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 $[0,1]$? $\endgroup$ Dec 6, 2019 at 3:33
  • $\begingroup$ Sorry, but it seems your attempt is just repeating the hint? $\endgroup$
    – Giskard
    Dec 6, 2019 at 3:43
  • $\begingroup$ Hi @Giskard, Honestly, I have no idea how to solve that question :( $\endgroup$
    – Akira
    Dec 6, 2019 at 8:08
  • $\begingroup$ Hi @corran_horn, each player choose value in the interval $[0,1]$. $\endgroup$
    – Akira
    Dec 6, 2019 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 $y \in (0,1)$ $\endgroup$ Dec 6, 2019 at 13:16

1 Answer 1


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 $2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.