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The Nash Equilibrium is a point where neither player can unilaterally improve their position by changing strategy (if their opponent stays the same). This is not the same as the best outcome. In the prisoner's dilemma, both confess is the Nash Equilibrium - neither player benefits by changing from confession to silence if their opponent stays on confession.
And it's actually rational for both of them to confess. And the confession is actually a Nash equilibrium. And we'll talk more about this, but a Nash equilibrium is where each party has picked a choice given the choices of the other party. So when we think of, or each party has to pick the optimal choice, given whatever choice the other party ...
Game theorists and social scientists have been fascinated by Prisoner's Dilemma, a two-by-two game (two players, each with two possible pure strategies) with a particular payoff matrix (Rapoport and Chammah; Poundstone). The game's nickname and the accompanying story were provided by A. W. Tucker. Suppose that two prisoners, accused of jointly ...
The term Nash-equilibrium applies to the set of strategies taken by all the players, not to any one player’s individual strategy. If a player can only do worse by deviating then the equilibrium is strict, if she can do just as well (but no better) then then the equilibrium is weak, and if she can do better, then it is not an equilibrium. The ...
Sep 4, 1997 · In a game like this, the notion of nash equilibrium loses some of its privileged status. Recall that a pair of moves is a nash equilibrium if each is a best reply to the other. Let us extend the notation used in the discussion of the asynchronous PD and let \(\bDu\) be the strategy that calls for defection at every node of an IPD.
Jun 16, 2024 · The Nash equilibrium in this example is for both players to betray one other, even though mutual cooperation leads to a better outcome for both players; however, if one prisoner chooses mutual ...
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If the iterated prisoner's dilemma is played a finite number of times and both players know this, then the dominant strategy and Nash equilibrium is to defect in all rounds. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn.
