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 Don't Be a Square (Posted on 2003-05-19)
Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within any common semicircle?

 See The Solution Submitted by DJ Rating: 4.4667 (15 votes)

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 The Solution & Generalization Comment 21 of 21 |
The solution posted is ingenious (and amazingly simple). In fact for all arc lengths less than 180 degrees similar formulae hold, such as the probability that all the points will lie in the same quadrant, the formula is n/4^(n-1), as the occurrences of all within a given clockwise arc from each point are mutually exclusive.

A more difficult generalization would be for arc lengths greater than 180 degrees. These are no longer mutually exclusive. For example if we seek the probability that all the points lie within a 270-degree arc (that is, the largest gap is at least 90 degrees), the events clockwise from a given point are not mutually exclusive as can be seen from three points separated by 120 degrees. All the points fit within a 270-degree arc going clockwise starting at any one of the points.

So what formula can fit the simulation results for 1,000,000 trials that come up as follows? (expressed as a fraction of the 1,000,000):
3 1
4 1
5 .996157
6 .974637
7 .926162
8 .852604
9 .761418
10 .663289
11 .565835

 Posted by Charlie on 2003-05-24 06:01:53

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