Consider a game with the following payoff-matrix
$\textbf{M}=\begin{array}{c|c c c c} \! & A & B & C & D \\ \hline A & 0 & 1 & -1 & 0 \\ B & -1 & 0 & 1 & 1 \\ C & 1 & -1 & 0 & -1 \\ D & 0 & -1 & 1 & 0 \end{array}$
It's easy to see that there are no saddle points or dominated strategies. So I tryed to find a mixed strategy envolving all possible strategies $A,B,C,D$, but I failed with this step. It means the game has a solution in a 3x3 or 2x2 subgame. I let work a computer (http://banach.lse.ac.uk/) and I obtained two optimal strategies, note them $\mathsf{S}=(A,B,C,D)=(1/3,1/3,1/3,0)$ (I guess obtained as a solution of $(A,B,C)\times(A,B,C)$ subgame) and $\mathsf{T}=(1/2,0,0,1/2)$ (which might be obtained as a solution of $(A,D)\times(A,D)$ subgame). If I understand Minimax Theorem (http://mathworld.wolfram.com/MinimaxTheorem.html) correctly, there will be infinitely many optimal strategies in form $\mathsf{O}=\lambda\mathsf{S}+(1-\lambda)\mathsf{T}=\left(\frac{3-\lambda}{6},\frac{\lambda}{3},\frac{\lambda}{3},\frac{1-\lambda}{2}\right)$ for $0\le\lambda\le1$.
Now a task. Prove explicitely, that every non-optimal strategy $\mathsf R=(\alpha,\beta,\gamma,\delta)\neq\mathsf O$ is at most as good as $\mathsf O$. My main problem here is to deal with expected payoff, which should be equaled to $0$. How can be then $\mathsf R$ worse? :)
It is also possible, that I misunderstood something, so everything written above may be completely wrong. The most problematic place for me is an acquisition of $\mathsf T$. Where did it come from? Is that really from subgame $(A,D)\times(A,D)$? If so, it si a little bit weird for me, because you have a "singular payoff" matrix and there every strategy $(x,1-x)$ is "optimal", or am I wrong? :)
And last few questions. Consider payoff matrix $\textbf{M}'=k\cdot\textbf{M}$ for some nonzero $k\in\mathbb R$ (every possible outcome is multiplied by $k$). Then expected payoffs and optimal strategy should stay without any changes, right? So, why? And consider matrix $\textbf{M}''=\textbf{M}+c=\left(a_{ij}+c\right)_{4\times4}$ (every outcome is raised by $c\in\mathbb R$). Then optimal strategies remain same and the expect payoff for row player will be $c$, ok? And again, why? :)
Thx a lot for any help.
Let $p=(p_A,p_B,p_C,1-p_A-p_B-p_C)$. Then if we look for an equilibrium where all strategies are player with positive prob we get:
$\begin{align*} &U_1(A,p)=p_B-p_C,\\ &U_1(B,p)=1-2p_A-p_B,\\ &U_1(C,p)=2p_A+p_C-1,\\ &U_1(D,p)=p_C-p_B. \end{align*}$
So $U_1(A,p)=U_1(D,p)\Leftrightarrow p_C=p_B\Rightarrow U_1(A,p)=0$. And so $U_1(C,p)=0\Rightarrow $p_A=(1-p_B)/2$.
You can see that $p_B=p_C=0$ also satisfies all below so you can also get an eq. where only $A$ and $D$ are played with positive prob. but you must then have $p_A=(1-p_B)/2=1/2$.