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Parity Tic-Tac-Toe

Two players play Tic-Tac-Toe, one playing "Evens" and the other playing "Odds".
On each turn, a player must place a 0 (zero) or a 1 (one) in any unoccupied
space. When all nine spaces of the the board are filled, each of the eight
rows of three spaces is summed (horizontally, vertically, and diagonally.)
The goal for each player is to have the most sums of their parity (most even or
odd sums.)If played randomly, is there an advantage to either player?
Is there an obvious strategy to play? Is there an advantage to picking a
parity or going first? Can you find final configurations for which the
"Odd" player has 8,7,6,5, and 4 sums?

Source: Original.
Sorry it doesn't have an obvious "puzzle answer," but I thought I'd share it
anyway.

Solution

Mail to Ken