Given a game with outcome matrix: $$Player 2$$ $$ Player 1 \begin{array}{c|lcr} & \text{A'} & \text{B'} \\ \hline A & (-1,1) & (1,-1) \\ B & (h,-h) & (-1,1) \\ \end{array} $$
where (B,A') continues the game to the next turn, with neither player winning or losing.
In a single game, A' can only be used 8 times. After those 8 times, Player 2 can only use B', which would mean given that the information is not hidden, Player 1 would always choose A to win.
I put (h,-h) there to illustrate that if Player 2 does waste an A', they have lost some advantage in the game, and therefore it is gained by Player 1. This means that the outcome should not be considered (0,0), and instead can be represented with the variable h, where -1 $\leq$ h $\leq$ 1. If this is incorrect please tell me.
My question is two-part: how do I calculate the value of h for the number of A' uses left, and further, what are the odds of Player 1 winning on average in a full game, given optimal play from both sides?
I tried using the mixed-strategy formula that was used for penalty kicks in soccer, but the finite number of A' threw me off. I apologize if the formatting is off, I couldn't figure out how to label it neatly.
Never saw anything of that kind, but I would consider going from the last step to the front one, like in the dynamic programming. Let me explain.
If $A'$ was already used $8$ times, the strategy is obviously $(A, B')$: P2 does not have any other choice and P1 chooses at her will. So, if $A'$ was used $8$ times, P2 knows that it gets $-1$ for sure.
Next, if $A$ was used only $7$ times, the payoff table becomes $$ \begin{array}{c|lcr} & A'& B' \\ \hline A & (-1,1) & (1,-1) \\ B & (h + 1,-h - 1) & (-1,1) \\ \end{array} $$ and you can solve it easily to get what are the optimal strategies and corresponding payoffs. Next, you just need to insert those payoff for the case $(B,A')$ to compute it for $A'$ being used $6$ times already, etc.