Bertrand's chord versus the triangle side

Question

A chord is drawn at random in a circle of radius $r$ .

What is the probability that the chord is longer than the side of the inscribed equilateral triangle?

This is the classic form of Bertrand's paradox. The answer depends on how the chord is drawn, so give it under the three standard interpretations and explain why they differ.

Accepted Answer

Take r = 1; the triangle side is \sqrt{3} \approx 1.732, so we want P(	ext{length} \sqrt3). Method 1, random endpoints. Fix one endpoint and choose the other uniformly on the circumference. With central angle 2\alpha, the length is 2\sin\alpha and \alpha is uniform on [0,\pi). Require 2\sin\alpha \sqrt3 \iff \sin\alpha 	frac{\sqrt3}{2} \iff \alpha \in \left(	frac{\pi}{3}, 	frac{2\pi}{3}ight). That interval has length 	frac{\pi}{3} out of \pi, so P1 = 	frac{1}{3}. Method 2, random radial midpoint. Pick a radius, then the midpoint uniformly along it, chord perpendicular. Then d is unifo…

Bertrand's chord versus the triangle side

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