Three random points, one semicircle

Question

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

Accepted Answer

For each point i, let Ai be the event that the semicircle beginning at point i and sweeping clockwise contains all three points. These events are mutually exclusive (only the clockwise-most point of the group can start such a semicircle). Given point i, the other 2 points each land in that half independently with probability 	frac12, so P(Ai) = \left(	frac12ight)^2. With 3 disjoint choices of starting point, P(	ext{all in a semicircle}) = 3 \cdot \left(	frac12ight)^{2} = \frac{3}{4} . Takeaway. The formula \dfrac{n}{2^{\,n-1}} gives 	frac{3}{4} for n = 3. Three random points usually …

Three random points, one semicircle

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