Six random points, one semicircle

Question

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

Accepted Answer

For each point i, let Ai be the event that the clockwise semicircle starting at point i holds all six points. These events are mutually exclusive, and given the start, the other 5 points each land in that half with probability 	frac12: P(Ai) = \left(	frac12ight)^5 . With 6 disjoint starting points, P(	ext{all in a semicircle}) = 6 \cdot \left(	frac12ight)^{5} = \frac{6}{32} = \frac{3}{16} . Takeaway. \dfrac{n}{2^{\,n-1}} at n = 6 gives 	frac{3}{16}. The disjoint-events trick makes every case of this family a one-line count.

Six random points, one semicircle

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