L | C | R | |

T | 2,2 | 2,2 | 2,2 |

U | 3,3 | 0,2 | 3,0 |

D | 0,0 | 3,2 | 0,3 |

Van Damme (1989) |

There are a continuum of Nash equilibria of this game in which player 1 plays T, in all of which player 2 plays C with positive probability; there are no equilibria in which player 1 plays D. With common knowledge of (near) rationality, then, it would seem that player 1 should never be expected to play D; if the possibility of D is excluded, then C is weakly dominated. Van Damme thus argues that (U,L) is the only "reasonable" equilibrium.

It occurs to me that the other example I wanted to present requires extensive form, and I don't feel like drawing game trees. I'll note that Mertens (1995) in Games and Economic Behavior 8:378–388 constructed two game trees with unique subgame perfect equilibria and the same reduced normal form such that the equilibria of the two games are different. Elmes and Reny had a 1994 JET paper improving on a classic result that claims to prove that any two extensive form games with the same reduced normal form can be converted to each other by means of three "inessential" transformations that should leave the strategic considerations unaffected. I have an inclination at some point to reconcile these results for myself.

While I'm here, I will write down one more normal form, not entirely unrelated to the one above:

L | C | R | |

U | 4,-4 | -4,4 | 1,1 |

M | -4,4 | 4,-4 | 1,1 |

D | 2,0 | 2,0 | 2,0 |

*does*play D, the choice between L and C doesn't matter. Similarly, player 1 can choose between U and M as though player 2 is definitely playing L or C. In the game in which choices R and D are disallowed, there's obviously a single Nash equilibrium; thus we should reasonably expect that player 1 will only play U and M with equal probability, and player 2 will only play L and C with equal probability. The game then reduces to

L/C | R | |

U/M | 0,0 | 1,1 |

D | 2,0 | 2,0 |

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