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Bertrand's Problem

What is the Probability that a Chord drawn at Random on a Circle of Radius $r$ has length $\geq r$? The answer, it turns out, depends on the interpretation of ``two points drawn at Random.'' In the usual interpretation that Angles $\theta_1$ and $\theta_2$ are picked at Random on the Circumference,

\begin{displaymath}
P={\pi-{\pi\over 3}\over \pi} ={2\over 3}.
\end{displaymath}

However, if a point is instead placed at Random on a Radius of the Circle and a Chord drawn Perpendicular to it,

\begin{displaymath}
P={{\sqrt{3}\over 2} r\over r} ={\sqrt{3}\over 2}.
\end{displaymath}

The latter interpretation is more satisfactory in the sense that the result remains the same for a rotated Circle, a slightly smaller Circle Inscribed in the first, or for a Circle of the same size but with its center slightly offset. Jaynes (1983) shows that the interpretation of ``Random'' as a continuous Uniform Distribution over the Radius is the only one possessing all these three invariances.


References

Bogomolny, A. ``Bertrand's Paradox.'' http://www.cut-the-knot.com/bertrand.html.

Jaynes, E. T. Papers on Probability, Statistics, and Statistical Physics. Dordrecht, Netherlands: Reidel, 1983.

Pickover, C. A. Keys to Infinity. New York: Wiley, pp. 42-45, 1995.




© 1996-9 Eric W. Weisstein
1999-05-26