This is the set up to Bertrand's Paradox:
Randomly choose two points on a circle. Construct a line segment (circle chord) between them. Construct an inscribed equilateral triangle within the circle. What is the probability that the (randomly chosen) chord is longer than one side of the triangle?
Three equally compelling approaches to this puzzle lead two three solutions: $\frac{1}{3}, \frac{1}{4}, \frac{1}{2}$ (I won't describe those solutions here. Grant Sanderson does so eloquently in this video by Numberphile.)
Monte Carlo simulations can be run which support each of the three solutions.
The paradox:
How can the puzzle have three mutually exclusive correct solutions?
I am inclined to think that the question, although it appears to be well-formed, is otherwise. The culprit is here:
Randomly choose....
The assumption is that we can choose randomly from an infinite set of points. The Axiom of Choice sounds like a trite assumption. Yet, we know it can lead to odd results (such as the Banach Tarski Paradox).
The question:
Am I correct in my assertion that Bertrand's Paradox relies on the Axiom of Choice?
We assume that a choice function exists to select from an infinite set of circle chords. Furthermore, we pit three such functions against each other, and receive different results.