Five random points, one semicircle

Question

Five points are placed independently and uniformly at random on the circumference of a circle. What is the probability that all five lie within some common semicircle?

Accepted Answer

For each point i, let Ai be the event that the clockwise semicircle starting at point i contains all five points. These events are mutually exclusive, and given the starting point, the other 4 points each fall in that half with probability 	frac12: P(Ai) = \left(	frac12ight)^4 . With 5 disjoint starting points, P(	ext{all in a semicircle}) = 5 \cdot \left(	frac12ight)^{4} = \frac{5}{16} . Takeaway. \dfrac{n}{2^{\,n-1}} falls fast: at n = 5 it is already 	frac{5}{16}. More points make a shared semicircle increasingly unlikely.

Five random points, one semicircle

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