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This question can be phrased in terms of (0,1)-matrix permanents. Player 1 places copies of the number 1 and player 2 places copies of the number 0, and empty cells count as 0. Player 1 wants to achieve a non-zero permanent.

Permuting the rows and columns, or taking the matrix transpose gives an equivalent game.

EDIT: I was actually quite surprised at how easy it is for the o's to make a losing mistake! Consider the position on the left in the diagram below (after 4 o's have been placed and 3 x's have been placed, with x to move). An equivalent position seems like a very natural position to arise. Surprisingly, x can force the win from here! I put in yellow the forced moves by o (otherwise x goes in that square and o loses).

Consequently, o has made a mistake in allowing the first position above, and can win instead as illustrated in the following sketch, for example.

1

This question can be phrased in terms of (0,1)-matrix permanents. Player 1 places copies of the number 1 and player 2 places copies of the number 0, and empty cells count as 0. Player 1 wants to achieve a non-zero permanent.

Permuting the rows and columns, or taking the matrix transpose gives an equivalent game.