Answers to PSNE and MSNE questions
Game Theory & Experiments, ECON 6206-001
January 22, 2013
Directions: Answer all questions as completely as possible. You may work in pairs on the assignment. If you do so, simply turn
in one assignment per pair with both names on the assignment.
1. Consider the game represented by the following matrix:
Player A
U
C
D
L
2; 1
1; 2
2; 2
Player B
M
R
1; 1
0; 0
3; 3
2; 1
1; 0
1; 1
Find all pure and mixed strategy Nash equilibria for this game.
Answer:
There are two PSNE. They are: (1) Player A choose U, Player B choose L and (2) Player A choose C, Player B choose M.
There is one MSNE, though it will involve one player choosing a pure strategy. Looking at the matrix we see that strategy R
is strictly dominated by strategy M, so Player B will never use strategy R. Also, strategy L is weakly dominated by strategy
M, so the only possible way for Player A to make Player B indi¤erent among his pure strategies is to choose strategy U with
100% probability.
Now, Player B will play strategies L and M with positive probability and so will need to choose probabilities for strategies R
and M to make Player A indi¤erent among strategies U, C, and D. However, once strategy R is removed for Player B, Player
A’s payo¤s to U and D are the same, so Player A is always indi¤erent between U and D. Now let p be the probability that
Player B chooses L and (1 p) be the probability Player 2 chooses M. To make Player A indi¤erent we have:
E [U ] = E [C]
2p + 1 (1 p) = 1p + 3 (1 p)
2p + 1 p = p + 3 3p
3p = 2
2
p =
3
So if Player B chooses to play L with probability 23 and M with probability 31 then Player A will be indi¤erent among strategies
U, C, and D (Player A always receives 53 on average from any pure strategy). So the MSNE to this game is: Player A choose
U with 100% probability, Player B choose L with probability 32 and M with probability 13 .
2. Suppose that there are 2 hunters, Fred and Barney. The hunters can go after big game or small game. If a hunter goes after
small game then he catches small game for a payo¤ of 1. If he goes after big game and he hunts alone he fails to catch anything,
for a payo¤ of 0. However, if a hunter hunts for big game and both hunters are hunting big game, then they have a hunting
party and catch the big game for a payo¤ of 3 each.
a Write down the normal form version of this 2-player game.
Answer:
Barney
Fred
Small game
Big game
Small game
1; 1
0; 1
Big game
1; 0
3; 3
b What is (are) the pure strategy Nash equilibria (PSNE) of this game?
Answer:
Using the normal form of the game from part a and the method of best responses we can see that there are two PSNE to this
game,
Barney
Fred
Small game
Big game
Small game
1;1
0; 1
Big game
1; 0
3;3
(1) Fred hunt small game and Barney hunt small game and (2) Fred hunt big game and Barney hunt big game.
c Is there a mixed strategy Nash equilibrium to this game? If so …nd it.
Answer:
Since this is a coordination game (like Boxing-Opera), the answer is yes, there will be a MSNE. To …nd Barney’s probabilities
we need to set the expected value of Fred choosing to hunt small game or big game equal:
EF [Small]
1
1
3 small
small
=
=
=
=
EF [Big]
0
small + 3 (1
3 3 small
2
2
=
3
small )
Note that I didn’t use the probabilities for EF [Small] because Fred always receives 1 if he hunts small regardless of what
Barney does. For Fred’s probabilities we will go through the exact same process because Barney has identical payo¤s to Fred
(EB [Small] = 1 and EB [Big] = 0
small + 3 (1
small )), so we have the following MSNE:
Fred chooses to hunt small game with probability 23 and big game with probability 31 , while Barney chooses to hunt small game
with probability 23 and big game with probability 13 .
Now suppose that there are 3 hunters: Fred, Barney, and Bamm-Bamm. If each goes after small game then each gets a payo¤
of 1. If one hunter goes after big game and the other two go after small game, then the small game hunters receive a payo¤
of 1 and the big game hunter receives a payo¤ of 0. If 2 hunters hunt big game and 1 hunter hunts small game, the 2 hunters
who hunted big game each receive a payo¤ of 3, while the hunter who hunted small game receives a payo¤ of 5 (there is some
big game left over and the big game hunters share it with the small game hunter). If all 3 hunters hunt big game, then all 3
hunters receive a payo¤ of 3.
d What is (are) the PSNE of this game?
Answer:
There are 4 PSNE to this game. (1) All hunt small, (2) Fred and Barney hunt big, Bamm-Bamm hunts small, (3) Fred and
Bamm-Bamm hunt big, Barney hunts small, and (4) Barney and Bamm-Bamm hunt big, Fred hunts small. Essentially, if the
other two are hunting big then the third wants to hunt small to receive a higher payo¤ (basically the third person is wasted
on hunting big game). The last three equilibria are of this type. However, if the other two are hunting small then the best
response is to hunt small as the third hunter cannot catch big game alone. The normal form of the game (with the best
responses marked) is:
Fred
SG
BG
Barney
SG
BG
1; 1;1
1; 0; 1
0; 1; 1
3; 3;5
-Small Game
Fred
Big Game%
SG
BG
Barney
SG
BG
1; 1; 0 5; 3;3
3; 5;3 3; 3; 3
Bamm-Bamm
e Suppose that there are N > 3 hunters. If they all hunt small then they all receive a payo¤ of 1. If at least 2 of the hunters
hunt big game, then those who hunt big game receive a payo¤ of 3 and the ones who hunt small game receive a payo¤ of 5.
If only 1 hunter hunts big game then that hunter receives a payo¤ of 0 while the other hunters (who are all hunting small
game) receive a payo¤ of 1. Qualitatively describe (how many hunt big game, how many hunt small game) the PSNE for
the game with an unknown number of hunters, N .
Answer:
This is similar to part d in that there is always a PSNE where all hunters choose small (since none can deviate and catch big
game alone). There are also many PSNE where 2 players hunt big and N 2 hunt small. If one of the big hunters chooses to
hunt small then his payo¤ drops from 3 to 1. If any of the players who are choosing small hunt big they are again essentially
wasting e¤ort and their payo¤ would drop from 5 to 3.
3. Find all pure and mixed strategy Nash equilibria to the following game. If there are none of either type explain why there are
none:
Player 2
F
G
H
I
J
A
18 ; 11
9 ; 12
6; 1
8; 0
7; 1
B
6; 6
7; 8
5; 7
4; 5
9 ; 11
Player 1 C
9; 0
4; 5
14 ; 4
4; 10
5; 16
D
3; 4
3; 6
2; 3
6; 7
1; 9
8;6
E
0; 0
4; 2
7; 1
7; 4
Answer:
The best responses are marked in the matrix above. This leads to three PSNE:
(1) P1 choose A, P2 choose G; (2) P1 choose B, P2 choose I; (3) P1 choose E, P2 choose J.
What you should notice is that strategies F and H are both strictly dominated by G, and strategy D is strictly dominated by
A, so all three of those can be removed leaving:
Player 1
A
B
C
E
G
9 ; 12
7; 8
4; 5
4; 2
Player 2
I
8; 0
9 ; 11
4; 10
7; 4
J
7; 1
4; 5
5; 16
8;6
Now we can see that strategy C, once H has been removed, is strictly dominated by A so it too can be removed leaving:
Player 1
A
B
E
G
9 ; 12
7; 8
4; 2
Player 2
I
8; 0
9 ; 11
7; 4
J
7; 1
4; 5
8;6
Now let’s try to …nd an MSNE. Let g, i, and j = 1
J respectively. We need:
g
i be the probabilities that Player 2 uses to choose strategies G, I, and
E [A] = E [B] = E [E]
Focusing on:
E [A]
9g + 8i + 7 (1 g i)
9g + 8i + 7 7g 7i
2g + i + 7
3 4i
=
=
=
=
=
E [B]
7g + 9i + 4 (1 g i)
7g + 9i + 4 4g 4i
3g + 5i + 4
g
E [B]
7g + 9i + 4 (1 g i)
7g + 9i + 4 4g 4i
3g + 5i + 4
7g + 6i
=
=
=
=
=
E [E]
4g + 7i + 8 (1 g i)
4g + 7i + 8 8g 8i
4g i + 8
4
Now focusing on:
Substituting we have:
7 (3
21
7g + 6i
4i) + 6i
28i + 6i
17
=
=
=
=
i =
4
4
4
22i
17
22
Then we have:
3
4i
17
3 4
22
66 68
22 22
2
22
= g
= g
= g
= g
2
7
At this point it looks like there is a problem because we are going to end up with: g = 22
, i = 17
22 , and j = 22 . This is a
problem because g < 0. The question is, do these probabilities make player 1 indi¤erent among the pure strategies he plays
with positive probabilities?
E [A]
= 9
E [B]
= 7
E [E]
= 4
2
22
2
22
2
22
+8
+9
+7
17
+7
22
17
+4
22
17
+8
22
7
167
=
22
22
7
167
=
22
22
7
167
=
22
22
They do –but since one of the probabilities violates the laws of probability, there cannot be an MSNE in the 3x3 game ... using
all 3 strategies.1
But there are actually two MSNE using only 2 strategies each from the 3x3 game. Consider the following 2x2 game:
Player 1
A
B
G
9 ; 12
7; 8
Player 2
I
8; 0
9 ; 11
Now for player 1 we have:
E [A]
9g + 8 (1 g)
9g + 8 8g
g+8
3g
g
So Player 2 would use G with probability
1
3
=
=
=
=
=
E [B]
7g + 9 (1 g)
7g + 9 9g
9 2g
1
1
=
3
and I with probability 23 . Now let’s look at ALL of Player 1’s expected values:
E [A]
= 9
E [B]
= 7
E [C]
= 4
E [D]
= 3
E [E]
= 4
1
3
1
3
1
3
1
3
1
3
+8
+9
+4
+6
+7
2
3
2
3
2
3
2
3
2
3
=
=
=
=
=
25
3
25
3
12
3
15
3
18
3
Note that the strategies that Player 1 uses with positive probability, A and B, have an expected payo¤ of
uses the MSNE 0; 13 ; 0; 23 ; 0 , while the other strategies Player 1 could use have an expected payo¤ less than
1 For
the record, if you found player 1’s probabilities …rst you would have found that a =
having an expected value of 562
for each pure strategy he plays with positive probability.
113
18
,
113
b=
26
,
113
and e =
69
.
113
25
3 when Player 2
25
3 . Now we need
These should lead to player 2
to …nd Player 1’s probabilities using:
E [G]
12a + 8 (1 a)
12a + 8 8a
4a + 8
15a
=
=
=
=
=
E [I]
0a + 11 (1
11 11a
11 11a
3
3
1
a =
=
15
5
a)
So if Player 1 uses A with probability 15 and B with probability 45 this will make Player 2 indi¤erent over strategies G and I.
Can Player 2 deviate to another pure strategy and receive a strictly higher payo¤?
E [F ]
E [G]
E [H]
E [I]
E [J]
4
35
=
5
5
4
44
= 12
=
5
5
1
4
29
= 1
+7
=
5
5
5
1
4
44
= 0
+ 11
=
5
5
5
1
4
21
= 1
+5
=
5
5
5
1
+6
5
1
+8
5
= 11
The answer is no, so there is an MSNE where: Player 1 chooses A with probability
chooses G with probability 13 and I with probability 23 .
1
5
and B with probability
4
5
while Player 2
Now consider the 2x2 game:
Player 2
G
J
Player 1 A
9 ; 12
7; 1
E
4; 2
8;6
Setting:
E [A]
9g + 7 (1 g)
9g + 7 7g
2g + 7
6g
g
So if Player 2 uses G with probability
1
6
=
=
=
=
=
E [E]
4g + 8 (1 g)
4g + 8 8g
8 4g
1
1
=
6
and J with probability
then Player 1’s expected values (for all 5 pure strategies) are:
1
5
44
+7
=
6
6
6
1
5
27
7
+4
=
6
6
6
5
24
1
4
+5
=
6
6
6
1
5
8
3
+1
=
6
6
6
1
5
44
4
+8
=
6
6
6
Now to …nd Player 1’s probabilities:
E [A]
= 9
E [B]
=
E [C]
=
E [D]
=
E [E]
=
so Player 1 would not want to switch to either B, C, or E.
5
6
E [G]
12a + 2 (1 a)
12a + 2 2a
10a + 2
15a
=
=
=
=
=
E [J]
1a + 6 (1 a)
1a + 6 6a
6 5a
4
4
a =
15
4
15
So if Player 1 uses A with probability
and E with probability
E [F ]
E [G]
E [H]
E [I]
E [J]
11
15
then Player 2’s expected values are:
4
+0
15
4
= 12
+2
15
4
= 1
+1
15
4
= 0
+4
15
4
= 1
+6
15
= 11
11
44
=
15
15
11
70
=
15
15
11
15
=
15
15
11
44
=
15
15
11
70
=
15
15
so that Player 2 would not want to switch to either F, H, or I. Thus, we have a second MSNE:
Player 1 chooses A with probability
probability 65 .
4
15
and E with probability
11
15
while Player 2 chooses G with probability
1
6
and J with
Now, let’s consider the last remaining 2x2:
Player 1
B
E
Player 2
I
9 ; 11
7; 4
J
4; 5
8;6
Setting:
E [B]
9i + 4 (1 i)
9i + 4 4i
5i + 4
6i
=
=
=
=
=
E [E]
7i + 8 (1 i)
7i + 8 8i
8 i
4
2
i =
3
Looking at Player 1’s expected values:
E [A]
= 8
E [B]
= 9
E [C]
= 4
E [D]
= 6
E [E]
= 7
2
3
2
3
2
3
2
3
2
3
+7
+4
+5
+1
+8
1
3
1
3
1
3
1
3
1
3
=
=
=
=
=
23
3
22
3
18
3
13
3
22
3
Now, what do we notice here? We notice that E [A] > E [B] = E [E]. Thus, while Player 2’s choice of I with probability 23
and J with probability 31 makes Player 1 indi¤erent between B and E, it doesn’t really matter because Player 1 would choose
A. This is the reason we do not have an MSNE for the 3x3 game. Thus, there is no MSNE for this 2x2 so there are 5 total
Nash equilibria.
3 PSNE:
(1) P1 choose A, P2 choose G
(2) P1 choose B, P2 choose I
(3) P1 choose E, P2 choose J.
2 MSNE:
(4) Player 1 chooses A with probability
probability 23 .
1
5
and B with probability
4
5
while Player 2 chooses G with probability
1
3
and I with
(5) Player 1 chooses A with probability
probability 56 .
4
15
and E with probability
11
15
while Player 2 chooses G with probability
1
6
and J with
4. Consider a three player game where the three players all simultaneously choose whether to use Beta or VHS. If all three choose
Beta, then all three receive a payo¤ of 1. If all three choose VHS, then all three receive a payo¤ of 1. However, if there is any
player who chooses di¤erently than the other two then all three players receive 0. A normal form representation of the game
is here:
Player 2
Player 2
Beta
VHS
Beta
VHS
Player 1 Beta 2; 2; 2 0; 0; 0
Player 1 Beta
0; 0; 0 0; 0; 0
VHS 0; 0; 0 0; 0; 0
VHS
0; 0; 0 1; 1; 1
-Beta
VHS%
- Player 3 %
Note that Player 3 is essentially choosing which of the two matrices to play, the one on the left (if he chooses Beta) and the
one on the right (if he chooses VHS). The payo¤s are listed as: row player, column player, matrix choice player (or Player 1,
Player 2, and Player 3 in this example).
a Find both pure strategy Nash equilibria in this game.
Answer:
We can go through and mark o¤ the best responses in the matrix or we can just reason through this.
If all players choose VHS does any player wish to deviate? No, because then that player would go from receiving 1 to receiving
0. So all players choosing VHS is a Nash equilibrium.
If all players choose Beta does any player wish to deviate? No, because then that player would go from receiving 2 to receiving
0. So all players choosing Beta is a Nash equilibrium.
If 2 players choose Beta and 1 chooses VHS, does any player wish to deviate? Yes, the one who chooses VHS can switch to
Beta and increase his payo¤ from 0 to 2. Note that the 2 players who choose Beta do NOT wish to deviate as if either one
switches from Beta to VHS that player’s payo¤ is still 0. A similar argument can be made for 2 players choosing VHS and 1
choosing Beta.
So the pure strategy Nash equilibria are: (1) Player’s 1, 2, and 3 all choose VHS and (2) Player’s 1, 2, and 3 all choose Beta.
b Find the mixed strategy Nash equilibrium to this game.
Answer:
In order to …nd the MSNE to this game, you follow the same process as in a 2 player game by setting the expected values of
each player’s strategies equal to each other. I’m going to use the following notation. Let 1V be the probability that Player 1
chooses VHS (so the probability he chooses Beta is given by 1
1V ). Let 2V and 1
2V be the probabilities that Player
2 chooses VHS and Beta, respectively, and let 3V and 1
3V be the probabilities that Player 3 chooses VHS and Beta,
respectively.
What is Player 1’s expected value of choosing VHS? He receives 1 if both Players 2 and 3 choose VHS, and 0 otherwise. The
probability that both Players 2 and 3 choose VHS is given by 2V
3V . So:
E1 [V HS] = 1
2V
3V
What is Player 1’s expected value of choosing Beta? He receives 2 if both Players 2 and 3 choose Beta, and 0 otherwise. The
probability that both Players 2 and 3 choose Beta is given by (1
2V ) (1
3V ). So:
E1 [Beta] = 2 (1
Setting these equal and solving for
3V
1
2V
2V
2V
2
2
2
2V
2V
2V
2V
3V
in terms of
2V
.
) (1
3V
)
we have:
E1 [V HS]
Now we know
2V
=
=
3V
=
3V
=
3V
0 =
2 =
2 =
2
=
2
E1 [Beta]
2 (1
2V ) (1
3V )
2 (1
2V
3V + 2V 3V )
2 2 2V 2 3V + 2 2V 3V
2 2 2V 2 3V + 2V 3V
2 3V
2V 3V
2)
3V ( 2V
3V
Now let’s work on setting E2 [V HS] = E2 [Beta]. If Player 2 chooses VHS he receives 1 with probability
chooses Beta he receives 2 with probability (1
1V ) (1
3V ). So:
E2 [V HS] =
1 1V 3V =
=
1V 3V
=
1V 3V
0 =
But we know
3V
in terms of
2V
0
E2 [Beta]
2 (1
1V ) (1
2 (1
1V
2 2 1V 2
2 2 1V 2
3V
3V
3V
3V
1V
3V
and if he
)
+ 1V
+ 2 1V
+ 1V
3V
)
3V
3V
so substituting in we have:
=
2
2
2
1V
2
2V
2V
2
2
2
+
2
2
2V
2 (2 2V 2) + 1V (2 2V
4 2V + 4 + 2 1V 2V 2
2V
1V
0 = 2 ( 2V 2) 2 1V ( 2V 2)
0 = 2 2V 4 2 1V 2V + 4 1V
0 = 2 1V 2 2V
=
2V
1V
2)
1V
So that is a nice result, that 2V = 1V . It will make things easier. Now let’s work on setting E3 [V HS] = E3 [Beta]. If Player
3 chooses VHS he receives 1 with probability 1V 2V and if he chooses Beta he receives 2 with probability (1
1V ) (1
2V ).
So:
E3 [V HS] =
1 1V 2V =
=
1V 3V
=
1V 2V
0 =
But we know that
1V
=
2V
E3 [Beta]
2 (1
1V ) (1
2 (1
1V
2 2 1V 2
2 2 1V 2
2V
2V
2V
2V
)
+ 1V
+ 2 1V
+ 1V
2V
)
2V
2V
, so:
0 = 2
0 = 2
0 = (
2 1V 2 2V +
2 1V 2 1V +
2
4 1V + 2
1V )
1V
2V
1V
1V
Note that you can’t FOIL this to …nd the solution, but you can use the quadratic formula. So we would have:
p
b
b2 4ac
=
1V
2a q
1V
=
4
1V
1V
1V
1V
So
1V
= 2+
p
2 or
1V
=2
p
2. Since 2 +
p
2
( 4)
( 4)
p
2 1
8
16
=
p2
4
8
=
2 p
4 2 2
=
2p
= 2
2
4 1 2
2 > 1 this cannot be a solution, which leaves
1V
=2
p
2. We already know
that
1V
=
2V
so
2V
=2
p
2 as well. As for
3V
2
, we know that
2V
2V
2
=
2
2 =
3V
2 2V
2)
3V ( 2V
p
p
2 2
2
2 =
2 2
3V 2
p
p
2
4 2 2 2 =
3V
p
p
2 2 2 =
2
3V
p
2 2 2
p
=
3V
2
23=2 21
=
3V
21=2
21 21=2 =
3V
p
2 =
2
3V
p
p
So all three players choose VHS with probability 2
2 and all three players will choose Beta with probability 1
2
2
p
p
2 1. While this is the correct MSNE, there is a problem with implementing it since 2
2 is not a rational number.
=
c Suppose now there are four players, and the payo¤s are such that if all 4 coordinate on Beta then they all receive 0, if all 4
coordinate on VHS then they all receive 2, and if there is any miscoordination then all players receive 0. What are the
pure strategy Nash equilibria in the four player game? Do they di¤er from those in the three player game (and if they
di¤er how do they di¤er)?
Answer:
There are more than just two pure strategy Nash equilibria now. The outcomes where all players choose VHS and all players
choose Beta are still Nash equilibria to this game (no one can unilaterally deviate and increase his or her payo¤). And the
case where 3 players choose VHS and 1 chooses Beta (or 3 choose Beta and 1 chooses VHS) are still NOT equilibria because
the one person who is choosing the opposite of the other three can switch and receive a higher payo¤. However, if 2 players
choose VHS and 2 players choose Beta, this is also an equilibrium to the game. Notice that no single player can switch his or
her strategy and receive a strictly higher payo¤, so this is a Nash equilibrium to the game. There are 6 such occurrences (1
and 2 choose Beta while 3 and 4 choose VHS; 1 and 3 choose Beta while 2 and 4 choose VHS; 1 and 4 choose Beta while 2 and
3 choose VHS; 2 and 3 choose Beta while 1 and 4 choose VHS; 2 and 4 choose Beta while 1 and 3 choose VHS; and 3 and 4
choose Beta while 1 and 2 choose VHS).
So the PSNE do di¤er between the 3 and 4 person games as in the 4 person game there are PSNE where miscoordination occurs,
while in the 3 player game there are no PSNE with miscoordination.
5. There are 5 players each with 5 tokens. The players will play a simultaneous game. Players can either keep the tokens for
themselves or contribute tokens to the pot. Any tokens kept by the player pay the player who kept the tokens, and only that
player, $1 each. So if a player keeps 4 tokens he receives $4, but the other players receive nothing from those 4 tokens. The
tokens in the pot are counted and every player in the group receives $0:5 for each token in the pot. So, if there are 6 tokens in
the pot then each player receives a payout of $3 from the pot, regardless of how many he or she contributed. Thus the player’s
total payo¤ is $1 for every token kept plus $0:5 times every token in the pot.
a Do the players have a strictly or weakly dominant strategy? Is so, what is the strictly or weakly dominant strategy? If not,
explain why not.
Answer:
Yes, the strategy of keeping all the tokens is a strictly dominant strategy. No matter what the other players do keeping all the
tokens is the best response. Consider Player 1’s decision –what matters to him is the total number of tokens contributed, not
who contributed. So there are 21 potential "states of the world", where there are between 0 and 20 tokens contributed.
The table below shows Player 1’s payo¤, and only Player 1’s payo¤, for each level of his own contribution given the total number
of tokens contributed by the other player:
P1#
0
1
2
3
4
5
Total contributions by other players
8
9
10 11
12
13
0
1
2
3
4
5
6
7
5
4.5
4
3.5
3
2.5
5.5
5
4.5
4
3.5
3
6
5.5
5
4.5
4
3.5
6.5
6
5.5
5
4.5
4
7
6.5
6
5.5
5
4.5
7.5
7
6.5
6
5.5
5
8
7.5
7
6.5
6
5.5
8.5
8
7.5
7
6.5
6
9
8.5
8
7.5
7
6.5
9.5
9
8.5
8
7.5
7
10
9.5
9
8.5
8
7.5
10.5
10
9.5
9
8.5
8
11
10.5
10
9.5
9
8.5
11.5
11
10.5
10
9.5
9
14
15
16
17
18
19
20
12
11.5
11
10.5
10
9.5
12.5
12
11.5
11
10.5
10
13
12.5
12
11.5
11
10.5
13.5
13
12.5
12
11.5
11
14
13.5
13
12.5
12
11.5
14.5
14
13.5
13
12.5
12
15
14.5
14
13.5
13
12.5
If the other 4 players contribute all 20 tokens then Player 1 will receive $15 if he contributes 0 tokens ($5 from keeping his own
tokens and $10 from the pot) and something less than $15 if he contributes any of his tokens. The same is true if Player 1
faces any other combination of tokens contributed by others –Player 1 always does better by keeping his tokens. Essentially,
each token contributed by a player will reduce his payo¤ by $0:5 , holding the number of tokens contributed by other player’s
constant.
b Find a pure strategy Nash Equilibrium to this game.
Answer:
The players all have a strictly dominant strategy of contributing 0, thus the only PSNE (and only NE) is for all players to
contribute zero. Note that this game is essentially a 5 player Prisoner’s Dilemma – all players have a strictly dominant
strategy (choosing to contribute 0 in this game, choosing Confess in the Prisoner’s Dilemma), but all would be better o¤ if
they cooperated (if everyone chose contribute all tokens in this game then all players would receive 12.5, rather than the 5 they
receive from playing the equilibrium strategies).
Now, consider the case of those 5 people, each with their 5 tokens, attempting to privately …nance a library. The pot can now
be thought of as the library fund. If the total contribution to the pot is greater than or equal to 10 tokens, then the library is
funded and if there are less than 10 tokens then the library is not funded. If the library is funded then each player receives a
bene…t of $15 from the library, regardless of how many tokens that player contributed. In addition to that $15, any remaining
tokens that they have also add $1 to their bene…t. So if the library is funded and a player kept 5 tokens, his payo¤ would be
$20. If the library is not funded then players only receive a bene…t equal to the amount of tokens that they kept.
c Find 2 pure strategy Nash equilibria to this game –one where the library is funded and another where it is not.
Answer:
The NE where the library is not funded is the same as the one in part b – all players keep all tokens. Notice that no player
can fund the library by herself, so if no one else contributes tokens the best response is to not contribute as well.
There are many NE where the library is funded. The key is that exactly 10 tokens are contributed. The distribution of the
10 tokens does not matter (for the determination of the NE – it does matter for the individual players). One NE is that all
players contribute exactly 2 tokens. In this case, all players receive a payo¤ of $18; they get $15 from the library being funded
and $3 from the 3 tokens they kept. Another NE would be one player contributes 5 tokens, a second contributes 4, and a third
contributes 1. In this case, the …rst player receives $15, the second $16, the third player $19, and the two players who did not
contribute receive $20. However, no player can deviate from this set of strategies and become better o¤, as the players who
contributed do not want to take back their contribution because then the library would not be funded and they would receive
a lower payo¤.
Notice that exactly 10 tokens need to be contributed –if there are more than 10 tokens then players could contribute less and
receive a larger payo¤ because the library would still be funded.