Bertrand's Paradox

Recently, I have been getting back into probability theory where I came across a seemingly simple problem but was surprised how some simple assumptions that we do not take into consideration could lead to some very interesting results. This problem while searching for it I found out was called “Bertrand’s Paradox”.

Today I will share my understanding of the different solutions that I found and my conclusion on what exactly the resolution for this paradox is. Let’s begin😁.

Problem Statement and its solutions

The problem statement is as follows:

Take a unit circle and draw a chord uniformly at random. What is the probability that the length of the “random chord“ is greater than \(\sqrt{3}\) ?

As we can see the problem statement seems pretty simple. Let’s try to solve this. I will try to explain my understanding of three different ways to solve this problem. Let’s begin:

1. Argument-1 (The random mid-point method): It is known that the perpendicular distance from the center of the circle to any chord of same length is same and the perpendicular intersects the chord at its mid-point. This implies that is a chord of length \(\sqrt{3}\), the perpendicular distance comes out to be \(\sqrt{1^2 - {\frac{\sqrt{3}}{2}}^2} = 0.5\). Now, if we draw a circle of radius \(0.5\) inside the unit circle, then any chord drawn at random which has length greater than \(\sqrt{3}\) will have its midpoint lying inside the inner circle. So, the probability of drawing a chord at random to have length greater than \(\sqrt{3}\) is

$$Pr(\mbox{Length of Random Chord} > \sqrt{3}) = \frac{\mbox{Area of inner circle}}{\mbox{Area of the unit circle}} = \frac{\pi(0.5)^2}{\pi(1^2)} = 0.25$$

2. Argument-2 (The random end-point method): If we choose a random point A on the unit circle and draw a tangent to the unit circle at that point then, the length of any drawn chord from that point A can only be greater than \(\sqrt{3}\), if and only if the angle \(\theta\) that the chord makes with the tangent lies in the range \([\frac{\pi}{3},\frac{2\pi}{3}]\). Since the chord is drawn at a random so the probability of its length be greater than \(\sqrt{3}\) is

$$Pr(\mbox{Length of Random Chord} > \sqrt{3}) = \frac{\frac{\pi}{3}}{\pi} = \frac{1}{3} \approx 0.33$$

3. Argument-3 (The random radial point method): We can also see from the first argument that perpendicular distance from the center of the circle to a chord of length \(\sqrt{3}\) is \(0.5\). So, for a chord drawn at random to have length greater than \(\sqrt{3}\), the perpendicular distance from the center of the circle has to be less than \(0.5\). So, the probability of the random chord to have its length greater than \(\sqrt{3}\) is

$$Pr(\mbox{Length of Random Chord} > \sqrt{3}) = \frac{0.5}{1} = 0.5$$

It seems quite strange that why do we get three different answers for a seemingly simple problem. This happens because the sample space under consideration in all three methods or arguments are different due to which it seems like we are solving the same problem stated above but in reality, these are three different solutions to three different problems. The difference in problems arises from the fact that how “randomness” is defined in this problem.

As stated in the problem “chord is drawn uniformly at random”, means that all the possible outcomes under consideration should be equally likely (The principal of indifference). But this paradox simply shows that the applicability of this principal of indifference needs to be thought of critically when the sample space under consideration becomes infinitely large.

This is just a small explanation from my side if you want to explore more on this, I recommend you to go through the below given link.

References

1. Drory, A. (2015). Failure and uses of jaynes’ principle of transformation groups. Foundations of Physics, 45(4), 439-460. https://doi.org/10.1007/s10701-015-9876-7

2. Bertrand paradox (probability) - Wikipedia

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