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From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re: circle probability (spoiler)
Message-ID: <33@ism780.UUCP>
Date: Fri, 16-Sep-83 15:40:00 EDT
Article-I.D.: ism780.33
Posted: Fri Sep 16 15:40:00 1983
Date-Received: Sat, 17-Sep-83 06:25:54 EDT
Lines: 24

For each of the n points, imagine a chord extending clockwise from that
point through the other n-1 points.  Since the n different chords include
all the ways that the points might be formed into a semicircle, and since
no two of the chords can both fall within a semicircle (with the
exception of singular cases), the odds that all the points fall within a
semicircle is the sum of the odds that any of the chords falls within a
semicircle; since the points are equivalent, p = n * p_one_chord.

Since, in order for one of the chords to fall within a semicircle, all of
the last n-1 points must be within 1/2 the circle from the first point,
p_one_chord is clearly (1/2) ** (n-1).  So, p = n/(2**(n-1)).

I actually derived the result empirically (Monte Carlo), and then found
an analytical explanation.  Many mathematical results have been derived
this way, despite the purely analytical form in which they are published.
Thus, the methods of mathematics are sometimes not as far from the
methods of science as is often believed.

I have no idea how to extend the result to n points within half an m-sphere,
or whether there are better and/or more formal proofs.

Jim Balter (decvax!yale-co!ima!jim), Interactive Systems Corp

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