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Let's consider this 3x3 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?

FluidMechanics Potential Flows
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    Depends on the conditions. There's a pretty standard assumption in game theory that players are rational, utility-maximizing, and don't mess up the execution of their strategy. In that type of standard situation, players won't deviate. If you assume that a player can mess up, there might be a way that one "equilibrium" is preferable to another, but I think you'd need to have some sort of a priori signal as to how they'll screw up execution. Also I'm not sure if it'd technically be an equilibrium in that type of situation, since deviation is possible (through poor execution). – Amaan M May 07 '24 at 22:51

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On your statement:

One should always go for "pure Nash Equilibria" (if there are some) to maximize their gain considering opponent might blunder

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.

Amir
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  • I see that what you say works but I'm still confused. We can find $(\frac{3}{5},\frac{2}{5})$ by saying I'm looking for $(p,1-p)$ such that the Player 2 would be indifferent whether they go Prize Fight or Ballet but it turns out that they end up not being indifferent since going to Prize Fight alone would benefit me. – FluidMechanics Potential Flows May 26 '24 at 09:23
  • Although, I guess in this scenario because it's not zero-sum, Player 2 doesn't mind that it would benefit me since their reward is not linked to the reward the opponent gets. – FluidMechanics Potential Flows May 26 '24 at 09:24
  • In the mixed-strategy equilibrium, it is true that Player 2 is indifferent whether she/he goes to Prize Fight or Ballet, but being indifferent does not mean Player 2 can go to Prize Fight with probability 1 (in fact, if Player 2 mistakenly goes for this, it will benefit Player 1, but when Player 2 goes to Prize Fight with probability $\frac25$, it will not benefit Player 1). Each player at the mixed-strategy equilibrium plays one of the two options stochastically based the obtained probabilities to control the other player's behavior, thought his/her gain is the same for each option. – Amir May 26 '24 at 16:39
  • I'm sorry what do you mean by "being indifferent does not mean Player 2 can go to Prize Fight with probability 1" – FluidMechanics Potential Flows May 27 '24 at 19:58
  • If $(p,1-p)$ and $(q,1-q)$ with $0<p,q<1$ are the mixed-strategies of the two players in a Nash equilibrium, it can be proven:

    $$U_1((1,0),(q,1-q))=U_1((0,1),(q,1-q)), U_2((p,1-p),(1,0))=U_2((p,1-p),(0,1))$$

    where $$U_i((p,1-p),(q,1-q))=pqu_i(A,A)+p(1-q)u_i(A,B)+(1-p)qu_i(B,A)+(1-p)(1-q)u_i(B,B).$$

    This helps us to find $(p,1-p)$ and $(q,1-q)$ more easily. The above is often interpreted as each player is indifferent between the two pure strategies A and B that she plays in the mixed-strategy Nash equilibrium, but it does not mean the player goes with probability 1 for A or B.

     – Amir May 28 '24 at 13:31
  • I see, and in no way when computing $p, q$ do we worry about minimising the other player's payoff. Which is just done naturally in zero-sum games. – FluidMechanics Potential Flows May 29 '24 at 19:00