Maxmin and minmax strategies

Question

I was solving for a stable equilibrium in the following 2 player zero sum game. I need to calculate the equilibrium using maxmin and minmax strategies. In this game they should come out to be identical and coincide with the mixed strategy Nash's equilibrium.

                        P2
               L                R
   L       (0.6,0.4)         (0.8,0.2)

P1

  R        (0.9,0.1)         (0.7,0.3)

If I solve it formally, the expected payoff of P1 will be the expression:

$E = 0.6(pq) + 0.8p(1-q) + 0.9(1-p)(q) + 0.7(1-p)(1-q)$

where p is probability of P1 playing L and q is the probability pf P2 playing L. Now, P2 is trying to minimize E keeping in mind that P1 is trying to maximize it

i.e. min max {E}

and P1 is trying to maximize E keeping in mind that P1 will try to minimize E i.e.

max min {E}

and vice versa. Essentially both strategies played simultaneously should give the same equilibrium. How to go about it, I am stuck in the maths part of the problem now.

Answer 1

The question was clarified in the comments as

How to calculate $\min\limits_q \max\limits_p E(p,q)$ for a given function $E(p,q)$?

What this notation means is that $p$ is chosen first, and $q$ is chosen afterwards.

Assume that $p$ is a given parameter. Then $E(p,\cdot)$ is a function over $q$ only. So you take the minimum of the function $E(p,\cdot)$. In economics this usually means taking the first derivative with respect to $q$. From this you get the optimal value of $q$, given $p$. This is basically a function $q^*(p)$. So you know how $q$ will be chosen given the choice of $p$. This gives you the function $E(q^*(p),p)$. You choose $p$ to maximize this function. Again in economics this usually means taking the first derivative w.r.t. $p$.

With $\max\limits_p \min\limits_q$ the procedure is essentially the same, but there you first derive the maximum for $p$ given $q$. This yields $p^*(q)$. Using this, you minimize $E(p^*(q),q)$ w.r.t. $q$.

A sidenote:
If $E(p,q)$ is linear in both $p$ and $q$ the functions $q^*(p)$ and $p^*(q)$ are usually not continuous. To be more precise they are not functions but mappings. As such, they upper hemicontinuous but that is very technical and you may not need to know what it means.