Game Theory
Solutions to Homework 1
For problems (1)-(3), the LPs are given on pages 3 and 4 of the Lecture 2 notes on the course home
page. The solutions are given below.
1. x = (4/7, 3/7, 0)0 , v = 22/7, y = (5/7, 0, 2/7, 0)0 .
2. x = (4/11, 50/143, 41/143)0 , v = 96/143, y = (4/13, 42/143, 57/143)0 .
3. x = (6/37, 20/37, 0, 11/37)0 , v = 121/37, y = (14/37, 4/37, 0, 19/37, 0)0 .
4. Since A = −AT , we have
xT Ay = y T AT x = −y T Ax.
Now, if x = y, then xT Ax = −xT Ax, which implies that xT Ax = 0. Thus, v̄ ≥ 0, and v ≤ 0, and
therefore, v = 0. Further, if Player 1’s optimal strategy is x, we have just showed that the value of
the game will be zero if Player 2 also plays x. This, if x is a saddle-point strategy for one of them,
then it is also a saddle-point strategy for the other.
5. In a 2 × 2, it is not difficult to see that either there exists a pure-strategy saddle-point or all
components of the mixed strategy probability distribution are non-zero for both players. Thus, if
Player 1 uses strategy x, then all components of xT A must be equal to v, the value of the game (if
one of the components is larger than the other, then Player 2 will only use that strategy and will
not randomize over its two choices). Thus,
xT A = 1 T v
Multiplying both sides by A∗ and using the fact that AA∗ = det(A)I, where I is the identity matrix,
we have
xT det(A) = 1T A∗ v.
Further, since x is a probability distribution,
1T x = 1.
Thus,
v=
det(A)
,
1T A ∗ 1
and the rest of the results follow.
6. The probabilities converge; see figures.
1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
100
200
300
400
500
600
700
800
900
1000
Figure 1: Maximizer’s probabilities
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
100
200
300
400
500
600
700
800
900
Figure 2: Minimizer’s probabilities
2
1000