## ECO290E: Game Theory

## Review of Lecture 4 ^{}

Spatial competition model has a unique NE in which both shops choose the middle point.

⇒_{ } Minimal differentiation!

Political platforms (left/right or liberal/ conservative)

TV shows (time slots)

Under the Bertrand model:

Price will be set at the level of marginal cost even when there are only two competitors.

⇒_{ } Huge jump from 1 (monopoly) to 2 (duopoly)!

Under the Cournot model:

Total output (price) is between monopoly level and that under perfect competition.

## Zero-Sum Game ^{}

Matching Pennies:^{
}

** Player 2 **
**Player 1 **

**Heads ** **Tails **

**Heads ** ** 1 **
** -1 **

** -1 **
** 1 **

**Tails ** ** -1 ** ** 1 **

## No Nash Equilibrium?

### The distinguishing feature of zero-sum game is that

### each player would like to outguess the other since

### there is No “win-win” situation.

### Examples are:

Porker: bluff or not

Battle (War): attack/defend by land or by sea

Tennis: left or right to serve/receive

### When each player would always like outguess the

### other(s), there is no Nash equilibrium.

### Is there no solution or stable outcome?

## Mixed Strategy

### A mixed strategy for a player is a probability

### distribution over some (or all) of her strategies.

### The strategy we have studied so far, i.e., taking some

### action for sure, is called a pure-strategy .

### When the outcome of the game is uncertain, we

### assume that each player maximizes expected value of

### her payoff.

⇒_{ }

### Expected utility theory (von Neumann and

### Morgenstern, 1944)

## Matching Pennies Again ^{}

Introducing mixed strategies:^{
}

** Player 2 **
**Player 1 **

**Heads (p) ** **Tails (1-p) **

**Heads **
**(q) **

** 1 **
** -1 **

** -1 **
** 1 **

**Tails **
**(1-q) **

** -1 **
** 1 **

** 1 **
** -1 **

## How to Find Equilibrium?

If a player takes both “Heads” and “Tails” with positive probability, she must be indifferent between these two pure strategies, i.e., the expected payoff derived by

choosing Heads must be equal to that by choosing Tails.

Otherwise, it is strictly better for her to choose either H or T for sure; no incentive to mix the strategies.

*-p + (1-p) = p - (1-p), hence p = 0.5. *

*q - (1-q) = -q + (1-q), hence q = 0.5. *

## How to Verify Equilibrium?

*Note that if p = 0.5, Player 1 does not have a strict *

*incentive to change her strategy from q = 0.5. *

### Similarly, Player 2 does not have a strict incentive to

*change his strategy from p = 0.5, if q = 0.5. *

⇒_{ }

* Therefore, p = q = 0.5 constitutes a mixed-strategy *

### equilibrium.

### Q: Is this equilibrium reasonable/stable?

### A: Yes (each player ends up randomizing two strategies

### equally if the rival is smart enough).

## Best Response Curves ^{}

Given the opponent mixed strategy, draw the best
response (mixed strategy) against it.^{
}

p

1

## BR2 ^{
}

## BR1 ^{
}

0.5^{
}

**NE** ^{
}

## Existence of NE

### Theorem (Nash, 1950)

### If a game has finite number of players and actions,

### then there exists at least one Nash equilibrium,

### possibly involving mixed strategies.

### Q: Are there any games which do not even have a

### mixed-strategy equilibrium?

### A: Yes. For example, integer game.

(Two person) Zero-sum game in which players independently choose a number simultaneously. The player who picked a larger number wins.

## Modified Matching Pennies ^{}

### Suppose the payoffs in the up-left cell changes as the

### following:

** Player 2 **
**Player 1 **

**Heads (p’) ** **Tails (1-p’) **

**Heads **
**(q’) **

** 2 **
** -2 **

** -1 **
** 1 **

**Tails ** ** -1 ** ** 1 **

## Indifference Property

Under mixed-strategy NE, Player 1 must be indifferent between choosing H and T:

⇒_{ }* -2p’ + (1-p’) = p’ - (1-p’), hence p’ = 0.4. *

Similarly, Player 2 must be indifferent between choosing H and T:

⇒_{ }* 2q’ - (1-q’) = -q’ + (1-q’), hence q’ = 0.4. *

*You can easily verify that (p’, q’) = (0.4, 0.4) indeed *
becomes a mixed-strategy NE.

Note that player 2 chooses H with less than a half chance, even if H looks “better” strategy than T.

## Multiple Nash Equilibria ^{}

** 2 **
**1 **

**D (x) ** **E (y) ** **F (1-x-y) **

**A **
**(a) **

** 6 **
** 7 **

** 8 **
** 5 **

** 0 **
** 0 **

**B **
**(b) **

** 8 **
** 5 **

** 6 **
** 7 **

** 0 **
** 0 **

**C ** ** 0 ** ** 0 ** ** 4 **

## Three Different Equilibria ^{}

Pure-strategy NE

^{(}^{C}^{, }** ^{F}**) is a unique pure-strategy NE.

Mixed-strategy NE (1)

P1 mixes A (q) and B (1-q); p2 mixes D (p) and E (1-p).

7p + 5(1-p) = 5p + 7(1-p) ^{⇒} p = 1/2.

6q + 8(1-q) = 8q + 6(1-q) ^{⇒} q = 1/2.

Mixed-strategy NE (2)

Both players use all three strategies.

A (a), B (b), C (1-a-b); D (x), E (y), F (1-x-y)

7x + 5y = 5x + 7y = 4(1-x-y) ^{⇒} x = y = 1/5.

## Example: Soccer (Penalty Kick) ^{}

** Kicker **
**GK **

**Left (x) ** **Middle (y) ** **Right (1-x-y) **

**Left **
**(p) **

** 40 **
** 60 **

** 100 **
** 0 **

** 80 **
** 20 **

**Middle **
**(q) **

** 80 **
** 20 **

** 0 **
**100 **

** 80 **
** 20 **

**Right ** ** 80 ** ** 100 ** ** 20 **

## Further Exercises ^{}

Construct a game which has both pure-strategy and mixed-strategy Nash equilibria.

Consider a modified matching penny game in which the payoff on (H, H) is changed from (-1, 1) to (-k, k) for k > 0. Then, derive a mixed-strategy Nash equilibrium.

Prove that a strategy which is erased by the iterated

elimination of strictly dominated strategies can never be
selected (with positive probability) in a mixed-strategy
Nash equilibrium. ^{
}