Misère Hex is like Hex, except that the winner and loser are swapped. Instead of trying to make a connection, you are trying to not make a connection.
One player will inevitably lose, since when the board is filled, there has to be either a red or blue connection.
Instead of saying that you lose when you make a connection, we'll instead say that it is illegal to make a connection, like how getting into check is illegal in chess. You lose if you have no more legal moves left. This definition makes Misère Hex suitable to analysis by Combinatorial game theory. In particular, for a given $m$ and $n$, we can find the game value $\operatorname {mhex}(m,n)$ corresponding to the game value of Misère Hex on a $m \times n$ board.
I conjecture that the value is of the form $x + *$ or $x$, for some dyadic fraction $x$, since most moves make the position worse for a player.
So, how do we calculate $\operatorname {mhex}(m,n)$ given $m$ and $n$?
EDIT: It appears that if generalize the game slightly, you can figure out values recursively (for example, if blue connects the top and bottom, the game is now the sum of two smaller Misère Hex games with irregular borders, and blue has extra moves).