality constraint on the remaining points. The interviewer is watching for clean use of continuous distributions on the circle, independence and exchangeability, recognition of mutually exclusive events across different reference arcs, and the ability to derive a compact closed-form expression rather than rely on brute-force integration or simulation.

What it tests

This problem class is governed by the principle of partitioning a symmetric space (like a circle) and analyzing the probability that all random points fall within a particular subregion (such as a semicircle). The key is that the circle's symmetry allows any point to serve as a reference, and the probability for all points to fall within a given arc is the same regardless of starting point. The events where all points are contained within an arc starting at each point are mutually exclusive, because the existence of a large gap (at least half the circle) between two points precludes any overlap between such arcs. This exclusivity means the total probability is simply the sum over all possible reference points, each with identical probability. The structure generalizes: for any arc length $x$ (with $x\le1/2$), the probability is $N x^{N-1}$, reflecting the independence and uniformity of placement, and the combinatorial count of possible reference points.