Answers to Problem Set 1
Economics 703
Spring 2016
1. For player 1, a2 strictly dominates a6 . No other strategy dominates (strictly or
otherwise) any other for either player. After eliminating a6 , b4 strictly dominates b2
for player 2. Again, no other dominance is present. After eliminating b2 , a3 strictly
dominates a2 for player 1. After eliminating a2 , b3 strictly dominates b4 for player 2.
After eliminating b4 , a5 strictly dominates a3 and a4 for 1. After eliminating a3 and a4 ,
b5 strictly dominates b3 . At the next step, a5 strictly dominates a1 . Hence b5 ends up
strictly dominating b1 . So this game is dominance–solvable and (a5 , b5 ) is the solution.
Hence this is the unique Nash equilibrium of the game.
2. (a) Let si denote the action of player i — that is, the quantity of his endowment he
hands to the other player. Then Si = [0, 1] for i = 1, 2. The payoff function for 1 is
u1 (s1 , s2 ) = min{1 − s1 , s2 }, while the payoff function for 2 is u2 (s1 , s2 ) = min{s1 , 1 − s2 }.
(b) This game has an enormous number of pure strategy Nash equilibria. To see this,
notice that the best reply for player 1 to some s2 is any s1 ≤ 1 − s2 . Reason: If the
amount 2 gives him is s2 , then his payoff is min{1 − s1 , s2 }. As long as 1 − s1 > s2 ,
the minimum is s2 and so increasing or decreasing s1 (i.e., giving the other player more
or less) does not affect 1’s utility. Certainly giving so much that 1 − s1 < s2 is not
optimal, since giving less would make player 1 strictly better off. Hence any s1 such that
1 − s1 ≥ s2 is optimal — that is, any s1 ≤ 1 − s2 . A similar argument shows that any
s2 ≤ 1 − s1 is a best reply for player 2 to s1 . But notice both inequalities say the same
thing: s1 + s2 ≤ 1. Hence the set of equilibria is the set of (s1 , s2 ) such that s1 + s2 ≤ 1.
Notice that there are some odd equilibria — such as s1 = 0 and s2 = 1.
(c) This game has a dominant strategy equilibrium: s1 = s2 = 0. Clearly, keeping
everything never hurts you and, if the opponent gives you everything, it’s the best strategy
for you.
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3. The normal form is
a
b
c
a 3, 2, 1 3, 2, 1 3, 2, 1
b 3, 2, 1 2, 1, 3 2, 1, 3
c 3, 2, 1 1, 3, 2 1, 3, 2
a
a
b
c
a 3, 2, 1 2, 1, 3 3, 2, 1
b 2, 1, 3 2, 1, 3 2, 1, 3
c 1, 3, 2 2, 1, 3 1, 3, 2
b
a
b
c
a 3, 2, 1 3, 2, 1 1, 3, 2
b 2, 1, 3 2, 1, 3 1, 3, 2
c 1, 3, 2 1, 3, 2 1, 3, 2
c
The pure strategy Nash equilibria are (a, a, a), (b, b, b), (c, c, c), (a, a, b), and (a, c, c).
The first round of elimination: For 1, voting for a dominates voting for b or c. For 2,
voting for c (his favorite outcome) dominates voting for b (his least favorite). However,
neither c nor a dominates the other. For example, if 1 votes b and 3 votes a, 2 prefers to
vote a. If 1 votes c and 3 votes a, 2 prefers c. Similarly, for 3, voting for b (his favorite)
dominates voting for a (his least favorite). However, neither b nor c dominates the other.
So we eliminate b and c for 1, c for 2, and a for 3.
On the second round of elimination, c dominates a for 2 and c dominates b for 3.
Hence we are left with (a, c, c) as the votes with c winning. Ironically, the outcome is the
worst outcome for player 1, the player who seems to have the most power in the voting.
4. (a) If si is a strictly dominant strategy for i, then it is necessarily i’s unique best
reply to any s∼i . Hence if every player has a strictly dominant strategy, the only Nash
equilibrium consists of the vector of these strategies. If si is a dominant strategy, it is a
best reply to any s−i . Hence if every player has a dominant strategy, the vector of these
actions must be a Nash equilibrium.
(b) Suppose that s∗ is the unique rationalizable strategy profile in the game. We have to
show two things: s∗ is a Nash equilibrium and there is no other Nash equilibrium. The
first part: Suppose not. Then there is some player who would gain by deviating. That
is, there must be some i and si ∈ Si with ui (si , s∗−i ) > ui (s∗ ). But si was eliminated by
successive elimination of strictly dominated strategies while s∗i was not. Suppose that
when si was eliminated, we had reduced the strategies of the other players to Ŝ−i . Then
there was some other action for i, say ŝi such that ui (ŝi , s−i ) > ui (si , s−i ) for all s−i ∈ Ŝ−i .
But since s∗−i never got eliminated, it must have been in Ŝ−i . So ui (ŝi , s∗−i ) > ui (si , s∗−i ) >
ui (s∗ ). Hence ŝi can’t be s∗i . Similarly, though, ŝi was eliminated at some point, so it was
strictly dominated by something else. Applying the same argument shows that whatever
eliminated it also could not have been s∗i or si since whatever eliminated ŝi must be
strictly better than s∗i or si against s∗−i . We could continue this to generate an arbitrarily
large collection of actions, each of which is better against s∗−i than all the preceding ones.
But the game is finite, so we have to run out of actions for i at some point, making this
impossible.
Now let’s show that there is no other Nash equilibrium. Suppose â is a different
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vector of strategies and is a Nash equilibrium. Of the set of i with ŝi 6= s∗i , fix any
player j whose strategy is eliminated earliest. (There may be more than one agent whose
strategy is eliminated in the same round before any of the other players had a strategy
eliminated. This lack of uniqueness doesn’t affect the argument to follow.) Since â is a
Nash equilibrium, we know that uj (ŝj , ŝ−j ) ≥ uj (sj , ŝ−j ) for all sj ∈ Sj . However, at the
round where ŝj was eliminated, none of the actions in ŝ−j had been eliminated. Hence
there could not have been some other action for j, say sj , with uj (sj , s−j ) > uj (ŝj , s−j )
for all the s−j ’s that had not yet been eliminated. This is a contradiction.
(c) Suppose G is dominance solvable with solution s∗ . An argument very similar to
the one in the first part of (b) establishes that this must be a Nash equilibrium. More
specifically, suppose s∗ is not a Nash equilibrium. Then there is some player who would
gain by deviating. That is, there must be some i and si ∈ Si with ui (si , s∗−i ) > ui (s∗ ).
But si was eliminated by successive elimination of dominated strategies while s∗i was
not. Suppose that when si was eliminated, we had reduced the strategies of the other
players to Ŝ−i . Then there was some other action for i, say ŝi such that ui (ŝi , s−i ) ≥
ui (si , s−i ) for all s−i ∈ Ŝ−i . But since s∗−i never got eliminated, it must have been in
Ŝ−i . So ui (ŝi , s∗−i ) ≥ ui (si , s∗−i ) > ui (s∗ ). Hence ŝi can’t be s∗i . Similarly, though, ŝi was
eliminated at some point, so it was dominated by something else. It is not hard to see
that if ŝi dominated si , the reverse could not be true. Thus whatever eliminated ŝi could
not have been s∗i or si . We could continue this to generate an arbitrarily large collection
of actions, each of which is better against s∗−i than all the preceding ones. But the game
is finite, so we have to run out of actions for i at some point, making this impossible.
A simple example showing that a dominance solvable game can have multiple Nash
equilibria:
b1
b2
a1
1,1
0,0
a2
0,0
0,0
This game is dominance solvable with solution (a1 , b1 ). However, (a2 , b2 ) is also a Nash
equilibrium.
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