14.12 Game Theory — Midterm I
10/19/2010
Prof. Muhamet Yildiz
Instructions.
This is an open book exam; you can use any written material. You have one
hour and 20 minutes. Each question is 25 points. Good luck!
1. Consider the following game.
(a) Using backward induction
fi
nd an equilibrium.
Answer:
At the last node, Player 1 chooses
; at the right node, Player 2 then
chooses
. At the left node, she chooses
. Hence, at the beginning, Player 1
chooses
. The equilibrium is
(
)
.
(b) Write the game in normal form.
Answer:
The game in normal form is
1
\
2
2,1
2,1
1,2
1,2
2,1
2,1
1,2
1,2
2,1
1,0
2,1
1,0
2,1
0,1
2,1
0,1
2. Compute a Nash equilibrium of the following game. (This is a version of RockScissors
Paper with preference for Paper.)
R
S
P
R
0
0
2
−
2
−
2
3
S
−
2
2
0
0
2
−
1
P
3
1
−
1
2
1
1
Answer:
(Because of a typo, the question became asymmetric, making the answer
longer. We gave nearly full credit to the students who thought the game was symmet
ric.)
1
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This game has a unique Nash equilibrium, which is in mixed strategies. Write
,
,
and
for the probabilities with which player
plays strategies R, S, and P, respectively.
Of course,
+
+
= 1
(1)
Now,
(
2
2
2
)
must make Player 1 indi
ff
erent between his strategies.
Indi
ff
erence
between R and S yields
2 (
2
−
2
) = 2 (
2
−
2
)
, i.e.,
2
2
=
2
+
2
Since
2
+
2
= 1
−
2
by (1), this yields
2
= 1
3
On the other hand, the indi
ff
erence between S and P yields
2 (
2
−
2
) = 1+2 (
2
−
2
)
.
Substituting
2
= 1
3
, we obtain
4
2
−
2
2
=
−
1
3
. Together with
2
+
2
= 2
3
, this
yields
2
=
1
6
2
=
1
2
Similarly, indi
ff
erence between R and S for player 2 yields
2 (
1
−
1
) = 2
1
−
1
, i.e.,
3
1
= 2 (
1
+
1
)
. Since
1
+
1
= 1
−
1
, this yields
1
= 2
5
The indi
ff
erence between S and P yields
2 (
1
−
1
) = 1+2 (
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