, 2021

Solution:

The main tools we will use are linearity of expectation and Pick's theorem. Note that the resulting polygon is a lattice polygon, and thus the expected area $A$ satisfies
$$A = I + \frac{B}{2} - 1$$
where $I$ is the expected number of interior points and $B$ is the expected number of boundary points. We may now use linearity of expectation to write this as
$$A = -1 + \sum_{p \in \{1,2, \ldots, 6\}^2} \mathbb{E}\left[X_p\right]$$
where $X_p$ is 1 if the point is inside the polygon, $1 / 2$ if the point is on the boundary, and 0 otherwise. Letting $f(p) = \mathbb{E}\left[X_p\right]$, we may write this by symmetry as
$$A = -1 + 4 f(1,1) + 8 f(1,2) + 8 f(1,3) + 4 f(2,2) + 8 f(2,3) + 4 f(3,3)$$
There are many ways to continue the estimation from here; we outline one approach. Since $X_{(1,1)}$ is $1 / 2$ if and only if $(1,1)$ is one of the selected points (and 0 otherwise), we see
$$f(1,1) = \frac{1}{12}$$
On the other hand, we may estimate that a central point is exceedingly likely to be within the polygon, and guess $f(3,3) \approx 1$. We may also estimate $f(1, y)$ for $y \in \{2,3\}$; such a point is on the boundary if and only if $(1, y)$ is selected or $(1, z)$ is selected for some $z<y$ and for some $z>y$. The first event happens with probability $1 / 6$, and the second event happens with some smaller probability that can be estimated by choosing the 6 points independently (without worrying about them being distinct); this works out to give the slight overestimate
$$f(1,2), f(1,3) \approx \frac{1}{8}$$
From here, it is not so clear how to estimate $f(2,2)$ and $f(2,3)$, but one way is to make $f(x, y)$ somewhat linear in each component; this works out to give
$$f(2,2) \approx \frac{1}{4}, \quad f(2,3) \approx \frac{1}{2}$$
(In actuality the estimates we'd get would be slightly higher, but each of our estimates for $f(x, y)$ up until this point have been slight overestimates.) Summing these up gives us an estimate of $A \approx \frac{31}{3}$ or $E=103333$, which earns 10 points. The actual value of $A$ is $10.4552776 \ldots$, and so $N=104552$.