PI found in Minesweeper - a new conjecture Playing with Monte Carlo simulations over Minesweeper, I happened by chance to find PI appearing in a pattern of the numbers relating to cells neighbouring mines on n*n grids with n mines.
My observations point to the following conjecture:

*

*We define $S_n$ as the average sum of all numbers on a quadratic Minesweeper board with dimensions $n\times n$, and a number of mines equal to $n$.

*We define $S_{n-1}$ as the average sum of all numbers on a quadratic Minesweeper board with dimensions $(n-1) \times (n-1)$, and a number of mines equal to $n-1$.

*Subtracting $S_{n-1}$ from $S_n$ yields approximately $7.85398 \approx 2.5\pi$.

The full conjecture is given in a PDF here: Github link
I, being a mere computer scientist, do not have enough insight into the topic, so I have discussed this with a professor of mathematics at my university. He thinks the edge effects are not that important, and suggests looking closer at the overlaps, further suggesting looking at the Minesweeper grid as a torus.
What are your thoughts on this curious occurence of PI? Do you find any flaws in my conjecture?
 A: To make the back-of-the-envelope computation in @Mindlack's comment exact:
With $n$ mines on an $n\times n$ board, the expected number of mines on an arbitrary field is exactly $\frac n{n^2}=\frac1n$. But if we know that a specific field is a mine, then for any other field, the expected number of mines on that field is $\frac{n-1}{n^2-1}=\frac1{n+1}$. Hence, if a field with $k$ neighbours is a mine, then the expected contribution to the sum by this mine is $k\cdot(1-\frac1{n+1})$ and therefore the expected contribution of that field (without already knowing that it is a mine) is
$$ \frac1n\cdot k\cdot\left(1-\frac1{n+1}\right)=\frac k{n+1}.$$
Now on our (non-toroidal) $n\times n$ board, we have $4$ corner fields with $3$ neighbours, $4n-8$ edge fields with $5$ neighbours, and $(n-2)^2$ inner fields with $8$ neighbours.
We conclude that
$$ S_n = 4\cdot \frac3{n+1}+(4n-8)\cdot\frac5{n+1}+(n-2)^2\frac8{n+1}=\frac{8n^2-12n+4}{n+1}=8n-20+\frac{24}{n+1}$$
exactly.
It follows that
$$ S_n-S_{n-1}=8-\frac{24}{n^2+n}$$
and
$$ \frac1nS_n=8-\frac{20}n+\frac{24}{n^2+n}.$$
It is no coincidence that the limit of these expressions as $n\to\infty$ is precisely $8$.
