Mixed strategy equilibrium of game

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Mixed strategy equilibrium of game in Figure 1:-

Show that the game in Figure 1 has no non degenerate mixed strategy equilibrium. Each action of each player is a best response to some action of the other player (for example, T is a best response of player 1 to R, M is a best response to C, and B is a best response to L).

Thus, setting Z1 = {T, M, B} and Z2 = {L, C, R} we see that every action of each player is rationalizable. In particular the actions T and B of player 1 are rationalizable, even though they are not used with positive probability in any Nash equilibrium. The argument for player 1's choosing T, for example, is that player 2 might choose R, which is rational for her if she thinks player 1 will choose B, and it is reasonable for player 2 to so think since B is optimal for player 1 if she thinks that player 2 will choose L, which in turn is rational for player 2 if she thinks that player 1 will choose T, and so on.

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Even in games in which every rationalizable action is used with positive probability in some Nash equilibrium, the predictions of the notion of rationalizability are weaker than those of Nash equilibrium. The reason is that the notion of Nash equilibrium makes a prediction about the profile of chosen actions, while the notion of rationalizability makes a prediction about the actions chosen by each player. In a game with more than one Nash equilibrium these two predictions may differ.

Consider, for example, the game in Figure 2. The notion of Nash equilibrium predicts that the outcome will be either (T, L) or (B, R) in this game, while the notion of rationalizability does not restrict the outcome at all: both T and B are rationalizable for player 1 and both L and R are rationalizable for player 2, so the outcome could be any of the four possible pairs of actions.

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Reference no: EM131252761

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