Econ 302
Assignment 2 – Solution
Problem 1
Part a) Note: I will explain everything only for question 1 part a. For the rest of
the answer key, I will give only short answers since they follow the same logic.
(i) ISD
Round 1 - B is strictly dominated by T
•
Given player 2 choose L, T give player 1 higher payoff than B (4 > 2)
•
Given player 2 choose C, T give player 1 higher payoff than B (2 > 1)
•
Given player 2 choose R, T give player 1 higher payoff than B (3 > 0)
L
T
M
B
C
R
4,0 2,1 3,2
2,2 3,4 0,1
2,3 1,2 0,3
Now Strategy B is out!
L
Round 2 - L is strictly dominated by C
•
Given player 1 choose T, C give player 2 higher payoff than L (1 > 0)
•
Given player 1 choose M, C give player 2 higher payoff than L (4 > 2)
T
M
B
C
R
4,0 2,1 3,2
2,2 3,4 0,1
2,3 1,2 0,3
Now Strategy L is out!
Note: that we don’t consider the case in which player 1 chooses B, because B has been
out since round 1.
Conclusion:
•
Player 1’s pure strategies that survive ISD is T and M
•
Player 2’s pure strategies that survive ISD is C and R
(ii) Find NE
Pure strategy NE
•
We don’t consider B and L since they are strictly dominated, so we can rewrite
that game as follow.
•
Circles represent player 1’s best responses.
•
o
If player 2 chooses C, player 1 choose M (since 3 > 2)
o
If player 2 chooses R, player 1 choose T (since 3 > 0)
Squares represent player 2’s best responses.
o
If player 1 chooses T, player 2 choose R (since 2 > 1)
o
If player 1 chooses M, player 2 choose C (since 4 > 1)
C
T 2 , 1
M 3 , 4
R
3 , 2
0 , 1
1
•
We can see that there exist two pure strategy NE: (T,R) and (M,C)
Mixed strategy NE
•
Again, we don’t consider B and L since they are strictly dominated.
•
Let player 1 plays T with probability p and plays M with probability 1-p
•
Let player 2 plays C with probability q and plays R with probability 1-q
Find p – player 1 choose p such that player 2’s expected payoff from choosing C is
equal to expected payoff from choosing R.
o
Player 2’s expected payoff form choosing C = p(1) + (1-p)(4) = 4 – 3p
o
Player 2’s expected payoff form choosing R = p(2) + (1-p)(1) = 1 + p
o
Since they must be equalized:
4 – 3p = 1 + p
•
Therefore, p = ¾
Find q – player 2 choose q such that player 1’s expected payoff from choosing T is
equal to expected payoff from choosing M.
o
Player 1’s expected payoff form choosing T = q(2) + (1-q)(3) = 3 – q
o
Player 1’s expected payoff form choosing M = q(3) + (1-q)(0) = 3q
o
Since they must be equalized:
3 – q = 3q
•
Therefore, q = ¾
•
There exists one mixed strategy NE: ( ¾T + ¼M, ¾C + ¼R)
Conclusion: There exist three NE.
•
Two pure strategy NE: (T,R) and (M,C)
•
One mixed strategy NE: ( ¾T + ¼M, ¾C + ¼R)
Part b)
(i) ISD
Round 1: A is strictly dominated by C
A is out!
Round 2: X is strictly dominated by Z
X is out!
Round 3: B is strictly dominated by C
B is out!
Round 4: Y is strictly dominated by Z
Y is out!
Round 5: C is strictly dominated by D
C is out!
Conclusion:
2
•
Player 1’s pure strategies that survive ISD is D
•
Player 2’s pure strategies that survive ISD is Z
(ii) NE
•
Since each player has only one strategy that survive ISD, there exists only one
NE, which is pure strategy NE: (D,Z)
Part c)
(i) ISD
Round 1: A is strictly dominated by (0.5B + 0.5C)
A is out.
•
Given player 2 choose X, (0.5B + 0.5C) give player 1 higher payoff than A (7.5 > 7)
•
Given player 2 choose Y, (0.5B + 0.5C) give player 1 higher payoff than A (4.5 > 4)
•
Given player 2 choose Z, (0.5B + 0.5C) give player 1 higher payoff than A (8.5 > 8)
Round 2: Z is strictly dominated by (0.6X + 0.4Y)
Z is out.
•
Given player 1 choose B, (0.6X + 0.4Y) give player 2 higher payoff than Z (6.2 > 6)
•
Given player 1 choose C, (0.6X + 0.4Y) give player 2 higher payoff than Z (1.2> 1)
Conclusion:
•
Player 1’s pure strategies that survive ISD is B and C
•
Player 2’s pure strategies that survive ISD is X and Y
(ii) NE
•
There exist three NE.
o
B
C
There exist two pure strategy NE: (B,X) and (C,Y)
X
Y
10 , 7
5 , 0
2 , 5
7 , 3
o
There exist one mixed strategy NE: (0.6B + 0.4C, 0.5X + 0.5Y)
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Problem 2
(i)
ISD
(ii)
NE
All strategies of each player survive ISD.
•
Obviously, there does not exist pure strategy NE. (Do it by yourself.)
•
Then we find mixed strategy NE.
•
Let player 1…
•
o
plays A with probability p1
o
plays B with probability p2
o
plays C with probability 1 – p1 - p2
Since we want to find symmetric NE, therefore player 2…
o
plays X with probability p1
o
plays Y with probability p2
o
plays Z with probability 1 – p1 - p2
Find p1 and p2
•
First, consider player 1. Player 1 choose p1 and p2 such that player 2’s expected
payoff from choosing X = expected payoff from choosing Y = expected payoff from
choosing Z.
o
We need that
p2 + 2(1 - p1 – p2) = 2p1 + (1 - p1 – p2) = p1 + 2p2
2 - 2p1 – p2 = p1– p2 + 1 = p1 + 2p2
o
•
Solving the above equation system, we get p1 = 1/3 , p2 = 1/3
Since the game is symmetric, we don’t need to consider player 2. However, by
yourself you can try to find p1 and p2 by considering player 2. You will get the
same answer.
Conclusion There exists one symmetric NE, which is a mixed strategy:
( 1/3A + 1/3B + 1/3C, 1/3X + 1/3Y + 1/3Z)
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Problem 3
•
Let me define the game this way:
o
There are 4 identical players in the game.
o
Each player has 2 pure strategies: hit on the blond (BL) and hit on a
brunette (BR)
Finding NE
1) Pure Strategy NE
Claim: There exist 4 pure NE: (BL,BR,BR,BR), (BR,BL,BR,BR) , (BR,BR,BL, BR) , (BR,
BR,BR,BL). In other words, one player hit on the blond girl and each of the rest hit on
a brunette girl)
Proof:
Main idea – show that no player has incentive to deviate.
•
•
•
•
•
•
For example, we want to prove that (BL,BR,BR,BR), (player 1 hit on the blond
girl and player 2, 3 and 4 hit on brunette girls) is a pure NE
Consider player 1, given player 2, 3 and 4 hit on a Brunette girl he will
successfully hit on the blond girl and receive payoff = a, but he receive payoff
only b if he deviate (hit on a brunette girl). Therefore, he has no incentive to
deviate.
Consider player 2, given player 1 hit on the blond girl, to hit on a brunette girl
gives him payoff = b, but he receive payoff = 0 if he deviate (hit on the blond
girl). Therefore he also has no incentive to deviate
The same logic also applies to player 3 and 4.
Since nobody has no incentive to deviate, (BL,BR,BR,BR) is a pure NE.
The same logic also applies to the other three pure NE.
Note: All players hit on brunette girls, i.e. (BR,BR,BR,BR), is not a NE. You can prove
it by yourself by showing that each player has incentive to deviate!
2) Mixed Strategy NE
Claim: There exist 3 classes of mixed NE:
1) A class in which all players play mixed strategies. There is one NE in this class:
(
∗
,
∗
,
∗
,
∗
) where
∗
= 1−
+
.
2) A class in which 3 players play mixed strategies and 1 player choose BR. There
are 4 NE in this class ( , ∗ , ∗ , ∗ ) , ( ∗ , , ∗ , ∗ ) , ( ∗ , ∗ , , ∗ ) and
(
∗
,
∗
,
∗
,
) where
∗
= 1−
+
.
5
3) A class in which 2 players play mixed strategies and 2 player choose BR. There
are 6 NE in this class ( , , ∗ , ∗ ) , ( , ∗ , , ∗ ) , ( , ∗ , ∗ , )
(
∗
,
,
,
∗
),(
∗
,
,
∗
,
) and (
∗
,
∗
,
,
) where
∗
= 1−
+
Proof:
1) Proof for the class in which all player play mixed strategy.
•
Since 4 players are identical, we focus on symmetric NE.
•
Let a player chooses BL with probability (1-p) and BR with probability p.
•
Remember: to find a mixed NE, a player will choose p such that his opponent’s
expected payoffs from choosing BL and BR are equalized.
•
Again, since this game is symmetric, it’s enough to consider only one player
since every player has exactly the same expected payoff.
•
Expected payoff for choosing BL
= Prob(none of the others choose BL)*a + (1- Prob(none of the others choose BL))*0
•
•
o Prob(none of the others choose BL) =
o Therefore, Expected payoff for choosing BL = a
Expected payoff for choosing BL = b
To find a mixed NE, we need that
Expected payoff for choosing BL = Expected payoff for choosing BR
a =b
•
Therefore,
∗
= 1−
=
, and (
∗
,
∗
,
∗
,
∗
) is a mixed (symmetric) NE where
+
2) Proof for the class in which 3 players play mixed strategy.
•
•
•
We follow almost exact the same idea that applies for the previous proof, except
now there’re only three players playing mixed strategy.
Consider a player who plays mixed strategy, his payoff for choosing BL is now
a , since there are only two opponents who may hit on the blond.
To find a mixed NE in this class, we need that a
∗
= 1−
+
= b. Therefore
=
and
6
•
The last thing we have to show is to show that the player who plays BR has no
incentive to deviate.
o Given there are three other opponents playing the mixed strategy, the
player’s payoff from choosing BL is a .
o Since
< 1, a < a . We know that a = b, therefore a < b.
o That means BR is strictly better than BL, and any mixed strategy
involving BL will never be chosen by the player. Done!
3) Proof for the class in which 2 players play mixed strategy. – Try it by yourself.
Follow the same idea as the previous proof.
Conclusion
•
There exist 15 NE in 4 classes.
• A class in which all players play mixed strategies and no player choose BR.
There is one NE in this class: (
•
∗
,
∗
,
∗
) where
∗
= 1−
+
.
∗
,
∗
,
∗
,
) where
∗
= 1−
+
.
A class in which 2 players play mixed strategies and 2 players choose BR.
There are 6 NE in this class: ( , , ∗ , ∗ ) , ( , ∗ , , ∗ ) , ( , ∗ , ∗ , )
(
•
,
A class in which 3 players play mixed strategies and 1 player chooses BR.
There are 4 NE in this class: ( , ∗ , ∗ , ∗ ) , ( ∗ , , ∗ , ∗ ) , ( ∗ , ∗ , , ∗ ) and
(
•
∗
∗
,
,
)
,
∗
) , (
∗
,
,
∗
,
) and (
∗
,
∗
,
,
) where
∗
= 1−
+
A class in which all player play pure strategy and 3 players choose BR. There
are 4 NE in this class: (BL,BR,BR,BR), (BR,BL,BR,BR) , (BR,BR,BL, BR) ,
(BR, BR,BR,BL)
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Problem 4
Here is the payoff matrix:
red 2
black 7
black Ace
−2, 2
1, −1
red 8
8, −8
−7, 7
It is easy to check that no strategy is strictly dominated, and there is no pure strategy.
Note that this is an asymmetric game: Player I and Player II have different cards, i.e.,
different actions.
Let us conjecture that in a mixed-strategy Nash equilibrium Player I plays black Ace
with probability p and plays red 8 with probability 1 − p; and Player II plays red 2 with
probability q and plays black 7 with probability 1 − q.
In a mixed strategy Nash equilibrium, both the black Ace and the red 8 must be best
responses of Player I to Player II’s mixed strategy, hence:
−2q + (1 − q) = 8q − 7(1 − q),
i.e.,
4
q= .
9
Likewise, both the red 2 and the black 7 must be best responses of Player II to Player
I’s mixed strategy, hence:
2p − 8(1 − p) = −p + 7(1 − p),
i.e.,
5
p= .
6
Thus, the mixed-strategy Nash equilibrium is that Player I plays black Ace with probability 5/6 and plays red 8 with probability 1/6; and Player II plays red 2 with probability
4/9 and plays black 7 with probability 5/9.
In this equilibrium, player I gets an expected payoff of −2q + (1 − q)|q=4/9 = −1/3; player
II gets an expected payoff of 2p − 8(1 − p)|p=5/6 = 1/3. Thus, Player II wins on average.
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