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.

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


  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.

  • $\begingroup$ I'm voting to close this question. Either I don't know what you are asking (is the question about calculating the maximum of a function given a parameter $q$?) or it as off-topic because it is a basic question and there is no effort shown at a solution. As the site is not aimed at solving homework problems we try to avoid these questions. If you write down what you have tried so far the question may be reopened and you will probably get helpful feedback. $\endgroup$
    – Giskard
    Oct 3 '15 at 13:15
  • $\begingroup$ I have calculated the solution to this problem. in mixed strategies but I am not getting how to solve this using the maxmin and minmax strategies. Please help $\endgroup$ Oct 3 '15 at 13:59
  • $\begingroup$ What exactly is the difference between mixed equilibrium strategies and minmax strategies? $\endgroup$
    – Giskard
    Oct 3 '15 at 14:09
  • $\begingroup$ I'll ask again what I asked in the first comment: Is your question: "How to calculate $\min\limits_q\max\limits_p E(p,q)$ for a given function $E(p,q)$?" $\endgroup$
    – Giskard
    Oct 3 '15 at 14:13
  • $\begingroup$ Yes I am interested in knowing that only. $\endgroup$ Oct 3 '15 at 14:21

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.


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