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From: levy@princeton.UUCP
Newsgroups: net.math
Subject: Re: circle probability
Message-ID: <64@princeton.UUCP>
Date: Sun, 18-Sep-83 00:02:43 EDT
Article-I.D.: princeto.64
Posted: Sun Sep 18 00:02:43 1983
Date-Received: Mon, 19-Sep-83 00:13:15 EDT
References: <704@vax2.fluke.UUCP>
Organization: Princeton University
Lines: 46


I consider Debbie Smit's response to the circle probability problem an
excellent example of the usefulness and of the pitfalls of heuristic
reasoning in mathematics.  The impression I have is that although she
does not believe in the existence of continuous distributions of probability,
this does not prevent her from following a water-tight argument for the
three-point problem.

Here's what you do when your space of possible events is a subset of the
reals (like [0, pi] in this case):  instead of considering the probability
of a *single* event (say angle between two points is x) you consider the
probability of an *interval* of events (angle between two points is
less than or equal to x).  This gives you a function called the (cumulative)
probability distribution; if this function is differentiable, its derivative
is called the density distribution and intuitively gives the probability
that the random variable takes the value "between x and x+dx".  In other
words, if the density is big the chance that the event will take place in
that region is bigger.

So Debbie essentially applied this reasoning (including one necessary
integration) to get the correct answer 3/4.

For more than three points however she oversimplified assuming the chance
that points 1,2,3 are in the same semicircle is independent of the chance
that points 1,2,4 are in the same semicircle.  So the answer (3/4)^(n-2)
is wrong.  To find the answer for n=4, for example, I reason as follows:

The distribution of probabilities for the length of the segment spanned
by the first two points is uniform in the interval [0, pi]; let's call this
length x.  Then the cumulative distribution for the length y of the segment
spanned by the first three points will depend on x in the following way:
for y < x, it is constant and equal to 0; for x <= y <= pi it is affine
linear, growing from x / 2pi to 1 - x / 2pi.  (y > pi is irrelevant, since 
the three points already won't be in the same semicircle.)
Now for a fixed y the probability of that value of y as a function of x starts 
at y / pi for x = 0, decreases linearly to y / 2pi for x = y, then stays
equal to 0 for y < x < pi.  Averaging out gives 3 y*y / 4pi for the cumulative
probability distribution of y, so the density is 3y/2pi.  Now for any value
of y the probability that the four points will be in a semicircle is 1 - y/2pi;
integrating this with the density above gives 3/4 - 1/4 = 1/2; this is the
probability that the four points will be in a semicircle.

I don't know the answer for arbitrary n, and suspect it must be obtainable
by a simpler reasoning than the above.

		-- Silvio Levy (should be writing my thesis...)