We have an $m\times n$ rectangle. Two players take turns placing down diameter-$1$ coins that don't overlap with previously placed coins. Whoever cannot place a coin loses. Who wins?
(Note: coins are not required to fit neatly on a grid, so most turns will have uncountably many options.)
I'm fairly certain we are guaranteed a winner, even though the game tree has infinitely many vertices, because it's not infinitely deep: there's an upper bound to the length of a game, and if I recall correctly there's a general theorem that implies one player must have a winning strategy in such a case. I don't know how to determine what that winning strategy is, though, or who has it. Circle packings are pretty messy, so I'm not quite sure where to begin.
(For definitiveness, I will say that coins meeting at only one point - coins that are tangent - are allowed, though I am curious how the game changes if this rule is changed.)
I realized this several minutes after posting this question. Player one wins. Player one plays in the center; forever after that, player one plays directly opposite player two.
EDIT: As Ivan Kaznacheyeu points out, this is assuming there is enough space for the first coin.