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Is there a totally asymmetric 2-player zero-sum game with all payoffs $\pm1$, with a unique Nash equilibrium which assigns positive probability to each strategy? By totally asymmetric, I mean that there is no nontrivial way to permute the rows and columns of the payoff matrix that leaves the payoffs invariant. For example, the zero-sum game with this payoff matrix: $\begin{pmatrix} 1 & -1 \\ -1 & 1 \\ \end{pmatrix}$ (only player 1's payoffs are shown) is not totally asymmetric, because swapping the columns and swapping the rows gives the same payoff, although it satisfies all the other requirements. The zero-sum game with this payoff matrix:$\begin{pmatrix} -1 & -1 \\ -1 & 1 \\ \end{pmatrix}$ is totally asymmetric, but it has a continuum of Nash equilibria, and the second column is not assigned positive probability in any of them.

What I've tried

If every Nash equilibrium of a game assigns positive probability to every strategy, none of the columns of the payoff matrix can be dominated. For $2\times n$ games this uniquely identifies $\begin{pmatrix} 1 & -1 \\ -1 & 1 \\ \end{pmatrix}$. For $3\times n$ games the options are $\begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{pmatrix}$ and its negation. For $4\times 4$ games the options are $\begin{pmatrix} 1 & -1 & -1&-1 \\ -1 & 1 & -1&-1 \\ -1 & -1 & 1&-1 \\ -1 & -1 & -1&1 \\ \end{pmatrix}$,$\begin{pmatrix} 1 & -1 & -1&-1 \\ -1 & 1 & -1& 1 \\ 1 & -1 & 1& 1 \\ -1 & 1 & 1&-1 \\ \end{pmatrix}$,$\begin{pmatrix} 1 & -1 & -1& 1 \\ -1 & 1 & -1& 1 \\ 1 & -1 & 1&-1 \\ -1 & 1 & 1&-1 \\ \end{pmatrix}$,$\begin{pmatrix} 1 & -1 & -1&-1 \\ -1 & -1 & 1& 1 \\ -1 & 1 & -1& 1 \\ -1 & 1 & 1&-1 \\ \end{pmatrix}$, and their negations. None of these are totally asymmetric.

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