NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/rec/puzzles/7467

Text File | 1992-11-23 | 2.0 KB | 43 lines

Newsgroups: rec.puzzles
Path: sparky!uunet!zaphod.mps.ohio-state.edu!cs.utexas.edu!swrinde!emory!athena.cs.uga.edu!groucho.dev.uga.edu!is
From: [e-mail] (Bob Stearns)
Subject: Re: Random Points on a Sphere
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: University of Georgia
References: <[e-mail]> <[e-mail]>
Date: Mon, 23 Nov 1992 22:34:53 GMT
Lines: 31

In article <[e-mail]> [e-mail] (Dilip V. Sarwate) writes:
>[e-mail] (Mark Isaak) writes:
>
>>Four points are randomly selected from the surface of a sphere.
>>What is the probability that all four lie in the same hemisphere?
>>How about for 5 points?  6?  more?
>
>One simple-minded solution (which doesn't worry about boundaries) is as follows.
>The first two points picked, together with the center of the sphere, define a
>plane that divides the sphere into two hemispheres.  The two points are on the
>boundary dividing the hemispheres.  With probability 1, the third point lies
>strictly in the interior of one of the hemispherical surfaces, and this becomes
>our chosen hemisphere (the first two points, being on the boundary, are included
>by courtesy in the chosen hemisphere.)  Thus, assuming that the points are
>picked with a uniform distribution over the surface of the sphere, the
>probability that the fourth point lies in the chosen hemisphere is 1/2, the
>probability that the fifth point also lies in the chosen hemisphere is 1/4, etc.

This makes the first two points in some way distinguished. As I
interpret the question, is there no way to chose one pair of the four to
use in the manner you outline such that the other two fall into one of
the two hemispheres created by the first chosen two. This makes the
problem much more complex. I have no simple solution to propose, but
merely wish to point out that your solution is too simple by far.


--
Bob Stearns
University of Georgia
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