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

Question

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 $1-\frac{n}{3}$.)

Answer 1

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:

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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*}