Chapter 15

Exercise 15.10.1

Bob’s strategies are functions from the card he observes into actions he takes, so

that he has four possible strategies: CC,CF, FC, FF (C stands for “Call” and F for

“Fold”, so that for example the strategy CF means call if he observes H and to fold

if he observes L, and the strategy FF instructs him to fold in either case). Alice’s

strategies are functions from observed Bob’s actions into her actions, so she has four

strategies cc, cf, fc, ff (e.g., strategy fc means to fold if call and call if fold).

To illustrate how to compute the payoffs in the strategic form, take for instance

the profile (FC, cf). With prob = 12

Bob will observe either of H and L. In case he

observes H he will fold, and Alice will then fold as well, resulting in the payoff of

0 to each. If he observes L he will call and Alice will call, so that Alice gets 1 and

Bob gets −1. So the payoffs for (FC, cf) are 12(0, 0) + 1

2(−1, 1) = (−1

2, 1

2). The other

payoffs are obtained similarly. The resulting strategic form of the game is

cc cf fc ff

CC0 0 0 0

0 0 0 0

CF−1

2−1

20 0

12

12

0 0

FC12

12

0 0

−12

−12

0 0

FF0 0 0 0

0 0 0 0

Nash equilibria are constituted by any strategy profile where Bob plays any of

CC,CF, FF , and Alice plays either fc or ff . Strategy CF weakly dominates all

the other strategies for Bob.

1

Chapter 15

If Bob observes the card only with probability p the situation doesn’t change

much. Then Bob has 8 strategies (i.e., strategy tells him what he does if he observes

H, observes L, or doesn’t observe anything), and as before we would compute the

strategic form. But when Bob calls if he doesn’t observe the card, his expected payoff

is 0 regardless of what Alice does, so that this case will never contribute anything

to Bob and Alice’s payoffs. Similarly, when Bob folds if he doesn’t observe, their

expected payoffs are 0. Therefore, the strategic form would now be just two payoff

matrices from the previous case, both multiplied by p, and stacked on top of each

other and the same conclusions hold as before.

2

Chapter 15

Exercise 15.10.2

As in exercise 15.10.1 we can compute the strategic form which is now

cc cf fc ff

CC0 0 −1 −1

0 0 1 1

CF−1

2−1

20 0

12

12

0 0

FC32

32

0 0

−32

−32

0 0

FF1 1 1 1

−1 −1 −1 −1

Nash equilibrium is given by Bob’s mixed strategy (13, 2

3, 0, 0), and Alice’s (β, 2

3−

β, α, 13− α), where α ∈ [0, 1

3], β ∈ [0, 2

3]. Clearly, Alice will sometimes bet when Bob

is sure to win.

3

Chapter 15

Exercise 15.10.3

Bob’s analysis is incorrect because he counts the money in the pot as being his

whereas it actually belongs to whomever will win the pot. That is, his contribution to

the pot is sunk. Bob incurs an opportunity cost because he foregoes the opportunity

of being a winner.

4

Chapter 15

Exercise 15.10.4

The strategic form is (B stands for Bet, and F stands for Fold, so that e.g., BF

stands for Bet if L and Fold if H),

bb bf fb ff

BB0 −3a− b b− a −a

0 3a+ b a− b a

BF3a+ b 0 a+ b −2a

−3a− b 0 −a− b 2a

FBa− b −a− b 0 −2a

b− a a+ b 0 2a

FFa 2a 2a 0

−a −2a −2a 0

After successive deletion of strongly dominated strategies we are left with BB,FB

and bb, fb, and the claim follows.

5

Chapter 15

Exercise 15.10.5

In the casting stage the Nature chooses which of the roles a player will get to play,

in the Quiche, there are 2 possible roles (Tough or Wimpy) for Bob and a single

role (Nasty) for Alice. Once the casting stage is over, we then fill in the appropriate

payoffs into the outcome nodes of the two possible games that the players play,

depending on what has happened in the casting stage. These two games are depicted

in figure 15.9. Together (adding the casting move by Nature, and appropriately

drawing Alice’s information sets) these two games constitute the game of imperfect

information that Alice and Bob end up playing.

6

Chapter 15

Exercise 15.10.6

The strategic form is (e.g., dh stands for dove if N and hawk if Y )

dd dh hd hh

DD5 13

292

6

5 52

52

0

DH52

174

114

92

132

174

174

2

HD52

174

114

92

92

114

114

1

HH0 2 1 3

6 92

92

3

Now hh strongly dominates hd and dd, and after eliminating these for both players,

hh strongly dominates dh. Hence the only rationalizable strategy profile is (HH, hh),

contrary to Rousseau.

7

Chapter 15

Exercise 15.10.7

We refer to figure 15.6(a) for the picture. When Player II gets to move, he compares

his expected payoff from firing with what he gets if he chickens out, which is b.

Thus, he will fire if and only if b ≤ 12. Given this strategy of II, Player I compares his

payoff of chickening out with his expected payoff if he fires. Denote by Pr(b ≤ 12|a)

the conditional probability that Player I assigns to the event that II will fire, given

that his own type is a. Now, the expected payoff of I if his type is a and he fires

is 23

(Pr(b ≤ 1

2|a)(1

21 + 1

2a) + Pr(b > 1

2|a)1

)+ 1

30. We have to consider 2 different

cases. First, if a ≤ 12, then Pr(b ≤ 12|a) = 1

212

+ 121 = 3

4(because half of the time b

is equal to a, and since a is assumed to be less than or equal 12, b will be as well, and

the other half of the time, b is independent, so the probability of it being less than

12

is then equal to 12). Thus, the expected payoff of firing for I is then 1

12(5 + 3a),

and he will fire if this is more than a, which is always the case for a ≤ 12. Similarly,

when a ≥ 12, I’s expected payoff from firing is 1

12(7 + a), so that he then fires if and

only if a ≤ 711

.

8

Chapter 15

Exercise 15.10.8

As in section 10.2.2, Alice’s and Bob’s profits are given by πA(a, b) = (K−A−a−b)a

and πB(a, b) = (K−B−a−b)b, except that A andB are now random variables. So the

expected profit of Alice if she is of type A is EπA(a, b) = E [(K − A− a− b)a] = (K−

A−a− b)a, because πA(a, b) is a linear function of b so that when we take expectation

over b, we can distribute the expectation over the product and the sum (in fancy

language, expectation is a linear operator). Here b is Bob’s average output, which is

at this point still unknown. Similarly, we have for Bob EπB(a, b) = (K−B− a− b)b.

Now we can compute Alice’s best-reply function by maximizing her expected profit

with respect to her output, while treating her type A as a constant (Alice knows

her type). This gives Alice’s output as a function of her type A and Bob’s average

output b, a = K−A−b2

. Now, b is a constant, so we can take the expectation over a to

obtain a = K−c1−b2

. Similarly, we get for Bob b = K−B−a2

and b = K−c2−a2

. Together,

the equations for a and b give a = K−2c1+c23

and b = K−2c2+c13

.

9

Chapter 15

Exercise 15.10.9

Notice that Pr(B = L|A = H) = 1 so that Alice knows for sure that Bob is of type

L. Bob, however, only thinks that Alice is of type H with Pr(A = H|B = L) = 19, so

that Alice knows Bob is badly mistaken. Alice can exploit this fact by producing more

than she would be able to produce otherwise. The reason is that Bob is convinced

she is of type L which produces more than type H, so Bob will produce less than he

would if he knew Alice was H. Alice, knowing this, will be able to cash in. Formal

argument needs a bit of accounting: different types have different beliefs about the

other player. Let aH denote Alice’s output when she is of type H, aL is her output

when she is type L, and a = 19aH + 8

9aL. Similarly for Bob. Then, from maximization

of Alice’s and Bob’s profit, we get the following equations for Alice’s output (and

symmetrically for Bob): aH = 12(K−H−bL), aL = 1

2(K−L− b), and a = 1

9aH + 8

9aL.

From these three equations and from the three symmetric equations for Bob, we get

aH = bH = 151

(17K + 9L− 26H), aL = bL = 151

(H + 17K − 18L). It is easy to check

that when A = H, Alice is playing a best reply to Bob’s production level, as opposed

to Bob, who is under-producing given Alice’s production.

10

Chapter 15

Exercise 15.10.10

We use the expressions for α(A) and β(B) derived in section 15.5.1. We need to

inductively compute ¯A, ¯B,¯A, and so on. Since A = cB, we have ¯A = cB = cdA.

Similarly, we get ¯B = cdB. Now we can compute¯A = cA = c2dB, whence we can

already observe the pattern. Next we plug this into the formulae for α(A) and β(B).

We show how to compute the expression for α(A). The expression is

α(A) =1

2(K − A)− 1

4(K − dA) +

1

8(K − cdA)− 1

16(K − cd2A) + ...

We rearrange this into

α(A) = K(1

2− 1

4+

1

8− ...)− A(

1

2− 1

4d+

1

8cd− 1

16cd2 + ...).

We already know from section 15.5.1 that the coefficient in front of K equals 13. To

compute the coefficient multiplying A denote it by x and observe that

x =1

2− 1

4d+

1

4cdx.

From here we can immediately compute x = 2−d4−cd

, so that α(A) = 13K − 2−d

4−cdA. The

expression for β(B) is computed in the same manner.

11

Chapter 15

Exercise 15.10.11

If their beliefs are consistent then what Bob believes about Alice must be the same

as what Bob believes about what Alice believes that Bob believes about Alice. That

is, it must be that¯A = A, which gives cd = 1.

12

Chapter 15

Exercise 15.10.12

We have that b = K−B−a2

, implying that b = 13(K − 2c2 + A). Since a = K−A−b

2, so

that a = K−A−b2

. Now we can insert the expression for b, to obtain a = 16(2K+ 2c2−

3A− A).

In case Alice is type L < c2 but A > c2 Bob will overproduce, and knowing that,

Alice will best-reply to his production level and produce less.

13

Chapter 15

Exercise 15.10.13

Let [L,H] be the interval of possible types for Alice, and let l be her type such that

she reveals her cost for all types A ≤ l, but doesn’t reveal her type for A > l. The

production level for type A < l is then a(A) = 13(K−2A+c2), and Bob’s production

is b(A) = 13(K−2c2+A). On the other hand, if Alice is of type A′ > l, her production

is given as in exercise 15.10.12, except that we need to take conditional expectation

over her types, A(l) = E[A | A > l]. Thus, α(A) = 16(2K + 2c2 − 3A − A(l)) and

b = 13(K − 2c2 + A(l)). Inserting this into the expression for Alice’s profit, it follows

that for A > l, close to l, Alice would be better off revealing her cost.

14

Chapter 15

Exercise 15.10.14

We have b = K−B−a2

, so that b = K−B−¯a2

. Next, a = K−A−b2

, so that a = K−A(B)−b2

,

and taking the expectation we obtain ¯a = K− ¯A−b2

. Using these to solve for α(A) and

β(B), we obtain the desired equations.

Another type B of Bob might hold a belief that A(B) is high so that ¯A would

then be high and Alice would produce less.

15

Chapter 15

Exercise 15.10.15

Pr(I = A | II = D) = 0 whereas Pr(I = A) = 0.01, so that choices aren’t indepen-

dent. Pr(I = B | II = C) = .9. A and D will know for sure.

16

Chapter 15

Exercise 15.10.16

(a) A,C: (d,R); A,D: (d, L); B,C: (u, L); B,D : (u,R).

(b) After successive deletion of dominated strategies, the solution is A plays u,

B plays u, C plays L, D plays R (note that everyone except actor A has a

strongly dominant strategy, given the probabilities in figure 15.11(b)).

(c) Probabilities, payoffs to actor C and nothing else, since all actors except actor

A are playing a strongly dominated strategy.

(d) See answer to part (c).

(e) The difference reflects the fact that the actors have to optimize against the

average opponent rather than against a specific opponent as in part (a). Given

that C plays the strategy which is best on average, A can get 8 instead of −4.

17

Chapter 15

Exercise 15.10.17

(a) From left to right: (8,−8), (−8, 8), (0, 0), (−4, 4); (−4, 4), (8,−8), (0, 0),

(12,−12); (−8, 8), (12,−12), (−12, 12), (16,−16); (4,−4), (−4, 4), (0, 0), (−8, 8).

(b) Von Neumann’s pure strategies are s1s1, s1s2, s2s1, s2s2, similarly for Morgen-

stern.

(c) For instance the payoff vector for the profile (s1s1, t1t2) is (8α−8β−4γ,−8α+

8β + 4γ), where α = 0.01, β = 0.09, γ = 0.9. Similarly, we compute the rest of

the strategic form. The resulting game is clearly zero sum.

(d) The saddle point is (s1s1, t1t2).

(e) t1t2 strongly dominates all the other strategies for II, after deleting these, s1s1

is optimal for I.

18

Chapter 15

Exercise 15.10.18

An actor always knows his identity but also knows that the other actors only know

his “expected identity”, meaning that as perceived by the actor (through his belief

and the belief about others’ beliefs) the game is not zero sum.

19

Chapter 15

Exercise 15.10.19

(a) It does not have to be modified.

(b) In the approach of exercise 15.10.16, each player would, after observing his type,

decide with what probability to play either of his two actions, and then throw

an appropriate die. In the approach of exercise 15.10.17, each player would

in advance decide the probability with which to play each of his 4 strategies,

then throw an appropriate die, and then observe his type and implement the

appropriate action.

20

Chapter 15

Exercise 15.10.20

(a) The belief of I is (12, 1

2) regardless of the actor, while the belief of II is either

(12, 1

2) or (0, 1). I’s beliefs imply that C and D would have to be equally likely

and Pr(I = A, II = B) > 0. However, belief of D implies that Pr(I = A, II =

B) = 0 so that these beliefs are inconsistent.

(b) The model is the same as figure 15.12, but the casting move has different

probabilities for each type.

(c) For example, the payoffs to the strategy profile (s1s1, t2t2) are (3.96, 3.4).

21

Chapter 15

Exercise 15.10.21

(a) Nash equilibria are (Pay, Free−ride), (Free−ride, Pay) and a mixed equilib-

rium given by (4−c24, 4−c1

4). If c1 = c2 = 1 then the public good will be provided

with probability 1516

; if c1 = c2 = 3 the probability is 716

.

(b) Compute the payoffs, Ui(Pay | ci) = p(4 − ci) + (1 − p)(4 − ci) = 4 − ci, and

Ui(Free− ride | ci) = 4(1− p). Given that p ∈ [14, 3

4] it is now immediate that

(Free− ride, Pay) is an equilibrium. The probability that the public good is

provided is 1− p2.

(c) In a symmetric equilibrium, one of the types, depending on p, now plays a

mixed strategy. If p ∈ [34, 1], denote α = Pr(Pay | High). Then, as before,

Ui(Pay | ci) = 4 − ci, and Ui(Free − ride | ci) = 4 (αp+ (1− p)). For High

(ci = 3) to be indifferent, 1 = 4 (αp+ (1− p)), so that α = 4p−34p

; note that

α ∈ [0, 14]. It is easy to verify that Low prefers Pay.

If p ∈ [0, 14] then denote β = Pr(Pay | Low), and as before, we obtain β =

34(1−p)

∈ [34, 1].

(d) Let α and β be as in (c). Again, Ui(Pay) = 4−ci, but Ui(Free−ride | High) =

4 (0.2α + 0.8β), and Ui(Free − ride | Low) = 4 (0.8α + 0.2β). If α, β ∈ (0, 1)

then both types must be indifferent, hence 14

= 0.2α+0.8β and 34

= 0.8α+0.2β.

Solving these we obtain α = 1112

and β = 112

, and the probability of providing

the good is 0.1× α2 + 2× 0.4× αβ + 0.1× β2 = 1172

.

22

Chapter 15

Exercise 15.10.22

Let KA and KE be the knowledge operators of Adam and Eve, respectively, and let

CK be the common knowledge operator. If n is even, then we have KA({n}) = {n−

1, n}, KE({n}) = {n, n + 1}, so that KA({n}) ∩KE({n}) = {n}, hence CK({n}) =

{n}. The event that there are finitely many periods is E = ∪∞n=1{n}, so that E =

CK(E), hence in every state of the world, E is common knowledge.

23

Chapter 15

Exercise 15.10.23

In order for grim-trigger to be an equilibrium it has to be the case that in each state

of the world, both players believe that it is twice as likely that there will be more

repetitions rather than less. In order to construct a common prior P which would

make these beliefs consistent, it would thus have to be the case that P ({n}) = q2n,

for some constant q > 0. Since P is a probability measure, we would need to pick q

such that∑∞

n=1 q2n = 1, but that is impossible since

∑∞n=1 2n =∞.

24

Chapter 15

Exercise 15.10.24

Consider an actor of type n. If both players follow the prescribed strategies, than

n’s payoff is 23(2(n− 2)− 1) + 1

3(2(n− 1) + 3) = 2(n− 1), if n deviates and instead

plays hawk in period n−2, then his payoff is 2(n−3)+3 = 2(n−1)−1. The actors’

beliefs now come from a common prior P (N) = 12N , i.e., actor n obtains a signal that

the state of the world is in {n, n+ 1}, so that after updating, he puts probability 23

on state n.

25