There is no answer to this problem, I think, unless you make the assumption that unmarked cards have to be played first before marked cards. If unmarked cards have to be played first, then odd can keep playing even numbers and the only way even can win is if even writes down an even number, but then eventually even will be left with only even cards whereas odd will have blank cards left, so at that point odd can keep writing down odd numbers and/or playing odd cards and win the rest of the hand. It's also easy to calculate the worst-case minimum number of moves for odd to win if they use this strategy. If players are allowed to play marked or unmarked cards, then consider the following. If even has an even or odd marked card in their hand, then they play that card with 50/50 probability, else they play the opposite card, possibly marking a card if necessary. Now if odd decides to always play even, and even has an even card, then even will win but even will still have unmarked cards so that they have the ability to play odd in the future. In essence, if one player decides to play even/odd with 50/50 probability using marked cards when they can, then I don't think you can predict the winner of the game, or how many moves it will take to win.