suppose that there are two players $a$ and $b$ playing two types of zero-sum game. Type 1 is expressed by:\begin{array}{*{20}{c}} {}&{{a_1}}&{{a_2}}\\ {{b_1}}&1&0\\ {{b_2}}&0&1 \end{array}
Type 2 is expressed by: \begin{array}{*{20}{c}} {}&{{a_1}}&{{a_2}}\\ {{b_1}}&1&{0.5}\\ {{b_2}}&{0.7}&1 \end{array}
where the payoff in each matrix is calculated from the view of $a$, the payoff of $b$ is the negative of it.
Suppose only $a$ knows which game he plays, but $b$ does not know (he only knows Type 1 is played with the probability of $p$ and Type 2 $1-p$). I want to know how to calculate the optimal strategies.
I check many documents for simultaneous move Bayesian game, but they all do with non-zero sum game, besides, in each type of game, there exists a dominant strategy, which makes the optimal strategy pure one.
Thank you very much
First look from $a$s point of view. He expects $b$ to play $b_1$ with probability $q$. If they are playing game $1$ he will play $a_1$ when $q \gt \frac 12$ and $a_2$ if $q \lt \frac 12$, winning $q$ or $1-q$ respectively. If they are playing game $2$ by playing $a_1$ he wins $q+0.7(1-q)=0.3q+0.7$ By playing $a_2$ he wins $0.5q+(1-q)=1-0.5q$. He should therefore play $a_1$ when $q \gt \frac 38$. Now $b$ can compute $a$s average win depending on $q$. If $q \gt \frac 12$, $a$ wins $pq+(1-p)(0.3q+0.7)=0.7-0.7p+0.3q+0.7pq$ You can repeat this calculation for $\frac 38 \lt q \lt \frac 12$ and for $q \lt \frac 38$, then find the minimum over $q$. As they will be linear in $q$ it must come at one of the breakpoints.