I am trying to prove the following: Let $X$ be a finite set and let $F$ be a family of subsets of $X$. Prove that if Maker has a winning strategy in the breaker-maker game $(X,F)$, then Maker also has a winning strategy in the maker-breaker game $(X, F )$. Show that the converse is false. That is, find $(X,F)$ such that Maker wins the maker- breaker game $(X, F )$, but Breaker wins the breaker-maker game $(X, F )$.
my thoughts are that if maker has a winning strategy in the breaker maker game then he can copy this strategy for the maker breaker game ignoring his first move and therefore win this game, also in a game of $3^2$ tic tac toe, this is a maker win in the maker breaker game but also a draw in the strong game, if it is a draw in the strong game then there must be strategy for player 2 to prevent player 1 from winning, therefore the breaker can also use this strategy to block the maker from winning in the breaker maker game. i believe this is a counterexample showing the converse is not correct. what is the best way to answer this question in the correct mathematical way?
Yes, strategy stealing is the way to go. To win the Maker-Breaker game, first play randomly, then respond using the winning strategy in the Breaker-Maker game, ignoring the first move. If this strategy every calls for you to play on your first random spot, then play randomly again, and ignore this new random play instead. You could state this more formally, but it would be a huge pain and not very enlightening.
The game $(\{1\},\{\{1\}\})$ is a counter-example to the converse; Maker wins if they go first, and same for Breaker. This represents a $1\times 1$ tic-tac-toe board where the Maker is trying to get one in a row.