# How to find mixed optimal strategies in this zero-sum game?

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 $$1$$ play $$x=0$$ 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!

• to clarify, does each player choose either $0$ or $1$ or is each player choosing a value between $[0,1]$? – corran_horn Dec 6 '19 at 3:33
• Sorry, but it seems your attempt is just repeating the hint? – Giskard Dec 6 '19 at 3:43
• Hi @Giskard, Honestly, I have no idea how to solve that question :( – LAD Dec 6 '19 at 8:08
• Hi @corran_horn, each player choose value in the interval $[0,1]$. – LAD Dec 6 '19 at 8:09
• I’ll update my answer, but think through the payoff structure. For any given $x \in [0,1]$, what is the optimal response for $Y$? Is it ever optimal to play $y \in (0,1)$ – corran_horn Dec 6 '19 at 13:16

Now, let's look at the one you're suggested to try for, where player $$1$$ chooses $$0$$ with $$p$$ and $$1$$ with $$1-p$$. Again let player $$2$$ choose $$G$$ with full support.
If player $$1$$ chooses $$0$$ and player $$2$$ chooses $$y$$, Pl $$1$$ gets $$\int_{0}^{1}yg(y)dy = \mu$$ and if he chooses $$1$$, he gets $$\int_{0}^{1}(1-2y)g(y)dy = 1 - 2\mu$$ where $$\mu := \mathbb{E}_{G}[y]$$ Hence, for indifference we need $$\mu = 1/3$$. Ah but we also need this to be robust to any other deviation. Namely, we need. $$\frac{1}{3} \geq x\int_{0}^{x}(1-2y)g(y) + (1-2x)\int_{x}^{1}yg(y)$$ for all $$x \in [0,1]$$.
Knowing that $$\mathbb{E}_{G}(y) = 1/3$$ for $$G(y) = \sqrt{y}$$, let's guess that that is the solution. The right-hand side reduces to $$\dfrac{2x^\frac{3}{2}-2x+1}{3}$$ which is maximized at $$x = 0$$ and $$x = 1$$ and equals $$1/3$$ there, as required.
Finally, for player $$2$$, she is indifferent over any $$y$$ if $$p(-y) - (1-p)(1-2y)$$ does not depend on $$y$$. This holds if $$p = 2/3$$.
Thus the equilibrium is given as follows: player $$1$$ chooses $$0$$ with probability $$2/3$$ and $$1$$ with probability $$1/3$$; and player $$2$$ chooses cdf $$G(y) = \sqrt{y}$$ on $$[0,1]$$. The value of the game is $$1/3$$ for player $$1$$ and $$-1/3$$ for player $$2$$.