Let's consider this 3×3 game:
\begin{matrix}
&A&B&C \\
A&1,1 & 10,0 & -10,1 \\
B&0,10 & 1,1 & 10,1 \\
C&1,-10 & 1,10 & 1,1
\end{matrix}
Player 1 is the row player, player 2 is the column player.
From my understanding (based on this Math SE answer), there is a pure strategy that is a Nash Equilibrium (which is (A,A)). Indeed, when player 1 plays A, player 2 would gain 1 if playing A, 0 if playing B, 1 if playing C and therefore would have no incentive to deviate from (A,A). Similarly, when player 2 plays A, player 1 has no incentive to deviate from (A,A).
There are also mixed Nash Equilibria.
To solve them, we can do:
Consider player 2. They play column A with probability $p$, B with probability $q$, and C with probability $1−p−q$. We need to find $p$,$q$ such that player 1 is indifferent between his pure strategies A,B,C.
Similarly for player 1 we could solve the same thing.
If now, I'm player 1, or player 2 and I try to maximise my gain. If I go for a mixed strategy, no MATTER WHAT my opponent will play, they cannot "blunder" / "make a bad decision".
However, if I go for the strategy related to the pure Nash Equilibrium, my opponent CAN "blunder" / "make a bad decision". E.g., if I'm player 2 and I play A (following the pure Nash Equilibrium (A,A)), my opponent can play B and 'blunder', leading me to gain 10.
I do understand that searching for a Nash Equilibrium "just" means searching for both players' strategies such that 'no player's expected outcome can be improved by changing one's own strategy' (ref) but in a "Game Theory" setting, does that mean that one should always go for "pure Nash Equilibria" (if there are some) to maximise their gain considering opponent might blunder? Since going for a "mixed Nash Equilibrium" means no matter what the opponent does, the payoff will be the same?
Would the answer to my question change if this was a zero-sum game?
Best Answer
On your statement:
Taking advantage of possible mistakes of the opponent is also possible when you go for a mixed-strategy Nash equilibrium, so the above argument is not necessarily correct. For example, in the battle-of-sexes game (see this version) if you (row player) go for $\left (\frac35,\frac25 \right )$, and your opponent instead of going for $\left (\frac25,\frac35 \right)$ mistakenly goes for $ (1,0)$ or $ (x,1-x)$ for any $x>\frac25$, it helps you to gain more.