Q: What is the expected area of a cyclic quadrilateral inscribed in a unit circle? Choose four points randomly on the circumference of a circle with radius $1$. Connect them to form a quadrilateral. What is the expected area of this quadrilateral? 
I have attempted to simulate to find an answer but not sure how to approach finding an exact value. The simulation fixes one of the points at $0$ and generates 3 other points uniformly around the circle between $0$ and $2\pi$. Then it orders the points and takes the differences between them to get the 4 central angles of the quadrilateral. From these 4 central angles it finds the length of each side $s_i$ using the formula $s_i=2sin\frac{\theta_i}{2}$. Once I have the four sides I can use Brahmagupta's Formula to find the area $K$ of the quadrilateral. I repeat this 100k times and take the average of $K$ and get $K\approx.96$.

A: The four central angles have the same distribution, and the expected area is $4$ times the expected area of one of the four triangles spanned by the central angles.
The probability density of the central angle is proportional to the volume it leaves to the remaining two points: $f_\alpha(\alpha)\propto(2\pi-\alpha)^2$. Normalization yields $f_\alpha(\alpha)=\frac3{(2\pi)^3}(2\pi-\alpha)^2$.
The area of the triangle spanned by the central angle $\alpha$ is $\frac12\sin\alpha$. Thus the expected area of the quadrilateral is
$$
4\int\limits_0^{2\pi}\frac3{(2\pi)^3}(2\pi-\alpha)^2\frac12\sin\alpha\,\mathrm d\alpha=\frac3\pi\approx0.955\;.
$$
