Prove or disprove: There is a way to choose independent random chords in a circle so that their intersections are uniformly distributed in the circle.

Prove or disprove:

There is a way to choose independent random chords in a circle so that
their intersection points (given that they exist) are uniformly distributed in the circle.

One common way to choose a random chord in a circle, is to connect two uniformly random points on the circle. Assume the circle has center at origin with radius $$1$$ : $$x^2+y^2=1$$.
Numerical investigation (using Excel) suggests that, when two such chords are chosen independently and they intersect, the probability that their intersection point lies inside the smaller circle having center at origin with radius $$1/2$$ : $$x^2+y^2=\frac{1}{4}$$, is $$\frac{1}{6}$$. But the smaller circle's area is $$\frac{1}{4}$$ the larger circle's area. So the intersection points are not uniformly distributed in the circle.

So I wonder if there is some method of choosing independent random chord, that yields intersection points that are uniformly distributed in the circle.

Context: I have been generally interested in questions about random chords, for example: cutting a pizza, a counter-intuitive result, and a possibly intuitive result.

• Interesting. In the usual discussions of the Bertrand Paradox at least three methods for choosing "random chords" are usually discussed. Do all three fail?
– lulu
Commented Nov 10, 2023 at 12:33
• @lulu "Random radial point" method works, as shown by SmileyCraft's answer. I think "random midpoint method" fails: numerical investigation suggests that the probability that an intersection (within the original circle) lies in $x^2+y^2=\frac14$, is $\frac{1}{16}$.
– Dan
Commented Nov 10, 2023 at 13:17
• "are uniformly distributed in the circle." << I'm a bit confused by this. Are the chords supposed to intersect at their endpoints on the circle? Or does "circle" mean "disc" in this case?
– Stef
Commented Nov 11, 2023 at 11:01
• @Stef The chords intersect inside the circle (not at the endspoints of the chords). Or equivalently, the chords intersect on the disk that is bounded by the circle.
– Dan
Commented Nov 11, 2023 at 11:06

The random radial point method should suffice: Choose a uniformly random diameter of the circle, choose a uniformly random point on the diameter and construct the chord through this point and perpendicular to the diameter.

Let $$p$$ be the probability two generated chords intersect.

Then for any circle of radius $$r<1$$ contained within the larger circle of radius $$1$$, any generated chord intersects the smaller circle with probability $$r$$, and the conditional distribution is equivalent to applying the random radial point method on the smaller circle.

It follows that two generated chords intersect within the smaller circle with probability $$pr^2$$ regardless of where the smaller circle lies within the larger circle.

• Just realize, you do not even need to choose the diameter uniformly. The same logic still applies for any arbitrary non-deterministic distribution of diameters. If you require the distribution of chords to be rotationally symmetric however, I believe the random radial point method is the only distribution giving a uniformly random intersection point. If $f$ is the probability density function of the distance of the chord to the center, you basically need $\int_0^\pi f(r\sin\theta)\ \mbox{d}\theta$ to be constant with respect to $r\in[0,1]$. Commented Nov 10, 2023 at 13:19

This is a comment on @SmileyCraft's nice answer.

The random radial point method which @SmileyCraft's answer refers to, can be thought of as first drawing $$m$$ regular polygons, each with $$n$$ sides, inside the circle, concentric with the the circle and with corresponding sides parallel, with radii that are in uniformly distributed (including a polygon with radius $$0$$, which is not visible).

Then we extend each side of each polygon, to produce chords of the circle.

Then we let $$n$$ and $$m$$ approach infinity (moving from discrete to continuous). It is intuitively obvious that the intersections of the chords are uniformly distributed throughout the circle.