# Nash equilbrium in a 5x5 game

Consider a zero-sum game with the following matrix for row player: $$A= \begin{pmatrix} 0 & 1 & -1& -1&1 \\ -1&0&1&-1&-1\\ 1&-1&0&-1&1\\ 1&1&1&0&-1\\ -1&1&-1&1&0 \end{pmatrix}$$

I wanted to compute nash equilbrium. It is clear that there is no pure nash equilbrium. So following the strategy given in many sources. Let $$(p_1,\ldots,p_5)$$ be the probabilities for the first player. Then the expected payoff if the column player plays the first strategy is $$-p_2+p_3+p_4-p_5$$, for the second strategy, it is $$p_1-p_3+p_4+p_5$$, for the third it is $$-p_1+p_2+p_4-p_5$$, fourth: $$-p_1-p_2-p_3+p_5$$ and finally fifth $$p_1-p_2+p_3-p_4$$. Now setting this all equal (i.e. first = second, second = third, etc.) gets me to: \begin{align*} -p_2+p_3+p_4-p_5 &= p_1-p_3+p_4+p_5 \\ p_1-p_3+p_4+p_5 &= -p_1+p_2+p_4-p_5 \\ -p_1+p_2+p_4-p_5 &= -p_1-p_2-p_3+p_5 \\ -p_1-p_2-p_3+p_5 &= p_1-p_2+p_3-p_4 \end{align*} and together with the fact that $$p$$'s are probabilities i.e. $$p_1+p_2+p_3+p_4+p_5=1$$ I should get the following augmented matrix / system of lin. equations: $$\begin{pmatrix} -1&-1&2&0&-2 \\ 2&-1&-1&0&2\\ 0&2&1&1&-2\\ -2&0&-2&1&1\\ 1&1&1&1&1 \end{pmatrix} \begin{pmatrix} p_1\\p_2\\p_3\\p_4\\p_5 \end{pmatrix} =\begin{pmatrix} 0\\0\\0\\0\\1 \end{pmatrix}$$ but this gives me result $$(p_1,p_2,p_3,p_4,p_5)=(-1/7,1/7,3/7,1/7,3/7)$$.

The equilbrium should be $$(0,0,1/3,1/3,1/3)$$. By looking at the matrix, we can observe that the second strategy is dominated by the fourth and then the first is dominated by the third. We are left with a submatrix rows 3,4,5, columns 3,4,5, which is a copy of the well known rock-paper-scissors game and there is an equilbrium $$(1/3,1/3,1/3)$$. But where is the problem with the first approach ? Is it wrong to argue with the equalities of expected payoffs when there are dominant strategies? Or is there a numeric error in my calculations?

In your example, since the equilibrium assigns zero probability to the first two pure strategies, these two strategies need not have the same payoff as the others. They do need to have lower payoffs, though, since otherwise one could gain by shifting probability to them – and indeed they do have a lower payoff ($$-\frac13$$) than the three pure strategies with non-zero probabilities ($$0$$).
This isn’t specific to dominance – there can be strategies that aren’t dominated as pure strategies and yet are assigned zero probability in a Nash equilibrium. In your case, you could slightly increase the $$1$$s in the weakly dominated strategies to make them non-dominated, and your Nash equilibrium would still be valid and these strategies would still be assigned zero probability in it.