Bertrand's Paradox Calculator
Bertrand's paradox is a famous problem in probability that shows how the phrase at random can hide more than one meaning. Draw a circle, then draw a chord across it at random. What is the chance the chord is longer than a chosen length, classically the side of the inscribed equilateral triangle? The unsettling answer is that it depends entirely on how you make the chord random, and three perfectly reasonable methods give three different probabilities. This calculator works all three at once. Enter the circle radius and a threshold chord length, and it returns the probability the chord beats that length under the random endpoints method, the random radius method and the random midpoint method, along with the triangle-side reference length so you can reproduce the classic one third, one half and one quarter. The random endpoints method fixes one end of the chord and spins the other uniformly around the circle. The random radius method slides a point uniformly along a radius and takes the perpendicular chord. The random midpoint method drops the chord's midpoint uniformly anywhere in the disk. Each is a valid uniform choice, yet they disagree, and that disagreement is the whole lesson: a probability question is only well posed once the sample space is pinned down. Use the default length, the square root of three times the radius, to see the textbook values, or set your own threshold to explore how each method responds.
The threshold cannot exceed the diameter, twice the radius, since no chord is longer than that. The default threshold equals the inscribed triangle side, giving the classic values. Teaching tool.
How it works
Let t be the threshold length divided by twice the radius. A chord beats the threshold when its perpendicular distance from the centre is small enough. The random endpoints method gives a probability of one minus two over pi times the arcsine of t. The random radius method gives the square root of one minus t squared, because the midpoint distance is uniform along the radius. The random midpoint method gives one minus t squared, because the midpoint is uniform over the disk, so its distance follows an area law. When the threshold equals the triangle side, the square root of three times the radius, t is the square root of three over two, and the three formulas collapse to one third, one half and one quarter.
Worked example
Take a radius of 1 and a threshold of 1.7320508, the square root of three, which is the side of the inscribed equilateral triangle. Then t is 1.7320508 divided by 2, about 0.8660. The random endpoints method gives one minus two over pi times arcsine of 0.8660, which is 33.33 percent. The random radius method gives the square root of one minus 0.8660 squared, which is 50.00 percent. The random midpoint method gives one minus 0.8660 squared, which is 25.00 percent. Three sound methods, three different answers, all for the same innocent-sounding random chord.
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