Lecture Notes 5: Mixed Strategies
All of our previous examples have involved pure strategies. That is, players in the equilibrium choose a single strategy for sure (with probability 1). But it is also possible for a player to use a
mixed strategy, in which he deliberately randomizes over multiple strategies.
Mixed strategies can be motivated in a variety of different ways, but a good example is to think about a tennis player. In equilibrium, he’s not always going to choose a particular kind of shot with probability 1. He’s more likely to keep his opponent guessing by randomizing over various shots that he can play.
In this section, we will extend our discussion of Nash Equilibrium to allow not only for pure strategy choices but also for mixed strategies.
Finding a Mixed Equilibrium
Consider the game below.
Y Z
A 1,4 5,2
B 7,0 3,4
Standard best-response analysis shows that there is no pure-strategy Nash Equilibrium. However, there is a mixed equilibrium.
The basic idea behind mixed strategies is that, in order for a player to be willing to use a mixed strategy, he must be indifferent over the strategies he is mixing. For example, if player 1 mixes between 𝐴𝐴 and 𝐵𝐵 in the game above, his expected payoff must be the same for both strategies. After all, if 𝐴𝐴 gave him a higher payoff, then he would select 𝐴𝐴 for sure – not mix 𝐴𝐴 and 𝐵𝐵. He is only willing to randomize if he is indifferent between 𝐴𝐴 and 𝐵𝐵.
Player 1 is indifferent between playing 𝐴𝐴 and 𝐵𝐵 when his expected payoff is the same from playing either strategy. Notice that player 1’s expected payoffs from 𝐴𝐴 or 𝐵𝐵 depend on the probability with which player 2 chooses 𝑌𝑌 or 𝑍𝑍. We will let 𝑝𝑝𝑌𝑌 and 𝑝𝑝𝑍𝑍 designate the probability with which player 2 chooses 𝑌𝑌 and 𝑍𝑍. Here are player 1’s expected payoffs.
Player 1 is only willing to randomize between 𝐴𝐴 and 𝐵𝐵 when these are equal:
Π𝐴𝐴 = Π𝐵𝐵
1𝑝𝑝𝑌𝑌+ 5𝑝𝑝𝑍𝑍 = 7𝑝𝑝𝑌𝑌+ 3𝑝𝑝𝑍𝑍
Since player 2 either uses strategy 𝑌𝑌 or 𝑍𝑍, we must have by definition that 𝑝𝑝𝑍𝑍 = 1− 𝑝𝑝𝑌𝑌.
1𝑝𝑝𝑌𝑌+ 5(1− 𝑝𝑝𝑌𝑌) = 7𝑝𝑝𝑌𝑌+ 3(1− 𝑝𝑝𝑌𝑌)
1𝑝𝑝𝑌𝑌+ 5−5𝑝𝑝𝑌𝑌 = 7𝑝𝑝𝑌𝑌+ 3−3𝑝𝑝𝑌𝑌 𝑝𝑝𝑌𝑌 = 14
Here, 𝑝𝑝𝑌𝑌 =1
4 implies that 𝑝𝑝𝑍𝑍 = 3
4. In summary, when player 2 randomizes, playing 𝑌𝑌 with
probability 1
4 and playing 𝑍𝑍 with probability 3
4, then player 1 is indifferent between choosing 𝐴𝐴 and
choosing 𝐵𝐵. Player 1’s expected payoff is the same for either 𝐴𝐴 or 𝐵𝐵. For any other strategy choice by player 2, player 1 would strictly prefer 𝐴𝐴 or 𝐵𝐵 and would not be willing to mix.
We also need to make sure that player 2 is actually willing to play a mixed strategy. Notice that player 2’s expected payoffs associated with playing 𝑌𝑌 or 𝑍𝑍 are a function of 𝑝𝑝𝐴𝐴 and 𝑝𝑝𝐵𝐵, the probabilities with which player 1 chooses 𝐴𝐴 or 𝐵𝐵. Player 2 is willing to play a mixed strategy when the expected payoffs from playing 𝑌𝑌 and 𝑍𝑍 are equal.
Π𝑌𝑌 = Π𝑍𝑍
4𝑝𝑝𝐴𝐴 + 0𝑝𝑝𝐵𝐵= 2𝑝𝑝𝐴𝐴+ 4𝑝𝑝𝐵𝐵
Since player 1 either uses strategy 𝐴𝐴 or 𝐵𝐵, we must have by definition that 𝑝𝑝𝐵𝐵 = 1− 𝑝𝑝𝐴𝐴.
4𝑝𝑝𝐴𝐴+ 0(1− 𝑝𝑝𝐴𝐴) = 2𝑝𝑝𝐴𝐴+ 4(1− 𝑝𝑝𝐴𝐴)
4𝑝𝑝𝐴𝐴 = 2𝑝𝑝𝐴𝐴+ 4−4𝑝𝑝𝐴𝐴 𝑝𝑝𝐴𝐴 = 23
That is, in order for player 2 to be willing to mix between 𝑌𝑌 and 𝑍𝑍, player 1 must mix between 𝐴𝐴 and 𝐵𝐵 with probabilities 𝑝𝑝𝐴𝐴 =2
3 and 𝑝𝑝𝐵𝐵 = 1 3.
Summarizing, player 1 is willing to mix when player 2 mixes using 𝑝𝑝𝑌𝑌 =1
4 and 𝑝𝑝𝑍𝑍= 3
4. Player 2
is willing to mix when player 1 mixes using 𝑝𝑝𝐴𝐴 =2
3 and 𝑝𝑝𝐵𝐵 = 1
other, each mixing in response to the other player mixing, and thus these strategies constitute a Nash Equilibrium. Formally, the Nash Equilibrium of this game is stated as follows
𝑁𝑁𝑁𝑁: �23𝐴𝐴+13𝐵𝐵,14𝑌𝑌+34𝑍𝑍�
Note that, by construction, player 1’s payoff in this equilibrium in this game is the same whether he chooses 𝐴𝐴 or 𝐵𝐵, given the mixing probabilities of player 2.
Π𝐴𝐴 =14(1) +34(5) = 4
Π𝐵𝐵 =14(7) +34(3) = 4
Similarly, player 2’s payoff in equilibrium is the same whether she chooses 𝑌𝑌 or 𝑍𝑍, given the mixture probabilities specified for player 1.
Π𝑌𝑌 =23(4) +13(0) = 83
Π𝑍𝑍 = 23(2) +13(4) = 83
Pure and Mixed Equilibria
Consider the game below.
Y Z
A 2,2 0,0
B 0,0 1,1
There are pure-strategy equilibria at (𝐴𝐴,𝑌𝑌) and (𝐵𝐵,𝑍𝑍). There is also a mixed-strategy equilibrium.
Player 1 is willing to mix when 𝐴𝐴 and 𝐵𝐵 generate the same expected payoff.
Π𝐴𝐴 = Π𝐵𝐵
2𝑝𝑝𝑌𝑌+ 0𝑝𝑝𝑍𝑍 = 0𝑝𝑝𝑌𝑌+ 1𝑝𝑝𝑍𝑍
2𝑝𝑝𝑌𝑌+ 0(1− 𝑝𝑝𝑌𝑌) = 0𝑝𝑝𝑌𝑌+ 1(1− 𝑝𝑝𝑌𝑌) 𝑝𝑝𝑌𝑌 = 13⇒ 𝑝𝑝𝑍𝑍 =23
Π𝑌𝑌 = Π𝑍𝑍
2𝑝𝑝𝐴𝐴 + 0𝑝𝑝𝐵𝐵= 0𝑝𝑝𝐴𝐴+ 1𝑝𝑝𝐵𝐵
2𝑝𝑝𝐴𝐴+ 0(1− 𝑝𝑝𝐴𝐴) = 0𝑝𝑝𝐴𝐴+ 1(1− 𝑝𝑝𝐴𝐴) 𝑝𝑝𝐴𝐴 = 13⇒ 𝑝𝑝𝐵𝐵= 23
These mixture probabilities imply that both players are willing to mix, so they are mutual best responses and this constitutes a Nash Equilibrium. This game has three Nash Equilibria. Two of them are pure-strategy equilibria and one is a mixed-strategy equilibrium. Formally, we can state the set of Nash Equilibria as follows.
𝑁𝑁𝑁𝑁: �(𝐴𝐴,𝑌𝑌), (𝐵𝐵,𝑍𝑍),�31𝐴𝐴+23𝐵𝐵,13𝑌𝑌+23𝑍𝑍��
The payoffs from (𝐴𝐴,𝑌𝑌) are (2,2). The payoffs from (𝐵𝐵,𝑍𝑍) are (1,1). The mixed equilibrium is
the worst of all. It gives payoffs �2
3, 2 3�.
Nonexistence of Mixed Equilibria
Not all games have mixed equilibria. Consider the following game.
L R
T 2,2 -1,4
B 4,-1 0,0
If you tried to set up the indifference condition for player 1 to mix between 𝑇𝑇 and 𝐵𝐵 you would quickly see the problem.
Π𝑇𝑇 =Π𝐵𝐵
2𝑝𝑝𝐿𝐿+−1𝑝𝑝𝑅𝑅 = 4𝑝𝑝𝐿𝐿+ 0𝑝𝑝𝑅𝑅
2𝑝𝑝𝐿𝐿+−1(1− 𝑝𝑝𝐿𝐿) = 4𝑝𝑝𝐿𝐿+ 0(1− 𝑝𝑝𝐿𝐿) 𝑝𝑝𝐿𝐿 =−1
is a dominant strategy for player 2. (𝐵𝐵,𝑅𝑅) is the only Nash Equilibrium of this game. There is no mixed-strategy equilibrium.
When both players have a strictly dominant strategy, there can be no mixed-strategy Nash Equilibrium. A strictly dominant strategy means that you want to play that strategy for sure, not mix.
Continuum of Equilibria
Sometimes games can have a whole continuum of Nash Equilibria. Consider the following game.
L R
T 2,1 1,2
B 0,1 1,3
Under what conditions is player 1 willing to play a mixed strategy?
Π𝑇𝑇 = Π𝐵𝐵
2𝑝𝑝𝐿𝐿+ 1𝑝𝑝𝑅𝑅 = 0𝑝𝑝𝐿𝐿+ 1𝑝𝑝𝑅𝑅
2𝑝𝑝𝐿𝐿+ 1(1− 𝑝𝑝𝐿𝐿) = 0𝑝𝑝𝐿𝐿+ 1(1− 𝑝𝑝𝐿𝐿) 𝑝𝑝𝐿𝐿 = 0
This makes sense from inspection of the game. 𝑇𝑇 is a weakly dominant strategy for player 1. If player 2 plays 𝐿𝐿 at all, then player 1 will prefer to play 𝑇𝑇 rather than 𝐵𝐵. But if player 2 plays 𝑅𝑅 for sure (a pure strategy), then player 1 is willing to mix between 𝑇𝑇 and 𝐵𝐵.
Player 2 will always play 𝑅𝑅 since 𝑅𝑅 is a strictly dominant strategy. Given this, player 1 can mix between 𝑇𝑇 and 𝐵𝐵 with any probabilities that he wants. Think about best responses – player 2 is always going to choose 𝑅𝑅 and then player 1 is indifferent as to what he does.
Formally, the Nash Equilibria of this game are stated as:
𝑁𝑁𝑁𝑁: (𝑝𝑝𝑇𝑇+ (1− 𝑝𝑝)𝐵𝐵,𝑅𝑅) for any 0≤ 𝑝𝑝 ≤1
Notice that this includes the pure equilibria (𝑇𝑇,𝑅𝑅) and (𝐵𝐵,𝑅𝑅) when 𝑝𝑝 = 1 and 𝑝𝑝= 0, respectively. This game has two pure equilibria and a whole continuum of mixed equilibria. Any mixing probability 0≤ 𝑝𝑝 ≤ 1 will work.
L R
T 2,1 1,2
B 0,4 1,3
This is a little bit more complicated. Player 1’s payoffs are the same. 𝑇𝑇 is a weakly dominant strategy and player 1 is willing to mix only when player 2 plays 𝑅𝑅 for sure.
In the previous example, 𝑅𝑅 was a dominant strategy and so player 2 was always willing to play 𝑅𝑅. In this example, player 2 is willing to use the pure strategy 𝑅𝑅 only when the expected payoff is higher than the expected payoff for 𝐿𝐿.
Π𝑅𝑅 ≥ Π𝐿𝐿
2𝑝𝑝𝑇𝑇+ 3𝑝𝑝𝑅𝑅 ≥ 1𝑝𝑝𝑇𝑇+ 4𝑝𝑝𝑅𝑅
2𝑝𝑝𝑇𝑇+ 3(1− 𝑝𝑝𝑇𝑇)≥ 1𝑝𝑝𝑇𝑇+ 4(1− 𝑝𝑝𝑇𝑇) 𝑝𝑝𝑇𝑇 ≥ 12
For this game, player 1 can’t mix between 𝑇𝑇 and 𝐵𝐵 with any probability he wants. He needs to play 𝑇𝑇 with 𝑝𝑝𝑇𝑇 ≥ 1
2. Otherwise, player 2 will not be willing to play 𝑅𝑅 and then player 1 will not be
willing to mix at all. Summarizing, this game has a pure equilibrium at (𝑇𝑇,𝑅𝑅) and also a continuum of mixed equilibria as follows:
𝑁𝑁𝑁𝑁: (𝑝𝑝𝑇𝑇+ (1− 𝑝𝑝)𝐵𝐵,𝑅𝑅) for any 𝑝𝑝 ≥12
Again, note that this includes the pure equilibrium (𝑇𝑇,𝑅𝑅) as the special case where 𝑝𝑝= 1.
Domination by a Mixed Strategy
Consider the following game.
X Y
A 5,2 1,1
B 1,3 5,2
C 2,1 2,4
Restricted to pure strategies, player 1 does not have any strictly dominated strategies. 𝐶𝐶 is not dominated by 𝐴𝐴 or 𝐵𝐵. However, notice that 𝐶𝐶 is strictly dominated by the mixed strategy 1
2𝐴𝐴+ 1 2𝐵𝐵
To show this, suppose that player 2 chooses 𝑋𝑋. Then player 1’s payoff from the mixed strategy
1 2𝐴𝐴+
1 2𝐵𝐵 is:
Π1 =12(5) +12(1) = 3
But if player 2 chooses 𝑌𝑌, then player 1’s payoff from the mixed strategy 1
2𝐴𝐴+ 1 2𝐵𝐵 is:
Π1 =12(1) +12(5) = 3
In other words, regardless of which strategy player 2 chooses, player 1 is always better off playing
the mixed strategy 1
2𝐴𝐴+ 1
2𝐵𝐵 than he is playing the pure strategy 𝐶𝐶. Thus, 𝐶𝐶 is a strictly dominated
strategy. It is not strictly dominated by any pure strategy, but it is strictly dominated by the mixed
strategy 1
2𝐴𝐴+ 1
2𝐵𝐵. If we eliminate 𝐶𝐶, the game is now dominance-solvable.
X Y
A 5,2 1,1
B 1,3 5,2
C 2,1 2,4
Strategy 𝑌𝑌 is now strictly dominated by 𝑋𝑋 for player 2, so we can eliminate it
X Y
A 5,2 1,1
B 1,3 5,2
C 2,1 2,4
Given what’s left, 𝐵𝐵 is strictly dominated by 𝐴𝐴 for player 1. This leaves (𝐴𝐴,𝑋𝑋) as the solution obtained through iterated removal of strictly dominated strategies. It is also the game’s only Nash Equilibrium.
X Y
A 5,2 1,1
B 1,3 5,2
Fully and Partially Mixed Equilibria
In games with more than two strategies, there is a possibility that a mixed strategy might involve all of a player’s strategies or might involve only some of the player’s strategies. Consider the following game.
X Y Z
A 4,3 10,1 0,2
B 1,7 4,3 9,4
C 3,0 7,5 4,3
This game has a pure-strategy equilibrium at (𝐴𝐴,𝑋𝑋).
There is a mixed-strategy equilibrium with partially mixed strategies where player 1 mixes 𝐴𝐴 and
𝐶𝐶 and where player 2 mixes 𝑋𝑋 and 𝑍𝑍. Given that player 2 uses only 𝑋𝑋 and 𝑍𝑍, player 1 is willing to mix 𝐴𝐴 and 𝐶𝐶 when:
Π𝐴𝐴 = Π𝐶𝐶
4𝑝𝑝𝑋𝑋+ 0𝑝𝑝𝑍𝑍 = 3𝑝𝑝𝑋𝑋+ 4𝑝𝑝𝑍𝑍
4𝑝𝑝𝑋𝑋+ 0(1− 𝑝𝑝𝑋𝑋) = 3𝑝𝑝𝑋𝑋+ 4(1− 𝑝𝑝𝑋𝑋) 𝑝𝑝𝑋𝑋= 45
Similarly, given that player 1 uses only 𝐴𝐴 and 𝐶𝐶 player 2 is willing to mix between 𝑋𝑋 and 𝑍𝑍 when:
Π𝑋𝑋 = Π𝑍𝑍
3𝑝𝑝𝐴𝐴+ 0𝑝𝑝𝐶𝐶 = 2𝑝𝑝𝐴𝐴+ 3𝑝𝑝𝐶𝐶
3𝑝𝑝𝐴𝐴+ 0(1− 𝑝𝑝𝐴𝐴) = 2𝑝𝑝𝐴𝐴+ 3(1− 𝑝𝑝𝐴𝐴) 𝑝𝑝𝐴𝐴 = 34
These are mutual best responses. Player 1 is willing to mix 𝐴𝐴 and 𝐶𝐶 when player 2 mixes with
4 5𝑋𝑋+
1
5𝑍𝑍. And player 2 is willing to mix 𝑋𝑋 and 𝑍𝑍 when player 1 mixes with 3 4𝐴𝐴+
1
4𝐶𝐶. Thus, this
strategy set comprises a Nash Equilibrium: �3
4𝐴𝐴+ 1 4𝐶𝐶,
4 5𝑋𝑋+
1
5𝑍𝑍�. This is called a partially mixed equilibrium because the mixtures involve only some of the strategies
There is also a fully mixed equilibrium in which all players use all strategies. In order for player 1 to be willing to mix over all strategies, we need the following conditions to hold. Payoffs for all three strategies must be equal, and the mixture probabilities must sum to 1.
Π𝐴𝐴 = Π𝐵𝐵 ⇒4𝑝𝑝𝑋𝑋+ 10𝑝𝑝𝑌𝑌+ 0𝑝𝑝𝑍𝑍 = 1𝑝𝑝𝑋𝑋+ 4𝑝𝑝𝑌𝑌+ 9𝑝𝑝𝑍𝑍 Π𝐴𝐴 =Π𝐶𝐶 ⇒4𝑝𝑝𝑋𝑋+ 10𝑝𝑝𝑌𝑌+ 0𝑝𝑝𝑍𝑍 = 3𝑝𝑝𝑋𝑋+ 7𝑝𝑝𝑌𝑌+ 4𝑝𝑝𝑍𝑍
𝑝𝑝𝑋𝑋+𝑝𝑝𝑌𝑌+𝑝𝑝𝑍𝑍 = 1
This is a system of 3 equations in 3 variables, which can be solved to give (𝑝𝑝𝑋𝑋,𝑝𝑝𝑌𝑌,𝑝𝑝𝑍𝑍) =�1
3, 1 3,
1 3�.
We also need to make sure that player 2 is willing to mix between all of her strategies. Similar to what we just did for player 1, the indifference conditions are as follows.
Π𝑋𝑋 =Π𝑌𝑌 ⇒3𝑝𝑝𝐴𝐴 + 7𝑝𝑝𝐵𝐵+ 0𝑝𝑝𝐶𝐶 = 1𝑝𝑝𝐴𝐴+ 3𝑝𝑝𝐵𝐵+ 5𝑝𝑝𝐶𝐶 Π𝑋𝑋 =Π𝑍𝑍 ⇒3𝑝𝑝𝐴𝐴 + 7𝑝𝑝𝐵𝐵+ 0𝑝𝑝𝐶𝐶 = 2𝑝𝑝𝐴𝐴+ 4𝑝𝑝𝐵𝐵+ 3𝑝𝑝𝐶𝐶
𝑝𝑝𝐴𝐴+𝑝𝑝𝐵𝐵+𝑝𝑝𝐶𝐶 = 1
This system can be solved to give (𝑝𝑝𝐴𝐴,𝑝𝑝𝐵𝐵,𝑝𝑝𝐶𝐶) =�1
2, 1 6,
1 3�
Thus, both players are indifferent between all 3 of their strategies and are willing to play fully mixed strategies when their opponents also play fully mixed strategies with the probabilities given.
These are mutual best responses and so �1
2𝐴𝐴+ 1 6𝐵𝐵+
1 3𝐶𝐶,
1 3𝑋𝑋+
1 3𝑌𝑌+
1
3𝑍𝑍� is a Nash Equilibrium
of this game.
All told, this game has three Nash Equilibria: a pure equilibrium, a partially mixed equilibrium and a fully-mixed equilibrium.
𝑁𝑁𝑁𝑁: �(𝐴𝐴,𝑋𝑋),�34𝐴𝐴+14𝐶𝐶,45𝑋𝑋+15𝑍𝑍�,�12𝐴𝐴+16𝐵𝐵+13𝐶𝐶,13𝑋𝑋+13𝑌𝑌+13𝑍𝑍��
Symmetric Mixed-Strategy Equilibria in Games with Many Players
In games with many players, it can be tricky to compute all the mixed equilibria. However, for games with a symmetry to the payoff structure, we can compute a symmetric mixed equilibrium. The example in this section illustrates the “bystander effect.”
In a symmetric mixed strategy equilibrium, each player reports the crime with probability 𝑝𝑝 and does not report it with probability 1− 𝑝𝑝. We say that the equilibrium is symmetric because we are computing a mixed equilibrium where all players use the same strategy. There may be other equilibria. We are computing an equilibrium specifically of this form.
Like any other mixed strategy, players are willing to mix when their expected payoffs from reporting and not reporting are equal. A player who reports the crime gets a certain payoff equal to Πreport= 𝑉𝑉 − 𝑐𝑐 since filing the report himself guarantees that the report is issued, but he has to bear the cost 𝑐𝑐 to make the report.
What about a player who does not report the crime? It is possible that none of the other players will report the crime. This happens with probability (1− 𝑝𝑝)𝑛𝑛−1 – when all the rest of the 𝑛𝑛 −1 players do not file a report. This leads to a payoff Π= 0. On the other hand, with probability
1−(1− 𝑝𝑝)𝑛𝑛−1, someone else will report the crime and then our player’s payoff is Π= 𝑉𝑉 since
the crime is reported but our player doesn’t have to bear the cost of filing the report. Together, the expected payoff associated with not filing a report is:
Πno report= (1− 𝑝𝑝)𝑛𝑛−1⋅0 + (1−(1− 𝑝𝑝)𝑛𝑛−1)⋅ 𝑉𝑉
Each witness is willing to use a mixed strategy when the two are equal to each other.
Πreport = Πno report
𝑉𝑉 − 𝑐𝑐= (1− 𝑝𝑝)𝑛𝑛−1⋅0 + (1−(1− 𝑝𝑝)𝑛𝑛−1)⋅ 𝑉𝑉 𝑉𝑉 − 𝑐𝑐
𝑉𝑉 = 1−(1− 𝑝𝑝)𝑛𝑛−1
(1− 𝑝𝑝)𝑛𝑛−1 = 𝑐𝑐 𝑉𝑉 𝑝𝑝= 1− �𝑉𝑉�𝑐𝑐
1 𝑛𝑛−1
To study this equilibrium a bit, suppose that 𝑉𝑉 = 12, 𝑐𝑐 = 1 and that there are 𝑛𝑛= 10 witnesses. In the symmetric mixed equilibrium, the probability with which each witness files a report is:
𝑝𝑝= 1− �121� 1 10−1
= 0.2413
That is, each witness files a report with probability 0.2413 and does not file a report with probability 0.7587. Thus, the probability that none of the 10 witnesses will report the crime is
What if the number of witnesses rises to 𝑛𝑛= 100? In the new mixed equilibrium, the probability with which each witness files a report is:
𝑝𝑝 = 1− �121� 1 100−1
= 0.0248
The probability that none of the 100 witnesses will report the crime is 0.9752100 = 0.0812.
Look at what happened. Increasing the number of witnesses from 10 to 100 actually increases the chance that the crime will go unreported! This is the well-known bystander effect. Each individual feels less compelled to intervene when there are more bystanders. Overall, having a whole bunch of witnesses might not help increase the chance of someone intervening. Everyone might just wait for someone else to get involved. So much for safety in numbers!
A Pirate Expedition
Here is another example of how to compute a symmetric mixed strategy equilibrium. Five pirates encounter a potential mission to capture a treasure. Going out on the mission costs 7. If all five pirates go, they can capture the treasure and each pirate gets a payoff of 20 from a successful mission. Each pirate decides simultaneously whether to go or not go. The treasure cannot be captured unless all five pirates go. Pirates who don’t go on the mission earn a payoff of 0.
First of all, the game has two equilibria where all players use pure strategies.
• If all five pirates go on the mission, this is a Nash Equilibrium. All pirates get a net payoff of 13. But, if one deviates, then he would get a payoff of zero. There is no incentive for a single pirate to deviate
• If all five pirates do not go on the mission, this is also a Nash Equilibrium. All pirates get a payoff of 0. But, if one pirate decides to deviate alone and go on the mission, the mission will be unsuccessful and the pirate’s payoff will be −7. There is no incentive for a single pirate to deviate.
There is also a mixed strategy Nash Equilibrium where all pirates go on the mission with probability 𝑝𝑝. Again, to compute a mixed equilibrium, we need to equate the expected payoff from going on the mission with the expected payoff from not going on the mission.
A pirate who goes on the mission gets a payoff of 13 if the mission is successful. The mission will only be successful if the other four pirates all go out on the mission as well, which happens with probability 𝑝𝑝4.
A pirate who goes on the mission gets a payoff of −7 if the mission is unsuccessful. The mission will be unsuccessful with probability 1− 𝑝𝑝4 – that is, if the other four pirates do not all agree to go out on the mission.
To compute the mixed strategy equilibrium, we equate the expected payoff from going on the mission with the expected payoff from not going on the mission.
Πgo =Πdon't go
𝑝𝑝4⋅13 + (1− 𝑝𝑝4)⋅ −7 = 0
13𝑝𝑝4 = 7⋅(1− 𝑝𝑝4)
20𝑝𝑝4 = 7
𝑝𝑝∗ =�7
20�
1 4⁄
= 0.7692
Problems
1. Find a mixed-strategy Nash equilibrium for the game below.
Y Z
A 8,3 1,0
B 2,1 4,7
2. Consider the game below.
a. What are the pure-strategy Nash equilibria?
b. Find an equilibrium where player 1 mixes A and B and player 2 mixes X and Z.
X Y Z
A 4,2 0,0 0,1
B 0,0 2,4 1,3
3. Consider the game below.
a. For 𝑥𝑥< 1, what is the Nash Equilibrium?
b. For 𝑥𝑥> 1, what are the Nash Equilibria? There are two pure equilibria and one mixed equilibrium.
c. Work out the equilibria for 𝑥𝑥= 1.
L R
T 𝑥𝑥,𝑥𝑥 0,1
B 1,0 1,1
4. Consider the game below.
a. What are the pure strategy Nash Equilibria?
b. Describe all equilibria where player 2 mixes between 𝑀𝑀 and 𝑅𝑅.
L M R
A 2,2 0,3 1,3
5. A group of 10 players each simultaneously chooses 𝐴𝐴 or 𝐵𝐵. Any player who chooses 𝐴𝐴 gets a payoff of 1. If a player chooses 𝐵𝐵, his payoff is 2 as long as no other player chooses 𝐵𝐵. However, the payoff from playing 𝐵𝐵 is 0 if any other player or players also chooses 𝐵𝐵. Find the symmetric, mixed-strategy equilibrium of this game.
