The game goes like this:
Before the (American) football game begins, $5$ players each choose $20$ squares on a $10 \times 10$ grid. After choosing, the numbers $0-9$ are chosen at random twice, once to label each row and once to label each column. At the end of the game, the last digit of each team's score is taken, and the player on the corresponding lattice wins.
For example, if rows represent the home team's final score and the columns represent the away team's score, then the winner is whoever chose the square indexed $(\text{home mod } 10, \text{away mod } 10)$. There are, indeed, most common NFL scores, so the scores themselves are not randomly distributed. My question is
Are there squares that increase the odds of winning?
I assume no, since the fact that the indices are randomized, the grid is also randomized, but I wanted to know if my intuition was wrong, and if so, why?
I would instinctively assume yes because of limited modulo residues, and in this case, that is correct.
From my non-American understanding of American Football scoring, you score 6 points for a touchdown, one for a post-touchdown extra point, 3 for a field goal, and 2 for a safety.
Let's make some assumptions to help.
With these assumptions in mind, we are most of the time scoring in $7$s and $3$s with $60:40$ probability. Conveniently, $7+3=10$, so we can identify the end digits quite easily. Let $\tau$ be the number of touchdowns, and $\phi$ the number of field goals a given team scores. Then, the end digit $d$ is found as: $$d=\begin{cases} ... \\ 1 & \tau=\phi+3 \\ 4 & \tau=\phi+2 \\ 7 & \tau=\phi+1 \\0&\tau=\phi \\ 3&\tau=\phi-1 \\ 6 &\tau=\phi-2 \\9 &\tau=\phi-3 \\... \end{cases}$$
Most likely, the digits are going to end in $0,3,4,7$ (bearing in mind the $60:40$ tilt towards getting more touchdowns), sometimes in $1,6$ and almost never in $2,5,8,9$.
The data sheet you provided seems to line up very nicely with this.