January 19, 2001 EXAMPLE WHERE ITERATED DOMINANCE DOES NOT LEAD TO THE UNIQUE NASH EQUILIBRIUM Conjecture 1: If a game has a unique Nash equilibrium in pure strategies, then that equilibrium can be reached by iterated dominance. Counterexample; COLUMN: LEFT CENTER RIGHT UP 50,1 1,35 1,1 ROW: MIDDLE 1,35 50,1 1,1 DOWN 1,1 1,1 10,10* The starred combination, DOWN, RIGHT, is Nash. Consider ROW. Is any strategy weakly dominant? UP is best, for 50, if COL picks LEFT. MIDDLE is best, for 50, if COL picks CENTER. DOWN is best, for 10, if COL picks RIGHT. Consider COL. Is any strategy weakly dominant? LEFT is best, for 35, if ROW picks MIDDLE. CENTER IS best, for 35, if ROW picks UP. RIGHT is best, for 10, if ROW picks DOWN. Thus, no strategy is dominated, and iterated dominance gets us nowhere. There is, to be sure, more than one equilibrium, but there is only one equilibrium in pure strategies. I think there are two other equilibria. One mixes UP, MIDDLE and LEFT, CENTER. The other mixes all three strategies for each player. (But I haven't checked these.) That suggests the conjecture: Conjecture 2: If a game has a unique Nash equilibrium, which is in pure strategies, then that equilibrium can be reached by iterated dominance. Proof idea: ... Two concepts of dominance: 1. Strategy X never does better than Y. (the standard definition) 2. Strategy X is not the best response to ANY pure strategy of the other player. An interesting point: a strategy might be best against a mix of two strategies of the other player, though it is not a best response to any single pure strategy. That willb e true whenever a straegy is dominated according to definition 2 but not by defintion 1.