Mixed Strategy Nash Equilibrium IV

Additional problem solving

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We will focus on solving more games that involve pure strategy Nash Equilibrium and mixed strategy Nash Equilibrium. It is important that we understand how to solve these games as it lays the foundation for further topics in game theory. This is especially the case if you are interested in advancing your studies. In these problems the goal is to identify the Nash Equilibrium and not necessarily the payoffs.

Let’s start with Problem #1:

q 1-q
1\2 Red Blue
p Red 1,4 2,3
1-p Blue 3,2 1,4

Player 1 chooses Blue when Player 2 chooses Red.

Player 1 chooses Red when Player 2 chooses Blue.

Player 2 chooses Red when Player 1 chooses Red.

Player 2 chooses Blue when Player 1 chooses Blue.

There is no pure strategy Nash Equilibrium in this game. However, let’s see if a mixed strategy Nash Equilibrium exists. Similar to previous examples, we need to solve for p, 1-p, q, and 1-q to determine the mixed strategies.

For Player 1:

4p + 2(1-p) = 3p + 4(1-p)

=> 4p + 2 – 2p = 3p + 4 – 4p

=> 2p + 2 = 4 – p

=> 3p = 2

=> p = 2/3, 1-p = (1 – 2/3) = 1/3

For Player 2:

1q + 2(1-q) = 3q + 1(1-q)

=> q + 2 -2q = 3q +1 – q

=> 2 – q = 2q + 1

=> 1 = 3q

=> q = 1/3, 1-q = (1 – 1/3) = 2/3

Summary: Pure strategy Nash Equilibrium does not exist. Mixed strategy for Player 1 (Red = 2/3, Blue = 1/3) and Player 2 (Red = 1/3, Blue = 2/3)

Problem #2

q 1-q
1\2 Left Right
p Up 5,6 4,1
1-p Down 3,2 5,3

Player 1 chooses Up when Player 2 chooses Left.

Player 1 chooses Down when Player 2 chooses Right.

Player 2 chooses Left when Player 1 chooses Up.

Player 2 chooses Right when Player 1 chooses Down.

The pure strategy Nash Equilibrium for this game is (Up, Left) and (Down, Right). Now let’s determine if a mixed strategy Nash Equilibrium exists.

For Player 1:

6p + 2(1-p)  = 1p + 3(1-p)

=> 6p + 2  – 2p = p + 3 – 3p

=> 4p + 2 = 3 – 2p

=> 6p = 1

=> p = 1/6,  1-p = (1 – 1/6) = 5/6

For Player 2:

5q + 4(1-q) = 3q + 5(1-q)

=> 5q + 4 – 4q = 3q + 5 – 5q

=> q + 4 = 5 – 2q

=> 3q = 1

=> q = 1/3, 1-q = (1-1/3) = 2/3

Summary: In this game we have both pure strategy Nash Equilibrium and a mixed strategy Nash equilibrium. Pure strategies are (Up, Left) and (Down, Right). Mixed strategies are Player 1 (Up = 1/6, Down = 5/6) and Player 2 (Left = 1/3, Right = 2/3).

Problem #3:

q 1-q
1\2 Yellow Green
p Yellow 5,4 2,4
1-p Green 1,4 3,3

Player 1 chooses Yellow when Player 2 chooses Yellow.

Player 1 chooses Green when Player 2 chooses Green.

Player 2 is indifferent between Yellow and Green when Player 1 chooses Yellow.

Player 2 chooses Yellow when Player 1 chooses Green.

There is only one pure strategy Nash Equilibrium in this game (Yellow, Yellow). Now for the mixed strategy Nash equilibrium:

For Player 1:

4p + 4(1-p) = 4p + 3(1-p)

=> 4p + 4 – 4p = 4p + 3 – 3p

=> 4 = 3 +p

=> p = 1, 1-p = (1 – 1) = 0

What does it mean for p = 1? Remember in a probability distribution the sum of the probabilities must equal 100%. This means there is no mixed strategy Nash Equilibrium for Player 1. So Player 1’s optimal choice is to always choose the pure strategy of Red.

For Player 2:

5q + 2(1-q) = 1q + 3(1-q)

=> 5q + 2 – 2q = q + 3 – 3q

=> 3q + 2 = 3 – 2q

=> 5q = 1

=> q = 1/5, 1-q = (1 – 1/5) = 4/5

Summary: The pure strategy Nash Equilibrium is (Yellow,Yellow). Even though Player 1 does not have a mixed strategy Nash Equilibrium, Player 2’s mixed strategy is (Yellow = 1/5, Green = 4/5)

For additional practice I recommend you look for available problems online. Feel free to leave comments or feedback below.

 

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