Mixed Strategy Nash Equilibrium III

Is a mixed strategy optimal?

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So far we have solved and analyzed games with only a mixed strategy Nash Equilibrium. Today we will analyze a scenario where both mixed and pure strategies exist. On occasion my wife and I are too lazy to cook and choose to go out to dinner. However, this is where the fun really begins. Like other couples, sometimes we don’t always agree on where to eat and have our own preferences.

This can be represented in the following game:

Me\Wife Mexican BBQ
Mexican 2,3 0,0
BBQ 0,0 3,2

On this night, I prefer to eat BBQ (love tri tip) over my wife’s preference of Mexican food. As you can see in this game, I would rather eat Mexican food over eating alone and my wife would rather have BBQ over eating alone. Eating alone represents a payoff of 0. In this game we have two pure strategy Nash Equilibrium represented as (Mexican, Mexican) and (BBQ, BBQ).

As we have seen in the past few blogs a mixed strategy Nash Equilibrium exists as well. Let’s solve for this equilibrium.

q 1-q
Me\Wife Mexican BBQ
p Mexican 2,3 0,0
1-p BBQ 0,0 3,2

Using the information above we will first solve for p and 1-p:

3(p) + 0(1-p) = 0(p) + 2(1-p)

=> 3p = 2 – 2p

=> 5p = 2

=> p = 2/5 and 1-p = 3/5

Now let’s solve for q and 1-q:

2(q) + 0(1-q) = 0(q) + 3(1-q)

=> 2q = 3 – 3q

=> 5q = 3

=> q = 3/5 and 1-q = 2/5

Here is an update of the game matrix with the mixed strategies.

q = 3/5 1-q = 2/5
Me\Wife Mexican BBQ
p = 2/5 Mexican 2,3 0,0
1-p = 3/5 BBQ 0,0 3,2

With the probabilities of each strategy being chosen solved, let’s determine the payouts for both my wife and I choosing a mixed strategy. We first need to determine the probability of us choosing:

(Mexican, Mexican) = (p)(q) = (2/5)(3/5) = 6/25

(Mexican, BBQ) = (p)(1-q) = (2/5)(2/5) = 4/25

(BBQ, Mexican) = (1-p)(q) = (3/5)(3/5) = 9/25

(BBQ, BBQ) = (1-p)(1-q) = (3/5)(2/5) = 6/25

Next we need to take summation of each individual player’s pure strategy payoff multiplied by the corresponding probability in that quadrant.

My payoff: (2)(6/25) + (0)(9/25) + (0)(4/25) + (3)(6/25) = 1.2

Wife’s payoff: (3)(6/25) + (0)(9/25) + (0)(4/25) + (2)(6/25) = 1.2

For this game my payoffs are (3, 2, and 1.2) for (BBQ, Mexican, and Mixed) and my wife’s are (3, 2, and 1.2) for (Mexican, BBQ, and Mixed). Looking at these payoffs, is a mixed strategy Nash Equilibrium optimal? We clearly don’t want to eat dinner alone (payoff of 0) and we run the risk of this occurring if we choose a mixed strategy. In addition, even though I favor BBQ over Mexican food, Mexican food with my wife is a better payoff than choosing a mix.

I can summarize my preferences as BBQ > Mexican > Mixed (3 > 2 > 1.2). Since we assume rational decisions, my wife will approach this game similarly. Her preferences are Mexican > BBQ > Mixed (3 > 2 > 1.2). It is not optimal for us to choose a mixed strategy. Plus, why would you want to choose a mixed strategy on your wife or husband in the first place? Are you trying to trick/confuse him or her? Sometimes it is best to concede to your spouse’s preference for the night and he or she will return the favor another night.

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