Nash equilibria of bimatrix games
0 6
A= 2 5
3 3
B=
2
1
4
Nash equilibrium =
pair of strategies x , y with
x best response to y and
y best response to x.
1
3
3
Mixed equilibria
0 6
A= 2 5
3 3
2/3
x = 1/3
0
4
Ay = 4
3
B=
2
1
4
1
3
3
xTB = 5/3 5/3
yT = 1/3 2/3
only pure best responses can have probability > 0
Best response condition
Let x and y be mixed strategies of player I and II, respectively.
Then x is a best response to y
⇐⇒ for all pure strategies i of player I:
xi > 0 =⇒ (Ay)i = u = max{(Ay)k | 1 ≤ k ≤ m}.
Here, (Ay)i is the i th component of Ay , which is the expected
payoff to player I when playing row i .
Proof.
m
m
xAy = ∑ xi (Ay)i = ∑ xi (u − (u − (Ay)i)
i=1
m
i=1
m
m
= ∑ xi u − ∑ xi (u − (Ay)i) = u − ∑ xi (u − (Ay)i) ≤ u,
i=1
i=1
i=1
because xi ≥ 0 and u − (Ay)i ≥ 0 for all i . Furthermore,
xAy = u ⇐⇒ xi > 0 implies (Ay)i = u , as claimed.
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1,0
0,1
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1,0
0,1
Best responses to mixed strategy of player 2
6
1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
0
1,0
0,1
Best responses to mixed strategy of player 2
5
2
2
1,0
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
3
3
1,0
3
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
6
1
3
3
5
2
3
2
0
1,0
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
1
3
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
best response polyhedron
1,0
0,1
Best responses to mixed strategy of player 2
5
4
1
3
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
best response polyhedron
with facet labels
1,0
0,1
Best responses to mixed strategy of player 2
5
4
1
3
5
3
2
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1
4
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
5
3
2
1
4
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
1
0
0
0
0
1
1
0
1
0
Best responses to mixed strategy of player 1
4
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
1
0
0
0
0
1
1
0
1
0
Best responses to mixed strategy of player 1
3
3
1
1
0
0
0
0
1
0
1
0
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
Best responses to mixed strategy of player 1
3
3
1
1
0
0
0
0
1
0
1
0
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
1
2
4
5
0
0
1
1
0
0
3
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
best response
polyhedron
with facet labels
(front)
0
1
0
Alternative view
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
4
5
3
1
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Constructing games using geometry
low dimension: 2, 3, (4) pure strategies:
subdivide mixed strategy simplex into
response regions, label suitably
high dimension:
use polytopes with known combinatorial structure
e.g. for constructing games with many equilibria,
or long Lemke-Howson computations
[Savani & von Stengel, FOCS 2004,
Econometrica 2006]
Construct isolated non−quasi−strict equilibrium
5
5
4
3
2
1
2
4
6
1
6
3
Construct isolated non−quasi−strict equilibrium
5
5
4
3
2
1
2
4
6
1
3
6
4
0 2 0
A= 2 0 0
1 1 1
1
2
3
5
6
1 0 0
B= 1 0 2
1 1 0
The Lemke−Howson algorithm
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
0
0
0
artificial equilibrium
0,0
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
0
0
0
artificial equilibrium
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
found label
4
5
3
2
Why Lemke-Howson works
LH finds at least one Nash equilibrium because
•
finitely many "vertices"
for nondegenerate (generic) games:
•
unique starting edge given missing label
•
unique continuation
precludes "coming back" like here:
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
Odd number of Nash equilibria!
start at Nash equilibrium
5
4
3
2
2
1
1
found label
4
5
3
2
Nondegenerate bimatrix games
Given:
m × n bimatrix game (A,B)
X = { x ∈ Rm | x ≥ 0, x1 + . . . + xm = 1 }
Y = { y ∈ Rn | y ≥ 0, y1 + . . . + yn = 1 }
supp(x) = { i | xi > 0 }
supp(y) = { j | yj > 0 }
(A,B) nondegenerate
⇔ ∀ x ∈X,
y ∈Y:
| { j | j best response to x } | ≤ | supp(x) |,
| { i | i best response to y } | ≤ | supp(y) |.
Nondegeneracy via labels
m × n bimatrix game (A,B) nondegenerate
⇔
no x  X has more than m labels,
no y  Y has more than n labels.
E.g. x with > m labels,
s labels from { 1 , . . . , m } ,
⇒ > m−s labels from { m+1 , . . . , m+n }
⇔ > |supp(x)| best responses to x.
⇒ degenerate.
Example of a degenerate game
4
5
1
2
1
2
1
3
3
2
1
4
5
3
4 4
=B
payoffs to
player II
Handling degenerate games
Lemke–Howson implemented by pivoting, i.e., changing from
one basic feasible solution of a linear system to another by choosing an entering and a leaving variable.
Choice of entering variable via complementarity (only difference
to simplex algorithm for linear programming).
Leaving variable is unique in nondegenerate games.
In degenerate games: perturb system by adding (ε, . . . , εn)> ,
creates nondegenerate system.
Implemented symbolically by lexicographic rule.