The game "Hex" is a simple game which apparently has been invented at least twice (Piet Hein and John Nash). The game consists of an n by n grid of hexagons, with two opposite sides marked as blue and the other pair of opposite sides marked as red. The red pair belongs to the red player, and the blue pair belongs to the blue player. Players alternate marking hexs either red or blue. The goal is to form a path of hexs of your color which connect one's two sides.
It is well known that there is a winning strategy for the first player on an $n$ by $n$ board. Proof sketch:
First show that in any completely filled board, there must be a winning path for at least one player.
Second, note that if the second player had a winning strategy, the first player could simply make some extra random move, and then play using the second player's strategy, with no penalty.
If one plays around on a few tiny boards with $n>2$ one will notice that edge moves are really bad first moves. In fact, a little work will show you that on a 4 by 4 grid, an initial move on an edge is a loser for the first player. The intuition is that an edge move does not give many options whereas a move in the middle of the board is much harder to block. Given this, two questions:
First, can we show that for any $n>2$, any edge move is a losing first move?
Second, suppose that the first player's move is randomly chosen; can we say anything about the proportion of those moves which lead to a winning board for the first player? My guess is that this proportion either goes to 0 or goes to 1 as $n$ goes to infinity, but my intuition on which seems to be oscillating violently.
