Three random points inside a third of the circle

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 arc of 120^\circ (a third of the circle)?

Accepted Answer

Let the arc length be the fraction \alpha = 	frac13 of the circle. For each point i, let Ai be the event that the arc of length \alpha starting at point i and sweeping clockwise contains all three points. Because \alpha \le 	frac12, at most one point can serve as the clockwise-most starting point, so the Ai are mutually exclusive. Given the starting point, the other 2 points each land in that arc with probability \alpha, so P(Ai) = \alpha^{2}. With 3 disjoint choices, P(	ext{all in a } 	frac13	ext{-arc}) = 3 \cdot \left(	frac13ight)^{2} = \frac{3}{9} = \frac13 . Takeaway. For n points…

Three random points inside a third of the circle

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