I'm trying to solve this problem from last year final exam in game theory:
Consider the zero-sum game $G=(X, Y, g)$ where $X=Y=[0,1]$, and $$\forall (x,y) \in X \times Y: g(x, y)=\max \{x(1-2 y), y(1-2 x)\}$$
Find a mixed optimal strategy for each player. (Hint: one can consider mixed strategies of player $1$ which plays $x=0$ with some probability and $x=1$ with the remaining probability).
My attempt:
Let $\sigma$ be a mixed optimal strategy in which player $1$ plays $x=0$ with probability $p$ and $x=1$ with probability $1-p$. A mixed optimal strategy $\tau$ of player $2$ is a probability measure on $[0,1]$.
Then I'm stuck to proceed. Could you please help me finish this exercise? Thank you so much!
If player $1$ chooses $\ x=0\ $, his or her payoff is $$ \max(0,y)=y\ . $$ If he or she chooses $\ x=1\ $, the payoff is $$ \max(1-2y,-y)=1-2y\ . $$ Thus, player $1$ can guarantee an expected payoff of $\ \frac{1}{3}\ $ by choosing $\ x=0\ $ with probability $\ \frac{2}{3}\ $ and $\ x=1\ $ with probability $\ \frac{1}{3}\ $, since the expected payoff from this strategy is $$ \frac{2}{3}y +\frac{1}{3}(1-2y)= \frac{1}{3} $$ for all $\ y\in[0,1]\ $.
On the other hand, if player $2$ chooses $\ y= \frac{1}{3}\ $ with probability $1$, then the payoff to player $1$ becomes \begin{align} \max\left(\frac{x}{3}, \frac{1-2x}{3}\right)\le \frac{1}{3} \end{align} for all $\ x\in[0,1]\ $. Thus, this (pure) strategy for player $2$ prevents player $1$ from obtaining an expected payoff of any more than $\ \frac{1}{3}\ $.
It follows from this that the strategies described above are optimal for their respective players, and the value of the game is $\ \frac{1}{3}\ $ (to player $1$).