I'm trying to solve a question from last year final exam in game theory:
I've solved question (3a), but it takes me a lot of time to compute because of many possible cases to consider. I think I can not complete the exam in time with my approach.
Could you please inform me a shorter approach on this problem?
My attempt:
Because $x(1-2y)$ and $y(1-2x)$ are continuous, $g(x,y) =\max \{x(1-2y),y(1-2x)\}$ is continuous. Moreover, $X$ and $Y$ are compact. It follows that $\alpha = \max_{x \in X} \min_{y \in Y} g(x,y)$ and $\beta = \min_{y \in Y} \max_{x \in X} g(x,y)$.
We have $$\begin{aligned} \min_{y \in [0,x]} g(x,y) &= \min_{y \in [0,x]} x(1-2y) = x(1-2x) \\ \min_{y \in [x,1]} g(x,y) &= \min_{y \in [x,1]} y(1-2x) = \begin{cases} 1-2x & \text{if} \quad x \ge 1/2 \\ x(1-2x) & \text{if} \quad x < 1/2 \\ \end{cases} \end{aligned}$$
Consequently, $$\min_{y \in Y} g(x,y) = \begin{cases} 1-2x & \text{if} \quad x \ge 1/2 \\ x(1-2x) & \text{if} \quad x < 1/2 \\ \end{cases}$$
Hence $$\alpha = \max_{x \in X} \min_{y \in Y} g(x,y) = \max_{x \in [0,\frac{1}{2}]} x(1-2x) = \frac{1}{8}$$
Similarly, we have $$\begin{aligned} \max_{x \in [0,y]} g(x,y) &= \max_{x \in [0,y]} y(1-2x) = y \\ \max_{x \in [y,1]} g(x,y) &= \max_{x \in [y,1]} x(1-2y) = \begin{cases} y(1-2y) & \text{if} \quad y \ge 1/2 \\ 1-2y & \text{if} \quad y < 1/2 \\ \end{cases} \end{aligned}$$
Consequently, $$\max_{x \in X} g(x,y) = \begin{cases} y & \text{if} \quad y \ge 1/3 \\ 1-2y & \text{if} \quad y \le 1/3 \\ \end{cases}$$
Hence $$\beta = \min_{y \in Y} \max_{x \in X} g(x,y) = \frac{1}{3}$$
Computing $\alpha$ involves first minimizing $g(x,y)$ with respect to $y$. This implies that we need to set $x(1-2y)=y(1-2x)$. To see this, note the following: If $x(1-2y)>y(1-2x)$, $g(x,y)=x(1-2y)$ and we can increase $y$ to decrease $g(x,y)$. If $y(1-2x)>x(1-2y)$, then $g(x,y)=y(1-2x)$ and we can decrease $y$ to decrease $g(x,y)$. Thus, if $x(1-2y)=y(1-2x)$ is false, we can increase or decrease $y$ to decrease $g(x,y)$. Hence, for fixed $x$, a minimum of $g(x,y)$ with respect to $y$ must be such that $x(1-2y)=y(1-2x)$. But $x(1-2y)=y(1-2x)$ implies $x=y$, so $\alpha=\max_x\min_yg(x,y)=\max_xg(x,x)=\max_xx(1-2x)=g(1/4,1/4)=1/8$.
Now, consider $\beta$. If $y>1-2y$, one would like to set $x=0$ for payoff $y$ when maximizing $g(x,y)$ with respect to $x$. Why? Well, to maximize $g(x,y)=\max(x(1-2y),y(1-2x))$ with respect to $x$, we should maximize $x(1-2y)$ or $y(1-2x)$. Thus, we should set $x=0$ or $x=1$. $x=0$ gives $y$, and $x=1$ gives $1-2y$. Since $y>1-2y$, we should set $x=0$. Similarly, if $y<1-2y$, one would like to set $x=1$ for payoff $1-2y$. Thus, if $y\neq 1-2y$, the other player would like to increase or decrease $y$ to decrease $\max_xg(x,y)$. Hence, the other player will set $y=1-2y$, i.e., $y=1/3$, so that $\beta=g(0,1/3)=g(1,1/3)=1/3$.