# Expected number of intersection points when $n$ random chords are drawn in a circle

There are $n$ random chords drawn in a circle. I am trying to determine the expected number of points in which the chords intersect within the circle.

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You need to specify the distribution the random chords are chosen with. There are several intutively "reasonable" distributions on the chords in a circle that give different answers; see Bertrand's paradox. – Henning Makholm Oct 16 '11 at 16:02

If the chords are identically distributed, this is $\frac12n(n-1)$ times the probability that two given chords meet. Two chords are determined by four points on the circle. Amongst the three possible pairings of these four points, one produces meeting chords and the two others do not (place without loss of generality the four points at North, South, East and West positions, then the winning pairing is N-S vs W-E).

Assume that by symmetry, each pairing of the four points is equally probable. Then the mean total number of intersection points is $\frac16n(n-1)$.

The condition for this result to hold is that, for every two chords $C=[A,B]$ and $C'=[A',B']$, the distribution of $\{A,B,A',B'\}$ is exchangeable. For example, one can choose each endpoint of every chord uniformly on the circle and independently of the others.

Other procedures to choose the chords may produce other values, for example if one chooses randomly uniformly the midpoint of every chord in the disc and if these $n$ choices are independent, the mean total number of intersection points is $\frac12pn(n-1)$ where $p$ is the probability that two chords chosen according to this procedure meet.

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Are you sure that all those $\frac12n(n-1)$ events are independent? Perhaps they are, but to me at least, it's not obvious. – TonyK Oct 16 '11 at 14:12
@TonyK, these are certainly not--and one does not care since the expectation of any sum, independent or not, is the sum of the expectations. – Did Oct 16 '11 at 14:20
Yes, you are right. – TonyK Oct 16 '11 at 15:55