Security level and equilibrium payoffs in $3-$person zero sum game.

Question

Let the security level ($p-$payoff, $M-$set of all strategies) $$B_i:=\sup_{\sigma_i \in M_i} \inf_{\sigma_{-i}\in M_{-i}}p_i(\sigma_{-i},\sigma_i)$$ Now I consider $3-$person zero sum game. The questions are:

1. When every player use strategy, which guarantee $B_i$ is that Nash equilibrium?
2. Is that true: $B_1+B_2+B_3=0$?

Both are true for $2$ person game. Does something change for $3$ players?

Thanks in advance

Answer 1

The answer is no, neither is true. Consider the following example, which is given in the paper "Zero-sum Polymatrix Games: A Generalization of Minmax". Player $A$ chooses matrix, strategy $m_1$ (he has actually only one choice), Player $B$ chooses row, strategies $(r_1,r_2)$and Player $C$ chooses column, strategies $(c_1,c_2)$ with payoffs given as follows $$\begin{pmatrix}(1,0,-1)&(1,-2,1)\\(-1,0,1)&(-1,2,-1)\end{pmatrix}$$

Obviously, player $A$ cannot do much, so $$B_1=-1$$ for the strategy $m_1$ if Player $B$ decides to "hit" him and choose his second row, strategy $r_2$. Next, Player $B$'s security level is $$B_2=0$$ when he chooses $r_2$ and Player $C$ decides to "hit" him and choose his first column, strategy $c_1$. Finally, $$B_3=0$$ since Player $C$ can mix uniformly his strategies, i.e. employ the mixed strategy $\frac12c_1+\frac12c_1$ to guarantee himself a payoff of $0$. But he cannot guarantee more than that. So $$B_1+B_2+B_3=-1<0$$ which answers negatively your second question. For the first question, consider the strategy profile $$(m_1,r_2,\frac12c_1+\frac12c_2)$$ which we saw that guarantees all three players their security levels. But, given $m_1, r_2$ Player $C$ can deviate to $c_1$ and improve his payoff to $1$, which makes the previous not a Nash equilibrium.