$2n$ points on a circle. What's the probability two strings will intersect? $2n$ points are placed on a circle with equal intervals between them. Assume $n>1$. We randomly pick two points and draw a straight line between them. Out of the remaining $2n-2$ points, we randomly pick another pair of points and again draw a straight line between them. What's the probability the lines will intersect?
 A: Hint: Pick four points from your set of $2n$. How many ways are there to divide them into two pairs to draw line segments between?  How many of these have the two line segments intersecting?
A: Because of the infinite symmetries of the circle and the similarity of all circles, we can just use the unit circle, and we can let the first point $P_1$ be fixed at $(0,1)$. 
Let $\theta$ be a random variable for the second point $P_2$, so that $\theta\in [0,2\pi]$ chooses another point on the unit circle. Again, because of the horizontal symmetry of the circle, we may instead choose $\theta$ in the interval $[0,\pi]$. Here's a picture of what the placement of the first two points might look like:

Now for our third point $P_3$, let $\phi$ be our random variable. There are two cases - one in which $P_3$ is inside the minor arc $P_1P_2$ (the one marked with a red angle in the picture), and the other in which $P_3$ ends up on the other side of the arc. Let $C_A$ be the probability that $\overline{P_1P_2}$ and $\overline{P_3P_4}$ intersect inside the unit circle when $P_3$ is in the minor arc, and let $C_B$ be the probability when $P_3$ is not in the minor arc. Then, if $A$ is the probability that $P_3$ is in the minor arc and $B$ is the probability that it is not in the minor arc, the probability of intersection (given the value of $\theta$) is
$$P(I|\theta)=AC_A+BC_B$$
And it is trivial to determine that
$$A=\frac{\theta}{2\pi}$$
$$B=1-\frac{\theta}{2\pi}$$
Furthermore, notice that intersection happens whenever $P_3$ and $P_4$ are in different arcs, so
$$C_A=1-\frac{\theta}{2\pi}$$
$$C_B=\frac{\theta}{2\pi}$$
So we have
$$P(I|\theta)=2\bigg(\frac{\theta}{2\pi}\bigg)\bigg(1-\frac{\theta}{2\pi}\bigg)$$
Now we need to integrate this probability over all possible values of theta, so we have that the final answer is
$$\frac{1}{\pi}\int_0^\pi 2\bigg(\frac{\theta}{2\pi}\bigg)\bigg(1-\frac{\theta}{2\pi}\bigg)d\theta$$
$$=\frac{1}{2\pi^3}\int_0^\pi 2\pi\theta-\theta^2 d\theta$$
$$=\frac{1}{2\pi^3} \bigg[\pi\theta^2-\frac{1}{3}\theta^3\bigg]_0^\pi$$
$$=\frac{1}{2\pi^3} \bigg[\pi^3-\frac{1}{3}\pi^3\bigg]$$
$$=\frac{1}{2\pi^3} \bigg[\frac{2}{3}\pi^3\bigg]$$
$$=\color{\green}{\frac{1}{3}}$$
