# Mixed Strategy Equilibrium in the Game of Morra

• chen0000
In summary, the two-player game "Morra" is a symmetric strictly competitive game where each player's equilibrium payoff is 0. In order to guarantee a payoff of at least 0, player 1 can use a mixed strategy of playing (1 and 2) with probability 3/5, (1 and 3) with probability 0, (2 and 3) with probability 0, and (2 and 4) with probability 2/5. Another option is to play (1 and 2) with probability 4/7, (1 and 3) with probability 0, (2 and 3) with probability 0, and (2 and 4) with probability 3
chen0000

## 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

I really don't know what else to do to approach the question...

## 1. What is the objective of the game?

The objective of Morra is to correctly guess the total number of fingers that both players have shown.

## 2. How do you play the game?

The game starts with both players showing a number of fingers (1-5) on each hand. They then simultaneously guess the total number of fingers shown by both players. If one player correctly guesses the total, they win. If both players guess correctly, it is a tie. If neither player guesses correctly, the player with the closest guess wins.

## 3. Are there any variations to the game?

Yes, there are variations to the game such as playing with more than two players, using different objects instead of fingers, or adding in additional rules such as penalties for guessing incorrectly.

## 4. Is there a strategy to winning the game?

There is no guaranteed strategy to winning Morra, as it is a game of chance. However, some players may try to predict their opponent's moves or use psychological tactics to influence their opponent's guesses.

## 5. What are the origins of the game?

Morra is believed to have originated in ancient Greece and was commonly played by soldiers to pass time during wars. It has since spread to various cultures and is now played as a popular game of chance around the world.

• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
29
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
938
• Calculus and Beyond Homework Help
Replies
2
Views
557
• Calculus and Beyond Homework Help
Replies
1
Views
542
• Calculus and Beyond Homework Help
Replies
2
Views
589
• Calculus and Beyond Homework Help
Replies
2
Views
715
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
1K