Intersecting Chords

To find the solution, first consider what the probability is that two random chords generated in the method of the problem intersect:

Drawing two random chords is the same thing as drawing 4 random points and grouping them into pairs. Doing it this way, there are three ways to group them, but only one way results in two crossed lines (when you connect the "diagonal" points), so the probability of choosing this pairing randomly is 1/3.

Now if there are 18 chords, there are 18 choose 2 ways to pair them up, or 153 different pairs. We previously calculated that each pair has a 1/3 chance of resulting in an intersection, so we should expect 153/3 = 51 of these pairs to intersect, so our answer is 51.

Let us consider only 2 chords A1A2 and B1B2, where A1 A2 B1 B2 are points on the circle.

Without lack of generality, let us fix A1 and let A2 move around the circle, at an angle X from A1. X varies from 0 to 2PI.

The probability of selecting X is (dX/2PI)

If the other chord intersects this chord, then : either B1 must be between 0 and X, and B2 must be between X and 2PI or B2 must be between 0 and X, and B1 must be between X and 2PI.

The probability of this is : ( X/2PI * (2PI -X)/2PI ) * 2

So total probability of this event is : (dX/2PI) * ( X/2PI * (2PI -X)/2PI ) * 2 Integrate this from X=0 to 2PI

We get X^2/4PI^2-X^3/12PI^3 between X=0 and X=2PI We get 1-2/3 We get (1/3)

Now for 18 chords, we have (18 17/2) pairs. The expected number of intersections will be (18 17/2) * (1/3) = 51

First we look into the probability whether two random chords intersect or not. We will have 4 points, and there are three ways to partition them in to 2 pairs. Only one partition will make the chords intersect, so the probability is $\frac{1}{3}$ . Now there are 18 chords here, with ${18\choose 2}$ pairs. Each pair has a probability of $\frac{1}{3}$ to intersect each other, so the answer is ({18\choose 2}\times \frac{1}{3}=51).

When we have 2 chords, let be $p$ the probability of they intersect each other. If we have 18 chords, the probability of two of them intersect is independent of other chords, then $X= \{$ Number of intersections $\}$ is a binomial distribution $X \in Bi(n,p)$ . So the expected number is $n\cdot p$

• We will calculate now the value of $n$ :

The first chord can't intersect, the second chord have a maximum of 1 intersection, the third chord have a maximum of 2 intersections, ... , the 18 chord have a maximum of 17 intersections. Then the number of maximum intersections is: $1 + 2 + \ldots + 17 = \frac{17 \cdot (17 +1)}{2} = 153$ . So we have $n = 153$ .

• We will calculate now the value of $p$ :

Let be the circle of radius 1 and centered at $(0,0)$ . It's allowed to assume that the first chord starts at $(1,0)$ and finishes at A, and the second one starts at B and finishes at C.

Considering that the points A, B and C can be written in the polar coordinate system, then A, B and C are independent uniformly distributed randoms variables on $[0, 2 \pi]$ .

Now the probability density function of (A,B,C) is $f(a,b,c)=\frac{1}{(2 \pi)^3 }$ if $0 < a,b,c < 2 \pi$ .

With the previous explanations, the event "two chords intersect each other" is done by the set $\{ (a,b,c) \in [0, 2\pi] \times [0,2\pi] \times [0,2\pi] / 0 < b < a < c < 2\pi \\ \textrm{ or } 0 < c < a < b < 2\pi \}$ Hence $p = 2 \int_0^{2 \pi} \int_a^{2 \pi} \int_0^a \frac{1}{(2\pi)^3} \partial b \partial c \partial a = \frac{1}{3}$

Then we have that expected value is $153 \cdot \frac{1}{3} = 51$

Any two chords have a probability 1/3 of intersecting. There are 18 choose 2 pairs of chords. Each has probability 1/3 of intersecting. The expected value is thus 18 choose 2 times 1/3, which is 51.

We may assume that all chords have distinct vertices, since this occurs with probability 1. Given any pair of chords $C_i, C_j$ , if we consider the 4 endpoints of the chords around the circle, then the probability that the pair belonging to one chord are opposite is $\frac{1}{3}$ , so the probability that any pair of chords intersects is $\frac{1}{3}$ . Since there are $18$ chords, there are $\binom{18}{2}$ pairs of chords, so the expected number of pairs of crossing chords is $\binom{18}{2}/3 = 51$ .