Bertrand Paradox 1

Compute the probability that a random chord is longer than the side of the inscribed equilateral triangle (length $\sqrt3$ ) under three different constructions.

Answer

Method 1: Two random endpoints. Choose two points independently and uniformly on the circumference (this defines a random cord). In this case, w.l.o.g., let the first point be whatever and let it define one of the vertices of the triangle (if it doesn't we would just rotate the circle so that it does). In this context it is clear that the probability that a chord exceeds $\sqrt3$ is

\frac{1}{3}.

because out of three equal arc lengths, only two satisfy the above condition.

Method 2: Random midpoint. Pick a point uniformly inside the circle to serve as the chord's midpoint (this defines a random chord). A chord is longer than $\sqrt3$ exactly when its midpoint lies within the concentric circle of radius $$1/2$$ . The probability is the area ratio

\frac{\pi(1/2)^2}{\pi(1)^2} = \frac{1}{4}.

Method 3: Random radial angle & perpendicular chord. Choose a random radius (angle uniformly in $[0,2\pi)$ ), then pick a point uniformly on that radius line $$[0,1]$$ and draw the chord perpendicular to the radius through that point (basically the choosen point acts as the midpoint). This defines a random chord. The chord is longer than $\sqrt3$ if the chosen point lies within distance $$1/2$$ of the center, giving probability

\frac{1}{2}.