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Consider the following 2-players game: the set of actions for the player 1 (P1) is $U_1$, for the player 2 (P2) is $U_2$. Utility functions are $g_1(u_1,u_2)$ and $g_2(u_1,u_2)$.

First, P1 decide about his action and say it to P2 - so P2 can make a decision using this knowledge. Whenever P1 makes an action $u_1\in U_1$ it's optimal for P2 to make a decision $u_2\in I(u_1)$ where the subset $I(u_1)\subset U_2$ is defined as $$ U_2^*(u_1) = Arg\max_{u_2\in U_2}g_2(u_1,u_2). $$ This fact is clear since P2 would like to maximize his utility $g_2$ provided an action $u_1$ of P1.

P1 knows that whenever he makes an action $u_1$, P2 will make an action $u_2\in U_2^*(u_1)$. If for some $u_1$ the set $I(u_1)$ contains more than one element, player P1 don't know exactly which action will perform $u_2$. Let us suppose here that P1 is considering then worst case scenario, then it's optimal for he to make the following action $$ u^*_1 \in U^*_1 = Arg\max_{u_1\in U_1}\left(\min\limits_{u_2\in U_2^*(u_1)}g_1(u_1,u_2)\right). $$

This is an optimal strategy for the both players. My question is the following: if for any $u^*_1\in U^*_1$ and $u^*_2\in U^*_2(u^*_1)$ the pair $(u^*_1,u^*_2)$ corresponds to the Nash equilibrium?

For the simplicity we can assume that $U_1$ and $U_2$ are finite.

P.S. I am not so experienced in the game theory, so this solution is open to critics about the solution itself and remarks on the terminology.

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up vote 1 down vote accepted

Single strategies will not necessarily lead to a Nash equilibrium. For example, consider a zero-sum game where the utilities to the two players are shown by $(x_1,x_2)$ based on the choices.

                         Player 2 chooses
                          Y           Z

Player 1   A           (+2, -2)    (-1, +1)
           B           (-2, +2)    (+1, -1)

Player 1 knows that with a choice of A, Player 2 will choose Z at a utility of $-1$ for Player 1; and with a choice of B, Player 2 will choose Y at a utility of $-2$ for Player 1. So in your argument Player 1 will choose A and Player 2 Z. But this is not a Nash equilibrium as Player 1 can benefit changing to B if Player 2 does not change from Z. Wikipedia says:

If each player has chosen a strategy and no player can benefit by changing his or her strategy while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitute a Nash equilibrium

The Nash equilibrium for this (and many other two player zero-sum games) is a mixed strategy, in this case where Player 1 chooses A and B randomly with equal probability and Player 2 chooses Z twice as often as choosing Y.

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Thank you for the answer. Maybe the difference is that in my example the 2nd player is in better position since he can decide based on the decision of the 1st player, while the Nash equilibrium is presented for the games where all players are on the same level. – Ilya Apr 13 '11 at 9:43
There is a difference between: (a) Player 1 announcing a mixed strategy and then Player 2 deciding what to do and only later discovering which choices they make in a particular example; and (b) Player 1 having a (not necessarily mixed) strategy and then annoucing which choice is made in a particular example before Player 2 has to do anything. I was dealing with (a) while you seem to be concentrating on (b). – Henry Apr 13 '11 at 9:58

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