Here’s a beautiful geometric probability puzzle:

Problem: Place \( N \) points independently and uniformly at random on the circumference of a circle.
What is the probability that all \( N \) points lie within some semicircle (i.e., an arc of length \(180^\circ\))?

Step 1: Understand the Event

We ask: What is the chance that all \( N \) random points can be “seen” within a 180° arc?

Visually, this means you could take a semicircle “window” and rotate it around the circle to include all the points.

Step 2: Strategy

We fix one of the points (say, point A) to be at angle \(0^\circ\) without loss of generality—because the circle is rotationally symmetric.

Then the rest of the \(N-1\) points are placed uniformly at random on the circle.

The configuration fits inside a semicircle if and only if all the other \(N-1\) points fall within an arc of \(180^\circ\) starting from point A.

We now ask: What’s the probability that the angular spread between the largest and smallest of the \(N\) points is \( \leq 180^\circ \)?

Step 3: Core Result

The probability that \( N \) points all lie in some semicircle is:

\[ P(N) = \frac{N}{2^{N-1}} \]

Examples:

  • \( P(2) = \frac{2}{2^1} = 1 \)
  • \( P(3) = \frac{3}{4} \)
  • \( P(4) = \frac{4}{8} = \frac{1}{2} \)
  • \( P(5) = \frac{5}{16} \)

This formula emerges from symmetry and combinatorics:
For each point, consider whether the other \(N - 1\) points fall within a half-circle arc originating at that point.

Final Answer

The probability that \( N \) random points lie within some semicircle is:

\[ \boxed{\frac{N}{2^{N-1}}} \]

Reference