Uniform Distribution of Chords

geometric-probabilitygeometryprobabilityprobability distributionsprobability theory

In the context of the Bertrand Paradox, I understand that different methods of choosing a chord lead to different probability density functions. For instance, selecting a chord based on a random central angle tends to favor longer chords since the derivative of chord length with respect to the chord angle is zero when the chord angle equals $\pi$ (or $180°)$.

However, I am looking for a method that ensures a uniform distribution of chords, where the probability density is equal for all chords. One approach I considered is randomly choosing two points on the circle's circumference, as it doesn't seem to favor any particular chord. However, I am unable to prove this. Is there a method to uniformly choose a chord, and if so, can you provide proof or explanation for its uniformity?

Best Answer

Considering Bertrand's "paradox":

I've made a hobby of collecting references to this problem and I have about 500!

Note part of the conundrum in Bertrand's paradox is figuring out "chord" and "secant".

Choosing a midpoint over the area is just wrong. Bertrand didn't say "circular area" he explicitly said circle. We all learned in grade school that a circle has an infinite number of diameters. So the center isn't just any point in the area, it is an asymptote. Thus at the center of the circle the density of points must go to infinity. aka Gabriel's horn, or Gabriel's wedding cake. (oddly I think just one of my references points out this fact.)

Now consider choosing two points randomly over the circumference of the circle. However let's add a twist. Let's use a measuring circle (radius r) concentric with the chosen circle R (radius R). The midpoint of any diameter longer than the side of the inscribed equilateral triangle triangle must be within the circle inscribed in the triangle. So for two randomly chosen points on our measuring circle then there are three different intervals to consider for $r$:

[0, 1/2R] the probability of a longer chord is 1
(1/2R, R] the probability of a chord decreases from 1 to a 1/3
(R, $\infty$] the probability rises from just over a 1/3 to 1/2

Note that splicing the various curves together does not result in a continuous function. In particular at r=R there is a discontinuity.

So there are two "magic" radii to consider. First when r=R, and also when r=$\infty$. Of course when counting chords for r>R one must also stipulate that the chord cuts circle R.

Obviously Bertrand's paradox relies on the symmetry of the circle. Now generalize the problem. Given an equilateral triangle of side $\sqrt{3}$ what is the probability that a random chord is longer than the radius of the inscribed circle? Now it isn't so easy to apply geometric probability when choosing two points on the circumference. It is just as easy for a Monte Carlo computer program.

Let's think deeper about this problem however. If the measuring circle has a radius of R, what happens at a circle R' 1 kilometer away? All of the random lines generated from the circle R when r=R will be essentially parallel at circle R'. So using two random points on the circumference of circle R doesn't generate a whole plane of random lines. However if we make our measuring circle 1 astronomical unit in diameter, the the lines are essentially isotropic and of the same density at both our circle R and at our circle R'.

Now let's also think of the problem another way. I have two sheets of rigid plastic. On one is a circle and on the other much larger one is a line. Is throwing the line on the circle different that throwing the circle on the line? For throwing a circle on a line the answer is unequivocally 1/2 withe condition that only throws where the circle intersects the line are counted. From Buffon's needle problem we can see that a large area covered with parallel lines space R distance apart will always yield one random chord.

The use of a measuring circle of infinite radius immediately follows from the work of Crofton who found that for for convex body the number of random lines of the plane cutting the convex body is equal to the perimeter of the body. Interestingly for a line segment the perimeter is two times the length - think about an infinitely thin rectangle.

Related Solutions

Probability – Solutions to Bertrand Paradox

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.

Probability chord of bigger circle intersects smaller circle

I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.

This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.

As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.

Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.

You can get $\frac{1}{2}$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $[0,r]$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $[0,2\pi)$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $<\frac{r}{2}$, which happens exactly $\boxed{\frac{1}{2}}$ of the time.

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Probability – Solutions to Bertrand Paradox

Probability chord of bigger circle intersects smaller circle

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