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Players’ Strategies

Strategies:

Elmer: Price war if Fred opens

No Price war if Fred opens.

Fred: Open

Don’t Open

Two Nash equilibrium

- Subgame perfect Nash equilibrium where Fred opens and Elmer does not start a price war.
- How about a Nash equilibrium that is not subgame perfect?

Why is these answers wrong?

- There is a Nash equilibrium that is not subgame perfect where Fred does not open and Elmer either starts a price war or doesn’t.
- Fred does not open a station.

What would make Fred stay out?

- This is a game.
- The payoff to each of Fred’s actions depends on what he believes Elmer will do.
- If Fred believes that Elmer won’t start a price war, Fred will start a gas station.
- If Fred believes that Elmer will start a price war, Fred will not start a gas station.

A correct answer

- There is a Nash equilibrium that is not subgame perfect in which Elmer’s strategy is to start a price war if Fred opens a station and in Fred’s strategy is to not open a station.
- In Nash equilibrium, each is doing a best response to the other’s strategy.
- Why is this Nash equilibrium not subgame perfect?

Strategic Form

Elmer

Fred

If Fred believes that Elmer will start a price war if he enters, Fred will stay out.

If Elmer believes that Fred will stay out, Elmer gets as high a payoff from the

Strategy “Start a price war if Elmer enters” than from “Don’t start a price war if Elmer enters.”

Lemons Problem

- Value of cars to current owners:
- Lemons $1,900
- Good cars $10,000
- Value of cars to potential buyers
- Lemons $2,000
- Good Cars $12,000
- Fraction of all cars that are lemons: q

Values and Beliefs

- If everybody believes that all used cars are on the market, what is the expected value of a used car to a buyer?
- $2000q+$12,000(1-q)
- If everybody believes that only the lemons come on the market, what is the expected value of a used car to a buyer?
- $2000

Which of these is true for the lemons market?

- in a pooling equilibrium, the lemons would be sold and the good cars would not be.
- In a separating equilibrium, the good cars would be sold and the lemons would not be.
- In a separating equilibrium, all the used cars would be sold.
- In a pooling equilibrium, all the used cars would be sold at the same price.
- In a pooling equliibrium, the lemons would sell for $2,000 and the good cars would sell for $12,000.

Pooling and separating equilibrium

- Pooling equilibrium if all types act the same.
- Separating if they act differently.

When will there be no pooling equilibrium?

- There will be no pooling equilibrium if, the price of used cars when all used cars reach the market is not high enough to induce all owners to sell their cars.
- Owners of good used cars will not sell their cars for less than $10,000
- Price of a used car if everybody expects them all to be on the market is 2,000q+12,000(1-q).
- Good cars are on market only if

2,000q+12,000(1-q)>10,000.

This is so if and only if q<1/5.

What if q>1/5?

- If q>1/5, we have seen that there is no pooling equilibrium.
- What about a separating equilibrium in which only the lemons are on the market.
- If all buyers believe that only lemons will be available, the expected value of a car to buyers will be $2,000.
- At this price, owners of lemons will want to sell. Owners of good cars will not want to sell.
- Beliefs of buyers are confirmed. This is an equilibrium.

What if q<1/5

- If q<1/5, then (as we showed) when all cars are available, if buyers believe that all used cars will reach the market, then the expected value of a used car to buyers will be greater than $10,000.
- If the price of used cars is greater than $10,000, all car owners, including both lemon owners and good car owners will want to sell.
- So there is an equilibrium in which buyers believe that all used cars are available and all owners want to sell their used cars.

A second equilibrium

- When q<1/5, we showed that there is a pooling equilibrium with all cars available.
- There is also a separating equilibrium.
- Suppose that buyers believe that only lemons will be on the market.
- Then the most a buyer will pay for a used car is $2000.
- At this price, only lemon owners will sell.
- The belief that only lemons will be available is confirmed.
- This is an equilibrium. Everybody maximizes payoff given beliefs and beliefs are confirmed by the outcome.

A signaling equilibrium

- Two types of workers, Beavers and Drones.
- Beavers’ productivity is $5,000 per month.
- Drones’ productivity is $3,000 per month.
- Employers can’t tell them apart or monitor their work.
- Listening to dull lectures is worse for drones than for beavers.
- Cost to beaver $20 per hour
- Cost to drone $100 per hour

Finding a separating equilibrium

- Workers can take a course of 15 hours of dull lectures at tuition price T.
- Employers expect only beavers to take the course and they pay $5,000 wage to those who take course and $3,000 to those who do not.
- Find T such that beavers will take the course and drones won’t take it.

What do drones do?

- If a drone takes the course, he will get wages of $5,000 and have costs of 15x100+T.
- If a drone doesn’t take the course, he gets wages of $3,000 and has no costs.
- A drone will not take the course if

5,000-1500-T>3,000 . This will be the case if

T>500.

What do beavers do?

- A beaver can earn $5,000, from taking the course and $3,000 from not taking the course.
- Cost of course to a beaver is 15x20+T.
- Beaver will take the course if

5,000-300-T>3,000. This means T<$1700

Separating equilibrium

- With tuition rate T
- Drones will not take course if T>500
- Beavers will take course if T<1700
- So there is a separating equilibrium for any tuition rate between 500 and 1700.
- The highest tuition the college could charge and get any students is just under $1700.

Easier course, higher tuition

- Suppose the course is shortened to 10 hours.
- Drones will not take course if

5000-10x100-T>3000, which implies T>1000.

- Beavers will take course if

5000-10x20-T>3000, which implies T>1800.

So now there is a separating equilibrium for any T between $1,000 and $1,800.

Alice, Bob, and Carol

Top number is Bob’s payoff, bottom number is Alice’s payoff.

Carol

Invites

Doesn’t invite

Bob

Bob

C

A

B

B

A

Alice

B

A

B

A

A

B

A

B

A

B

2

3

1

1

2.5

0

3

2

2.5

1

1

1

0

0

2

3

0

0

3

2

Strategic form

- Bob has 6 possible strategies, Alice has 2.
- Write x/y to mean Bob goes to Movie x if Carol invites him and y if she does not. In each box, payoff for Alice is written above payoff for Bob.
- Note that payoffs depend on c when Bob’s action depends on what Carol does.

What are the pure strategy Nash equilibria?

- Since Alice has only two possible strategies, it is probably a good idea to start by looking for Nash equilibria among her possible choices.

Is there a Nash equilibrium where Alice goes to B?

- Yes. In this equlibriumBob goes to C if Carol invites him and to B if she does not.
- Yes, in this equilibrium Bob goes to A if Carol invites him and to B if she does not.
- Yes, in this equilibrium, Bob always goes to B.
- Yes, in this equilibrium, Bob always goes to A.
- No

Is there a Nash equilibrium in which Alice goes to A?

- Yes, in this equilibrium, Bob always goes to A.
- Yes, in this equilibrium, Bob goes to C if Carol invites him and he goes to A if Carol does not.
- Yes, in this equilibrium, Bob always goes to B.
- Yes, in this equilibrium, Bob goes to C if Carol invites him and to B if Carol does not.
- No

The card game

Let x/y denote the strategy for Player 1, x if high, y if low

This game

- Has no regular, proper subgames
- Has no pure strategy Nash equilibrium.

Mixed strategy Nash equilibrium and “bluffing”.

- Suppose that Player 1 always plays Raise if she has a high card, but plays Raise with probability q when she has a low card.
- In such an equilibrium, Player 1 must be indifferent between playing Raise or See if she has a low card.
- Let’s find mixed strategy for Player 2 that makes Player 1 indifferent when she has a low card.

Mixed strategy for 2

- If player 2 plays Pass with probability p,

then for Player 1, if she has a low card

- Expected Payoff from playing See is -1.
- Expected Payoff from playing Raise is

p-2(1-p).

- Player 1 is indifferent if p-2(1-p)=-1
- This is equivalent to 3p=1, or p=1/3.
- So in a mixed strategy equilibrium, Player 2 must play Pass with probability 1/3 and Meet with probability 2/3.

Mixed strategy for 1

- In a mixed strategy equilibrium, if whenever Player 1 has a low card, she plays Raise with probability q, it must be that Player 2 is indifferent between Pass and Meet.
- Let P(H|R) be the probability that Player 1 has a high card, given that she plays Raise.
- If Player 1 plays Raise:
- payoff to Player 2 from Pass is -1.
- Payoff to Player 2 from Raise is

2(1-P(H|R)) -2P(H|R)

2(1-P(H|R)) -2P(H|R)=-1

- But what is P(H|R).
- Try Bayes’ Law: P(H|R)=P(H and R)/ P(R).
- P(H and R)=P(H)=1/2
- P(R)=1/2+1/2(q)=1/2(1+q)
- So P(H|R)= 1/(1+q)

Solving

- So P(H|R)=1/(1+q)
- Expected payoff to 2 from Meet is
- 2(1-P(H|R)) -2P(H|R)=2(q-1)/(1+q)
- Expected payoff to 2 from Pass is -1
- Solve the equation 2(q-1)/(1+q)=-1
- Answer is q=1/3.
- So in the mixed strategy Nash equilibrium, Player 1 always plays Raise when she has a high card and “bluffs” 1/3 of the time when she has a low card.
- Player 2 plays “meet” 2/3 of the time.

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