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.
Best Answer
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.