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Bertrand's Paradox

Date: 05/09/99 at 23:00:45
From: David Krasik
Subject: Bertrand's Paradox

Dear Dr. Math,

I've visited several math sites on Bertrand's Paradox and I've been to 
the library to look for information on it, but I haven't found 
anything that has helped me to clearly understand it. My teacher 
showed it to my probability and statistics class basically to confuse 
us and my job is to unconfuse us. Hope you can help a little.

Thank you,
David Krasik

Date: 05/10/99 at 08:27:59
From: Doctor Anthony
Subject: Re: Bertrand's Paradox

Bertrand's Paradox shows that you can get 3 different answers to the 
question.  What is the probability that a random chord drawn in a 
circle will be longer than the side of the inscribed equilateral 
triangle. The three answers are 1/4, 1/3 and 1/2 and they differ 
depending on the way you define 'random chord'.

Method (1)
You get a random chord by choosing a point at random in the circle and 
letting this point be the midpoint of the random chord.

The distance from the centre to a side of an inscribed equilateral 
triangle is  a.sin(30)  = a/2  where a = radius of the circle.

The area of the circle within this distance from the centre is 
therefore     pi.(a^2/4)

So the probability that a random chord exceeds the length of the side 
of the triangle is this area divided by the area of the circle pi.a^2 
and it follows that the required probability is 
(pi.a^2/4)/(pi.a^2) = 1/4

Method (2)
You draw a chord parallel to a given line with the position of the 
chord uniformly distributed along a diameter which is perpendicular to 
the given line. The distance of the midpoint of a side of the 
equilateral triangle from the centre of the circle is a.sin(30) = a/2.  
If x = distance of mid-point of chord from the centre, then the length 
of the chord is greater than the side of the triangle if x < a/2. So 
the probability that a chord is greater than the side of the triangle 
is the probability that -a/2 < x < a/2, and this is a/(2a) = 1/2 .

Method (3)
A point is chosen at random on the circumference of the circle and a 
tangent drawn at this point. Then a chord is drawn such that the angle 
between the chord and tangent is uniformly distributed between 0 and 
180. If this angle exceeds 60 degrees and is less than 120 degrees 
then the chord is longer that the side of the equilateral triangle.  
So the probability that chord exceeds the side of the triangle is

       120 - 60      60
       --------  =  -----  =  1/3
          180        180 

And so we get 3 perfectly correct but different answers to the 
question. This example illustrates the very important fact that there 
may be equally valid but different ways of defining 'randomness' and 
in some situations it is not at all clear which is the truly 'random' 

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Probability

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