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## Homework Statement

In the two-player game "Morra", the players simultaneously hold up some fingers and each guesses the total number of fingers help up.

If exactly one player guesses correctly, then the other player pays her the amount of her guess(in dollars, say). If either both players guess correctly or neither does, then no payments are made.

Consider a version of the game in which the number of fingers each player may hold up is restricted to either one or two.

a. Given the symmetry of the game, each player's equilibrium payoffs is 0 by the result from (some exercise that says: show that in any symmetric strictly competitive game in which U2 = -U1, where Ui is player i's expected payoff function for i = 1,2, each player's payoff in every mixed strategy Nash equilibrium is 0.). Find the mixed strategies of player 1 that guarantee that her payoff is at least 0 (i.e. the strategies such that her payoff is at least 0 for each pure strategy of player 2) and hence find all the mixed strategy equilibria of the game.

b. Find the rationalizable actions of each player in this game.

## Homework Equations

Maxminimization is relevant.

## The Attempt at a Solution

Player 1 plays (1 and 2) with probability a, (1 and 3) with probability b, (2 and 3) with probability c, and (2 and 4) with probability (1 – a – b – c).

EPayoff1[player 1 mixes, (1 and 2)] = (0)a + (-2)b + (3)c + (0)(1 – a – b – c) = -2b + 3c≥0

EPayoff1[player 1 mixes, (1 and 3)] = (2)a + (0)b + (0)c + (-3)(1 – a – b – c) = 2a – 3 + 3a + 3b + 3c = 5a + 3b + 3c – 3 ≥ 0

EPayoff1[player 1 mixes, (2 and 3)] = (-3)a + (0)b + (0)c + (4)(1 – a – b – c) = -3a + 4 – 4a – 4b – 4c = 4 – 7a – 4b – 4c ≥ 0

EPayoff1[player 1 mixes, (2 and 4)] = (0)a + (3)b + (-4)c + (0)(1 – a – b – c) = 3b – 4c ≥0

5a + 5b ≥ 3 7a + 7b ≥ 4 doesn’t make sense remove one of 2’s pure strategy

Player 1 plays (1 and 2) with probability a, (1 and 3) with probability b, (2 and 3) with probability (1 – a – b)

EPayoff1[player 1 mixes, (1 and 2)] = (0)a + (-2)b + (3)(1 – a – b) = -2b + 3 – 3a – 3b = 3 – 3a – 5b ≥ 0 3 – 3(0) – 5b ≥ 0 3 ≥ 5b 3/5 ≥ b

EPayoff1[player 1 mixes, (1 and 3)] = (2)a + (0)b + (0)(1 – a – b) = 2a ≥ 0 a ≥ 0

EPayoff1[player 1 mixes, (2 and 3)] = (-3)a + (0)b + (0)(1 – a – b) = -3a≥ 0 0 ≥ a

EPayoff1[player 1 mixes, (2 and 4)] = (0)a + (3)b + (-4)(1 – a – b) = 3b – 4 + 4a + 4b = 7b + 4a – 4 ≥ 0 7b + 4(0) – 4 ≥ 0 7b ≥ 4 b ≥ 4/7

All ≥ 0, satisfies the conditions. a = 0 3/5≥ b (1 – a – b) ≥ 2/5

Player 1 plays (1 and 2) with probability a, (1 and 3) with probability b, (2 and 4) with probability (1 – a – b)

EPayoff1[player 1 mixes, (1 and 2)] = (0)a + (-2)b + (0)(1 – a – b) = -2b ≥ 0

EPayoff1[player 1 mixes, (1 and 3)] = (2)a + (0)b + (-3)(1 – a – b) = 2a – 3 +3a + 3b = 5a + 3b – 3 ≥ 0 5a + 3(0) – 3 ≥ 0 5a – 3 ≥ 0 a ≥ 3/5

EPayoff1[player 1 mixes, (2 and 3)] = (-3)a + (0)b + (4)(1 – a – b) = -3a + 4 – 4a – 4b = 4 – 7a – 4b ≥ 0 4 – 7a ≥ 0 4 ≥ 7a 4/7 ≥ a

EPayoff1[player 1 mixes, (2 and 4)] = (0)a + (3)b + (0)(1 – a – b) = 3b ≥ 0

All ≥ 0, satisfies conditions. 4/7 ≥ a ≥ 3/5 b = 0 2/5 ≥ (1 – a – b)

Player 1 plays (1 and 2) with probability a, (2 and 3) with probability b, (2 and 4) with probability (1 – a – b)

EPayoff1[player 1 mixes, (1 and 2)] = (0)a + (3)b + (0)(1 – a – b) = 3b ≥ 0

EPayoff1[player 1 mixes, (1 and 3)] = (2)a + (0)b + (-3)(1 – a – b) = 2a – 3 + 3a + 3b = 5a + 3b – 3 ≥ 0 5a + 3(0) – 3 ≥ 0 5a ≥ 3 a ≥ 3/5

EPayoff1[player 1 mixes, (2 and 3)] = (-3)a + (0)b + (4)(1 – a – b) = -3a + 4 – 4a – 4b = 4 – 7a – 4b ≥ 0 4 – 7a – 4(0) = 4 – 7a ≥ 0 4 ≥ 7a 4/7 ≥ a

Payoff1[player 1 mixes, (2 and 4)] = (0)a + (-4)b + (0)(1 – a – b) = -4b ≥ 0

All ≥ 0, satisfies conditions 4/7 ≥ a ≥ 3/5 b = 0 2/5 ≥ (1 – a – b)

Player 1 plays (1 and 3) with probability a, (2 and 3) with probability b, (2 and 4) with probability (1 – a – b)

EPayoff1[player 1 mixes, (1 and 3)] = (0)a + (0)b + (-3)(1 – a – b) = 3a + 3b – 3 ≥ 0

4b + 3b – 3 ≥ 0 7b ≥ 3 b ≥ 3/7

EPayoff1[player 1 mixes, (2 and 3)] = (0)a + (0)b + (4)(1 – a – b) = 4 – 4a – 4b ≥ 0

4 – 4a – 3a = 4 – 7a 4 ≥ 7a 4/7 ≥ a