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

For all \( N \) points to fit inside a semicircle, exactly one of the points must act as the “starting” or “leading” edge of that semicircle (e.g., the counter-clockwise most point).

Let’s calculate the probability that a specific point, say point A, is this leading edge. For A to be the leading edge, all the other \(N-1\) points must fall within the \(180^\circ\) arc immediately following point A.

Since the points are placed independently and uniformly, the probability of any given point landing in that specific \(180^\circ\) arc is \(1/2\). Therefore, the probability that all \(N-1\) remaining points fall into this arc is \((1/2)^{N-1}\).

Step 3: Core Result

Since any of the \( N \) points could be the leading edge, and these \( N \) events are mutually exclusive (probability of a tie is 0), we can add their probabilities together.

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

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

Examples:

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

Final Answer

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

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

Reference