How to interpret negative probability for a strategy in mixed nash equilibrium? I am trying to get the mixed strategy in Nash equilibrium for the following matrix.
$$\begin{pmatrix}
0 & 3  & 4  & 5  & 6 \\ 
3 & 0 & 5 & 6 & 7 \\ 
4 & 5 & 0 & 7 & 8 \\ 
5 & 6 & 7 & 0 & 9 \\ 
6 & 7 & 8 & 9 & 0
\end{pmatrix}$$
When I solve the problem by assigning probabilities to each column, and solving system of equations $row_1 = row_2$, $row_2 = row_3$ and so on, I got the following probabilities:
[-0.15693431  0.17153285  0.2810219   0.33576642  0.36861314]

Can anyone explain me how to interpret such result?
 A: Recall what you are doing when you are setting those equations. You are finding probabilities that one player must assign to his actions such that the other player is indifferent across his actions.
If you solve for a negative probability, something has gone wrong. Not with your algebra, necessarily, but more likely with your search space. Remember that a rational player will never put positive weight on a strictly dominated action, so there can be no mixed Nash equilibria where he does. Your solved probabilities suggest that P2 would need to be putting a negative weight on action 1 to make P1 indifferent. Since this is impossible, that implies that there is no way to make P1 indifferent between his five actions.
Action 1 is, in fact, weakly dominated for P1 by a uniform mix over the other four actions. That is, action 1 is weakly dominated by $\left(0,\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$. There's slack everywhere but in the fifth column. So we can take some $\epsilon>0$ from action 5 and put it on action 4, and then the resulting mix $\left(0,\frac{1}{4},\frac{1}{4},\frac{1}{4}+\epsilon,\frac{1}{4}-\epsilon\right)$ strictly dominates action 1.
