Sample Problem Solutions Economics 2020b Spring 2012

Sample Problem Solutions
Economics 2020b
Spring 2012
1.1 Converting Normal Form Games to Extensive Form
(See Problem Set 1 Solutions)
1.2 Imperfect Memory
(a) In this extensive form, nature moves first, then either player 1a or player 2 has a
choice of “Continue” or “Stop”. After “Continue” by either player, player 1b has the
choice to “Exchange” or “Keep” the cards. Since player 1b does not observe any of the
details of play prior to this choice, player 1b has a single information set that includes
both possible paths leading to his move.
(b) Players 1a and 1b can coordinate their actions in advance – specifying one of two
possible actions at each choice of move. Thus, there are four pure strategies for players
1a and 1b:
R1 = 1a chooses “Continue”, 1b chooses “Exchange”
R2 = 1a chooses “Continue”, 1b chooses “Keep”
R3 = 1a chooses “Stop”, 1b chooses “Exchange”
R4 = 1a chooses “Stop”, 1b chooses “Keep”
(c) The normal form is a 4x2 table with strategies R1 through R4 for the combination of
players 1a and 1b and strategies “Continue” and “Stop” for player 2. To compute the
payoffs for each cell in the table, first identify the specific outcomes corresponding to
each combination of strategies and choice by nature.
Case 1: NATURE gives player 1a card H.
Note that player 2’s strategy doesn’t matter in this case.
The payoff is (1, -1) from “Stop” by player 1a (R3 or R4);
The payoff is (-2, 2) from “Continue” by player 1a, “Exchange” by player 1b (R1);
The payoff of (2, -2) from “Continue” by player 1a, “Keep” by player 1b (R2).
Case 2: NATURE gives player 1a card “L”.
The payoff is (-1, 1) from “Stop” by player 2 (player 1’s strategy doesn’t matter);
The payoff is (2, -2) from “Continue” by player 2, “Exchange” by player 1b (R1 or R3);
The payoff of (-2, 2) from “Continue” by player 2, “Keep” by player 1b (R2 or R4).
Case 1 Payoffs:
R1
R2
R3
R4
“Continue”
(-2, 2)
(2, -2)
(1, -1)
(1, -1)
“Stop”
(-2, 2)
(2, -2)
(1, -1)
(1, -1)
Case 2 Payoffs:
R1
R2
R3
R4
“Continue”
(2, -2)
(-2, 2)
(2, -2)
(-2, 2)
“Stop”
(-1, 1)
(-1, 1)
(-1, 1)
(-1, 1)
Combined Payoffs (probability ½ of Case 1 and probability ½ of Case 2)
“Continue”
“Stop”
R1
(0, 0)
(-1.5, 1.5)
R2
(0, 0)
(0.5, -0.5)
R3
(1.5, -1.5)
(0, 0)
R4
(-0.5, 0.5)
(0, 0)
Note that since nature randomly assigns a card to player 1, the payoffs in the “Combined
Payoff” matrix are still expected utilities (even though these payoffs correspond to pure
strategy choices by the players).
(d) If there is just a single player 1 with bad memory, the extensive form in (a) would
apply under the assumption that this player can plan out his strategy for the game in
advance and write it out on a card – with specific instructions to himself about what to do
if he receives card H and what to do if he gets the choice to “Exchange” or “Keep” the
cards.
But if there is just a single player 1, player 2 might not understand that player 1 has poor
memory. If player 2 thinks that player 1 will remember his card, then player 2 perceives
the game to be one where there are two distinct information sets rather than just one
information set (for player 1b) after “Continue”. Then player 2 would never choose
“Continue”, which would have secondary implications for player 1’s choice of actions.
2.1 Levels of Affection (See Problem Set 2 Solutions)
2.2. Iterated Elimination of Dominated Strategies
1 2
, . After
3 3
1 2
deleting R, D is dominated by (p, 1-p, 0) probabilities of (U,M,D), for any p
, .
3 3
The remaining strategies form a good basis for prediction of how players might play the
game if we assume common knowledge of rationality (and common knowledge of
payoffs). If player 2 is rational, and player 1 knows player 2 is rational, then we should
never see player 2 play R or player 1 play D. (Note that we don’t need to extend the chain
of common knowledge beyond this: we don’t need player 2 to know player 1 is rational,
for instance.)
(a) R is dominated by (p, 1-p, 0) probabilities of (L,C,R), for any p
(b) There are two natural possible orders of elimination: R, M, U, L and M, R, U, L. Note
that there are other possibilities, but regardless of the order, only (D, C) survives iterated
elimination.
(c) There are two natural possible orders of elimination leading to different predictions of
play: M, R, U and U, L, M. In the first case, we end up with (D, L), whereas in the
second case we end up with (D, R). (Again, these are not the only two possibilities.)
2.3 Limited Memory and Dominated Strategies
(a) The normal form takes the following form:
R1
R2
R3
R4
“Continue”
(0, 0)
(0, 0)
(1.5, -1.5)
(-0.5, 0.5)
“Stop”
(-1.5, 1.5)
(0.5, -0.5)
(0, 0)
(0, 0)
For player 2, “Continue” does better than “Stop” against R2 and R4, but “Stop” does
better than “Continue” against R1 and R3. So neither of player 2’s pure strategies strictly
dominates or weakly dominates the other.
For player 1, strategies R2 and R3 guarantee weakly positive payoffs, whereas R1 and R4
never produce a positive payoff. This suggests that R1 and R4 are dominated strategies,
and in fact,
(1) R3 strictly dominates R1; (2) R2 strictly dominates R4.
However, even after the elimination of R1 and R4, there are no dominated strategies for
Player 2, so the game is not dominance solvable.
R2
R3
“Continue”
(0, 0)
(1.5, -1.5)
“Stop”
(0.5, -0.5)
(0, 0)
(b) The strategies are defined as follows:
R1 = 1a chooses “Continue”, 1b chooses “Exchange”;
R2 = 1a chooses “Continue”, 1b chooses “Keep”;
R3 = 1a chooses “Stop”, 1b chooses “Exchange”;
R4 = 1a chooses “Stop”, 1b chooses “Keep”.
We found in (a) that R1 strictly dominates R3. In each of these strategies, player 1b
chooses “Exchange”. Intuitively, if player 1b chooses “Exchange”, then when player 1a
gets card H, players 1a and 1b will get a higher payoff (payoff of +1) if player 1a chooses
“Stop” than if player 1a chooses “Continue” (payoff of -2). That is, the choice by 1a to
“Stop” rather than “Continue” given that 1b will “Exchange” increases payoff by 3 when
player 1a has card H, which translates into an increase of 3 * ½ = 1.5 in expected payoff
since player 1a gets card H with probability ½. Thus, for any strategy by player 2, R3
produces expected payoff 1.5 higher than R1, so R3 strictly dominates R3.
Similarly, we found in (a) that R2 strictly dominates R4. In each of these strategies,
player 1b chooses “Keep”. Intuitively, if player 1b chooses “Keep”, then when player 1a
gets card H, players 1a and 1b will get a higher payoff (payoff of +2) if player 1a chooses
“Continue” than if player 1a chooses “Stop” (payoff of +1). That is, the choice by 1a to
“Continue” rather than “Stop” increases payoff by 1 when player 1a has card H, which
translates into an increase of ½ in expected payoff since player 1a gets card H with
probability ½. Thus, for any strategy by player 2, R2 produces expected payoff 0.5
higher than R4, so R2 strictly dominates R4.
(c) From the perspective of player 1, if player 2 chooses “Stop”, then player 1’s actions
only affect player 1’s payoff of player 1a gets card H. Thus, player 1 has a clear
incentive to choose R2 (“Continue”, “Keep”), which achieves maximum payoff when
player 1a gets card H.
But if player 2 chooses “Continue”, then R2 is a winning strategy when player 1a gets
card H and a losing strategy when player 2 gets card H. Since each player gets card H
with probability ½, these positive and negative outcomes offset in expectation, producing
an overall expected payoff of 0 from R2. In contrast, if player 2 chooses “Continue”,
then R3 (“Stop”, “Exchange”) produces a positive payoff for player 1 no matter who gets
card H. Therefore, if player 2 chooses “Continue”, then player 1 has clear incentive to
choose R3.
Combining these observations, R2 does not dominate R3 and R3 does not dominate R2.
Similarly, neither of player 2’s strategies dominates the other when R2 and R3 are
possible choices for player 1. Therefore, the game cannot be solved by iterated
dominance.
3.1. Mixed Strategy Equilibria
(a) Best responses for each player are listed in bold in the table. There are three pure
strategy equilibria: (U, C), (M, L) and (D, R).
2
L
1
C
R
U
0,0
2,4
1,1
M
4,2
1,1
0,0
D
1,1
0,0
3,3
(b) There are three combinations of two pure strategies for each player:
(U, M), (U, D), and (M, D) for player 1.
(L, C), (L, R), and (C, R) for player 2.
Combining these possibilities, there are 3x3 = 9 possible combinations of pairs of two
strategies for player 1 together with pairs of strategies for player 2.
Possibility 1: (U, M) for player 1 and (L, C) for player 2
Possibility 2: (U, M) for player 1 and (L, R) for player 2
Possibility 3: (U, M) for player 1 and (C, R) for player 2
Possibility 4: (U, D) for player 1 and (L, C) for player 2
Possibility 5: (U, D) for player 1 and (L, R) for player 2
Possibility 6: (U, D) for player 1 and (C, R) for player 2
Possibility 7: (M, D) for player 1 and (L, C) for player 2
Possibility 8: (M, D) for player 1 and (L, R) for player 2
Possibility 9: (M, D) for player 1 and (C, R) for player 2
Some of these possibilities can be ruled out by inspection and strict dominance based on
comparisons of payoffs for player 2:
If player 1 plays only U and M, then C strictly dominates R for player 2. This
rules out possibilities 2 and 3.
If player 1 plays only U and D, then R strictly dominates L for player 2.
This rules out possibilities 4 and 5.
If player 1 plays only M and D, then L strictly dominates C for player 2. This
rules out possibilities 7 and 9.
We can perform similar analysis for player 1’s strategies:
If player 2 plays only L and C, then M strictly dominates D for player 1. This
rules out possibilities 4 and 7.
If player 1 plays only L and R, then D strictly dominates U for player 1.
This rules out possibilities 2 and 5.
If player 1 plays only C and R, then U strictly dominates M for player 1. This
rules out possibilities 3 and 9.
Interestingly, each of these two separate approaches rules out the same set of
possibilities: 2, 3, 4, 5, 7 and 9. There are three remaining possibilities:
Possibility 1: (U, M) for player 1 and (L, C) for player 2
Possibility 6: (U, D) for player 1 and (C, R) for player 2
Possibility 8: (M, D) for player 1 and (L, R) for player 2
(c) This is possibility 1 from (b), which we cannot rule out by dominance. So look for
specific probabilities that would produce a mixed strategy equilibrium.
If player 1 plays U with probability p and M with probability (1-p), then the expected
payoffs for player 2 from L is 2(1-p) = 2 – 2p and the expected payoff for player 2 from
C is 4p + (1-p) = 1 + 3p. These payoffs are equal if p = 1/5, so this is the candidate
strategy for player 1 in a mixed strategy equilibrium.
If player 2 plays L with probability q and C with probability (1-q), then the expected
payoffs for player 1 from U is 2(1-q) = 2 – 2q and the expected payoff for player 1 from
M is 4q + (1-q) = 1 + 3q. These payoffs are equal if q = 1/5, so this is the candidate
strategy for player 2 in a mixed strategy equilibrium.
These candidate strategies produce a mixed strategy equilibrium in the 2x2 game where
player 1 is restricted to U and M and player 2 is restricted to L and C. They will also
produce a mixed strategy equilibrium in the full 3x3 game if
(1) player 1 does not get higher utility from D than from U and M when player 2
plays L with probability 1/5 and C with probability 4/5;
(2) player 2 does not get higher utility from R than from L and C when player 1 plays
U with probability 1/5 and M with probability 4/5.
To check this, note that when q = 1/5, player 1’s expected utility from D is 1/5 while
player 1’s expected utility from U and M is 8/5, so player 1 cannot gain by playing D
instead of U and M. Similarly, when p = 1/5, player 2’s expected utility from R is 1/5
while player 2’s expected utility from L and C is 8/5, so player 2 cannot gain by playing
R instead of L and C. This verifies that we have found a mixed strategy equilibrium for
the full game. (We could also have reached this conclusion from dominance since we
know that if player 1 plays only U and M, then C strictly dominates R for player 2 and
similarly that if player 2 plays only L and C, then M strictly dominates D for player 1.)
(d) and (e) We ruled out both of these possibilities in the analysis of (b).
(f) Suppose that player 2 plays L with probability q1, M with probability q2 and R with
the remaining probability 1 – q1 – q2.
Then player 1’s payoffs from each strategy are as follows:
U:
2q2 + 1*(1 - q1 – q2) = 1 - q1 + q2;
M:
4q1 + 1*q2
= 4q1 + q2;
D:
1q1 + 3*(1 - q1 – q2) = 3 - 2q1 + 3q2.
For player 1 to play all three strategies with positive probability, these strategies must all
yield the same expected payoff. This produces a system of three equations in two
unknowns, but the third equation must hold by transitivity if the first two equations hold
(if (1) expected payoff from U equals expected payoff from M and (2) expected payoff
from U equals expected payoff from D, then expected payoff from M equals expected
payoff from D by transitivity.)
Solving the system of equations:
1 - q1 + q2
= 4q1 + q2;
1 - q1 + q2
= 3 - 2q1 + 3q2.
The first equation can be solved directly for q1: q1 = 1/5. Given q1 = 1/5, the second
equation can be solved directly for q2: q2 = 9/20. Both of these are valid probabilities
(because they are between 0 and 1). The probability of R is 1 – q1 – q2 = 7/20, which is
also a valid probability.
The game is symmetric, so the same set of probabilities for player 1 will make player 2
indifferent between each of her three strategies.
Therefore, there is a unique mixed strategy equilibrium where each player plays all three
pure strategies with positive probability:
Player 1 plays U with probability 1/5, M with probability 9/20
and R with probability 7/20;
Player 2 plays L with probability 1/5, C with probability 9/20
and D with probability 7/20.
3.2 Nash Equilibrium in a 2x3 Game (See Solutions to Problem Set 2)
(a) Best responses are shown as bold payoffs in the normal form representation above.
There is no pure strategy equilibrium because the pure strategy best responses for the two
players never coincide.
U
D
L
4, 10
8, 2
M
6, 6
6, 6
R
8, 1
4, 9
(b) Suppose that player 1 plays U with probability p and D with probability (1-p). Then
the expected payoffs for player 2 are as follows:
L: 10p + 2 (1-p) = 8p + 2.
M: 6p + 6 (1-p) = 6.
R: p + 9 (1-p) = 9 - 8p.
For player 2 to choose both L and R with positive probability, it must be that the payoffs
from these strategies are equal, meaning 8p + 2 = 9 - 8p or p = 7/16. Then the payoffs to
L and R are each 5.5. But then player 2 would prefer M to either L or R -- meaning that
there is no mixed strategy equilibrium where player 2 randomizes between L and R.
(c) If player 2 never plays R, then D weakly dominates U for player 1. If player 2 plays L
and M with strictly positive probabilities (and doesn't play R), then player 1 has a strict
preference for D, which in turn means that player 2 would have an incentive to shift to R.
Once again, there is no mixed strategy equilibrium of this sort.
(d) If player 2 plays M, then player 1 is indifferent between U and D. So player 1 can
play any mixed strategy and the only question is whether there is a mixing probability p
so that M is player 2's best response.
M is a best response when two conditions are satisified:
6 > 8p + 2 (so that M is preferred to L);
6 > 9 - 8p (so that M is preferred to R).
These conditions are satisfied simultaneously if 3/8 < p < 1/2.
So, there are a range of mixed strategy equlibria of the form [ pU + (1-p)D; M ], where p
is any of the values in this range, 3/8 < p < 1/2.
(e) Player 2 is indifferent between L and M if p = 7/16, as shown in part (b). Player 2 is
indifferent between L and R if 6 = 8p + 2 or p = 1/2. But it is not possible for player 2 to
be indifferent among all three strategies at once, meaning that there is no mixed strategy
equilibrium where player 2 plays all three pure strategies with positive probabilities.
3.3 Nash Equilibrium in an Unusual 2x3 Game
(a) Player 2’s best response to “Up” is “Right” and player 2’s best response to “Down” is
“Left”. But player 1’s best response to “Right” is “Down” and player 1’s best response
to “Left” is “Up”. Therefore, there is no pure strategy equilibrium of this game.
(b) Solving for player 2’s expected payoffs as a function of p:
E(Left)
= 1p + 7(1-p) = 7 – 6p
E(Middle)
= 5p + 5(1-p) = 5
E(Right)
= 7p + 4(1-p) = 4 + 3p
Left gives a higher expected payoff than Middle if 7 – 6p > 5, or p < 1/3
Left gives a higher expected payoff than Right if 7 – 6p > 4 + 3p or p < 1/3
Middle gives a higher expected payoff than Right if 5 > 4 + 3p or p < 1/3.
That is, if p < 1/3, Left gives a higher expected payoff than the other pure strategies,
while if p > 1/3, Right gives a higher expected payoff than the other pure strategies.
Thus, if p < 1/3, Left is the best response for player 2, while
if p > 1/3, Right is the best response for player 2.
Finally, if p = 1/3, all three strategies for player 2 give the same expected payoff and
therefore any feasible mixed or pure strategy for player 2 is a best response.
(c) From the answer to (b), player 1 must use the strategy with p = P(Up) = 1/3 in any
mixed strategy Nash equilibrium.
In a mixed strategy equilibrium, player 1’s expected payoff is the same for both pure
strategies:
6 P(Left) + P(Middle) + 4 P(Right) = 5 P(Left) + P(Middle) + 7 P(Right)
OR
P(Left) = 3 P(Right).
Assuming P(Middle) = q, then we have three equations in three unknowns:
P(Left) + P(Middle) + P(Right)
=1
P(Left)
= 3 P(Right)
P(Middle)
=q
Substituting the last two equations into the first one gives
3 P(Right) + q + P(Right)
=1
OR
P(Right)
= (1-q) / 4.
Thus, any mixed strategy equilibrium where player 2 plays all three pure strategies with
positive probability takes the form (1/3, 2/3) for player 1 and
[3(1-q) / 4, q, (1-q) / 4] for player 2.
(d) As shown in part (c), the lines for player 2’s expected payoffs as a function of p
intersect in a single point. (Ordinarily, three separate lines have three separate pairwise
intersection points.)
(e) In any 2x2 game with Middle and one other strategy for player 2, player 1 will have a
weakly dominated strategy, since player 1’s payoff is the same value (1) for either
strategy when player 2 plays “Middle”.
Specifically, if player 2 only plays “Middle” and “Left” with positive probability, then
“Up” weakly dominates “Down” for player 1.
Similarly if player 2 only plays “Middle” and “Right” with positive probability, then
“Down” weakly dominates “Right” for player 1.
In either case, if player 2 plays a mixed strategy with positive probability of “Middle”
and one other strategy, then player 1’s best response is the weakly dominant pure strategy
– implying that there cannot be a mixed strategy equilibrium.
(f) From part (c), player 2 is only indifferent between “Left” and “Right” if player 1 plays
a mixed strategy with P(Up) = 1/3.
If player 2 plays “Left” with probability x and “Right” with probability 1-x, then player
1’s expected payoff from “Up” is 6x + 4(1-x) = 4 + 2x, and player 1’s expected payoff
from “Down” is 5x + 7(1-x) = 7 – 2x. Setting these expected payoffs equal yields the
solution x = 3/4.
That is, there is a unique mixed strategy equilibrium of this form where player 1’s
strategy is (1/3, 2/3) and player 2’s strategy is (3/4, 0, 1/4).
3.4 Complementary Monopoly
(a) The firm solves the following profit-maximization problem:
Max(Pv) : (100 - Pv) * Pv - (Ch + Ca)*(100 - Pv)
FOC: 100 - 2Pv + Ch + Ca = 0
Pv = (100 + Ch + Ca)/2
Q = (100 - Ch - Ca)/2
Profit = Q *(Pv - Ch - Ca) = 1/4 * (100 - Ch - Ca)
2
(b) The airline division treats the hotel’s price as fixed, and solves its own profit
maximization problem, based on the total demand (which depends on the hotel price):
Max(Pa) : [100 - (Pa + Ph)] * (Pa) - Ca*[100 - (Pa + Ph)]
FOC: 100 - 2Pa - Ph + Ca = 0
Pa = (100 - Ph + Ca)/2
(c) We solve the hotel division’s profit maximization, when it treats the airline price as fixed.
Due to the symmetry of the division profit function, we can just take the solution from b and
make the appropriate substitutions:
Ph = (100 - Pa + Ch)/2
The two functions we solved above are the best response functions for each
division: it specifies the optimal price for the division, given what price the other division
has chosen. So the Nash Equilibrium occurs when each division’s price is its best response
to the other division’s price:
Ph = [100 - (100 - Ph + Ca)/2 + Ch]/2
Ph = (100- Ca + 2Ch)/3
By symmetry, Pa = (100- Ch + 2Ca)/3
Given the prices that occur at equilibrium, we solve for the quantities and profit:
Qa = Qh = (100 - Ch - Ca)/3
Total Profit I = (200 + Ch + Ca)/3 * (100 - Ch - Ca)/3 - (Ch + Ca)(100 - Ch - Ca)/3
= 2/9 * (100 - Ch - Ca)
2
(d) The total profit in part c, when the divisions make separate pricing decisions, is less than
the single firm in part b (2/9 < 1/4). It resembles the Prisoner’s Dilemma because when each
firm makes its own optimal (but selfish) decision, both end up worse off than if they could
cooperate. If we start at the single firm’s monopoly price, each division manager has an
incentive to raise her price. But the effect of both divisions’ increased prices is lower total
profits. The mechanism for this outcome is an externality in pricing: If the hotel division
raises its price, it will increase its revenues while driving down the total demand. But the
losses due to decreased demand are shared by both divisions, while the hotel alone enjoys the
benefit of its own price increase. The single firm internalizes the full costs and benefits of
increasing the price of either the hotel or airline, so it is able to obtain a higher total profit.
3.5. Existence and Oddness of Nash Equilibrium (See Solutions to Problem Set 2)
3.6 Tic-Tac-Toe (Omitted, but should be straightforward)
3.7 Final Jeopardy
(a) The best way to produce an accurate representation of the game is to be
comprehensive and laborious. Let’s first list the payoffs for all four combinations of
right and wrong answers by the players:
Case 1: Player 1 is correct, Player 2 is correct (probability ¾ * ¾ = 9/16);
Case 2: Player 1 is correct, Player 2 is incorrect (probability ¾ * 1/4 = 3/16);
Case 3: Player 1 is incorrect, Player 2 is correct (probability ¾ * 1/4 = 3/16);
Case 4: Player 1 is incorrect, Player 2 is incorrect (probability ¼ * ¼ = 1/16).
In Case 1, player 1 wins if she bets $6,000 or if she bets $2,000 and player 2 bets
anything other than $8,000.
Case 1 Payoffs
S3: Bet $8,000
S4: Bet $4,000
S5: Bet $0
S1: Bet $6,000
16,000, 0
16,000, 0
16,000, 0
S2: Bet $2,000
0, 16,000
12,000, 0
12,000, 0
In Case 2, player 1 always wins and receives her initial balance of $10,000 plus whatever
she chooses to bet.
S1: Bet $6,000
S2: Bet $2,000
Case 2 Payoffs
S3: Bet $8,000
S4: Bet $4,000
16,000, 0
16,000, 0
12,000, 0
12,000, 0
S5: Bet $0
16,000, 0
12,000, 0
In Case 3, player 2 always wins unless he bets 0 and player 1 bets only $2,000.
S1: Bet $6,000
S2: Bet $2,000
Case 3 Payoffs
S3: Bet $8,000
S4: Bet $4,000
0, 16,000
0, 12,000
0, 16,000
0, 12,000
S5: Bet $0
0, 8,000
8,000, 0
In Case 4, player 1 always wins unless he bets $6,000 and player 2 bets $0.
S1: Bet $6,000
S2: Bet $2,000
Case 4 Payoffs
S3: Bet $8,000
S4: Bet $4,000
4,000, 0
4,000, 0
8,000, 0
8,000, 0
S5: Bet $0
0, 8,000
8,000, 0
The normal form for the game is simply a probability-weighted sum of the four separate
payoff matrices for these individual cases, with weight 9/16 on Case 1 payoffs, weight
3/16 on Case 2 payoffs, weight 3/16 on Case 3 payoffs and weight 1/16 on Case 4
payoffs. NOTE: It may be helpful to use an Excel spreadsheet to compute these
probability-weighted payoffs, since this will facilitate further calculations for later parts
of the problem.
NORMAL FORM COMBINING CASE 1 THROUGH CASE 4 PAYOFFS
S3: Bet $8,000
S4: Bet $4,000
S5: Bet $0
S1: Bet $6,000
12,250, 3,000
12,250, 2,250
12,000, 2,000
S2: Bet $2,000
2,750, 12,000
9,500, 2,250
11,000, 0
(b) For player 1, strategy S1 strictly dominates S2. For player 2, strategy S3 strictly
dominates S4 and strictly dominates S5. There is a Nash equilibrium in strictly
dominating strategies where player 1 plays S1 and player 2 plays S3.
(c) This simply changes the weights on the payoffs from the Normal Form game in (b) to
¾ on Case 1 and ¼ on Case 4. Under these conditions, the Normal Form is given by:
NORMAL FORM COMBINING CASE 1 THROUGH CASE 4 PAYOFFS
S3: Bet $8,000
S4: Bet $4,000
S5: Bet $0
S1: Bet $6,000
13,000, 0
13,000, 0
12,000, 2,000
S2: Bet $2,000
2,000, 12,000
11,000, 0
11,000, 0
Now S2 is still strictly dominated, but player 2’s best response to S1 is S5. The Nash
equilibrium is (S1, S5).
(d) Explain the intuition for the equilibrium outcomes in (b) and (c), highlighting the
explanation for any differences between these outcomes.
In both cases, player 1 has an optimal strategy of betting $6,000. This large bet has two
virtues. First, it ensures that player 1 wins the game with a correct answer, regardless of
whether player 2’s answer is correct or incorrect. Second, it maximizes player 1’s
winnings when correct. Given the high probability, ¾, of a correct answer, these virtues
outweigh the cost of the large bet – namely that player 1 tends to lose (or not win very
much money) when answering incorrectly.
Player 2’s best response to a bet of $6,000 by player 1 is to make the largest possible bet
in (b), but the smallest possible bet in (c). Given player 1’s bet of $6,000, player 2 can
only win if player 1 is incorrect.
In (c), when player 1 is incorrect, player 2 is also incorrect. Therefore, player 2’s goal
should be to minimize losses from a wrong answer when player 1 is incorrect – i.e. player
2 should bet the minimum amount of $0 when the answers are perfectly correlated.
By contrast, in (b) when player 1 is incorrect, player 2 still has probability ¾ of
answering correctly. So player 2 is both likely to win the game and to make money from
her bet on the final question. For this reason, player 2’s best response is to bet as much
as possible even though this reduces his chance of winning the game:
Bet $8,000
– win $16,000 with probability 3/16. EV = $3,000;
Bet $0
- win $8,000 with probability ¼.
EV = $2,000.
(e) In the Nash equilibrium from (a), player 2 bets her entire balance of $8,000 and player
1 bets $6,000.
Player 2 Best Response
When player 1 bets $6,000 (or more), player 2 cannot win if player 1’s answer is correct.
So player 2 should assume that player 1 will be incorrect, leaving player 1 with a balance
of $10,000 – bet = $4,000 and respond accordingly.
If player 2 bets $3,999 or less, then player 2 will always win when player 1 is incorrect,
for a total value of initial balance = $8,000 + the expected value of player 2’s bet = ¾ *
bet – ¼ * bet = ½ * bet. In this range of strategies, player 2’s expected value is
increasing in bet size, so $3,999 is the best choice. Here player 2’s expected value is ¼ *
(8,000 + bet / 2) < 2,500. This is less than the expected value from betting $8,000, so it is
not a best response for player 2.
If player 2 bets $4,000 or more, then player 2 will only win when player 1 is incorrect
and player 2 is correct, which occurs with probability 3/16. The expected value for
player 2 is then 3/16 * (8,000 + bet), which is increasing in bet size. So, player 2’s best
response is still to bet the entire balance of $8,000.
Player 1 Best Response
When player 2 bets her entire balance, player 1 wins if
(1) player 2 is incorrect
(2) player 2 is correct, player 1 is correct and player 1 bets at least $6,000
If player 2 is incorrect (probability ¼), player 1 wins a total of initial balance ($10,000)
plus the expected value from the bet = ¾ * bet – ¼ * (bet) = bet / 2.
If player 2 is correct and player 1 is also correct (probability 9/16), player 1 wins a total
of initial balance plus bet so long as bet is at least $6,000.
In either case, player 1’s expected payoff is increasing in bet, so in fact, player 1 should
increase her bet to her entire balance of $10,000.
So, the equilibrium from (b) is not an equilibrium of the more general game with all
possible choices of bets available because player 1 would want to increase her bet from
$8,000 to $10,000.
3.8 Colonel Blotto
(a) A pure strategy is a vector with three components adding to 10, where the ith
component represents the number of divisions devoted to country i. Suppose that country
1 chooses pure strategy (x1, x2, x3) and country 2 chooses pure strategy
(y1, y2, y3).
If one country wins less than 1.5 territories with these strategies, then that country could
improve to winning 1.5 territories by simply copying the other country’s strategy, so
there cannot be a pure strategy Nash equilibrium where one country wins more and one
country wins less than half the territories.
So suppose that each country chooses a pure strategy and wins 1.5 territories. In this
case country 2 can improve to winning 2 territories by placing 0 units in the territory
where country 1 places most units. Suppose without loss of generality that x1 > x2 > x3.
Then x2, x3 < 4, so country 2 wins 2 territories against (x1, x2, x3) with the pure strategy
(0, 5, 5). Thus, there also cannot be a pure strategy Nash equilibrium where each country
wins an equal number of territories.
(b) Note that it is never possible for one player to win more than two territories in this
game. (If country 1 wins territories 1 and 2, for example, then country 2 has more
divisions left to deploy and must win territory 3.) So any strategy that wins two
territories must be a best response for player 1. In this case, any strategy that puts 5 units
in each of two territories (e.g. (5, 5, 0)), wins two territories and is a best response for
country 1.
(c) Country 2’s strategy has probability 2/3 of putting 5 units in any specific territory and
probability 1/3 of putting 0 units in any specific territory.
So country 1’s expected payoff for number of divisions placed in a given territory are as
follows:
0 units:
tie with probability 1/3, lose with probability 2/3
Expected Wins: 1/3 * ½ + 2/3 * 0 = 1/6.
1 to 4 units: win territory with probability 1/3, lose with probability 2/3
Expected Wins: 2/3 * 0 + 1/3 * 1 = 1/3.
5 units:
win territory with probability 1/3, tie with probability 2/3
Expected Wins: 1/3 * 1 + 2/3 * ½ = 2/3.
6-10 units:
win territory with probability 1
Expected Wins: 1 * 1 = 1.
Given country 2’s mixed strategy, there is no difference in outcome between 1, 2, 3 and 4
divisions in a territory or between 6, 7, 8, 9, and 10 divisions in a territory. So country 2
should only consider strategies that are restricted to 0, 1, 5, and 6 divisions in each
territory.
There are only five distinct strategies to consider (all other combinations of 0, 1, 5, 6
divisions adding to 10 are equivalent in outcome to one of these five choices)
(5, 5, 0)
(5, 4, 1)
(6, 4, 0)
(6, 3, 1)
(4, 4, 2)
Expected wins = 2/3 + 2/3 + 1/6
Expected wins = 2/3 + 1/3 + 1/3
Expected wins = 1 + 1/3 + 1/6
Expected wins = 1 + 1/3 + 1/3
Expected wins = 1/3 + 1/3 + 1/3
= 1.5 territories;
= 4/3 territories;
= 1.5 territories;
= 5/3 territories;
= 1 territory.
The best result is from strategy (6, 3, 1) and so this (and other equivalent strategies) is
country 1’s best response to country 2’s mixed strategy.
There is a natural mathematical intuition for this result. Country 1 starts with no
divisions in any territory and an expected number of 1/6 wins per territory for ½ expected
wins overall.
The marginal value of the first division placed in any territory is 1/6, since this improves
the number of expected wins in that territory from 1/6 to 1/3.
There is no change in expected wins from the second through fourth units, but the fifth
unit placed in a territory increases the expected number of wins in that territory from 1/3
to 2/3. So it is appropriate to think of the marginal value for a block of four divisions
added to the first division in a territory – this has marginal value of (2/3 – 1/3) / 4 = 1/12
per division.
The marginal value of the sixth division placed in any territory is 1/3, since this increases
expected wins in a given territory from 2/3 to 1. But this calculation is only relevant after
5 units have already been assigned to a given territory. So it is also appropriate to think
of the marginal value for a block of five divisions added to the first division in a territory
– this has marginal value (1 – 1/3) / 5 = 2/15 per division.
Summarizing these calculations:
Marginal Value: increase from 0 to 1 division
1/6 expected win per division
(increase from 1/6 expected wins to 1/3 expected wins in the territory with 1 division)
Marginal Value: increase from 1 to 6 divisions
2/15 expected wins per division
(increase from 1/3 expected wins to 1 expected win in the territory with 6 divisions)
Marginal Value: increase from 1 to 5 divisions
1/12 expected wins per division
(increase from 1/3 expected wins to 2/3 expected wins in the territory with 5 divisions)
A simple comparison suggests that country 1 should
Start with allocation (0, 0, 0), which has expected payoff of ½ win overall.
Then increase allocations to 1 division per territory, which increases expected
payoff by 1/6 expected win per division allocated. The allocation (1, 1, 1) has
expected payoff of 1 win overall.
Then increase allocation to 6 divisions in one territory, which increases expected
payoff by 2/15 expected wins for each of the five divisions just allocated. The
allocation (6, 1, 1) has expected payoff of 5/3 expected wins overall.
At this point, country 1 has allocated 8 divisions and has a total of 5/3 expected wins.
Unfortunately, there is no way to improve expected wins from an assignment of
(6, 1, 1) given just two additional divisions to deploy. An alternate possibility of
(5, 5, 0) utilizes each division productively, but yields fewer expected wins (since the
biggest gain is 1/3 expected wins from moving from 5 divisions to 6 divisions in a
particular territory). Therefore, the best response for player 1 to player 2’s mixed
strategy is (6, 3, 1).
(d) Either country can achieve at least an expected value of 1.5 territories won by
adopting exactly the same strategy as the other country. So in any combination of
strategies where one country wins fewer than 1.5 territories, the losing country must not
be playing a best response. Therefore, any mixed strategy equilibrium must produce an
expected value of 1.5 territories for each country.
(e) The Colonel Blotto game can be viewed as a competition between candidates, with
each candidate trying to outspend the other on a district by district level. Political
scientists have used the framework of this game to study the campaign strategies in
different elections. For example, Jennifer Merolla, Michael Munger, and Michael Tofias
studied the United States Presidential Election of 2000 in this context:
https://pantherfile.uwm.edu/tofias/www/papers/mmt.apsa2006.pdf
One obvious difficulty in applying Colonel Blotto to political campaigns is that
candidates can observe each other’s strategies as they evolve and respond accordingly.
The simultaneous move rule of the Colonel Blotto campaign does not allow for these
dynamic adjustments to strategies during the course of the campaign.
3.9 Priority Voting
(a) Voting for the least favorite candidate is a weakly dominated strategy for each voter;
if voting for a least favorite candidate has any effect on the election outcome, it can only
lead to the worst outcome for a particular voter. So we can eliminate all strategies where
any player votes for her least favorite candidate.
From the perspective of voter 1, there are two possible outcomes for the votes of the
other players. First, it is possible that both of them vote for C, in which case person 1’s
vote has no effect on the outcome of the election. Otherwise, person 2 votes for one
candidate and person 3 votes for a different candidate. In this case, candidate A will win
the election so long as person 1 votes for A. That is, person 1 strictly prefers to vote for
A if voters 2 and 3 split their votes and weakly prefers to vote for A if voters 2 and 3 both
vote for the same candidate. So we can eliminate person 1’s option to vote for B and
assume that person 1 will vote for A. (In fact, voting for A weakly dominates voting for
B or for C for person 1 even if we allow all possible voting strategies for 2 and 3).
If person 1 is known to vote for A and person 2 is known to vote for C or for A, then B
cannot be elected. Therefore, voting for C weakly dominates voting for B for person 3,
so we can eliminate the possibility that person 3 votes for C.
Finally, if person 1 is known to vote for A and person 3 is known to vote for C, then
voting for C weakly dominates voting for A for person 2, so we can eliminate the
possibility that person 2 votes for A.
That is, the solution by iterated weak dominance is (A, C, C).
Interestingly, person 1 is harmed by her priority in the voting system – the iterated weak
dominance outcome causes person 1’s least favorite candidate to win the election.
Intuitively, voters 2 and 3 recognize that unless they can coordinate their votes to
counteract person 1’s priority in case of a tie, then candidate A will be elected. Since
person 3 recognizes that it is impossible to coordinate with person 2 in voting for B, the
logical result is for the two of them to coordinate in voting for C.
(b) First, observe that there are three pure strategy outcomes where all three players vote
for the same candidate. In each of these outcomes, one player is following a weakly
dominated strategy by voting for her least favorite candidate, but this is still a best
response given that the outcome of the election is already decided by the common votes
of the other two players.
Second, consider the possibility that two players vote for the same candidate and one
player votes for a different candidate.
If player 1 votes for one candidate, while players 2 and 3 vote for the same
different candidate, then player 1’s vote cannot affect the outcome of the election,
so player 1 is playing a best response. Players 2 and 3 could each change the
outcome of the election by changing votes to vote for the candidate chosen by
player 1. So this could only be a Nash equilibrium if players 2 and 3 both prefer
the candidate they are voting for to the candidate that player 1 votes for. Players
2 and 3 have just one pairwise preference in common – they both prefer C to A.
So there is a Nash equilibrium where player 1 votes for A and players 2 and 3
vote for C. But there is no other Nash equilibrium where player 1 votes for one
candidate and players 2 and 3 each vote for the same other candidate.
If player 2 votes for one candidate, while players 1 and 3 vote for the same
different candidate, then player 2’s vote cannot affect the outcome of the election,
so player 2 is playing a best response. Player 3 could change the outcome of the
election by changing votes to player 2’s candidate; player 1 could change the
outcome of the election by voting for either of the other two candidates. So the
proposed strategies can only be a Nash equilibrium if player 1 is voting for his
most preferred candidate and if player 3 prefers this candidate to the candidate
that player 2 is voting for. This is impossible because player 1’s most preferred
candidate is A, but player 3’s least preferred candidate is A. So there is no Nash
equilibrium where player 2 votes for one candidate and players 1 and 3 each vote
for the same other candidate.
If player 3 votes for one candidate, while players 1 and 2 vote for the same
different candidate, then player 3’s vote cannot affect the outcome of the election,
so player 3 is playing a best response. By similar analysis to the case where
player 2 votes for one candidate and players 1 and 3 vote for the same different
candidate, this can only be a Nash equilibrium if player 1 is voting for his most
preferred candidate and if player 2 prefers this candidate to the candidate that
player 3 is voting for. Players 1 and 2 have just one pairwise preference in
common – they both prefer A to B. So there is a Nash equilibrium where player 3
votes for B and players 1 and 2 vote for A. But there is no other Nash equilibrium
where player 3 votes for one candidate and players 1 and 2 each vote for the same
other candidate.
Third, consider the possibility that each player votes for a different candidate. Then
player 1’s vote takes priority over the others and so the candidate that player 1 votes for
must win the election. Further, given that players 2 and 3 vote for two different
candidates, player 1 can cause any candidate to win the election by voting for that
candidate. So in a Nash equilibrium where each candidate gets exactly one vote, player 1
must be voting for A. One of the other voters is not voting for C, but both players 2 and
3 prefer C to A. Therefore, the voter who is not voting for C is not playing a best
response. So there is no Nash equilibrium where each player votes for a different
candidate.
Thus, the set of pure strategy Nash equilibria includes three cases of unanimity
(1. all three players voting for A;
2. all three players voting for B;
3. all three players voting for C)
and two cases with 2 to 1 votes
(4. players 2 and 3 vote for C, player 1 votes for A;
5. players 1 and 2 vote for A, player votes for C).
(c) Iterated weak dominance selects one of the five pure strategy Nash equilibria, so a
natural way to interpret the question is to ask whether any of these other four pure
strategy Nash equilibria seem like reasonable predictions for the outcome of the election.
The three cases of unanimity all require one player to vote for the least preferred
candidate, which seems unreasonable. (This would obviously fail to satisfy the
trembling-hand perfection refinement, for example).
This leaves the Nash equilibrium where players 1 and 2 vote for A, while player 3 votes
for B. Iterated weak dominance ruled out this possibility because player 3 could
recognize that voting for B is a weakly dominated strategy for both players 1 and 2. In
practice, would player 3 operate with this level of rationality? It seems possible to argue
that player 3 would do so – in which case, iterated weak dominance is the right solution
concept for this game, but it also seems possible to argue that player 3 might vote for B in
the hope that either player 1 or 2 would make a “mistake” and vote for B as well.
(Alternatively, player 3 might attach positive probability to the possibility that either
player 1 or 2 actually prefers B to A and would vote for B for that reason.) So there is no
clear cut answer about the applicability of iterated weak dominance for this game.
3.10 Nash Equilibrium in a 3x3 Game
U
M
D
L
1, 6
5, 10
3, 14
C
9, 2
7, 12
11, 16
R
17, 4
15, 8
13, 18
We highlight pure strategy best responses in the normal form above. As we found in
class, there are no pure strategy Nash equilibria. There are also no Nash equilibrium
where one player plays a pure strategy and the other plays a mixed strategy because the
strict best response to a pure strategy by one player is a pure strategy by the other player.
First consider the possibility of a mixed strategy equilibrium where both players play all
three strategies. Suppose that player 1 plays U with probability p1, M with probability p2
and D with probability p3. Similarly suppose that player 2 plays L with probability q1,
C with probability q2 and R with probability q3.
Player 1’s expected utilities from U, M, and D are as follows:
EU1(U)
= q1 + 9q2 + 17 q3
EU1(M)
= 5q1 + 7q2 + 15 q3
EU1(D)
= 3q1 + 11q2 + 13 q3
For player 1 to play all three pure strategies with positive probability, it must be that each
pure strategy gives the same expected utility:
q1 + 9q2 + 17 q3
5q1 + 7q2 + 15 q3
=
=
5q1 + 7q2 + 15 q3
3q1 + 11q2 + 13 q3
(1)
(2)
Note that the two equations combine to produce the third equality EU1(U) = EU1(D). In
addition, player 2’s three probabilities must add to 1.
q1 + q2 + q3
=
1.
(3)
That is, we have three linear equations in three unknowns (q1, q2, q3) so (assuming that
these equations are jointly independent), there should be a unique solution. So long as
the unique solution produces values of q1, q2, q3 between 0 and 1 so that they are valid
probabilities, this solution will identify the probabilities required for player 2 so that
player 1 will be indifferent between all three of her pure strategies.
To solve this system of equations, first substitute (3) into (2). In (3), q1 = 1 - q2 + q3 , so
we can rewrite (2) as
5(1 - q2 - q3) + 7q2 + 15 q3
OR
2(1 - q2 - q3) + 2 q3
OR
2
OR
q2
=
=
=
=
3(1 - q2 - q3) + 11q2 + 13 q3
4q2
6q2
1/3
Now substituting (3) in (1), we have
q1 + 9q2 + 17 q3
OR
(1 - q2 - q3) + 9 q2 + 17 q3
OR
2 q2 + 2 q3
OR
6q2 + 6q3
=
=
=
=
5q1 + 7q2 + 15 q3
5(1 - q2 - q3) + 7 q2 + 15 q3
4(1 - q2 - q3)
4
Since we found q2 = 1/3 above, this gives q3 = 1/3 as well, so q1 = 1/3. That is, player 1
can only be indifferent between U, M, and D if player 2 plays the particular mixed
strategy (1/3, 1/3, 1/3).
A similar analysis shows that player 2 can only be indifferent between L, C, and R if
player 1 plays the particular mixed strategy (1/3, 1/3, 1/3).
We conclude that [(1/3, 1/3, 1/3), (1/3, 1/3, 1/3)] is the only mixed strategy equilibrium
where each player plays all three strategies with positive probability. In addition, this
analysis rules out the possibility of a mixed strategy equilibrium where one player plays
two pure strategies with positive probability and the other player plays three pure
strategies with positive probability – for either player to play a mixed strategy with all
three pure strategies, the other player also has to play a mixed strategy with all three pure
strategies.
Second, consider the possibility of a mixed strategy equilibrium where player 1 does not
play D (i.e. player 1 plays D with probability 0).
Excluding D, the normal form is as follows:
U
M
L
1, 6
5, 10
C
9, 2
7, 12
R
17, 4
15, 8
Now L strictly dominates R for player 2, so we can reduce the game to a 2x2 normal
form.
U
M
L
1, 6
5, 10
C
9, 2
7, 12
Assuming that player 1 plays U with probability p and player 2 plays L with probability
q, a mixed strategy equilibrium of this 2x2 game is given by the equations
EU1(U) = q + 9(1-q)
= EU1(M) = 5q + 7(1-q)
EU2(L) = 6p + 10(1-p)
= EU2(C) = 2p + 12(1-p)
These equations have the solutions p = 1/3, q = 1/3 corresponding to mixed strategies in
the 3x3 game [(1/3, 2/3, 0), (1/3, 2/3, 0)]. This combination of strategies is a candidate
mixed strategy equilibrium for the whole game because we “guessed” that the
equilibrium would not include strategy D for player 1 and now have to “verify” that D is
not a best response in this candidate equilibrium.
In this candidate equilibrium,
EU1(U) = 19/3; EU1(M) = 19/3; EU1(D) = 25/3.
EU2(L) = 26/3; EU1(C) = 26/3; EU1(R) = 20/3.
So player 2’s mixed strategy is a best response to (1/3, 2/3, 0) by player 1, but player 1’s
mixed strategy is not a best response to (1/3, 2/3, 0) by player 2. So we conclude that
there is no mixed strategy equilibrium where player 1 plays only the strategies U and M.
Third, consider the possibility of a mixed strategy equilibrium where player 1 does not
play M (i.e. player 1 plays M with probability 0).
Excluding M, the normal form is as follows:
L
C
1, 6
9, 2
U
3,
14
11,
16
D
R
17, 4
13, 18
Now R strictly dominates C for player 2, so we can reduce the game to a 2x2 normal
form.
L
R
1, 6
17, 4
U
3, 14
13, 18
D
Assuming that player 1 plays U with probability p and player 2 plays L with probability
q, a mixed strategy equilibrium of this 2x2 game is given by the equations
EU1(U) = q + 17(1-q)
= EU1(D) = 3q + 13(1-q)
EU2(L) = 6p + 14(1-p)
= EU2(R) = 4p + 18(1-p)
These equations have the solutions p = 2/3, q = 2/3 corresponding to mixed strategies in
the 3x3 game [(2/3, 0, 1/3), (2/3, 0, 1/3)].
In this candidate equilibrium,
EU1(U) = 19/3; EU1(M) = 25/3; EU1(D) = 19/3.
EU2(L) = 26/3; EU1(C) = 20/3; EU1(R) = 26/3.
Once again, player 2’s mixed strategy is a best response to (2/3, 0, 1/3) by player 1, but
player 1’s mixed strategy is not a best response to (2/3, 0, 1/3) by player 2. So we
conclude that there is no mixed strategy equilibrium where player 1 plays only the
strategies U and D.
Fourth, consider the possibility of a mixed strategy equilibrium where player 1 does not
play U (i.e. player 1 plays U with probability 0).
Excluding U, the normal form is as follows:
M
D
L
5, 10
3, 14
C
7, 12
11, 16
R
15, 8
13, 18
Now C strictly dominates L for player 2, so we can reduce the game to a 2x2 normal
form.
C
R
7, 12
15, 8
M
11, 16
13, 18
D
Assuming that player 1 plays M with probability p and player 2 plays C with probability
q, a mixed strategy equilibrium of this 2x2 game is given by the equations
EU1(U) = 7q + 15(1-q)
= EU1(D) = 11q + 13(1-q)
EU2(L) = 12p + 16(1-p)
= EU2(R) = 8p + 18(1-p)
These equations have the solutions p = 1/3, q = 1/3 corresponding to mixed strategies in
the 3x3 game [(0, 1/3, 2/3), (0, 1/3, 2/3)].
In this candidate equilibrium,
EU1(U) = 43/3; EU1(M) = 37/3; EU1(D) = 37/3.
EU2(L) = 38/3; EU1(C) = 44/3; EU1(R) = 44/3.
Once again, player 2’s mixed strategy is a best response to (0, 1/3, 2/3) by player 1, but
player 1’s mixed strategy is not a best response to (0, 1/3, 2/3) by player 2. So we
conclude that there is no mixed strategy equilibrium where player 1 plays only the
strategies M and D.
We have now exhausted the set of possibilities for mixed strategy equilibria and can
conclude that there is a unique Nash equilibrium to the original 3x3 game and that it is
given by [(1/3, 1/3, 1/3), (1/3, 1/3, 1/3)].
4.1 Extensive Form for Cournot and Stackelberg Games
(a) Each firm has two choices of quantity and the firms make these choices
simultaneously. So the game can be represented in extensive form with either player
moving first, using information sets to demonstrate that the second player in the extensive
form must choose quantity without knowledge of the action of the other player.
(b) With just one exception, this extensive form game is identical to the extensive form
for the Cournot game in (a) with player 1 listed first in the game tree. The only
difference in these extensive forms is that the information set for player 2 is removed in
the Stackelberg game tree.
(c) This is analogous to the simultaneous game based on the information available to
player 2 at the time of her decision. The choice of representation in the extensive form
turns on the information available to each player rather than the exact timing of moves.
4.2. Cournot Competition and Dominated Strateges (See Solutions to Problem Set 2)
4.3 The Herfindahl Index
(a) In order to compute the index we first need to find the optimal production levels for
firms in each scenario. You can do this by brute force, or here is one short way to do it.
Start by setting up the problem and finding one firm’s best response function for the N=5
case, because you can always set some of the firms’ quantities to zero to see what would
happen if N=4,3,2,1:
Max
p(Q)q1 (120 q1 q2 q3 q4 q5 )q1
q1
This gives the FOC (interior case):
120 2q1 q2 q3 q4 q5
0
120 q2 q3 q4 q5
2
Since the firms are all identical, in equilibrium they produce identical quantities.
So all of the q’s in the above equation can be replaced with q1*.
120 4q1*
q1*
2
Stop here for a moment and notice that we can generalize this equation for the Nfirm case. However many firms are in Cournot competition, we will have:
120 ( N 1)q1*
q1*
2
*
2q1 120 Nq1* q1*
q1*
( N 1)q1* 120
120
N 1
That gives the best response function for an individual firm in the N-firm Cournot
case. We can now set up a little chart to compute the Herfindahl index for any number of
firms:
q1*
Number of firms
1
2
3
4
5
qi (each firm’s q)
120/2=60
120/3=40
120/4=30
120/5=24
120/6=20
Q (total quantity)
60*1=60
40*2=80
30*3=90
24*4=96
20*5=100
Herfindahl index
[(60/60)^2]*1=1.0
[(40/80)^2]*2=0.5
[(30/90)^2]*3=0.33
[(24/96)^2]*4=0.25
[(20/100)^2]*5=0.20
(b)
i. Total Q=100.
40
H
100
ii. Total Q=100.
H
60
100
2
2
30
100
20
100
2
2
30
100
20
100
2
2
16
9
9
100 100 100
0.34
36
4
4
100 100 100
0.44
(c) Maybe, but not necessarily. The Herfindahl index is simply a measure of market
concentration – it is not necessarily a measure of market efficiency nor of consumer
surplus. Within a Cournot market, market concentration falls as the number of firms
increases (and as prices thus fall) – and thus H can be an indicator of efficiency in this
setting.
However, it may not be useful to compare across different types of markets,
especially if we don’t have a good understanding of where equilibrium production levels
come from in these markets. To see this, note that the markets in part (b) have strictly
higher Herfindahl indices than the Cournot markets in part (a) with 3, 4, or 5 firms. But
this is not necessarily bad, since the markets in part (b) also produce higher total quantity,
and thus provide consumers with lower prices than these Cournot markets with lower
indices (prices and quantity are equal in the N=5 case).
To make it even more clear, say the government only cared about lowering this
measure of market concentration. If we start from a market such as the one in (b)(i), the
government could tell firm 1 to cut its production to 30. Assuming the other firms’
production levels are unchanged, this would lower the Herfindahl index (from 0.34 to
0.33) but how could it possibly be good for consumers?
Understanding where the production levels in (b)(i) come from may help guide
whether or not to pay attention to the Herfindahl index. Clearly, these production levels
are not the result of Cournot competition with identical firms (unless firm 1 is not good at
profit maximizing). Perhaps the firms are not identical – perhaps firm 1 has lower
marginal costs? In this case, it may make sense to ignore the market concentration index.
But if we think firms are really identical, then why might one firm produce more than
another? Perhaps firm 1 is a bully, somehow scaring other firms out of the market
(perhaps by strategically committing early to producing a high quantity, and then letting
other firms respond). Then, it is possible that by placing caps on firm 1’s quantity, the
quantities produced by other firms may rise. Say firm 1 was reduced to 35, and firms 2
and 3 increased to 35 each. Then H=0.33 and total Q=105 – so we would have the nice
result that the H index falls and quantity increases as compared with (b)(i).
Practically speaking, the HHI is frequently used as a first-cut test in anti-trust
matters. Markets with low HHI are considered OK, while markets with high HHI are
considered potential problems that bear further investigation.
4.4 Monopoly vs. Cournot Part 1
(a) The monopolist solves profit-maximization problem
MaxQ Q [P(Q) – c],
with first-order condition
P(Q) + Q P’(Q) – c = 0.
With P(Q) = 20 – Q, the first-order condition simplifies to
20 – 2Q – c = 0.
(b) Let Q = Q1 + Q2, where Qj is the quantity produced by firm j.
Firm 1 faces profit-maximization problem
MaxQ1 Q1 [P(Q) – c]
with first-order condition
P(Q) + Q1 P’(Q) – c = 0.
Note that P’(Q) is the same for parts (a) and (b) under the assumption that firm 2’s
quantity in Cournot competition is fixed from the perspective of Firm 1.
With P(Q) = 20 – Q, this first-order condition simplifies to
20 – Q – Q1 – c = 0 OR
20 – 2Q1 – Q2 – c = 0
(c1) Monopoly vs. Perfect Competition
The first-order condition for perfect competition is simply P(Q) – c = 0. There is an
additional negative term in the first-order condition of the monopolist’s profitmaximization problem (1’). Thus, when FOC holds for monopoly, the left-hand side of
(1’) will be negative. Since (1’) can be interpreted as MR – MC or net profit per unit
produced, this indicates that marginal profit is negative for a monopolist at the
equilibrium quantity for perfect competition. This indicates that the monopolist will wish
to reduce quantity from the perfect competition equilibrium quantity, thereby increasing
price. This implies lower quantities and higher prices for monopoly than perfect
competition.
(c2) Monopoly vs. Cournot
Part (a) yields FOC
20 – 2Q – c
or
P(Q) – Q – c
=
=
0
0
(1)
(1’)
whereas part (b) yields FOC
20 – 2Q1 – Q2 - c
=
or
20 – 2Q – c + Q2
=
0
0
(2)
(2’)
The ranking is clearest from a comparison of (1) and (2’). (In the case of Cournot
competition in (2’), Q = Q1 + Q2.)
20 – 2 Qmonopoly – c
20 – 2 QCournot – c + Q2
=
=
0
0
(1)
(2’)
When Q = Qmonopoly, then equation (1) holds with equality, while the left-hand side of (2’)
is Q2. That is, (2’) does not hold at Q = Qmonopoly unless Q2 = 0 (i.e. a Cournot case that is
really a monopoly). Further if Q1 + Q2 = Qmonopoly, then the left-hand side of (2’) is
positive, indicating that marginal revenue for firm 1 is greater than marginal cost at the
total monopoly quantity. Thus, when firm 1 (and firm 2) set MR = MC in Cournot
competition, it must be that QCournot > Qmonopoly.
This implies higher quantities for Cournot than monopoly and lower prices for Cournot
than monopoly.
Taken together, (c1) and (c2) provide the desired rankings for equilibrium outcomes of
these three possibilities.
(d) The difference in FOC between monopoly and two-firm Cournot is that each Cournot
firm ignores the effect of a change in its quantity on the profits of the other firm. Once
again, comparing (1) and (2’), there is an additional positive Q2 term in the left-hand side
of (2’). This represents an effect on total profits from increase in Q1 that firm 1 neglects
in its incentives. That is, an increase in firm 1’s quantity reduces firm 2’s profit per unit
sold (at rate equal to firm 2’s quantity, Q2), but firm 1 does not account for this in its
profit maximization. This is a negative externality for the two firms from the perspective
of joint profit-maximization.
Thus, the firms overproduces in Cournot competition relative to the monopolist’s profitmaximizing quantity. But if they accounted for the negative externality in their
computations, then the FOC as represented by (2) in part (c) would be transformed into
(1) in part (c) and would lead to monopoly quantities and prices.
(e) Let Q*2 be total quantity with two Cournot firms and Q*3 be total quantity with three
Cournot firms. Similarly, let Qj represent firm j’s quantity produced.
In Cournot competition with three firms, firm 1 faces profit-maximization problem
MaxQ1 Q1 [P(Q) – c]
with first-order condition
P(Q) + Q1 P’(Q) – c = 0.
(3)
With P(Q) = 20 – Q, this first-order condition simplifies to
20 – Q*3 – Q1 – c = 0
OR
20 – 2Q1 – Q2 – Q3 - c = 0
(3’)
(3’’)
By comparison, with two Cournot firms, firm 1’s first-order condition is
20 – 2Q1 – Q2 – c = 0
(4)
OR
20 – Q*2 – Q1 – c = 0
(4’)
Here conditions (3’) and (4’) are identical, except that (3’) refers to total quantity with
three firms and (4’) refers to total quantity with two firms. If Q*2 = Q*3, then both (3’)
and (4’) would hold simultaneously if Q1 is the same in both cases.
Assuming a symmetric equilibrium in both cases (Q1 = Q2 = Q3), however, firm 1 would
produce less in the three-firm case than in the two-firm case if total quantities are the
same in both cases. That is, if Q*2 = Q*3 and (4’) holds, then the left-hand side of (3’) is
greater than the left-hand side of (4’), once again implying that MR > MC in (3’).
In words, firm 1’s effect on total price is smaller for a given total quantity as the number
of firms increases. Therefore, when Q*2 = Q*3 = Q*, firm 1 has greater marginal revenue
with three firms (when its production is Q*/3) than with two firms (when its production is
Q*/2). As a result, firm 1 (and firms 2 and 3) will increase production from Q*/3 in the
three-firm case, implying Q*3 > Q*2.
4.5 Cournot vs. Monopoly, Part 2
(b) Firm 1 solves:
Max
q1
p(Q)q1 c(q1 ) ( A Bq1 Bq 2 )q1
1
2
q12
Aq1 Bq12
1
2
Bq1q2
q12
So the FOC (interior solution) is:
A 2 Bq1* Bq 2
q1* 0
A Bq 2
2B
This represents firm 1’s best response function – it tells us what firm 1 would do
given any quantity produced by firm 2.
q1*
Firm 2 solves the maximization problem:
Max
p (Q)q 2 c(q 2 ) ( A Bq1
q2
Bq 2 )q 2
cq 2
Aq 2
Bq 22
Giving the FOC (interior solution):
A Bq1 2 Bq 2 c 0
q2*
A Bq1 c
2B
[Note: you are not asked to solve for the equilibrium, just to find the first order
Bq1q 2
cq 2
conditions.]
(c) As a monopolist, firm 1 would choose quantity by maximizing profits:
2
2
Max 1 p(q1 )q1 c1 (q1 ) Aq 1 Bq1 12 q1
q1
So the FOC (interior solution) is:
A 2 Bq1*
q1* 0
q1*
A
2B
Note that this corresponds to q1* in Cournot, where q2 = 0.
Using the first-order conditions from part (b) and substituting for firm 1’s Cournot
quantity:
A Bq 2
Bq 2
(
2 B)q 2
(
B)q 2
A
A
q2
2B
2B
2B
2B
2B
2B
Because of the additional term in the sum, this total Cournot quantity is greater
than firm 1’s monopoly quantity. And since we are told to assume that firm 1’s
monopoly quantity is bigger than firm 2’s monopoly quantity, it must be the case that the
total Cournot quantity is greater than either firm would produce as a monopolist (you
could also show this by solving for Firm 2’s monopoly quantity, (A-c)/2B, and then
showing that q1+q2 = (A–Bq1–c)/2B + q1 is larger than firm 2’s monopoly quantity).
q1
q2
4.6 Bertrand with Quadratic Demand