Two pairs of points are randomly chosen on a circle. Find the probability that the line joining the two points in one pair intersects that in the other pair.

I've been thinking over this problem, assuming one pair and finding that the other pair has to be entirely in one of the two arcs of the circle that the first pair of points divides it into.

But I've not been able to find an explicit answer. If I assume one point to be (a,b), I'm not able to manage the other three points.

Please help me out. Thank you.

  • 2
    $\begingroup$ This question strikes a chord with me. $\endgroup$ – David Oct 18 '14 at 13:37

For any configuration of the four points, there are three ways they can be divided into pairs. That means that as you go through all possible placements of the two chords, each configuration of points, regardless of which are paired with which will appear three times. Only one of those ways will give intersection.

Extra intuition on why the answer is lower than $0.5$: If you choose one pair first, and that pair's chord happens to be a diameter, then the probability that the other two points are on opposite sides of that diameter is $0.5$. If the first chord is not a diameter, then the probability of intersection is less than $0.5$.

  • $\begingroup$ So the probability is 1/3 then? $\endgroup$ – pkwssis Oct 18 '14 at 13:12
  • $\begingroup$ Yes, it is. $ $ $\endgroup$ – Arthur Oct 18 '14 at 15:09

The probability that any points overlap is $0$, so we can come up with an "ordering" of points with probability $1$. With pairs $(p, q)$ and $(r, s)$, let $p$ be the "smallest" point and define the "size" of a point as the clockwise angular distance from $p$. Since points are selected uniformly, it is the case that each unique "ordering" of points has equal probability. (e.g. $p<q<r<s$ has equal probability to $p<r<q<s$). Pairs of points intersect in two of these cases, and fail to intersect in four.

So the desired probability is $\frac{1}{3}$.

  • $\begingroup$ Why one case? The line segments fail to intersect in 4 out of 6 cases. So the required probability is $\dfrac{2}{6}=\dfrac{1}{3}$, isn't it? $\endgroup$ – pkwssis Oct 18 '14 at 13:11
  • $\begingroup$ @FenfyangWang You've forgotten four cases: $$p<q<s<r\\ p<r<s<q\\ p<s<r<q\\ p<s<q<r$$ $\endgroup$ – Arthur Oct 18 '14 at 15:12

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