Game Development Reference
In-Depth Information
By definition, no player achieves a higher payoff by switching from a
dominant strategy to another strategy. This explains the following obvious
Note 1.1 (Dominant Strategy) Consider a strategic game G. Suppose that
s is a joint strategy such that each s i is a dominant strategy. Then it is a
Nash equilibrium of G.
In particular, the conclusion of the lemma holds if each s i is a strictly
or a weakly dominant strategy. In the former case, when the game is finite,
we can additionally assert (see the IESDS Theorem 1.3 below) that s is a
unique Nash equilibrium of G . This stronger claim does not hold if each s i is
a weakly dominant strategy. Indeed, consider the game
T 1 , 1 1 , 1
B 1 , 1 0 , 0
Here T is a weakly dominant strategy for the player 1, L is a weakly dominant
strategy for player 2 and, as prescribed by the above Note, ( T,L ), is a
Nash equilibrium. However, this game has two other Nash equilibria, ( T,R )
and ( B, L ).
The converse of the above Note of course is not true. Indeed, there are
games in which no strategy is dominant, and yet they have a Nash equilibrium.
An example is the Battle of the Sexes game that has two Nash equilibria,
but no dominant strategy.
So to find a Nash equilibrium (or Nash equilibria) of a game it does not
su ce to check whether a dominant strategy exists. In what follows we
investigate whether iterated elimination of strategies can be of help.
1.3 Iterated elimination of strategies I
1.3.1 Elimination of strictly dominated strategies
We assumed that each player is rational. So when searching for an outcome
that is optimal for all players we can safely remove strategies that are strictly
dominated by some other strategy. This can be done in a number of ways.
For example, we could remove all or some strictly dominated strategies
simultaneously, or start removing them in a round robin fashion starting
with, say, player 1. To discuss this matter more rigorously we introduce the
notion of a restriction of a game.
Given a game G := ( S 1 ,...,S n ,p 1 ,...,p n ) and (possibly empty) sets
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