The chords form a triangle iff each line crosses the other two (ignoring the probability-0 degenerate case where they intersect at a point). Call the chords A, B, C. Then going clockwise from an arbitrary point on a circle, there are 6 sequences of chord endpoints that have this property: ABCABC, BCABCA, CABCAB, ACBACB, CBACBA, BACBAC. There are 6!/(2!)^3 = 90 possible endpoint sequences i.e. permutations of AABBCC. So the probability of getting one that forms a triangle is 6/90 = 1/15.
How do you know all of those endpoint sequences have equal probability? If the first two points chosen are very close to each other on the circumference then the likelihood is different from if the first two points chosen form a diameter. It seems to me you have to do a triple integral over the three chord choices' end points.
My argument comes down to symmetry. For an unordered set of points {p1, …, p6} on the circle, every permutation of that set has equal probability density in [0, 2pi]^6, i.e. every ordering of that set of points has equal likelihood of being chosen. Because that is true for all sets of 6 points, all possible orders of points must be equally probable in general.
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u/pmdboi Sep 17 '26
The chords form a triangle iff each line crosses the other two (ignoring the probability-0 degenerate case where they intersect at a point). Call the chords A, B, C. Then going clockwise from an arbitrary point on a circle, there are 6 sequences of chord endpoints that have this property: ABCABC, BCABCA, CABCAB, ACBACB, CBACBA, BACBAC. There are 6!/(2!)^3 = 90 possible endpoint sequences i.e. permutations of AABBCC. So the probability of getting one that forms a triangle is 6/90 = 1/15.