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Path: utzoo!linus!decvax!cca!ima!ism780!jim
From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re(2): circle probability (spoiler)
Message-ID: <38@ism780.UUCP>
Date: Mon, 19-Sep-83 16:07:00 EDT
Article-I.D.: ism780.38
Posted: Mon Sep 19 16:07:00 1983
Date-Received: Tue, 20-Sep-83 21:06:56 EDT
Lines: 26

re:
Jim Balter's solution to the circle problem is wrong.  I have run the
simulation and the result for n=4 agrees with my result, 1/2.  Moreover
I have an inductive proof that the probability for arbitrary n is equal
to n/(2^(n-1)).
---
Either my message got garbled in transmission, or you didn't read it properly.
My result agreed with yours (although it was arrived at in a significantly
different way).  To repeat (with the proper term "arc" substituted for the
improper "chord" in my original sending):

For each of the n points, imagine an arc extending clockwise from that
point through the other n-1 points.  Since the n different arcs include
all the ways that the points might be formed into a semicircle, and since
no two of the arcs 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 arcs falls within a
semicircle; since the points are equivalent, p = n * p_one_arc.

Since, in order for one of the arcs 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_arc is clearly (1/2) ** (n-1).  So, p = n/(2**(n-1)).

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

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