Consider the payoff matrix,
$$\begin{bmatrix} (0,0) & (-1,1)\\
(1,-1) & (-10,-10)
\end{bmatrix}$$
Assume players want to maximize payoff.
If the row player choses strategy $(1,0)$ i.e., strategy 1. Then the column player must choose strategy $(0,1)$ to maximize his payoff ($0$ vs $1$)
But then the row player wants to change to strategy $2$ to change his payoff from $-1$ to $1$. At this point, the column player can only change to strategy $1$ to avoid a payoff of $-10$.
This means the pure Nash of this game is $\{(0,1), (1,0)\}$ and the solution $\{(1,0), (0,1)\}$ will be forever unstable (one player will always want to change).
Why is that $\{(1,0), (0,1)\}$ is a solution when computed? https://intranet.csc.liv.ac.uk/cgi-bin/cgiwrap/rahul/input.py
Is the algorithm for computing pure NE:
- First consider case when row player move first?
- Next consider case when column player move first?
I think this 2 is what I am missing
Best Answer
If the row player chooses strategy 1, the column player can choose between payoff 0 or 1, and will therefore go with 1 (i.e. strategy 2). If the column player chooses strategy 2, the row player can choose between payoff -1 and -10, and will therefore choose -1 (i.e. strategy 1)
So row 1, column 2 is a Nash equilibrium. As is row 2, column 1 for analogous reasons.