Suppose $ n $ chords are uniformly chosen on a circle. What will be the expected number of pairs of intersecting chords?

From this discussion (Expected number of intersection points when $n$ random chords are drawn in a circle ) I found out that the expected number of points of intersection is $ n(n-1)/6 $. But what will be the expected number of pairs of intersecting chords? For example, if $5$ chords meet at a point, then the number of pairs of intersecting chords will be ${5 \choose 2}$.

The only approach I can think is that we can assume that the endpoints of the chords are distinct, since we are talking about geometric probability here and the probabiliy of this happening is $1$.

  • 1
    $\begingroup$ You need to define what "n chords are uniformly chosen on a circle" means $\endgroup$ – leonbloy May 10 '13 at 15:18
  • 1
    $\begingroup$ Anyway, the "number of pairs of intersection chords" equals the number of intersection points, except for the zero-measure event of more than two chords intersecting on a single point, hence the answer seems trivial. $\endgroup$ – leonbloy May 10 '13 at 15:24
  • $\begingroup$ @leonbloy It means two points are uniformly and independently chosen on the circumference of the circle and a chord is drawn joining these two points. This process is repeated $n$ times to draw $n$ chords. $\endgroup$ – user77204 May 10 '13 at 15:32
  • $\begingroup$ @leonbloy I disagree with your second comment. Please read the question carefully. $\endgroup$ – user77204 May 10 '13 at 15:33
  • 1
    $\begingroup$ @user77204 What leonbloy wantetd to point out is that the answer to your question must also be $n(n-1)/6$. The probability of more than two chords intersecting in one point is zero. $\endgroup$ – Hagen von Eitzen May 10 '13 at 15:53

Community wiki answer so the question can be marked as answered:

As discussed in the comments, the expected number of intersecting pairs and the expected number of intersection points is the same, since the probability for two intersection points to coincide is zero.

| cite | improve this answer | |
  • $\begingroup$ @bof: I'm not sure I understand what you mean. As I understood the question, the OP was worried that some intersection points might coincide, thus making the number of intersection points lower than the number of intersecting pairs. The probability for this to happen is $0$. I don't understand how your comment relates to this. $\endgroup$ – joriki Jun 18 '16 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.