Answer: $2+4 \ln (2)$
A vertex $P_{i}$ is part of the convex hull if and only if it is not contained in the triangle formed by the origin and the two adjacent vertices. Let the probability that a given vertex is contained in the aforementioned triangle be $p$. By linearity of expectation, our answer is simply $6(1-p)$. Say $\left|P_{0}\right|=a,\left|P_{2}\right|=b$. Stewart's Theorem and the Law of Cosines give that $p$ is equal to the probability that $\left|P_{1}\right|<\sqrt{a b-a b \frac{a^{2}+b^{2}+a b}{(a+b)^{2}}}=\frac{a b}{a+b}$; alternatively this is easy to derive using coordinate methods. The corresponding double integral evaluates to $p=\frac{2}{3}(1-\ln (2))$, thus telling us our answer.
8. For $i=0,1, \ldots, 5$ let $l_{i}$ be the ray on the Cartesian plane starting at the origin, an angle $\theta=i \frac{\pi}{3}$ counterclockwise from the positive $x$-axis. For each $i$, point $P_{i}$ is chosen uniformly at random from the intersection of $l_{i}$ with the unit disk. Consider the convex hull of the points $P_{i}$, which will (with probability 1) be a convex polygon with $n$ vertices for some $n$. What is the expected value of $n$ ?