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This is a crosspost from Math stackexchange as I left the question open a while and bountied it but received no answers.

If the game of Hex is played on an asymmetric board (where the hexes are arranged in a $k\times (k+1)$ parallelogram), the player who wants to connect the closer pair of sides can force a win, regardless of who goes first. The reflection strategy is detailed here.

I was wondering if this advantage conferred to the player connecting the closer pair of sides remains if each player plays $2$ stones per turn, rather than $1$. In this scenario the reflection-style proof seems to break down as a player can place in both a hex and its paired hex on the same turn.

We assume the first player is the one who needs to connect the farther pair of sides (as it is easy to see first player wins if they need to connect the closer pair). From playing around with a pencil and paper, I can tell that player $1$ wins for $k=1,3$, and player $2$ wins for $k=2$. It seems like player $2$ may also win for $k=4$ though I haven't worked through every possibility and fully convinced myself of that.

Who is the winner for large $k$? Will it always depend only on the parity of $k$? Or for sufficiently large $k$, will one of the two advantages (first move vs. shorter side) always win?

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    $\begingroup$ You could also consider "balanced" doublemove play, where the first player places only one stone on his first turn, and the "double moves" begin with the first turn of the second player. This is the usual rule in chess variants where the players make two moves in a turn. $\endgroup$
    – bof
    Commented May 13 at 6:55

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