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## Game Theory Coursera Quiz Answers

### Game Theory Week 01 Quiz Answers

Q1.

1 \ 2 | X | y | z |

a | 1,2 | 2,2 | 5,1 |

b | 4,1 | 3,5 | 3,3 |

c | 5,2 | 4,4 | 7,0 |

d | 2,3 | 0,4 | 3,0 |

Find the strictly dominant strategy:

- a;
- b;
**c;**- d;
- x;
- y;
- z

Q2.

1 \ 2 | X | y | z |

A | 1,2 | 2,2 | 5,1 |

B | 4,1 | 3,5 | 3,3 |

C | 5,2 | 4,4 | 7,0 |

D | 2,3 | 0,4 | 3,0 |

Find a very weakly dominant strategy that is not strictly dominant.

- c;
- x;
- d;
- a;
- y;
- b;
**z**

Q3.

1 \ 2 | x | y | z |

a | 1,2 | 2,2 | 5,1 |

b | 4,1 | 3,5 | 3,3 |

c | 5,2 | 4,4 | 7,0 |

d | 2,3 | 0,4 | 3,0 |

When player 1 plays d, what is player 2’s best response:

- Only x
**Only y**- Only z
- Both y and z

Q4.

1 \ 2 | x | Y | z |

A | 1,2 | 2,2 | 5,1 |

B | 4,1 | 3,5 | 3,3 |

C | 5,2 | 4,4 | 7,0 |

D | 2,3 | 0,4 | 3,0 |

Find all strategy profiles that form pure strategy Nash equilibria (there may be more than one, or none):

- (b, y);
- (d, y);
- (b, z);
- (a, y);
- (c, x);
**(c, y);**- (a, x);
- (b, x);
- (a, z);
- (d, z).
- (c, z);
- (d, x);

Q5. There are 2 players who have to decide how to split one dollar. The bargaining process works as follows. Players simultaneously announce the share they would like to receive s_1*s*1 and s_2*s*2, with 0≤s_10≤*s*1, s_2≤1*s*2≤1. If s_1+s_2≤1*s*1+*s*2≤1, then the players receive the shares they named and if s_1+s_2>1*s*1+*s*2>1, then both players fail to achieve an agreement and receive zero. This game is known as `Nash Bargaining’.

Which of the following is a strictly dominant strategy?

- 1;
- 0.5;
- 0;
**None of the above.**

Q6. There are 2 players who have to decide how to split one dollar. The bargaining process works as follows. Players simultaneously announce the share they would like to receive s_1*s*1 and s_2*s*2, with 0≤s_10≤*s*1, s_2≤1*s*2≤1. If s_1+s_2≤1*s*1+*s*2≤1, then the players receive the shares they named and if s_1+s_2>1*s*1+*s*2>1, then both players fail to achieve an agreement and receive zero.

Which of the following strategy profiles is a pure strategy Nash equilibrium?

- (0.3, 0.7);
- (0.5, 0.5);
- (1.0, 1.0);
**All of the above**

Q7. Two firms produce identical goods, with a production cost of c>0*c*>0 per unit.

Each firm sets a nonnegative price (p_1*p*1 and p_2*p*2).

All consumers buy from the firm with the lower price, if p_1≠p_2*p*1=*p*2. Half of the consumers buy from each firm if p_1=p_2*p*1=*p*2.

D is the total demand.

Profit of firm i*i* is:

- 0 if p_i>p_j
*pi*>*pj* (no one buys from firm i*i*); - D\frac{p_i−c}{2}
*D*2*pi*−*c* if p_i=p_j*pi*=*pj*(Half of customers buy from firm i*i*); - D(p_i−c)
*D*(*pi*−*c*) if p_i<p_j*pi*<*pj* (All customers buy from firm i*i*)

Find the pure strategy Nash equilibrium:

- Both firms set p=0
*p*=0. - Firm 1 sets p=0
*p*=0, and firm 2 sets p=c*p*=*c*. **Both firms set p=c***p*=*c*.- No pure strategy Nash equilibrium exists.

Q8.

- Three voters vote over two candidates (A and B), and each voter has two pure strategies: vote for A and vote for B.
- When A wins, voter 1 gets a payoff of 1, and 2 and 3 get payoffs of 0; when B wins, 1 gets 0 and 2 and 3 get 1. Thus, 1 prefers A, and 2 and 3 prefer B.
- The candidate getting 2 or more votes is the winner (majority rule).

Find all *very weakly***dominant** strategies (click all that apply: there may be more than one, or none).

**Voter 1 voting for A.**- Voter 1 voting for B.
- Voter 2 (or 3) voting for A.
**Voter 2 (or 3) voting for B.**

Q9.

- Three voters vote over two candidates (A and B), and each voter has two pure strategies: vote for A and vote for B.
- When A wins, voter 1 gets a payoff of 1, and 2 and 3 get payoffs of 0; when B wins, 1 gets 0 and 2 and 3 get 1. Thus, 1 prefers A, and 2 and 3 prefer B.
- The candidate getting 2 or more votes is the winner (majority rule).

Find **all** pure strategy Nash equilibria (click all that apply)? Hint: there are three.

**1 voting for A, and 2 and 3 voting for B.****All voting for A.****All voting for B.**- 1 and 2 voting for A, and 3 voting for B.

### Game Theory Week 02 Quiz Answers

Q1.

1 \ 2 | Left | Right |

Left | 4,2 | 5,1 |

Right | 6,0 | 3,3 |

Find a mixed strategy Nash equilibrium where player 1 randomizes over the pure strategy Left and Right with probability p*p* for Left. What is p*p*?

- 1/4
**3/4**- 1/2
- 2/3

Q2.

1 \ 2 | Left | Right |

Left | X X ,2 |
0,0 |

Right | 0,0 | 2,2 |

In a mixed strategy Nash equilibrium where player 1 plays Left with probability p*p* and player 2 plays Left with probability q*q*. How do p*p* and q*q* change as X*X* is increased (X>1*X*>1)?

*p*is the same,*q*decreases.*p*increases,*q*increases.*p*decreases,*q*decreases.*p*is the same,*q*increases.

Q3.

- There are 2 firms, each advertising an available job opening.
- Firms offer different wages: Firm 1 offers w_1=4
*w*1=4 and 2 offers w_2=6*w*2=6. - There are two unemployed workers looking for jobs. They simultaneously apply to either of the firms.
- If only one worker applies to a firm, then he/she gets the job
- If both workers apply to the same firm, the firm hires a worker at random and the other worker remains unemployed (and receives a payoff of 0).

Find a mixed strategy Nash Equilibrium where p*p* is the probability that worker 1 applies to firm 1 and q*q* is the probability that worker 2 applies to firm 1.

- p=q=1/4
*p*=*q*=1/4; - p=q=1/2
*p*=*q*=1/2; - p=q=1/5
*p*=*q*=1/5. **p=q=1/3***p*=*q*=1/3;

Q4.

- A king is deciding where to hide his treasure, while a pirate is deciding where to look for the treasure.
- The payoff to the king from successfully hiding the treasure is 5 and from having it found is 2.
- The payoff to the pirate from finding the treasure is 9 and from not finding it is 4.
- The king can hide it in location X, Y or Z.

Suppose the pirate has two pure strategies: inspect both X and Y (they are close together), or just inspect Z (it is far away). Find a mixed strategy Nash equilibrium where p*p* is the probability the treasure is hidden in X or Y and 1-p1−*p* that it is hidden in Z (treat the king as having two strategies) and q*q* is the probability that the pirate inspects X and Y:

**p=1/2***p*=1/2, q=1/2*q*=1/2;- p=4/9
*p*=4/9, q=2/5*q*=2/5; - p=5/9
*p*=5/9, q=3/5*q*=3/5; - p=2/5
*p*=2/5, q=4/9*q*=4/9;

Q5.

- A king is deciding where to hide his treasure, while a pirate is deciding where to look for the treasure.
- The payoff to the king from successfully hiding the treasure is 5 and from having it found is 2.
- The payoff to the pirate from finding the treasure is 9 and from not finding it is 4.
- The king can hide it in location X, Y or Z.

Suppose that the pirate can investigate any two locations, so has three pure strategies: inspect XY or YZ or XZ. Find a mixed strategy Nash equilibrium where the king mixes over three locations (X, Y, Z) and the pirate mixes over (XY, YZ, XZ). The following probabilities (king), (pirate) form an equilibrium:

- (1/3, 1/3, 1/3), (4/9, 4/9, 1/9);
- (4/9, 4/9, 1/9), (1/3, 1/3, 1/3);
- (1/3, 1/3, 1/3), (2/5, 2/5, 1/5);
**(1/3, 1/3, 1/3), (1/3, 1/3, 1/3);**

### Game Theory Week 03 Quiz Answers

Q1.

1 \ 2 | L | M | R |

U | 3,8 | 2,0 | 1,2 |

D | 0,0 | 1,7 | 8,2 |

We say that a game is *dominance solvable*, if iterative deletion of strictly dominated strategies yields a unique outcome. True or false: The above game si dominance solvable. Hint: Consider both pure strategies and mixed strategies to do your domination.

**True;**- False.

Q2. In order to illustrate the problem that arises when iteratively eliminating **weakly** dominated strategies, consider the following game:

1 \ 2 | L | M | R |

U | 4,3 | 3,5 | 3,5 |

D | 3,4 | 5,3 | 3,4 |

True or false: in the above game the order of elimination of **weakly** dominated strategies does not matter (that is, the final outcome is the same regardless of the order in which weakly dominated strategies are eliminated.).

[Hint: start the process of iterative elimination of **weakly** dominated strategies by eliminating different strategies at the beginning of the process.]

- True;
**False.**

Q3. Consider the matching pennies game:

1 \ 2 | Left | Right |

Left | 2,-2 | -2,2 |

Right | -2,2 | 2,-2 |

Which is a maxmin strategy for player 1:

- Play Left.
- Play Right.
**Play Left and Right with probability 1/2.**- It doesn’t exist.

Q4. Consider the matching pennies game:

1 \ 2 | Left | Right |

Left | 2,-2 | -2,2 |

Right | -2,2 | 2,-2 |

Apply the Minimax theorem presented in lecture 3-4 to find the payoff that any player must receive in any Nash Equilibrium:

- 2;
- -2;
- 1;
**0**.

Q5.

1 \ 2 | B | F |

B | 2,1 | 0,0 |

F | 0,0 | 1,2 |

Consider the following assignment device (for example a fair coin):

- With probability 1/2 it tells players 1 and 2 to play B, and with probability 1/2 it tells them to play F.
- Both players know that the device will follow this rule.

What is the expected payoff of each player when both players follow the recommendations made by the device? If one of players follows the recommendation, does the other player have an incentive to follow the recommendation as well?

- Expected payoff =2=2; player has an incentive to follow the recommendation.
- Expected payoff =1=1; player does not an incentive to follow the recommendation.
**Expected payoff =1.5=1.5; player has an incentive to follow the recommendation.**- Expected payoff =1.5=1.5; player does not have an incentive to follow the recommendation.

### Game Theory Week 03 Quiz Answers

Q1.

- Two players have to share 50 coins (of equal value).
- Players’ payoffs are the number of coins they each get
- First, player 1 splits the coins into 2 piles.
- Second, player 2 chooses one pile for him/herself and gives the other pile to player 1

What is agent 1’s strategy in a backward induction solution?

- Splitting coins into 1/49.
**Splitting coins into 25/25.**- Splitting coins into 15/35.
- Splitting coins into 0/50.

Q2. Find all of the pure strategy **Nash** Equilibria of this game. There can be more than one equilibrium. [Here ((Not Play,Steal),(Trust)) indicates that player 1 chooses Not Play at the first decision node and Steal at the second decision node, and 2 chooses Trust at his unique decision node.]

- ((Play, Share), (Trust))
**((Not play, Share), (Distrust))****((Not play, Steal),(Distrust))**- ((Not play, Steal), (Trust))
- ((Play, Steal), (Distrust))

Q3. Which is the Subgame Perfect Equilibrium of this game? [Here ((Not Play,Steal),(Trust)) indicates that player 1 chooses Not Play at the first decision node and Steal at the second decision node, and 2 chooses Trust at his unique decision node.]

**((Not play, Steal),(Distrust))**- ((Not play, Share), (Distrust))
- ((Not play, Steal), (Trust))
- ((Play, Steal), (Distrust))
- ((Play, Share), (Trust))

Q4.

- There are 2 investors. Each has deposited $10 in the same bank.
- The bank invested both deposits in a single long-term project.
- If the bank wants to end the project before its completion, a total of $12 can be recovered (out of the $20 invested).
- If the bank waits until the project is completed, it will receive a total of $30.
- Investors can withdraw money from their bank accounts at only 2 periods: before the project is completed and after.

The extensive from representation of the game between both investors is depicted below:

In order to find the subgame perfect Nash equilibrium of the whole game first focus on the subgame that starts with investor 1’s second decision node:

Which is a pure strategy Nash equilibrium of this subgame?

**(Withdraw, Withdraw)**- (Withdraw, Not)
- (Not, Withdraw)
- (Not, Not)

Q5.

- There are 2 investors. Each has deposited $10 in the same bank.
- The bank invested both deposits in a single long-term project.
- If the bank wants to end the project before its completion, a total of $12 can be recovered (out of the $20 invested).
- If the bank waits until the project is completed, it will receive a total of $30.
- Investors can withdraw money from their bank accounts at only 2 periods: before the project is completed and after.

The extensive from representation of the game between both investors is depicted below:

In order to find the subgame perfect Nash equilibrium of the whole game first focus on the subgame that starts with investor 1’s second decision node:

What are the subgame perfect Nash equilibria of the whole game? There might be more than one. [Hint: ((Withdraw, Not),(Not, Withdraw)) are the first and second investors’ strategies in their first and second decision nodes, respectively. So, ((Withdraw, Not),(Not, Withdraw)) indicates that the first investor withdraws at her first decision node but not at her second, while the second invest does not withdraw at his first decision node but does at his second decision node.]

**((Not, Withdraw), (Not, Withdraw))**- ((Not, Withdraw), (Withdraw, Withdraw))
- ((Withdraw, Withdraw), (Not, Withdraw))
**((Withdraw, Withdraw), (Withdraw, Withdraw))**

Q6.

- Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain) each wanting to maximize the number of coins that he gets.
- It is always the captain who proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed.
- If the captain fails to obtain support of at least half his crew (which includes himself), all pirates turn against him and make him walk the plank. The pirates then start over again with the next most senior pirate as captain (the pirates have a strict order of seniority denoted by A, B, C, D and E).
- Pirates’ preferences are ordered in the following way. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins he receives. Finally, each pirate would prefer to throw another overboard in the case of indifference.

What is the maximum number of coins that the original captain gets to keep across all subgame perfect equilibria of this game?

(Hint, work by backward induction to reason what the split will be if three captains have been forced to walk the plank and there are only two pirates left. Just one vote is enough to approve the split among the two pirates. Then use that to solve for what happens when two have walked the plank and there are three pirates left, and so forth.)

- 100;
- 0;
- 50;
**98**;

### Game Theory Week 05 Quiz Answers

Q1. Two players play the following normal form game.

1 \ 2 | Left | Middle | Right |

Left | 4,2 | 3,3 | 1,2 |

Middle | 3,3 | 5,5 | 2,6 |

Right | 2,1 | 6,2 | 3,3 |

Which is the pure strategy Nash equilibrium of this stage game (if it is played only once)?

- (Left, Left);
- (Left, Middle);
- (Left, Right);
- (Middle, Left);
- (Middle, Middle);
**(Middle, Right);**- (Right, Left);
- (Right, Middle);
- (Right, Right).

Q2. Two players play the following normal form game.

1 \ 2 | Left | Middle | Right |

Left | 4,2 | 3,3 | 1,2 |

Middle | 3,3 | 5,5 | 2,6 |

Right | 2,1 | 6,2 | 3,3 |

Suppose that the game is repeated for two periods. What is the outcome from the subgame perfect Nash equilibrium of the whole game:

- (Left, Left) is played in both periods.
**(Right, Right) is played in both periods.**- (Middle, Middle) is played in the first period, followed by (Left, Left)
- (Middle, Middle) is played in the first period, followed by (Right, Right)

Q3. Two players play the following normal form game.

1 \ 2 | Left | Middle | Right |

Left | 4,2 | 3,3 | 1,2 |

Middle | 3,3 | 5,5 | 2,6 |

Right | 2,1 | 6,2 | 3,3 |

Suppose that there is a probability p*p* that the game continues next period and a probability (1-p)(1−*p*) that it ends. What is the threshold p^**p*∗ such that when p \geq p^**p*≥*p*∗ (Middle, Middle) is sustainable as a subgame perfect equilibrium by grim trigger strategies, but when p < p^**p*<*p*∗ playing Middle in all periods is not a best response? [Here the grim strategy is: play Middle if the play in all previous periods was (Middle, Middle); play Right otherwise.]

- 1/2;
**1/3;**- 1/4;
- 2/5.

Q4. Consider the following game:

1 \ 2 | Left | Middle | Right |

Left | 1,1 | 5,0 | 0,0 |

Middle | 0,5 | 4,4 | 0,0 |

Right | 0,0 | 0,0 | 3,3 |

Which are the pure strategy Nash equilibria of this stage game? There can be more than one.

- (Left, Right);
**(Left, Left);**- (Left, Middle);
- (Middle, Right);
- (Middle, Left);
**(Right, Right).**- (Right, Middle);
- (Right, Left);
- (Middle, Middle);

Q5. Consider the following game:

1 \ 2 | Left | Middle | Right |

Left | 1,1 | 5,0 | 0,0 |

Middle | 0,5 | 4,4 | 0,0 |

Right | 0,0 | 0,0 | 3,3 |

Suppose that the game is repeated for two periods. Which of the following outcomes could occur in some subgame perfect equilibrium? (There might be more than one).

**(Middle, Middle) is played in the first period, followed by (Right, Right)****(Left, Left) is played in both periods.****(Right, Right) is played in both periods.**

Q6. Consider the following trust game:

There is a probability p*p* that the game continues next period and a probability (1-p)(1−*p*) that it ends. The game is repeated indefinitely. Which statement is true? [Grim trigger in (c) and (d) is player 1 playing Not play and player 2 playing Distrust forever after a deviation from ((Play,Share), (Trust)).]eval(ez_write_tag([[300,250],’networkingfunda_com-sky-4′,’ezslot_30′,708,’0′,’0′]));

- There exists a pure strategy Nash equilibrium in the one-shot game with player 2 playing Trust.
- There exists a pure strategy subgame perfect equilibrium with player 2 playing Trust in any period in the finitely repeated game.
**((Play,Share), (Trust)) is sustainable as a subgame perfect equilibrium by grim trigger in the indefinitely repeated game with a probability of continuation of p\geq5/9***p*≥5/9.

Q7. In an infinitely repeated Prisoner’s Dilemma, a version of what is known as a “tit for tat” strategy of a player i*i* is described as follows:

- There are two “statuses” that player
*i*might be in during any period: “normal” and “revenge”; - In a normal status player
*i*cooperates; - In a revenge status player
*i*defects; - From a normal status, player
*i*switches to the revenge status in the next period only if the other player defects in this period; - From a revenge status player
*i*automatically switches back to the normal status in the next period regardless of the other player’s action in this period.

Consider an infinitely repeated game so that with probability p*p* that the game continues to the next period and with probability (1−p)(1−*p*) it ends.

Cooperate (C) | Defect (D) | |

Cooperate (C) | 4,4 | 0,5 |

Defect (D) | 5,0 | 1,1 |

True or False:

When player 1 uses the above-described “tit for tat” strategy and starts the first period in a revenge status (thus plays defect for sure), in any infinite payoff maximizing strategy, player 2 plays defect in the first periodeval;

**True**.- False.

Q8. In an infinitely repeated Prisoner’s Dilemma, a version of what is known as a “tit for tat” strategy of a player i*i* is described as follows:

- There are two “statuses” that player
*i*might be in during any period: “normal” and “revenge”; - In a normal status player
*i*cooperates; - In a revenge status player
*i*defects; - From a normal status, player
*i*switches to the revenge status in the next period only if the other player defects in this period; - From a revenge status player
*i*automatically switches back to the normal status in the next period regardless of the other player’s action in this period.

Consider an infinitely repeated game so that with probability p*p* that the game continues to the next period and with probability (1−p)(1−*p*) it ends.

Cooperate (C) | Defect (D) | |

Cooperate (C) | 4,4 | 0,5 |

Defect (D) | 5,0 | 1,1 |

What is the payoff for player 2 from always cooperating when player 1 uses this tit for tat strategy and begins in a normal status? How about always defecting when 1 begins in a normal status?

- 4+4p+4p^2+4p^3+\ldots4+4
*p*+4*p*2+4*p*3+… ; 5+p+p^2+p^3+\ldots5+*p*+*p*2+*p*3+… **4+4p+4p^2+4p^3+\ldots4+4***p*+4*p*2+4*p*3+… ; 5+p+5p^2+p^3+\ldots5+*p*+5*p*2+*p*3+…- 5+4p+4p^2+4p^3+\ldots5+4
*p*+4*p*2+4*p*3+… ; 4+4p+4p^2+4p^3+\ldots4+4*p*+4*p*2+4*p*3+… - 5+4p+4p^2+4p^3+\ldots5+4
*p*+4*p*2+4*p*3+… ; - 5+p+p^2+p^3+\ldots5+
*p*+*p*2+*p*3+…

Q9. In an infinitely repeated Prisoner’s Dilemma, a version of what is known as a “tit for tat” strategy of a player i*i* is described as follows:

- There are two “statuses” that player
*i*might be in during any period: “normal” and “revenge”; - In a normal status player
*i*cooperates; - In a revenge status player
*i*defects; - From a normal status, player
*i*switches to the revenge status in the next period only if the other player defects in this period; - From a revenge status player
*i*automatically switches back to the normal status in the next period regardless of the other player’s action in this period.

Consider an infinitely repeated game so that with probability p*p* that the game continues to the next period and with probability (1−p)(1−*p*) it ends.

Cooperate (C) | Defect (D) | |

Cooperate (C) | 4,4 | 0,5 |

Defect (D) | 5,0 | 1,1 |

What is the threshold p^**p*∗ such that when p\geq p^**p*≥*p*∗ always cooperating by player 2 is a best response to player 1 playing tit for tat and starting in a normal status, but when p<p^**p*<*p*∗ always cooperating is not a best response?

- 1/2
**1/3**- 1/4
- 1/5eval(ez_write_tag([[250,250],’networkingfunda_com-mobile-leaderboard-2′,’ezslot_20′,650,’0′,’0′]));

### Game Theory Week 06 Quiz Answers

Q1. **War Game**

- Two opposed armies are poised to seize an island.
- Each army can either “attack” or “not-attack”.
- Also, Army 1 is either “weak” or “strong” with probability
*p*and (1−*p*), respectively. Army 2 is always “weak”. - Army’s 1 type is known only to its general.
- An army can capture the island either by attacking when its opponent does not or by attacking when its rival does if it is strong and its rival is weak. If two armies of equal strength both attack, neither captures the island.
- The payoffs are as follows
- The island is worth
*M*if captured. - An army has a “cost” of fighting, which is equal to
*s*>0 if it is strong and*w*>0 if it is weak (where*s*<*w*<*M*). - There is no cost of attacking if its rival does not attack.
- These payoffs are pictured in the payoff matrices below:

Weak | ||

1 \ 2 | Attack | Not-Attack |

Attack | -w,-w | M,0 |

Not-Attack | 0,M | 0,0 |

with probability p*p*

Strong | ||

1 \ 2 | Attack | Not-Attack |

Attack | M-s,-w | M,0 |

Not-Attack | 0,M | 0,0 |

with probability 1-p1−*p*.

When p=1/2*p*=1/2, which is a pure strategy Bayesian equilibrium (there could be other equilibria that are not listed as one of the options):

Strategies listed in format: (1’s type – 1’s strategy; 2’s strategy)

**(Weak – Not-Attack, Strong – Attack; Attack);**- (Weak – Not-Attack, Strong – Attack; Not-Attack);
- (Weak – Attack, Strong – Attack; Attack);
- It does not exist.

Q2. Consider the following variation to the Rock (R), Paper (P),Scissors (S) game:

- Suppose that with probability p player 1 faces a Normal opponent and with probability 1−p, he faces a Simple opponent that will always play P.
- Player 2 knows whether he is Normal or Simple, but player 1 does not.
- The payoffs are pictured in the payoff matrices below:

Normal | |||

1 \ 2 | R | P | S |

R | 0,0 | -1,1 | 1,-1 |

P | 1,-1 | 0,0 | -1,1 |

S | -1,1 | 1,-1 | 0,0 |

with probability p

Simple | |

1 \ 2 | P |

R | -1,1 |

P | 0,0 |

S | 1,-1 |

with probability 1−p1−*p*.

Suppose p = 1/3*p*=1/3, select all pure strategy Bayesian equilibria (there may be more than one):

(Form: 1’s strategy; 2’s type – 2’s strategy)

- (S; Normal – P, Simple – P)
- (R; Normal – P, Simple – P)
**(S; Normal – R, Simple – P)**- (P; Normal – P, Simple – P)

Q3. Consider the following variation to the Rock (R), Paper (P),Scissors (S) game:

- Suppose that with probability p player 1 faces a Normal opponent and with probability 1−p, he faces a Simple opponent that will always play P.
- Player 2 knows whether he is Normal or Simple, but player 1 does not.
- The payoffs are pictured in the payoff matrices below:

Normal | |||

1 \ 2 | R | P | S |

R | 0,0 | -1,1 | 1,-1 |

P | 1,-1 | 0,0 | -1,1 |

S | -1,1 | 1,-1 | 0,0 |

with probability p

Simple | |

1 \ 2 | P |

R | -1,1 |

P | 0,0 |

S | 1,-1 |

with probability 1−p1−*p*.

Suppose p = 2/3*p*=2/3, select all pure strategy Bayesian equilibria (there may be more than one):

(Form: 1’s strategy; 2’s type – 2’s strategy)

**none**- (R; Normal – P, Simple – P)
- (P; Normal – S, Simple – P)eval(ez_write_tag([[300,250],’networkingfunda_com-small-square-1′,’ezslot_34′,713,’0′,’0′]));
- (S; Normal – R, Simple – P)

Q4.

- An engineer has a talent t in {1,2} with equal probability (prob=1/2), and the value of t is private information to the engineer.
- The engineer’s pure strategies are applying for a job or being an entrepreneur and doing a startup.
- The company’s pure strategies are either hiring or not hiring the engineer.
*If the engineer applies for the job and the company does not hire, then the engineer becomes an entrepreneur and does a startup.*- The utility of the engineer is t (talent) from being an entrepreneur, and w (wage) from being hired.
- The utility of the company is (t−w) from hiring the engineer and 0 otherwise.
- These are pictured in the payoff matrices below, with the engineer being the row player and the company being the column player.

t=2 | Hire | Not |

Startup | 2,0 | 2,0 |

Work | w,2-w | 2,0 |

t=1 | Hire | Not |

Startup | 1,0 | 1,0 |

Work | w,1-w | 1,0 |

Suppose w=2*w*=2, which of the below are pure strategy Bayesian equilibria, there may be more than one and check all that apply. (Form: Engineer’s strategy, company’s strategy)

**(t=2***t*=2 Work, t=1*t*=1 Work, Not);- (t=2
*t*=2 Work, t=1*t*=1 Work, Hire); **(t=2***t*=2 Startup, t=1*t*=1 Work, Not);- (t=2
*t*=2 Startup, t=1*t*=1 Work, Hire);

Q5.

- An engineer has a talent t in {1,2} with equal probability (prob=1/2), and the value of t is private information to the engineer.
- The engineer’s pure strategies are applying for a job or being an entrepreneur and doing a startup.
- The company’s pure strategies are either hiring or not hiring the engineer.
*If the engineer applies for the job and the company does not hire, then the engineer becomes an entrepreneur and does a startup.*- The utility of the engineer is t (talent) from being an entrepreneur, and w (wage) from being hired.
- The utility of the company is (t−w) from hiring the engineer and 0 otherwise.
- These are pictured in the payoff matrices below, with the engineer being the row player and the company being the column player.

t=2 | Hire | Not |

Startup | 2,0 | 2,0 |

Work | w,2-w | 2,0 |

t=1 | Hire | Not |

Startup | 1,0 | 1,0 |

Work | w,1-w | 1,0 |

Suppose w=1*w*=1, which of the below are pure strategy Bayesian equilibria, there may be more than one and check all that apply.

(Form: Engineer’s strategy, company’s strategy)

**(t=2***t*=2 Work, t=1*t*=1 Startup, Hire);- (t=2
*t*=2 Startup, t=1*t*=1 Work, Hire); **(t=2***t*=2 Startup, t=1*t*=1 Work, Not);- (t=2
*t*=2 Work, t=1*t*=1 Startup, Not);

Q6. Change the Battle of Sexes to have incomplete information:

There are two possible types of player 2 (column):

- “Meet” player 2 wishes to be at the same movie as player 1, just as in the usual game. (This type has probability p
*p*) - “Avoid” 2 wishes to avoid player 1 and go to the other movie. (This type has probability 1−p1−
*p*)

2 knows her type, and 1 does not.

They simultaneously choose P or L.

These payoffs are shown in the matrices below.

Meet | ||

1 \ 2 | L | P |

L | 2,1 | 0,0 |

P | 0,1 | 1,0 |

with probability p*p*

Avoid | ||

1 \ 2 | L | P |

L | 2,0 | 0,2 |

P | 0,1 | 1,0 |

with probability 1−p1−*p*.

When p=1/2*p*=1/2, which is a pure strategy Bayesian equilibrium:

(1’s strategy; 2’s type – 2’s strategy)

**(L; Meet – L, Avoid – P);**- (P; Meet – P, Avoid – L);
- (L; Meet – P, Avoid – P);
- It does not exist.

Q7. Modify the Battle of Sexes to have incomplete information:

There are two possible types of player 2 (column):

- “Meet” player 2 wishes to be at the same movie as player 1, just as in the usual game. (This type has probability
*p*) - “Avoid” 2 wishes to avoid player 1 and go to the other movie. (This type has probability 1−
*p*)

2 knows her type, and 1 does not.

They simultaneously choose P or L.

These payoffs are shown in the matrices below.

Meet | ||

1 \ 2 | L | P |

L | 2,1 | 0,0 |

P | 0,1 | 1,0 |

with probability p*p*

Avoid | ||

1 \ 2 | L | P |

L | 2,0 | 0,2 |

P | 0,1 | 1,0 |

with probability 1−p1−*p*.

When p=1/4*p*=1/4, which is a pure strategy Bayesian equilibrium :

(1’s strategy; 2’s type – 2’s strategy)

**(L; Meet – L, Avoid – P);**- (P; Meet – P, Avoid – L);
- (L; Meet – P, Avoid – P);
- It does not exist.

### Game Theory Week 07 Quiz Answers

Q1.

- Three players together can obtain 11 to share, any two players can obtain 0.80.8, and one player by herself can obtain zero.
- Then, N=3
*N*=3 and v({1})=v({2})=v({3})=0*v*(1)=*v*(2)=*v*(3)=0, v({1,2})=v({2,3})=v({3,1})=0.8*v*(1,2)=*v*(2,3)=*v*(3,1)=0.8, v({1,2,3})=1*v*(1,2,3)=1.

Which allocation is in the core of this coalitional game?

- (0,0,0);
- (0.4, 0.4, 0);
- (1/3, 1/3, 1/3);
**The core is empty;**

Q2.

- There is a market for an indivisible good with
*B*buyers and*S*sellers. - Each seller has only one unit of the good and has a reservation price of 0.
- Each buyer wants to buy only one unit of the good and has a reservation price of 1.
- Thus
*v*(*C*)=*min*(*BC*,*SC*) where*BC*and*SC*are the number of buyers and sellers in coalition*C*(and so, for instance,*v*(*i*)=0 for any single player, and*v*(*i*,*j*)=1 if*i*,*j*are a pair of a buyer and seller).

If the number of buyers and sellers is B=2*B*=2 and S=1*S*=1, respectively, which allocations are in the core? [There might be more than one]

**Each seller receives 1 and each buyer receives 0.**- Each seller receives 0 and each buyer receives 1.
- Each seller receives 1/2 and each buyer receives 1/2.

Q3.

- There is a market for an indivisible good with
*B*buyers and*S*sellers. - Each seller has only one unit of the good and has a reservation price of 0.
- Each buyer wants to buy only one unit of the good and has a reservation price of 1.
- Thus
*v*(*C*)=*min*(*BC*,*SC*) where*BC*and*SC*are the number of buyers and sellers in coalition*C*(and so, for instance,*v*(*i*)=0 for any single player, and*v*(*i*,*j*)=1 if*i*,*j*are a pair of a buyer and seller).

Now assume that we increase the number of sellers so that B=2*B*=2 and S=2*S*=2. Which allocations are in the core? [There might be more than one]

**Each seller receives 1 and each buyer receives 0.****Each seller receives 0 and each buyer receives 1.****Each seller receives 1/2 and each buyer receives 1/2.**

Q4.

- The instructor of a class allows the students to collaborate and write up together a particular problem in the homework assignment.
- Points earned by a collaborating team are divided among the students in any way they agree on.
- There are exactly three students taking the course, all equally talented, and they need to decide which of them if any should collaborate.
- The problem is so hard that none of them working alone would score any points. Any two of them can score 4 points together. If all three collaborate, they can score 6 points.

Which allocation is in the core of this coalitional game?

**(0,0,0);**- (2, 2, 0);
- (2, 2, 2);
- The core is empty;

Q5.

- The instructor of a class allows the students to collaborate and write up together a particular problem in the homework assignment.
- Points earned by a collaborating team are divided among the students in any way they agree on.
- There are exactly three students taking the course, all equally talented, and they need to decide which of them if any should collaborate.
- The problem is so hard that none of them working alone would score any points. Any two of them can score 4 points together. If all three collaborate, they can score 6 points.

What is the Shapley value of each player?

- \phi=(0,0,0)
*ϕ*=(0,0,0) - \phi=(2,0,2)
*ϕ*=(2,0,2) **\phi=(1/3,1/3,1/3)***ϕ*=(1/3,1/3,1/3)- \phi=(2,2,2)
*ϕ*=(2,2,2)

Q6. There is a single capitalist (c*c*) and a group of 2 workers (w1*w*1 and w2*w*2).

The production function is such that total output is 0 if the firm (coalition) is composed only of the capitalist or of the workers (a coalition between the capitalist and a worker is required to produce positive output).

The production function satisfies:

*F*(*c*∪*w*1)=*F*(*c*∪*w*2)=3*F*(*c*∪*w*1∪*w*2)=4

Which allocations are in the core of this coalitional game? [There might be more than one]

**x_c=2***xc*=2, x_{w1}=1*xw*1=1, x_{w2}=1*xw*2=1;**x_c=2.5***xc*=2.5, x_{w1}=0.5*xw*1=0.5, x_{w2}=1*xw*2=1;**x_c=4***xc*=4, x_{w1}=0*xw*1=0, x_{w2}=0*xw*2=0;

Q7. There is a single capitalist (c*c*) and a group of 2 workers (w1*w*1 and w2*w*2).

The production function is such that total output is 0 if the firm (coalition) is composed only of the capitalist or of the workers (a coalition between the capitalist and a worker is required to produce positive output).

The production function satisfies:

*F*(*c*∪*w*1)=*F*(*c*∪*w*2)=3*F*(*c*∪*w*1∪*w*2)=4

What is the Shapley value of the capitalist?

- 3;
- 4;
**7/3;**- 7;

Q8. There is a single capitalist (c*c*) and a group of 2 workers (w1*w*1 and w2*w*2).

The production function is such that total output is 0 if the firm (coalition) is composed only of the capitalist or of the workers (a coalition between the capitalist and a worker is required to produce positive output).

The production function satisfies:

*F*(*c*∪*w*1)=*F*(*c*∪*w*2)=3*F*(*c*∪*w*1∪*w*2)=4

What is the Shapley value of each worker?

- 1;
**5/6;**- 3/4;
- 1/2;

Q9. There is a single capitalist (c*c*) and a group of 2 workers (w1*w*1 and w2*w*2).

The production function satisfies:

*F*(*c*∪*w*1)=*F*(*c*∪*w*2)=3*F*(*c*∪*w*1∪*w*2)=4

True or False: If there was an additional 3^{rd}3*rd* worker that is completely useless (i.e., his marginal contribution is 0 in every coalition), then the sum of the Shapley Values of the capitalist and the first two workers will remain unchanged.

**True;**- False;

### Game Theory Week 08 Quiz Answers

Q1.

1\ 2 | X | y | z |

a | 2,5 | 2,1 | 0,1 |

b | 3,2 | 4,4 | 1,1 |

c | 1,0 | 1,1 | 1,2 |

Find the strictly dominant strategies (click all that apply: there may be zero, one or more and remember the difference between strictly dominant and strictly dominated):

- c;
- a;
- x;
- y;
- z;
- b;
**none**

Q2.

1\ 2 | X | y | z |

a | 2,5 | 2,1 | 0,1 |

b | 3,2 | 4,4 | 1,1 |

c | 1,0 | 1,1 | 1,2 |

Find the weakly dominated strategies (click all that apply: there may be zero, one or more):

- b;
**x;**- z;
**c;**- y;
- a;

Q3.

1\ 2 | X | y | z |

a | 2,5 | 2,1 | 0,1 |

b | 3,2 | 4,4 | 1,1 |

c | 1,0 | 1,1 | 1,2 |

Which strategies survive the process of iterative removal of strictly dominated strategies (click all that apply: there may be zero, one or more)?

- b;
**y;****c;****x;**- a;
**z;**

Q4.

1\ 2 | X | y | z |

a | 2,5 | 2,1 | 0,1 |

b | 3,2 | 4,4 | 1,1 |

c | 1,0 | 1,1 | 1,2 |

Find all strategy profiles that form pure strategy Nash equilibria (click all that apply: there may be zero, one or more):

- (a, x);
- (b, x);
- (b, z);
- (a, z);
**(b, y);**- (a, y);
- (c, y);
**(c, z).**- (c, x);

Q5.

1\ 2 | Y | z |

b | 4,4 | 1,1 |

c | 1,1 | 2,2 |

Which of the following strategies form a mixed strategy Nash equilibrium? (p*p* corresponds to the probability of 1 playing \color{red}{\verb|b|}b and 1-p1−*p* to the probability of playing \color{red}{\verb|c|}c; q*q* corresponds to the probability of 2 playing y*y* and 1-q1−*q* to the probability of playing z*z*).

- p=1/3
*p*=1/3, q=1/3*q*=1/3; - p=1/3
*p*=1/3, q=1/4*q*=1/4; **p=1/4***p*=1/4, q=1/4*q*=1/4;- p=2/3
*p*=2/3, q=1/4*q*=1/4;

Q6.

- One island is occupied by Army 2, and there is a bridge connecting the island to the mainland through which Army 2 could retreat.
- Stage 1: Army 2 could choose to burn the bridge or not in the very beginning.
- Stage 2: Army 1 then could choose to attack the island or not.
- Stage 3: Army 2 could then choose to fight or retreat if the bridge was not burned, and has to fight if the bridge was burned.

First, consider the blue subgame. What is a subgame perfect equilibrium of the

- blue subgame?
**(Attack, Retreat).**- (Attack, Fight).
- (Not, Fight).
- (Not, Retreat).

Q7.

- One island is occupied by Army 2, and there is a bridge connecting the island to the mainland through which Army 2 could retreat.
- Stage 1: Army 2 could choose to burn the bridge or not in the very beginning.
- Stage 2: Army 1 then could choose to attack the island or not.
- Stage 3: Army 2 could then choose to fight or retreat if the bridge was not burned, and has to fight if the bridge was burned.

What is the outcome of a subgame perfect equilibrium of the whole game?

- Bridge is burned, 1 attacks and 2 fights.
- Bridge is not burned, 1 attacks and 2 retreats.
**Bridge is burned, 1 does not attack.**- Bridge is not burned, 1 does not attack.

Q8. Consider an infinitely repeated game where the game in each period is depicted in the picture.

There is a probability p*p* that the game continues next period and a probability (1-p)(1−*p*) that it ends. What is the threshold p^**p*∗ such that when p\geq p^**p*≥*p*∗ ((Play,Share), (Trust)) is sustainable as a subgame perfect equilibrium by a grim trigger strategy, but when p<p^**p*<*p*∗ ((Play,Share), (Trust)) can’t be sustained as a subgame perfect equilibrium?

[Here a trigger strategy is: player 1 playing Not play and player 2 playing Distrust forever after a deviation from ((Play,Share), (Trust)).]

**1/3;**- 2/3;
- 1/4.
- 1/2;

Q9.

- There are two players.
- The payoffs to player 2 depend on whether 2 is a friendly player (with probability
*p*) or a foe (with probability 1−*p*). - Player 2 knows if he/she is a friend or a foe, but player 1 doesn’t know.

See the following payoff matrices for details.

Friend | Left | Right |

Left | 3,1 | 0,0 |

Right | 2,1 | 1,0 |

with probability p*p*

Foe | Left | Right |

Left | 3,0 | 0,1 |

Right | 2,0 | 1,1 |

with probability 1−p1−*p*

When p=1/4*p*=1/4, which is a pure strategy Bayesian equilibrium:

(1’s strategy; 2’s type – 2’s strategy)

- (Left ; Friend – Left, Foe – Right);
- (Left ; Friend – Left, Foe – Left);
- (Right ; Friend – Right, Foe – Right);
**(Right ; Friend – Left, Foe – Right);**

Q10. Player 1 is a company choosing whether to enter a market or stay out;

- If 1 stays out, the payoff to both players is (0, 3).

Player 2 is already in the market and chooses (simultaneously) whether to fight

player 1 if there is entry

- The payoffs to player 2 depend on whether 2 is a normal player (with prob 1-p1−
*p*) or an aggressive player (with prob p*p*).

See the following payoff matrices for details.

Aggressive | Fight | Not |

Enter | -1,2 | 1,-2 |

Out | 0,3 | 0,3 |

with probability p*p*

Normal | Fight | Not |

Enter | -1,0 | 1,2 |

Out | 0,3 | 0,3 |

with probability 1−p1−*p*

Player 2 knows if he/she is normal or aggressive, and player 1 doesn’t know. Which are true (click all that apply, there may be zero, one or more):

- When p<1/2
*p*<1/2, it is a Bayesian equilibrium for 1 to enter, 2 to fight when aggressive and not when normal. - When p=1/2
*p*=1/2, it is a Bayesian equilibrium for 1 to enter, 2 to fight when aggressive and not when normal; **When p>1/2***p*>1/2, it is a Bayesian equilibrium for 1 to stay out, 2 to fight when aggressive and not when normal;**When p=1/2***p*=1/2, it is a Bayesian equilibrium for 1 to stay out, 2 to fight when aggressive and not when normal;

**Review: **

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