# Value of a zero sum game

I've been looking some around on the net for some info on zero-sum games, But I don't think I fully understand the principle; If we consider the (simple) matrix: $$\begin{pmatrix}\pi&0 \\ 0&e \end{pmatrix}$$

How can I determine the value of this zero-sum game?

Is this the possible correct answer regarding this example?

Let us just call first row R1, first column C1 and so on.

Then, the expectation for R1 is $\pi \cdot p_1$. Then, the expectation for R2 is $e \cdot p_2$.

Now I have to maximize: $min(E(R1),E(R2))$, which implies $p_1=e:p_2=\pi$.

So we see $p_1=0.464$ and $p_2=0.536$

We can do the same for C1,C2, which will give the same numbers.

Therefore I think the value of this game is $0.464 \cdot \pi=1.457= 0.536 \cdot e$

Is this ok? Thanks for checking!

-

We will denote the mixed strategies of the first player by $(p, 1- p)$ and of the second player by $(q, 1 - q)$ and the payoff function determined by your matrix by $g$. For the first player, we have to calculate $$V = \max_{0 \leq p \leq 1} \min_{0 \leq q \leq 1} g((p,1-p),(q,1-q)) = \max_{0 \leq p \leq 1} \min_{q \in \{0,1\}} g((p,1-p),(q,1-q)).$$ That is, $$V = \max_{0 \leq p \leq 1} \min \{ p\pi, (1-p)e \}.$$ Plotting it, we have
So we see that the max-min is attained only at the intersection of $p\pi$ and $(1-p)e$, that is, when $p = \frac{e}{\pi + e} \approx 0.464$ and the value of the game is the payment at $(p,1-p)$ so $V = \frac{\pi e}{\pi + e} \approx 1.457$.
@levap: You've got an error in the equation: $\max \min=\min \max$, instead of $\max \min = \max \min$. – amorfati Jun 23 at 10:49
@amorfati: There is no error. I've only calculated $\max \min$. It is equal to $\min \max$ but there is no need to treat it since this is guaranteed... – levap Jun 23 at 11:40