<p>A two-playerr <em>zero-sum game</em> in <em>normal form</em> is expressed as:
<span class="math-container">$g_1\left(s_{1,i}, s_{2,j}\right) + g_2\left(s_{1,i}, s_{2,j}\right) =0 \forall s_{1,I } \in S_1,\forall s_{2,j} \in S_2 $</span> where <span class="math-container">$ S_1=\left\{s_{1,1},..., s_{1,m}\right\} $</span> and <span class="math-container">$ S_2=\left\{s_{2,1},..., s_{2,n}\right\} $</span> are the strategies.
The payment functions <span class="math-container">$ g_1 $</span> and <span class="math-container">$ g_2 $</span> can be represented by means of a matrix consisting of <span class="math-container">$m$</span> rows and <span class="math-container">$n$</span> columns. The players 1 and player 2 know the contents of the payment matrix: the game consists for each k-th player in choosing a strategy <span class="math-container">$ s_{k,i} \in S_k $</span> without knowing the choice of the other.
With regards to to the payout matrix for player 1 we set
<span class="math-container">$ A_1\left(s_{1,i}, s_{2,j}\right) = A \left(s_{1,i}, s_{2,j}\right) $</span>
and <span class="math-container">$ A_2\left(s_{1,i}, s_{2,j}\right)  = - A\left(s_{1,i}, s_{2,j}\right) $</span>
The choice of a strategy by player 1 corresponds to the choice of a row, while for player 2 it corresponds to the choice of a column. Player 2 will pay player 1 the sum indicated at the intersection of the row chosen by player 1 with the column chosen by player 2.</p>
<p>The attention of the two players is directed to the same object: the matrix <span class="math-container">$A$</span>.
Player 1 tries to maximize <span class="math-container">$A$</span>, but his control is limited only to choosing one row. Ditto for player 2 who tries to minimize <span class="math-container">$A$</span>: his control is limited only to the choice of a column.</p>
<p><strong>What will be the criterion for choosing the two players?</strong></p>
<p>John von Neumann answered this question using the following criterion: player 1 first detects the smallest number in each row and decides to choose the row that contains the largest of the monetary values considered: in this way he is certain that his winnings will not be less than the value
<span class="math-container">$ max_i \left(min_j a_{i,j} \right) $</span>
whatever player 2 chooses.
Player 2, for his part, considers the largest number in each column and decides to choose the column that contains the smallest of the numbers considered: thus, he is certain that his loss will not be greater than the value
<span class="math-container">$ min_j \left(max_i a_{i,j} \right) $</span>
whatever player 1 chooses.
What has just been described represents the definition of <strong>rational behavior</strong> given by John von Neumann and is commonly referred to as the <em>minimax decision rule</em>.</p>
<p>The pair of strategies <span class="math-container">$ s_1^e $</span>, <span class="math-container">$s_2^e$</span> constitute an equilibrium point, <strong>saddle point</strong> if the two conditions hold:</p>
<ol>
<li><p><span class="math-container">$g\left(s_1^e, s_2^e \right) \ge \ g\left(s_{1,i}, s_2^e \right) \forall s_{1,i} \in S_1$</span></p>
</li>
<li><p><span class="math-container">$g\left(s_1^e, s_2^e \right) \le \ g\left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$</span></p>
</li>
</ol>
<p>1-st Condition refers to the <strong>maximizing player</strong>, 2-nd condition to the <strong>minimizing player</strong>.</p>
<p>The definition of equilibrium according to Von Neumann given in zero-sum games is the &quot;translation&quot; of the concept of equilibrium in the Nash sense given for non-cooperative games. Below we illustrate how to move from the concept of <strong>Von Neumann equilibrium</strong> to the concept of <strong>Nash equilibrium</strong>: just remember that for a zero-sum game we have: <span class="math-container">$g_1$</span> = <span class="math-container">$ g$</span> and <span class="math-container">$ g_2$</span> = <span class="math-container">$ -g$</span>.
It then becomes legitimate to write:
<span class="math-container">$g_1 \left(s_1^e, s_2^e \right) \ge \ g_1 \left(s_{1,i}, s_2^e \right) \forall s_{1,i} \in S_1$</span></p>
<p><strong>The first equilibrium condition in the Von Neumann sense is therefore the first equilibrium condition in the Nash sense referring to player 1.</strong></p>
<p>For player 2 we have:
<span class="math-container">$-g_2 \left(s_1^e, s_2^e \right) \le \ -g_2 \left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$</span></p>
<p>that is</p>
<p><span class="math-container">$g_2\left(s_1^e, s_2^e \right) \ge \ g_2\left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$</span>.</p>
<p><strong>Minimax Theorem</strong>
Let <span class="math-container">$ {\displaystyle g:\mathit{K_{1}}\times \mathit{K_{2}} \to \mathbb{R}} $</span> be a continuous function where <span class="math-container">$ \mathit{K_{1}} \subset \mathbb{R}^m $</span> and <span class="math-container">$ \mathit{K_{2}} \subset \mathbb{R}^n $</span> are compact convex set.
If  <span class="math-container">$g$</span></p>
<p><span class="math-container">$ g(\cdot, s_2):\mathit{K_{1}} \to \mathbb{R}$</span> is concave for fixed <span class="math-container">$s_2$</span> and</p>
<p><span class="math-container">$ g(s_1,\cdot):\mathit{K_{2}}\to \mathbb{R} $</span> is  convex for fixed <span class="math-container">$s_1$</span></p>
<p>then <span class="math-container">$\min_{s_1\in \mathit{K_{1}}}\max_{s_2\in \mathit{K_{2}}} g(s_1,s_2)=\max_{s_1\in \mathit{K_{2}}}\min_{s_1\in \mathit{K_{1}}}g(s_1,s_2) $</span></p>
<p>It is easy to think of the payment function <span class="math-container">$g:S_1 \times S_2 \rightarrow \mathbb{R} $</span> as an <em>alternating bilinear form</em> whose alternation property <span class="math-container">$g(s_1,s_2) = - g( s_2,s_1) $</span> mathematizes the total opposition of the two players, i.e. the fact that the sum of the payments is zero.
The two sets <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> appear to be compact since they are finite and therefore have no accumulation points. The payment functions <span class="math-container">$g_i$</span> are defined on sets without accumulation points, sets therefore consisting only of isolated points, i.e. points in which the two functions <span class="math-container">$g_i$</span> are continuous.
The reader should regard to <span class="math-container">$s_1$</span> as <em>mixed strategies</em>:  a non-negative m-tuple of numbers <span class="math-container">$(\tilde s_1,\ldots, \tilde s_m) $</span> such that <span class="math-container">$ \tilde s_1+...+\tilde s_m = 1 $</span>. The same for <span class="math-container">$s_2$</span>.
The domain of variation of mixed strategies is more extensive than that of pure strategies: for the latter, in fact, the domain is a simple finite set of <span class="math-container">$m$</span> ordered m-nuples, while mixed strategies are defined on a subset of the vector space of size <span class="math-container">$m$</span> consisting of all the <strong>convex linear combinations</strong> of the <span class="math-container">$m$</span> pure strategies. The same for <span class="math-container">$s_2$</span>.
The convex hull <span class="math-container">$K_i$</span> is compact because it is built on the compact set of pure strategies. If <span class="math-container">$ g_1(\cdot,\tilde s_2)$</span> is convex with respect to <span class="math-container">$\tilde s_1$</span> for a fixed <span class="math-container">$\tilde s_2$</span> and since <span class="math-container">$g_1=-g_2$</span>, then <span class="math-container">$g_2(\tilde s_1,\cdot)$</span> is concave with respect to <span class="math-container">$\tilde s_2$</span> for a fixed <span class="math-container">$\tilde s_1$</span>.</p>
<p>As a result the conditions of the minimax theorem are satisfied.</p>
<p>To conclude in the field of mathematical programming, remember that a two-player zero-sum game can always be turned into a pair of mutually dual linear programs. The optimal value
of the two problems corresponds to the value of the game <span class="math-container">$v$</span>. We identify the generic mixed strategy with <span class="math-container">$ \mathbf x$</span> where a k pure strategy is a vector of the type <span class="math-container">$(0, \cdots, , 0, 1, 0, \cdots, 0)$</span> with <span class="math-container">$x_k=1$</span>.</p>
<p><strong>Linear programming formulation</strong>
For the maximizing player we have the following primal problem:</p>
<p><span class="math-container">$\max_{x\in \mathit{K_{1}}} v $</span></p>
<p>Subject to</p>
<p><span class="math-container">$ \left\{ \begin{array}{l}
\begin{matrix} \sum_{j=1}^n a_{i,j} \cdot x_{j} \end{matrix} \ge v \quad \forall i=1,...,m \\
\begin{matrix} \sum_{j=1}^n x_{j} \end{matrix} = 1 \\
x_{j} \ge 0 \quad \forall j=1,...,n \\
\end{array} \right.$</span></p>
<p><strong>Example</strong>
Reader can easily fix the first matrix of the example (l,m) to <span class="math-container">$A$</span> and solve the Linear Programme.
For example, the (l,m) matrix
<span class="math-container">$$\begin{matrix} 
\begin{array}{c|c|c}
&amp;a^1 &amp;a^2 \\ \hline
a_1 &amp; 2 &amp; 1 \\ \hline
a_2 &amp; 0 &amp; 4
\end{array}
\end{matrix}
$$</span>
has the optimal strategy <span class="math-container">$=(s_{1,1}^e, s_{1,2}^e)=(0.8 , 0.2)$</span> with regards to maximizing player. The value of the game is <span class="math-container">$1.6$</span>. The point <span class="math-container">$(0.8 , 0.2)$</span> is the intersection of <span class="math-container">$ x_{1,1}+ x_{1,2}=1$</span> with <span class="math-container">$1.6= 1x_{1,1}+ 4x_{1,2}$</span> and with <span class="math-container">$1.6= 2x_{1,1}+ 0x_{1,2}$</span>.</p>

CC BY-SA 4.0
2024-04-15T20:13:09.233Z
Thank you for this answer, i upvoted as i appreciate the effort. it did clarify to me that for example if i write a Matrix A with positive payoffs, P1 has a max min problem and P2 has a Min Max problem, but if i write a seperate Matrix -A for player 2, then they also have a max-min problem (i.e. picking the most negative number in each column, and then choosing the largest (smallest in absolute value) of these. However your answer hasn't addressed the heart of my concerns regarding the mixed equilibrium etc.
CormJack
CC BY-SA 4.0
2024-04-15T20:14:20.317Z
To add more clarity to my questions, with regards to the specific example i have provided:
1) How do the Mixed Nash Equilibriums relate to Player 1's Min-Max and Max-Min Strategy`
2) How do the Mixed Nash Equilibriums relate to Player 2's Min-Max and Max-Min Strategy
3) How does this game illustrate the relationship found in zero sum games, between players Min-Max and Max-Min Strategy.
4) How does this game illustrate Von Neumann's Minimax theorem.
CormJack
CC BY-SA 4.0
2024-04-16T07:15:00.063Z
I hope the additional part clarifies
marco tognoli
CC BY-SA 4.0
2024-04-16T10:54:40.663Z
Hey, thanks again i mean it's helpful! But It would help me parse the information you have provided if it was related to the example i posted. For instance, I think that the lower envelope that corresponds to the payoff 8/11, represents for player, one a Max-Min solution, because the envelope represents the set of minimum playoffs to P1, and the equilibriums is the maximum point. And vice Versa given P2 has costs, it represents Min-max payoff to P2 because it represents her lowest cost envelope, but the largest cost point on it. And then these are reversed for payoff point 5/2. So...
CormJack
CC BY-SA 4.0
2024-04-16T10:58:27.877Z
So what we say is that for both points to be equilibriums in this Zero Sum Game, each point has to be a Max-min for one player and a min-Max for the other. And that is the point that Von Neumann was making (?). It's not that we necessarily require a unique equilibrium, just that the equilibriums in a zero sum game, master is a bit this property.... I think what I have said is true, but this is the kind of insight I was looking for with reference to the specific example with two equilibriums that I provided?
CormJack
CC BY-SA 4.0
2024-04-16T10:59:41.790Z
I really appreciate the time you put into your post, but I was finding some of the technical language and the general nature a slight barrier to understanding in this context.... Also i meant 8/5 not 8/11 above, thanks!
CormJack
CC BY-SA 4.0
2024-04-15T19:49:50.380Z
2024-04-17T11:15:13.767Z
2024-04-17T11:15:13.767Z
marco tognoli
<p>In the attached Zero Sum Game, i have solved for two Mixed Nash Equilibrium, <span class="math-container">$(l,m)$</span> and <span class="math-container">$(m,r.)$</span></p>
<ul>
<li><p>In <span class="math-container">$(l,m)$</span> the payoff is <span class="math-container">$8/5$</span> to <span class="math-container">$P_1$</span>, and <span class="math-container">$-8/5$</span> to <span class="math-container">$P_2$</span>. Here  <span class="math-container">$P_1$</span> mixes between Top and Bottom with probability <span class="math-container">$4/5$</span> and <span class="math-container">$1/5$</span> respectively. <span class="math-container">$P_2$</span> mixes between left and middle with probability <span class="math-container">$3/5$</span> and <span class="math-container">$2/5$</span> respectively.</p>
</li>
<li><p>In <span class="math-container">$(m,r)$</span> the payoff is <span class="math-container">$5/2$</span> to <span class="math-container">$P_1$</span> and <span class="math-container">$-5/2$</span> to <span class="math-container">$P_2$</span>. Here  player 1 mixes between Top and Bottom with probability <span class="math-container">$1/2$</span> each. And Player two mixes between middle and right with probability <span class="math-container">$1/4$</span> and <span class="math-container">$3/4$</span> respectively.</p>
</li>
</ul>
<p>I am confused about how this equilibrium relate to each players max-min and min-max strategy and Von Neumann's Minimax theorem. And the theorems of Zero-sum games more generally.</p>
<ul>
<li>For Example: I was under the impression there zero sum games have a unique mixed Nash Equilibrium which would involve player 1 &quot;maximising over the lower envelope.&quot; I.e.The 8/5 payoff representing some kind of Security Strategy in terms of Mixed Strategies. I thought this was meant to be the same as player two's Min-Max Strategy, but i feel like that would refer to the other equilibrium on the upper envelope.</li>
</ul>
<p><a href="https://i.sstatic.net/1pky3.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1pky3.jpg" alt="enter image description here" /></a></p>

<microeconomics><game-theory>
Zero-Sum Game with two Mixed Equilibrium Interpretation
CormJack
# Zero-Sum Game with two Mixed Equilibrium Interpretation

(Asked by: username_0 on 2024-04-15)

In the attached Zero Sum Game, i have solved for two Mixed Nash Equilibrium, $(l,m)$ and $(m,r.)$

* In $(l,m)$ the payoff is $8/5$ to $P\_1$, and $-8/5$ to $P\_2$. Here $P\_1$ mixes between Top and Bottom with probability $4/5$ and $1/5$ respectively. $P\_2$ mixes between left and middle with probability $3/5$ and $2/5$ respectively.
* In $(m,r)$ the payoff is $5/2$ to $P\_1$ and $-5/2$ to $P\_2$. Here player 1 mixes between Top and Bottom with probability $1/2$ each. And Player two mixes between middle and right with probability $1/4$ and $3/4$ respectively.

I am confused about how this equilibrium relate to each players max-min and min-max strategy and Von Neumann's Minimax theorem. And the theorems of Zero-sum games more generally.

* For Example: I was under the impression there zero sum games have a unique mixed Nash Equilibrium which would involve player 1 "maximising over the lower envelope." I.e.The 8/5 payoff representing some kind of Security Strategy in terms of Mixed Strategies. I thought this was meant to be the same as player two's Min-Max Strategy, but i feel like that would refer to the other equilibrium on the upper envelope.

[![enter image description here](https://i.sstatic.net/1pky3.jpg)](https://i.sstatic.net/1pky3.jpg)

**Tags:** microeconomics, game-theory

---

## 1 answers

### Answer by username_1 on 2024-04-15. Score: 1

A two-playerr *zero-sum game* in *normal form* is expressed as:
$g\_1\left(s\_{1,i}, s\_{2,j}\right) + g\_2\left(s\_{1,i}, s\_{2,j}\right) =0 \forall s\_{1,I } \in S\_1,\forall s\_{2,j} \in S\_2 $ where $ S\_1=\left\{s\_{1,1},..., s\_{1,m}\right\} $ and $ S\_2=\left\{s\_{2,1},..., s\_{2,n}\right\} $ are the strategies.
The payment functions $ g\_1 $ and $ g\_2 $ can be represented by means of a matrix consisting of $m$ rows and $n$ columns. The players 1 and player 2 know the contents of the payment matrix: the game consists for each k-th player in choosing a strategy $ s\_{k,i} \in S\_k $ without knowing the choice of the other.
With regards to to the payout matrix for player 1 we set
$ A\_1\left(s\_{1,i}, s\_{2,j}\right) = A \left(s\_{1,i}, s\_{2,j}\right) $
and $ A\_2\left(s\_{1,i}, s\_{2,j}\right) = - A\left(s\_{1,i}, s\_{2,j}\right) $
The choice of a strategy by player 1 corresponds to the choice of a row, while for player 2 it corresponds to the choice of a column. Player 2 will pay player 1 the sum indicated at the intersection of the row chosen by player 1 with the column chosen by player 2.

The attention of the two players is directed to the same object: the matrix $A$.
Player 1 tries to maximize $A$, but his control is limited only to choosing one row. Ditto for player 2 who tries to minimize $A$: his control is limited only to the choice of a column.

**What will be the criterion for choosing the two players?**

John von Neumann answered this question using the following criterion: player 1 first detects the smallest number in each row and decides to choose the row that contains the largest of the monetary values considered: in this way he is certain that his winnings will not be less than the value
$ max\_i \left(min\_j a\_{i,j} \right) $
whatever player 2 chooses.
Player 2, for his part, considers the largest number in each column and decides to choose the column that contains the smallest of the numbers considered: thus, he is certain that his loss will not be greater than the value
$ min\_j \left(max\_i a\_{i,j} \right) $
whatever player 1 chooses.
What has just been described represents the definition of **rational behavior** given by John von Neumann and is commonly referred to as the *minimax decision rule*.

The pair of strategies $ s\_1^e $, $s\_2^e$ constitute an equilibrium point, **saddle point** if the two conditions hold:

1. $g\left(s\_1^e, s\_2^e \right) \ge \ g\left(s\_{1,i}, s\_2^e \right) \forall s\_{1,i} \in S\_1$
2. $g\left(s\_1^e, s\_2^e \right) \le \ g\left(s\_1^e, s\_{2,j} \right) \forall s\_{2,j} \in S\_2$

1-st Condition refers to the **maximizing player**, 2-nd condition to the **minimizing player**.

The definition of equilibrium according to Von Neumann given in zero-sum games is the "translation" of the concept of equilibrium in the Nash sense given for non-cooperative games. Below we illustrate how to move from the concept of **Von Neumann equilibrium** to the concept of **Nash equilibrium**: just remember that for a zero-sum game we have: $g\_1$ = $ g$ and $ g\_2$ = $ -g$.
It then becomes legitimate to write:
$g\_1 \left(s\_1^e, s\_2^e \right) \ge \ g\_1 \left(s\_{1,i}, s\_2^e \right) \forall s\_{1,i} \in S\_1$

**The first equilibrium condition in the Von Neumann sense is therefore the first equilibrium condition in the Nash sense referring to player 1.**

For player 2 we have:
$-g\_2 \left(s\_1^e, s\_2^e \right) \le \ -g\_2 \left(s\_1^e, s\_{2,j} \right) \forall s\_{2,j} \in S\_2$

that is

$g\_2\left(s\_1^e, s\_2^e \right) \ge \ g\_2\left(s\_1^e, s\_{2,j} \right) \forall s\_{2,j} \in S\_2$.

**Minimax Theorem**
Let $ {\displaystyle g:\mathit{K\_{1}}\times \mathit{K\_{2}} \to \mathbb{R}} $ be a continuous function where $ \mathit{K\_{1}} \subset \mathbb{R}^m $ and $ \mathit{K\_{2}} \subset \mathbb{R}^n $ are compact convex set.
If $g$

$ g(\cdot, s\_2):\mathit{K\_{1}} \to \mathbb{R}$ is concave for fixed $s\_2$ and

$ g(s\_1,\cdot):\mathit{K\_{2}}\to \mathbb{R} $ is convex for fixed $s\_1$

then $\min\_{s\_1\in \mathit{K\_{1}}}\max\_{s\_2\in \mathit{K\_{2}}} g(s\_1,s\_2)=\max\_{s\_1\in \mathit{K\_{2}}}\min\_{s\_1\in \mathit{K\_{1}}}g(s\_1,s\_2) $

It is easy to think of the payment function $g:S\_1 \times S\_2 \rightarrow \mathbb{R} $ as an *alternating bilinear form* whose alternation property $g(s\_1,s\_2) = - g( s\_2,s\_1) $ mathematizes the total opposition of the two players, i.e. the fact that the sum of the payments is zero.
The two sets $S\_1$ and $S\_2$ appear to be compact since they are finite and therefore have no accumulation points. The payment functions $g\_i$ are defined on sets without accumulation points, sets therefore consisting only of isolated points, i.e. points in which the two functions $g\_i$ are continuous.
The reader should regard to $s\_1$ as *mixed strategies*: a non-negative m-tuple of numbers $(\tilde s\_1,\ldots, \tilde s\_m) $ such that $ \tilde s\_1+...+\tilde s\_m = 1 $. The same for $s\_2$.
The domain of variation of mixed strategies is more extensive than that of pure strategies: for the latter, in fact, the domain is a simple finite set of $m$ ordered m-nuples, while mixed strategies are defined on a subset of the vector space of size $m$ consisting of all the **convex linear combinations** of the $m$ pure strategies. The same for $s\_2$.
The convex hull $K\_i$ is compact because it is built on the compact set of pure strategies. If $ g\_1(\cdot,\tilde s\_2)$ is convex with respect to $\tilde s\_1$ for a fixed $\tilde s\_2$ and since $g\_1=-g\_2$, then $g\_2(\tilde s\_1,\cdot)$ is concave with respect to $\tilde s\_2$ for a fixed $\tilde s\_1$.

As a result the conditions of the minimax theorem are satisfied.

To conclude in the field of mathematical programming, remember that a two-player zero-sum game can always be turned into a pair of mutually dual linear programs. The optimal value
of the two problems corresponds to the value of the game $v$. We identify the generic mixed strategy with $ \mathbf x$ where a k pure strategy is a vector of the type $(0, \cdots, , 0, 1, 0, \cdots, 0)$ with $x\_k=1$.

**Linear programming formulation**
For the maximizing player we have the following primal problem:

$\max\_{x\in \mathit{K\_{1}}} v $

Subject to

$ \left\{ \begin{array}{l}
\begin{matrix} \sum\_{j=1}^n a\_{i,j} \cdot x\_{j} \end{matrix} \ge v \quad \forall i=1,...,m \\
\begin{matrix} \sum\_{j=1}^n x\_{j} \end{matrix} = 1 \\
x\_{j} \ge 0 \quad \forall j=1,...,n \\
\end{array} \right.$

**Example**
Reader can easily fix the first matrix of the example (l,m) to $A$ and solve the Linear Programme.
For example, the (l,m) matrix
$$\begin{matrix}
\begin{array}{c|c|c}
&a^1 &a^2 \\ \hline
a\_1 & 2 & 1 \\ \hline
a\_2 & 0 & 4
\end{array}
\end{matrix}
$$
has the optimal strategy $=(s\_{1,1}^e, s\_{1,2}^e)=(0.8 , 0.2)$ with regards to maximizing player. The value of the game is $1.6$. The point $(0.8 , 0.2)$ is the intersection of $ x\_{1,1}+ x\_{1,2}=1$ with $1.6= 1x\_{1,1}+ 4x\_{1,2}$ and with $1.6= 2x\_{1,1}+ 0x\_{1,2}$.

> **Comments:**
> **username_0**: Thank you for this answer, i upvoted as i appreciate the effort. it did clarify to me that for example if i write a Matrix A with positive payoffs, P1 has a max min problem and P2 has a Min Max problem, but if i write a seperate Matrix -A for player 2, then they also have a max-min problem (i.e. picking the most negative number in each column, and then choosing the largest (smallest in absolute value) of these. However your answer hasn't addressed the heart of my concerns regarding the mixed equilibrium etc.  
> **username_0**: To add more clarity to my questions, with regards to the specific example i have provided: 1) How do the Mixed Nash Equilibriums relate to Player 1's Min-Max and Max-Min Strategy` 2) How do the Mixed Nash Equilibriums relate to Player 2's Min-Max and Max-Min Strategy 3) How does this game illustrate the relationship found in zero sum games, between players Min-Max and Max-Min Strategy. 4) How does this game illustrate Von Neumann's Minimax theorem.  
> **username_1**: I hope the additional part clarifies  
> **username_0**: Hey, thanks again i mean it's helpful! But It would help me parse the information you have provided if it was related to the example i posted. For instance, I think that the lower envelope that corresponds to the payoff 8/11, represents for player, one a Max-Min solution, because the envelope represents the set of minimum playoffs to P1, and the equilibriums is the maximum point. And vice Versa given P2 has costs, it represents Min-max payoff to P2 because it represents her lowest cost envelope, but the largest cost point on it. And then these are reversed for payoff point 5/2. So...  
> **username_0**: So what we say is that for both points to be equilibriums in this Zero Sum Game, each point has to be a Max-min for one player and a min-Max for the other. And that is the point that Von Neumann was making (?). It's not that we necessarily require a unique equilibrium, just that the equilibriums in a zero sum game, master is a bit this property.... I think what I have said is true, but this is the kind of insight I was looking for with reference to the specific example with two equilibriums that I provided?  
> **username_0**: I really appreciate the time you put into your post, but I was finding some of the technical language and the general nature a slight barrier to understanding in this context.... Also i meant 8/5 not 8/11 above, thanks!  

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