# Bertrand paradox solutions

Wikipedia states the problem as follows: "The Bertrand paradox goes as follows: Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?"

solution being 1/2 seems to be the only solution as parallel chords have 1/2 probability to be longer than a side of the triangle.

I don't understand other solutions. Am I missing something?

To be more specific as requested, other solutions are based on circle arc and area (both are 2-dimensional).

The question is: why the use of 2-dimensional structures is appropriate/correct when solving Bertrand paradox?

Please delete the duplicate: https://math.stackexchange.com/questions/1509549/bertrand-paradox-solution which I posted as a guest.

It really isn't a paradox. It is more of a warning that statistics is a quirky area of mathematics.

It hinges on how a "random chord" is chosen. With no process defined, then the solver of the problem is allowed to use whatever process she wants. Choosing different processes gives different probabilities.

But let's turn the problem on its head.

Let's define a random chord as the chord between two points chosen randomly from the circumference of a "measuring circle." We'll let the measuring circle be concentric with the circle of interest. Now let's see what happens as we increase the diameter of the measuring circle.

   Diameter of measuring circle      probability
< 0.25                     1
>0.25  to <1                drops from 1 to 1/3
1                       1/3
100,000                   almost 1/2


Bertrand's 1/4 solution is "wrong" for a circle. It requires that the midpoint of the chord is evenly distributed over the area of the circle. He just hand waves away the center of the circle since it is just one point and one point has no area. But by definition the center of the circle has an infinite number of diameters passing through it, not 1. So the center of the circle must be an asymptote. Hence "circleness" prevents this solution.

There is also a symmetry wrinkle pointed out by Poincare. Consider the flip problem. We throw a circle of fixed diameter onto a line. Consider a hoop being tossed as a line on the floor. Given that the hoop intersects the line, then the only probability is 1/2.

• Thank you for your time and for your answer. From my perspective, I find your confirmation about the question not being an actual paradox really important. Information in your answer extends beyond "Wikipedia" and gives directions for further research. As for the process of choosing a random variable, I see it as a "reference axis" for probability distribution and might post a question regarding this topic after more research. Answer accepted. Nov 4, 2015 at 19:17
• The research problem here is whether the problem is "well defined" or not. That is to say does it have an unique solution. Jaynes' position is that it does. bayes.wustl.edu/etj/articles/well.pdf It really is an interesting problem to delve into the mysteries of probability.
– MaxW
Nov 4, 2015 at 19:26
• thank you for further reference. Nov 4, 2015 at 19:30

I came accross this paradox recently and have spent some time on it. IMHO the following holds: Bertrand's proposed method 2 gives a probability of 1/2, based on the chords of the radius that bisects the side of the triangle. This reasoning is valid: For every Radius there is only one chord which bisets it a right angles, it has the length of the side of the triangle. There exists an infinite number of chords on the "inside half" (towards the center) as well on the "outside half" (towards the circumference), the outer half being shorter, the inner half being longer then the side. As any radius (including its associated chords) can be obtained from any other by simple rotation, the ratio of longer to shorter chords is 1:1 for all possible chords in the circle, and the probability of a truely randomly selcted chord being longer then the side is therfore 1/2. When analysing selection method one as given by Bertrand, one finds that this selection method generates a subset of all existing chords (only chords on radii pointing to the arc running from 270 through 0 to 90 degrees), and only the chords bisected at right angles by the radii conncting to the arcs between 60 and 90, and 270 and 300 respectively, are longer then the side. This gives twice as many shorter then longer chords, and the probability 1/3 is correct with respect to the biased subset, but not with respect to all existing chords. Considering method three remember that all chords are bisected at right angles by a radius, and every radius bisects in this way an infinite number of chords. Furthermore, the circles are concentric (the have the same midpoint). It also holds that a radius runs from the centre of the circle to a point on the circumference. As a consequence, all radii of the outer circle pass throug the inner circle, and their "inner halfs" are a radius of the inner circle. The argument that a bigger arae implies more chords is clearly false, if correct it would imply that there are more radii existing in the ring, which is not possible as every radius has to connect to the center, and every additional radius would increase the number of radii in the inner circle, and consequently the number of chords. To clarify, I'm not a mathematician, and the above is more probable wrong then correct, but I can't find fault with the reasoning. Further reading on this problem quickly exceeded my mathematics, and it will take some time to understand the beginning of the complications that according to the professionals do exist, if ever. :-)