# Why is the probability that two random chords intersect $2/3$?

In a circle, a random chord is drawn by connecting a uniformly random point on the circle, with a uniformly random point on the opposite semicircle. (So the chord's length is at least $$\sqrt2$$ times the radius.)

What is the probability that two such independent random chords intersect?

I used Excel to run $$10^6$$ trials, and the proportion of intersecting chords was $$0.6665$$, suggesting that the probability is $$2/3$$. I am seeking an intuitive explanation for this result.

## Examples of intuitive explanations

Here is an example of an intuitive explanation. Suppose, instead, a random chord is drawn by connecting two uniformly random points on the circle, and we want to find the probability that two such chords intersect. First pick four random points on the circle, labelled $$A,B,C,D$$ going clockwise. Then pair them randomly, and connect each pair to form two chords. The chords intersect if and only if $$A$$ is connected to $$C$$, which has a probability of $$1/3$$.)

Here and here are other examples of intuitive explanations.

## Context

I have been thinking about probability questions that have simple answers, but seem to resist intuitive explanation, for example this one. I am also interested in questions about random chords, for example this one.

## EDIT

If a random chord is drawn by connecting a uniformly random point on the circle, with a uniformly random point chosen from the arc with central angle $$n\pi$$ directly opposite the first point, then what is the probability that two such chords intersect?

Well, if $$n=0$$, then the probability is obviously $$1$$. And if $$n=2$$, then the probability is $$1/3$$, as explained above. So, assuming that the relationship between $$n$$ and the probability is linear, then it follows that if $$n=1$$ (which is the OP) then the probability is indeed $$2/3$$. But is it intuitively obvious that the relationship is linear?

(In fact, numerical investigation shows that the relationship is linear, i.e. the probability $$1-\frac{n}{3}$$.)

• I simulated $10^6$ samples using Mathematica, and it seems that the estimated probability of two such chords intersecting is actually $\frac{2}{3}$. Perhaps can you share a bit about your method of simulation? (My understanding is that a chord $AB$ is sampled by (1) choosing $A$ uniformly at random from the circle, and (2) choosing $B$ uniformly at random from the semicircle which is "centered" at the antipodal of $A$.) Commented Nov 9, 2023 at 8:44
• I understood the problem in the same way as Sangchul Lee.
– Plop
Commented Nov 9, 2023 at 8:51
• @SangchulLee Your understanding is correct. Oh, you're right! I made a mistake in my Excel sheet, and after correcting it, I also get extremely close to $2/3$. I will edit. Fortunately the probability is still "nice".
– Dan
Commented Nov 9, 2023 at 8:57
• Calculating it with integrals, I also get $\frac{2}{3}$. Desmos link (beware, it takes some time to compute): desmos.com/calculator/99lib0j2pa I think one could simplify it and get rid of the indicator functions (see what length intervals are good). This doesn't give much(?) intuition though. Commented Nov 9, 2023 at 9:36
• I don't have a solution, but it may be helpful to observe that the following are equivalent: (1) choose a point $A$ at random, and choose a point $B$ at random on the semicircle opposite $A$ (2) choose a point $A$ at random, choose an antipodal pair of points $B_1,B_2$ at random, and connect $A$ to whichever of $B_1$ and $B_2$ is farther away from $A$. Using this fact, there may be an intuitive explanation available by starting with four pairs $A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2$ of antipodal points and somehow considering the symmetry. Commented Nov 21, 2023 at 22:26

Long Comment. Although I don't have a fully intuitive explanation, at least I can prove that the desired probability is $$\frac{2}{3}$$ as suggested by the simulation:

The key observation is as follows:

Observation. The following distributions are all identical:

1. The distribution of a chord sampled as in OP's scheme.
2. The conditional distribution of the chord joining two uniformly chosen points on the circle, given that the central angle of the chord is obtuse.
3. The distribution of the chord whose "direction" (the direction of the vector from the center of the circle to the midpoint of the chord) is uniformly distributed and the central angle is uniformly distributed between $$90^{\circ}$$ and $$180^{\circ}$$.

Now let $$A$$ and $$B$$ be i.i.d. chords sampled as in OP's scheme. Given that the central angle of $$A$$ and $$B$$ are $$\alpha$$ and $$\beta$$, respectively, where $$\alpha, \beta \in [\frac{\pi}{2}, \pi]$$, the probability that $$A$$ and $$B$$ intersect is

$$\mathbf{P}(\text{A and B intersect} \mid \alpha, \beta) = \frac{2\min\{\alpha,\beta\}}{2\pi}.$$

Below is an animation visually proving the above equality:

Since the central angle of $$A$$ and $$B$$ are independent and uniformly distributed over $$[\frac{\pi}{2}, \pi]$$, it follows that

\begin{align*} \mathbf{P}(\text{A and B intersect}) &= \int_{\frac{\pi}{2}}^{\pi}\int_{\frac{\pi}{2}}^{\pi} \mathbf{P}(\text{A and B intersect} \mid \alpha, \beta) \, \frac{\mathrm{d}\alpha}{\pi/2}\frac{\mathrm{d}\beta}{\pi/2} \\ &= \frac{2}{3}. \end{align*}

• Respected Mr Lee, can you return your previous pictures, dedicated to this problem, and add their to this animation? Commented Nov 9, 2023 at 10:50
• @VadimChernetsov Previous pictures were imgur nct5d.png and imgur 2x6PO.png, but I don't think restoring them in the answer would be useful without restoring the whole text they were illustrating. Commented Nov 9, 2023 at 13:18
• @CiaPan, Thank you, respected Mr CiaPan. Commented Nov 9, 2023 at 13:23
• How does the probability depend from the length of chords? Commented Nov 9, 2023 at 14:31