[math-fun] unbalanced hex

Here's a cute question, which may be trivial: Hex is of course a win for the first player. Let's give the second player a small advantage: every million moves (starting with the millionth) the 2nd player can place two stones instead of one. I conjecture that for all sufficiently large boards, the second player now has a winning strategy. Is this obvious? I know that the second player wins if we provide her with some other kinds of advantage, like removing one row of the board, but I don't see any obvious way to reduce this to that case. - Cris p.s. Is there anyone out there who sells a nice wooden Hex board?