The original form of Bertrand's paradox deals with random chords of a circle. Joseph Bertrand, a french mathematician poses this problem in Calcul des Probabilities. The problem is as follows. Given a circle of radius 1, inscribe an equilateral triangle within it. What is the probability, p, that a random chord of the circle will be longer than a side of the triangle, namely √3?

After Bertrand poses this problem, he proceeds to solve it in three different ways each yeilding a different answer, hence the paradox.

One argument suggests p=1/4.
Let O be the centre of the circle and choose a random point within it, M. One can create a chord by drawing a line perpendicular to OM. Thus each possible chord of the circle can be represented by each possible random point M in the disk. Upon inspection, one can see that the chord will be longer than √3 so long as M is chosen to be within the smaller disk centered at O with radius 1/2. Since this smaller disk represents 1/4 the area of the larger, one concludes that p=1/4.

Another proposed solution is p=1/3.
Let one end of the chord, P, remain fixed while the other end, Q, varies about the perimeter of the circle. Clearly the angle, θ, between PQ and the tangent to the circle at P is uniformly distributed distributed uniformly between 0 and π. For the chord PQ to be be longer than √3, θ must lie in the interval [π/3, 2π/3]. This is 1/3 of the possible range of θ, so this argument indicates p=1/3.

Another discussion points to p=1/2.
Consider that the direction of all the chords is fixed, say vertical. This does not affect the outcome, by the properties of symmetry. Consider also the line AB which passes through O perpendicular to the chords. AB forms the diameter of our unit disk which is 1 unit in length. For ease of calculation, lets say AB=[-1,1]. Now, each possible chord must intersect with AB, but note that only the chords that pass through the interval [-1/2, 1/2] will actually be greater than √3 in length. The length of this interval is only 1/2 the length of AB, so this argument shows that p=1/2.

J. Bertrand, Calcul des Probabilitiés
J. Holbrook, Sets and the Senses


Probability theory is a great field for "paradoxes": the Monty Hall problem, the two envelope paradox, Bertrand's paradox, Pascal's wager, and many others. The nice thing about probability is that it's so easy to compute. While there's a formal basis (involving measure theory, some analysis, and for geometric problems like this one even some geometry), you don't really need it to understand what's going on.

Even worse, it doesn't really help!

As always, the problem is defining what "at random" means. In everday speech, we expect it to mean "at random, chosen according to the uniform distribution over all outcomes". And unfortunately, even that's not enough. As this example shows, there's more than one way to pick "the" uniform distribution for the problem. And that makes all the difference -- just like you'd be willing to bet equal odds on "heads" in a coin toss, but not if you knew the coin was weighted to come up "tails" 75% of the time.

Betrand gives 3 different distributions for a "random chord" -- and comes up with 3 different probabilities for it being longer than the chord of an inscribed equilateral triangle. Here are the distributions, matching each of the 3 calculations above:

  1. Fix a diameter MOM' of the circle; pick a point inside the circle with a uniform distribution; the chord is the perpendicular to MOM' through the point. The choice of M (and its antipode M') makes no difference to the calculated result; the choice of distribution does.
  2. Fix one end P of the chord, and pick the other end Q at random along the perimeter, with a uniform distribution. Here, the choice of P makes no difference; again, the choice of distribution does.
  3. Finally, fix again a diameter AB of the circle; pick a point uniformly along AB, and extend the perpendicular chord through that point. Choice of A (and B) makes no difference; the choice of distribution does.

None of the 3 choices is an obvious choice to get a "uniform distribution" among all chords. We have no clear intuition for distributions along chords; which one we pick dictates the result (just as which set of loaded dice you pick dictates the probabilty of winning at craps). We do have a clear intuition (which has a firm formal basis) for a "uniform distribution" among points -- and in each of the 3 proposed distributions for chords, we make us of a uniform distribution, but among different points.

Bertrand calculates 3 different probabilities. And the results are 3 different numbers...

Just to show how easy it is, here's how to compute another number; this one is (hopefully) less justified than any of the others.

Pick one end A of the chord. Now pick a number x uniformly on the interval [-2,2]; |x| will be the length of the chord. Pick point B on the perimeter, at distance |x| from A, going clockwise if x<0 and counterclockwise otherwise. Clearly, the probability of |AB|=|x|>sqrt(3) is precisely 2-sqrt(3).

Note that a similar construction lets us get any value for the probability: just pick x according to some arbitrary distribution which gives the desired result!

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