SMU H3 Map

• Content map: SMU H3 Game Theory Map

Golden Balls

Players

• 2 players
• Both players have the same strategies (options)

Information Structure

• Simultaneous Game
• Both players decide on their strategy and make it known to the other player at the same time

Strategy

• 2 Players can choose to either split or steal
• $\{split, steal\}$

Outcomes

• If the players choose to split, both players get half the prise each
• If one player chooses to steal and the other player chooses to split, the player who chooses to steal receives all of the prise
• If both players choose to steal, both players get nothing

Payoffs

Summary

$\text{let } p \text{ be the total prise available to both players}$

Payoff Matrix

Outcomes

• When the other player chooses split, choosing steal is better than split
• When the other player chooses steal, choosing steal is as equally good/bad as split

Conclusions

• Steal is never worse, but sometimes better than split
• Steal is a weakly dominant strategy
• Split is a weakly dominated strategy

Black or Red

Instructions

• You will be asked to choose between red or black.
• Your earnings are determined by the colour that you choose and by the colour chosen by the person matched with you.
  • If you each play red, you will each earn 2 participation points.
  • If you each play black, you will each earn 3 participation points.
  • If you play black and the other person plays red, then you earn zero and the other person earns 5 participation points.
  • If you play red and the other person plays black, you earn 5 participations points, and the other person earns zero.

Players

• 2 players
• Both players have the same strategies (options)

Information Structure

• Simultaneous Game
• Both players decide on their strategy and make it known to the other player at the same time

Strategy

• 2 players can choose to either red or black

Payoff Matrix

Dominant and Dominated Strategy

Dominant Strategy

• A strategy that does better than all other strategies no matter the other player’s actions
• This means the strategy is independent of what the other player(s) perform
• A rational player will always choose a dominant strategy

Justification for Dominant Strategy

• Given common knowledge, since the player knows that the other player will play a dominant strategy, the current player will change their strategy to outperform him
• In many cases, this would be to not play a dominated strategy, and play a dominant strategy

Existence of Dominated Strategies

• In many games, there might not be a dominant strategy
• However, there exists dominated strategy which consistently performs worse than other possible strategy

Implications of Dominated Strategies

• We can delete dominated strategies from the game
• If new dominated strategies arise from this, we can delete them too

Iterative Deletion of Dominated Strategies

Initial Payoff

Iteration

• From the original case, D is dominated by C

• Z is dominated by W

• W is dominated by Y

• C is dominated by B

• Y is dominated by X

• B is dominated by A

• It can be concluded that the optimal strategy is for P1 to select A, P2 to select X
• This is unconditionally the best option for both players

Nash Equilibria

Definition

• Nash equilibria refers to the state which is the best response given the response of other players
• In a Nash equilibrium state, no player can benefit by changing their strategy alone, meaning that each player’s strategy is the best response to others’ strategies

Methods of Analysis

(1) Best Response Analysis

• Iterate through every possible strategy
• Identify the best response of both players given the choice made by the opposing player
• Record these responses in the payoff matrix
• If both payoffs are recorded, this strategy is a Nash equilibrium

(2) Guess and Verify

• Give a random strategy for both players
• Identify whether this is an equilibria using best response
• If this is not the best response to either (or both), this is not a Nash equilibrium

Notes

• Nash Equilibrium is a pair of strategies, NOT payoffs
• A dominant strategy is the Nash Equilibrium

Best Response

Initial State

Conditional Strategy

Player 1

• If P2 chooses L, the best strategy for P1 is M for a payoff of 2
• If P2 chooses C, the best strategy for P1 is T for a payoff of -1
• If P2 chooses R, the best strategy for P1 is T for a payoff of 2

Player 2

• If P1 chooses T, the best strategy for P2 is C for a payoff of 2
• If P1 chooses M, the best strategy for P2 is L for a payoff of 2
• If P1 chooses B, the best strategy for P2 is B for a payoff of 8

Analysis

• For $(T, C)$ and $(M, L)$: this is the best strategy for P2 given what P1 does AND for P1 given what P2 does
• Neither player has an incentive to change their strategy given what the other player does

First-Price Sealed Bid Auction

Terminology

• Bidders: $P_1, P_2$
• Bids: $b_1, b_2$
• Willingness to Pay: $V_1, V_2$

Players

• 2 Bidders, P1 and P2
• $V_1 = 100$ is the willingness to pay for the object by P1
• $V_2 = 250$ is the willingness to pay for the object by P2

Information Structure

• Players simultaneously submit bids in the form of $0, 0.1, 0.2, ...$
• Both players know their own and each others’ willingness to pay
• The highest bidder wins the object and pays their bid, the lower bidders do not get the object and do not pay
• In case of a tie, a winner is selected at random

Payoffs

• Payoffs are represented by the consumer surplus derived form consuming the object

Guess and Verify

Guess #1

• $(100, 100.1)$

Outcome #1

• P2 Wins
• $\pi_1 = 0$
• $\pi_2 = 250 - 100.1 = 149.9$

Verify #1

• P1 can choose three options:

  • $b_1 < 100$: P2 still wins, so $\pi_1 = 0$
  • $b_1 = b_2 = 100.1$: 50/50 chance of winning, so $\pi_1 = \frac{100 - 100.1}{2} < 0$
  • $b_1 > 100.1$: P1 wins, so $\pi_1 = 100 - b_1 < 0 \quad (\because b_1 > 100.1)$

• P2 can choose three options:

  • $b_2 < 100$: P1 wins, so $\pi_2 = 0$
  • $b_2 = b_1 = 100$: 50/50 chance of winning, so $\pi_1 = \frac{250 - 100}{2} = 75 < 149.9$
  • $b_2 > 100.1$: P1 wins, so $\pi_2 = 250 - b_2 < 149.9 \quad (\because b_2 > 100.1)$

• $(100, 100.1)$ is a Nash Equilibrium

Guess #2

• $(99.9, 100)$

Outcome #2

• P2 Wins
• $\pi_1 = 0.01$
• $\pi_2 = 250 - 100 = 150$

Verify #2

• P1 can choose three options:

  • $b_1 < 99.9$: P2 still wins, so $\pi_1 = 0$
  • $b_1 = b_2 = 100$: 50/50 chance of winning, so $\pi_1 = \frac{100 - 100}{2} = 0$
  • $b_1 > 100$: P1 wins, so $\pi_1 = 100 - b_1 < 0$

• P2 can choose three options:

  • $b_2 < 99.9$: P1 wins, so $\pi_2 = 0$
  • $b_2 = b_1 = 99.9$: 50/50 chance of winning, so $\pi_1 = \frac{250 - 99.9}{2} = 75.05 < 149.9$
  • $b_2 > 100$: P2 wins, so $\pi_2 = 250 - b_2 < 150$

• $(99.9, 100)$ is a Nash Equilibrium

Take-Home Problems

Take-Home Problem 1

”For the beauty contest, is the choice of 100 dominated by 99, 99 dominated by 98, 98 dominated by 97, 97 dominated by 96 and so on?”

Guess and Verify

$\text{for simplicity, let there be 2 players in the game}$

$\text{let } c_1 \text{ and } c_2 \text{ be the numbers chosen by P1 and P2 respectively}$

$\text{let } m \text{ be the mean of } c_1 \text{ and } c_2$

$\text{let } d_1 \text{ and } d_2 \text{ be the difference between } m \text{ and } c_1 \text{ and } c_2 \text{ respectively}$

• Winning conditions are as follows:

• $c_1 = 100$ is dominated by $c_1 \in \{0, 1, 2, ..., 99\}$
• Since you are comparing $c_1$ and $c_2$ with $\frac{m}{2}$, any $c \text{ s.t. } c = max(c_1, c_2)$ is a dominated strategy
• Choosing a lower number where $c > \frac{m}{2}$ will decrease the distance of $c$ from $\frac{m}{2}$

Macroscopic Perspective

• If players in the beauty contest chose numbers at random, the mean $m = \frac{1}{n} \sum^{n}_{i=0} c_i = 50$, with the target $\frac{m}{2} = 50$
• This means all strategies where $c > 50$ are strictly dominated
• Repeating this process:
  • Means will be $m = 25, 12.5, 7.25, 3.625, 1.8125, ...$
  • Targets will be $12.5, 7.25, 3.625, 1.8125, 0.90625, ...$
• At every step, $c > m$ are strictly dominated
• Since the sequence of $m$ converges at $0$, all strategies where $c > 0$ are dominated
• Thus, the strategy $c = 0$ is weakly dominant and is a Nash equilibrium

Take-Home Problem 2

For the First-Price Sealed Problem, show that bidding at $b \ge V$ is a dominated strategy

$\text{let } V_1 = 100, V_2 = 250$

• It can be concluded that $\pi \le 0$ for $b >$
• This holds true for $b \ge V$

For the First-Price Sealed Problem, show that any bid $(b, b + 0.1)$ where $100 \le b < 250$ is also a Nash Equilibrium

$\text{let } V_1 = 100, V_2 = 250$

• P1 has no incentive to change his strategy as the $\pi_1 = 0$, and all other choices result in either the same or worse payoffs
• P2 has no incentive to change his strategy as $\pi_2 = 250 - b$ is the maximum payoff given the bid of $(b, b + 0.1)$

For the Second-Price Sealed Problem, using $V_2 = 250$, show that $b_2 = 250$ is weakly dominant

$\text{let } V_1 = 100, V_2 = 250$

• Player 2’s payoff can be represented as:

• To guess and verify, we do the following:

• Therefore, $b_2 = V_2$ is never worse but sometimes better than $b_2 = x$

Notice how much the seller earns in both cases

$\text{let } V_1 = 100, V_2 = 250$

• First-Price Sealed Auction: Nash Equilbrium Strategy is $(99.9, 100)$
• Second-Price Sealed Auction: Nash Equilibrium Strategy is $(100, 250)$
• In both cases, the seller receives $100$