Probability That N Random Points Lie in Some Semicircle

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?

Interactive Visualization

Checking configuration...
Exact Probability
31.25%
Simulated Probability
—

Exact Answer

P = 5 / 24 = 31.25%
In general,

P(N) = N / 2N-1

Why is this true?

  1. If all points fit in some semicircle, then we can rotate that semicircle until its left endpoint lands on one of the points.
  2. So it is enough to check N possible anchor points, one for each point.
  3. Fix one anchor point. Starting from it, consider the clockwise semicircle of length 180°.
  4. Each of the other N − 1 points has probability 1/2 of falling into that semicircle.
  5. Since the points are independent, the probability for this anchor is (1/2)N-1.
  6. There are N possible anchors, and except for probability-zero boundary ties, these anchor events do not overlap.
  7. Therefore, P = N · (1/2)N-1 = N / 2N-1.

Key Interview Insight

The hard part is reducing an “infinite search over all semicircles” to a finite one. The trick is: if a covering semicircle exists, slide/rotate it until one boundary hits a point. That means every successful configuration is captured by one of the N anchor semicircles.