# mixed equilibria

I'm doing practice problems for an exam in a game theory course and I want solutions by some one more knowledgeable, so that I could see if I did them right. I need all the problems in the file "supplement" done. I also need problems 5 and 9 from the file "problems 5 and 9", problem 3 from from file "problem 3" and problem 8 from file "problem 8".

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Hello!

Here are your answers.

Supplement Q1

a. Let's call R to Rowena and C to Colin. The payoff matrix of this bimatrix game they are playing is:

R = [90 20]

[30 60]

C = [10 80]

[70 40]

Notice that for each possible outcome, the probabilities of winning add up to 100: clearly, when R has a 90% chance of winning the point, then C has a 10% chance of doing it.

b. The best response functions (BR) are:

Rowena: BR(Forehand) = Forehand

BR(Backhand) = Backhand

Colin: BR(Forehand) = Backhand

BR(Backhand) = Forehand

From the best response functions, it's clear that there are no Nash equilibria in pure strategies. Let's find the mixed-strategy equilibrium using the regular procedure.

Let's assume Colin plays F with probability q and B with probability (1-q). Now we compute the expected payoff for Rowena for each of her strategies:

If Rowena plays F, exp. payoff is q*90 + (1-q)*20

If Rowena plays B, exp. payoff is q*30 + (1-q)*60

We now have to equate these payoffs in order to find the q that makes her indifferent between her actions (if she weren't indifferent, then there would be no incentives to play mixed strategies):

q*90 + (1-q)*20 = q*30 + (1-q)*60

60q = 40(1-q)

q = 2/5

So in the mixed strategy equilibrium, q=2/5 and (1-q)=3/5. In order to find Rowena's probabilities, we repeat exactly the same procedure, from the "point of view" of Colin. Let's call p and 1-p to the probabilities that Rowena plays F or B. Then, Colin's expected payoffs are:

If he plays F: p*10 + (1-p)*70

If he plays B: p*80 + (1-p)*40

After equating and solving for p, we find that p=3/10, so 1-p=7/10. Therefore, the mixed-strategies NE happens when Rowena plays F with probability 3/10 and B with probability 7/10; and Colin plays F with probability 2/5 and B with probability 3/5.

c. i. We repeat the same procedure as before, but now the payoff when both players play B is 65 for Rowena and 35 for Colin. Let's find p and (1-p) again to see what happens with Rowena's probabilities:

p*10 + (1-p)*70 = p*80 + (1-p)*35

[notice that this is exactly the sa,e as the last equation from the last question, but using 35 instead of 40 for Colin's payoff for playing B]

Solving for p, we find that p=1/3. So the probability of using Backhand (1-p) decreases from 7/10 to 2/3 at the new Nash equilibrium.

c. ii. Without the need for doing any calculations, it's clear that her NE probability of winning must increase. There's no way her expected payoff can decrease when she has a "new" strategy (returning anticipated B with 65% probability instead of 60%) that is strictly better than the previous one.

Anyway, if you're interested in finding this analytically, you can easily do so. The NE probability of winning would simply be the probability of each outcome multiplied by the payoff of that outcome. For example, the probability of being in [F, F] is p*q, the probability of being in [F, B] is p*(1-q), etc. Therefore, the expected payoff (which is the probability of winning the point) is p*q*90 + p*(1-q)*20 + ... If you want to do this, you would first have to recalculate q for this case.

Supplement Q2

a. Strategy P for the column player is strictly dominated by a mixed strategy, in which this player plays R with probability 1/2 and S with probability 1/2. Let's see why:

Payoffs of playing P (column player):

If row plays R: 2

If row plays P: 1

If row plays S: ...

#### Solution Summary

Questions in game theory are assessed.