Semi-circle covering n points : Puzzle

Problem : (Posed by Saurabh Joshi, IIT Kanpur in his blog)

Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within any common semicircle?

Solution :
Highlight the part between the * symbols for the answer.
*
If a semi-circle covering all n points, indeed exists, then, a semi-circle covering all n points and starting from one of the points in a clock-wise direction also exists.

So, given a semi-circle which starts at one of the point in clock-wise direction. The probability that the rest of the n-1 points will be in that semi-circle is 1/(2^(n-1)). So for n such semi-circle, the probability will be n/(2^(n-1)) *

:)

Comments

  1. timequalsmoney9:38 AM, January 07, 2020

    Understand that given that the semicircle starts at one of the points the probability is p=(0.5)^(n-1). Don't understand the n circle argument. It doesn't seem necessary that the semicircle starts from a point only. It could start somewhere between two points as well. Trying to understand the exact math here.

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  2. timequalsmoney9:42 AM, January 07, 2020

    Actually i take that back. Semicircle cannot start from between two points.

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  3. Pratik Poddar12:50 PM, September 08, 2026

    Fair question - here is the exact argument. For each point i, let E_i be the event "all n points lie in the clockwise semicircle starting at point i". P(E_i) = (1/2)^(n-1) since each of the other n-1 points independently has a 1/2 chance of falling in that semicircle. Two observations make the count work: (1) if the points fit in ANY semicircle, slide that semicircle counterclockwise until its starting edge hits one of the points - the points still fit, so some E_i occurs; (2) the events E_i are mutually exclusive (two different points cannot both be the clockwise-first point of a containing semicircle). So the total probability is the sum, n/(2^(n-1)). (replied using AI)

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  4. Pratik Poddar12:50 PM, September 08, 2026

    Right - see my reply to your earlier comment for the full argument; the key move is that any containing semicircle can be rotated until it starts exactly at one of the points, so considering only those n semicircles loses nothing. (replied using AI)

    ReplyDelete

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