Tetragon for Two

There is a $\frac{1}{4}$ chance that the points are on the same side, a $\frac{1}{2}$ chance that they are on adjacent sides, and a $\frac{1}{4}$ chance that they are on opposite sides.

Case 1 : The points are on the same side.

Let $x$ be the distance from a vertex to the first point, and let $y$ be the distance from the same vertex to the second point.

Then the expected value of the distance between the points is:

$\begin{aligned} \int_{0}^{1} \int_{0}^{1} |y-x|\ dy\ dx &= 2\int_{0}^{1} \int_{0}^{x} (y-x)\ dy\ dx \\ &= \frac{1}{3} \end{aligned}$

Case 2 : The points are on adjacent sides.

Let $x$ be the distance from a vertex to the first point, and let $y$ be the distance from the same vertex to the second point.

Then the expected value of the distance between the points is:

$\int_{0}^{1} \int_{0}^{1} \sqrt{x^2+y^2}\ dy\ dx = \frac{1}{3}\left(\sqrt{2}+\ln \left(1+\sqrt{2} \right) \right)$

Case 3 : The points are on opposite sides.

Let $x$ be the distance from a vertex to the first point, and let $y$ be the distance from the adjacent vertex to the second point.

Then the expected value of the distance between the points is:

$\int_{0}^{1} \int_{0}^{1} \sqrt{1+(y-x)^2}\ dy\ dx = \frac{2-\sqrt{2}}{3} + \ln \left(1+\sqrt{2} \right)$

Weighting each of these expected values by their respective probabilities gives an expected value of approximately $\boxed{0.735}.$