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