In my prior blog I challenged readers to solve the game displayed in the title image. Before we move forward, let’s solve that game by finding the Nash Equilibrium.

1\2 | L | C | R |

T | 2,0 | 1,1 | 4,2 |

M | 3,4 | 1,2 | 2,3 |

B | 1,3 | 0,2 | 3,0 |

First we want to look for any dominated strategies. It appears that strategy T strictly dominates strategy B. Since we assume both players are rational decision makers we can eliminate that strategy since Player 1 will not choose strategy B. In addition, with B being eliminated, strategy R strictly dominates strategy C for Player 2 so strategy C can be removed. We are now left with the following game:

1\2 | L | R |

T | 2,0 | 4,2 |

M | 3,4 | 2,3 |

We can now solve for Nash Equilibrium. If Player 2 chooses L, Player 1 prefers M over T since 3>2. If Player 2 chooses R, Player 1 prefers T over M since 4>2. In addition, Player 2 prefers R over L when Player 1 chooses T and L over R when Player 1 chooses M. As a result, we end up with the following:

1\2 | L | R |

T | 2,0 | 4,2 |

M | 3,4 | 2,3 |

Wait, we have more than one Nash Equilibrium? The answer is yes! In fact, many games can have multiple Nash Equilibrium (ex. Battle of the Sexes). The next game we will analyze could have one or more Nash Equilibrium. This game will also involve three players instead of two. Don’t worry, we will work through this together.

In this game we have three college students who have finished school for the day and are debating whether they should go to the local bar or not (…should be studying). The problem is that the bar can be overcrowded (or empty) so the overall utility of going to the bar drops when too many people go (or not enough people go). The third value represents the utility (payoff) for Student 3.

Let’s look at this **step by step**:

Both Students 2 & 3 choose Go: Student 1 prefers to stay (1>-2)

Student 2 chooses Stay, Student 3 chooses Go: Student 1 prefers to go (2>1)

Student 2 chooses Go, Student 3 chooses Stay: Student 1 prefers to go (2>1)

Both Students 2 & 3 choose Stay: Student 1 prefers to stay (1>0)

Students 1 & 3 choose Go: Student 2 prefers to stay (1>-2)

Student 1 chooses Stay, Student 3 chooses Go: Student 2 prefers to go (2>1)

Student 1 chooses Go, Student 3 chooses Stay: Student 2 prefers to go (2>1)

Both Students 1 & 3 choose Stay: Student 2 prefers to stay (1>0)

Both Students 1 & 2 choose Go: Student 3 prefers to stay (1>-2)

Student 1 chooses Go, Student 2 chooses Stay: Student 3 prefers to go (2>1)

Student 1 chooses Stay, Student 2 chooses Go: Student 3 prefers to go (2>1)

Both Students 1 & 2 choose Stay: Student 3 prefers to stay (1>0)

In this game we end up with a total of four Nash Equilibrium. So it appears that each student prefers that only 2/3 students go to the bar otherwise they would rather stay home. Throughout these examples we have solved simultaneous games that involved pure strategy Nash Equilibrium. We have discovered that games could have one or many. But what happens if we come across a game where there is no pure strategy Nash Equilibrium? I will cover this occurrence in the next blog.

I appreciate your feedback in the comments below.