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Suppose $A$ is the $m \times n$ game matrix for a two-person zero sum game. Suppose row player uses the strategy $x\in P^m$, what is the condition for the strategy for the column player to be unique?

I know col would use $j$ -th strategy if $a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m$ is minimum among $j=1,2,\cdots,n$. But how to proceed?

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2 Answers 2

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The column player's strategy will be unique if and only if there is a unique $j$ that minimizes this quantity. In other words, for all $k \neq j$, we have $a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m < a_{1k}x_1+a_{2k}x_2+\cdots+a_{mk}x_m$.

If there are multiple $j$ that minimize this quantity, then the column player could use a mixed strategy, choosing among the minimizing $j$ with certain probabilities. The exact probabilities would depend on the specifics of the game and the row player's strategy.

So, the condition for the column player's strategy to be unique is that there is a unique column $j$ in the game matrix $A$ that minimizes the quantity $a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m$ for the given row player's strategy $x$. If this condition is not met, the column player may not have a unique strategy.

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  • $\begingroup$ it doesn't answer my question thoroughly, I want the condition ON THE MATRIX A $\endgroup$
    – user1324054
    Commented May 19 at 10:37
  • $\begingroup$ The condition on the matrix $A$ for the column player's strategy to be unique is that no two columns of $A$ are identical when weighted by the row player's strategy $x$. Formally, for all distinct $j$ and $k$, we have: $$a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m \neq a_{1k}x_1+a_{2k}x_2+\cdots+a_{mk}x_m$$ $\endgroup$
    – user1319982
    Commented May 19 at 10:39
  • $\begingroup$ Does that mean the rank of A has constraint? $\endgroup$
    – user1324054
    Commented May 19 at 10:47
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The rank of the matrix $A$ itself doesn't directly determine the uniqueness of the column player's strategy. The rank of a matrix is the maximum number of linearly independent columns (or rows). A higher rank indicates more linearly independent columns, but it doesn't necessarily mean that there is a unique column that minimizes the quantity $a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m$ for a given row player's strategy $x$.

However, if the rank of the matrix $A$ is equal to the number of columns $n$, then all columns are linearly independent. In this case, it is less likely (but still possible) that there will be multiple columns that minimize the quantity $a_{1j}x_1+a_{2j}x_2+\cdots+a_{mj}x_m$ for a given row player's strategy $x$. Therefore, a full-rank matrix $A$ might increase the likelihood of the column player's strategy being unique, but it does not guarantee it.

The uniqueness of the column player's strategy is determined by the specific values in the matrix $A$ and the row player's strategy $x$, not just the properties of the matrix $A$ itself. So, while the rank of $A$ can provide some insight, it doesn't give a complete picture. The condition I mentioned earlier about no two columns of $A$ being identical when weighted by the row player's strategy $x$ is a more direct condition for the uniqueness of the column player's strategy.

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