Calculating safety levels for Zero Sum games I am having trouble understanding safety levels for general sum games.
For a simple example,
$$
    \begin{matrix}
    (2,0) & (1,3)   \\
    (0,1) & (3,2)   \\
    \end{matrix}
$$
where player I's matrix A is,
$$
    A = \begin{matrix}
    2 & 1   \\
    0 & 3   \\
    \end{matrix}
$$
I know we have to find $$ max min((2p+0(1-p)),(1p+3(1-p)) $$
so we find the mixed probability strategy $$P=(p,1-p)$$
$$ 2p+0(1-p)=1p+3(1-p)$$
and solve which $$ P=(3/4,1/4)$$
therefore $$ min((2p+0(1-p)),(1p+3(1-p))=min(1.5,1.75)=1.5  $$
and then find the value that maximizes that result, but if there is only one value that comes from the minimizing function, so
$$max(1.5)=1.5$$
My question is in which cases does that maximum function actually maximizes more than one value?
 A: If I understand your question correctly, I suspect that it would be helpful to look at what your calculation is really intended to accomplish. What we are actually calculating is the maximum value of the function
$$f(p)=\min\{2p+0(1-p),p+3(1-p)\}$$
as $p$ ranges over the interval $[0,1]$.
First simplify: $2p+0(1-p)=2p$, and $p+3(1-p)=3-2p$. Now $2p\le 3-2p$ if and only if $4p\le 3$, or $p\le\frac34$, so
$$f(p)=\min\{2p,3-2p\}=\begin{cases}
2p,&\text{if }0\le p\le\frac34\\
3-2p,&\text{if }\frac34<p\le 1\,.
\end{cases}\tag{1}$$
This is a function of $p$, and we want to know its maximum value for $0\le p\le 1$. This function of $p$ is clearly increasing for $0\le p\le\frac34$, so its maximum value on that interval is $2\cdot\frac34=\frac32$. It is decreasing on the interval $\frac34\le p\le 1$, so its maximum value on that interval is $3-2\cdot\frac34=3-\frac32=\frac32$. Thus, the maximum value of the function $(1)$ on the interval $0\le p\le 1$ is $\frac32$, and it occurs at $p=\frac34$.
In fact as $p$ ranges over the interval $[0,1]$, $f(p)$ takes on every value between $0$ and $\frac32$, inclusive, so we are in fact finding the maximum of an infinite set of real numbers. Setting $2p=3-2p$ and solving for $p$ as you did in order to find $P$ is simply a shortcut; it simplifies the calculation at the cost of obscuring what you’re really doing.
(Note that you have an arithmetic error after you find that $p=\frac34$: when $p=\frac34$, $p+3(1-p)=1.5$, not $1.75$.)
