Convex Hull Problem from Monthly

From: Bill97 (nospam)
Date: 12/16/04 

Date: Wed, 15 Dec 2004 20:03:36 -0500

This concerns problem #11120 given in the current (December, 2004) copy of
the American Mathematical Monthly, on page 915.

It is asked to show that the expected number of vertices of the convex hull
of n points sampled from the standard normal distribution in the plane is
2*sqrt(2*pi*ln n). (The probability density function for the normal
distribution is 1/2pi * e^(-(x^2+y^2)/2).

A couple of my colleagues were discussiong this problem today. Let us just
consider the case n=3 right now:

It seems intuitive to me that the probability of 3 sampled points being
collinear is 0, so that the expected number of vertices when n=3 would be 3.
Moreover, the formula yields the unlikely value 2*sqrt(2*pi*ln 3) = 5.25!

Perhaps someone would be willing to help me to clarify my understanding of
the problem.

Thanks,
Bill

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