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!

  • $\begingroup$ to clarify, does each player choose either $0$ or $1$ or is each player choosing a value between $[0,1]$? $\endgroup$ – corran_horn Dec 6 '19 at 3:33
  • $\begingroup$ Sorry, but it seems your attempt is just repeating the hint? $\endgroup$ – Giskard Dec 6 '19 at 3:43
  • $\begingroup$ Hi @Giskard, Honestly, I have no idea how to solve that question :( $\endgroup$ – LAD Dec 6 '19 at 8:08
  • $\begingroup$ Hi @corran_horn, each player choose value in the interval $[0,1]$. $\endgroup$ – LAD Dec 6 '19 at 8:09
  • $\begingroup$ 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 $y \in (0,1)$ $\endgroup$ – corran_horn Dec 6 '19 at 13:16

First, a caveat: I'm on the job market this year in the midst of the couple weeks when calls are rolling in. Hence, this seemed like a nice way to kill some time (semi-)productively. This is also a disclaimer in case I've made an error :)

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 $2$.

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