Guillermo Sabbioni
Game Theory (ECO4400)
Book: Games of Strategy, by Dixit & Skeath
Chapter 7: Simultaneous-move games with mixed strategies: zero-sum games.
Randomization of strategies: extension of the equilibrium and strategy concepts
(especially) for games with no pure strategies equilibrium.
Its need arises when one player prefers coincidence of actions and the other does not
(direct conflict: zero-sum games for ex).
1. What is a mixed strategy?
Mixed strategy: it is a mixture between pure strategies. Choose one pure strategy with
some probability. Tennis point example: choose DL with probability p (as a result, she is
choosing CC with probability 1-p).
Payoffs are defined as the corresponding probability-weighted averages of the payoffs
from its constituent pure strategies. Example: if Evert picks (75% DL, 25% CC), she gets
0.75*50+0.25*90=60 against Navratilova’s DL.
That is her expected payoff. The NE concepts applies here: it is a list of (mixed)
strategies that gives each player her highest payoff given that the others are playing the
(mixed) strategies specified by the NE list.
Back to the continuous analysis, to be shown graphing BR. Nash showed that there is
(“almost”) always one equilibrium in mixed strategies, even if not in pure strategies.
2. Mixing moves to keep the opponent guessing
Back to the tennis point example, with no equilibrium in pure strategies.
a. Best response analysis
BR analysis is possible, like in continuous games (p is continuous). Evert chooses DL
with prob p and CC with prob 1-p. If p=0 or 1, back to pure strategies.
For Navratilova: DL with prob q and CC with prob 1-q. See fig 7.1. A new (mixed)
strategy is available for both players. Easy to calculate payoffs when one mixes against a
pure strategy of the other.
To find the equilibrium: use BR. Evert chooses p as a response to q from Navratilova.
Navratilova picks the best q as a BR to p from Evert.
See fig 7.2. This shows how each pure strategy for Navratilova performs against a p-mix
from Evert. Setting equal the expected payoffs for both pure strategies gives p=0.7. So if
p=0.7, Navratilova gets the same payoff (38) by using DL or CC.
For example, if p>0.7 (Evert picks DL with high prob), Navratilova is better off using
DL.
Formal BR for Navratilova:
 If p<0.7, choose pure CC (q=0)
 If p=0.7, choose anything (q=0 or q=1 or 0<q<1)
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Guillermo Sabbioni
 If p>0.7, choose pure DL (q=1).
Another way: CC is better than DL if 20p+80(1-p) > 50p+10(1-p); result: p<0.7.
Since this is a zero-sum game, we could also find the BR of Navratilova by using Evert’s
payoffs. See fig 7.3.
Now find Evert’s BR: how each of her pure strategies performs against a q-mix from
Navratilova. One way (not in the book):
Her DL does better if 50q+80(1-q) > 90q+20(1-q).
Doing algebra: if 50q+80-80q > 90q+20-20q
Or -30q+80 > 70q+20
Then 60 > 100q so… DL is better for Evert if q<0.6. Only need to do this.
Then, Evert’s BR is:
 If q<0.6, choose DL (p=1).
 If q=0.6, choose any p
 If q>0.6, choose CC (p=0).
Both BR can be graphed: see fig 7.4. Notice the “indifference” segments of both BR.
Reflect one of them and find the intersection (unique). That is the NE in mixed strategies:
(p=0.7, q=0.6).
For almost all values of p and q, the BR of the other is a pure strategy. Intersection: when
both players are at the “indifference” portion of their BR. Each player’s equilibrium
mixture is such that the rival is indifferent among all her mixes (opponent’s indifference
property).
BR graphs are complete: they can show all the equilibria (even in pure strategies). Also
notice: no need to graph. Algebraically we had found the equilibrium at the indifference
points of p and q.
b. Minimax method
In zero-sum games, players mix to prevent the opponent from taking advantage of
predictability of moves. See fig 7.5. Maximin and minimax payoffs are not the same: no
equilibrium in pure strategies.
See fig 7.6, incorporating mixed strategies. What is the minimum for the p-mix row for
Evert? It depends on p. See fig 7.7 showing this. At p=0.7 both expected payoffs for
Evert are equal (62). If Evert does not pick p=0.7, Navratilova exploits this. Navratilova
picks CC if p<0.7 (lower segment in fig 7.7).
The inverted V shows the maximin function: max at p=0.7. The vertex depicts (for Evert)
the max of all her minima (for diff p that she can choose). She gets 62, better than 50 and
20 (from pure strategies). For Navratilova: minimizing her maxima also yields 62
(coincidence) at the optimum: q=0.6. Recall: only valid for zero-sum games.
3. NE as a system of beliefs and responses
Recall about pure strategies: in a NE each player responds to a belief, and those beliefs
are correct. In mixed strategies: you cannot be sure about what the other chooses, but you
can have correct beliefs about the exact mixture the rival is playing.
Then: in a NE each player forms (correct) beliefs about the mixtures the others are
choosing and responds to that.
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Guillermo Sabbioni
4. Mixing when one player has 3 or more strategies
We consider only zero-sum games here. Typically, if a player has more than 2 strategies
but the other has two, the one with many strategies uses only 2 in equilibrium. Which
ones? Add one strategy for Evert, for example.
a. General case
See fig 7.8. Evert’s p-mix is defined: p1 is the prob of DL, p2 the prob of CC and 1-p1-p2
is the prob of Lob. How to solve?
See fig 7.9: Evert’s payoffs from playing each of her pure strategies, as a function of q
(by Navratilova). Then, Navratilova wants to minimize that function. The function shows
how Evert would respond to the q Navratilova picks.
Find the kinks: q=0.5 makes payoffs from DL and Lob equal, and q=2/3 makes payoffs
from Lob and CC equal.
Evert’s BR:
 DL if q<0.5
 Lob when 0.5<q<2/3
 CC when q>2/3.
Optimum for Navratilova (since zero-sum game) is the minimum of this curve (Evert’s
payoff): choose q=0.5. Evert gets 65 (hence Navratilova 35). At q=0.5, Evert is indiff
between DL and Lob. Hence, she does not use CC at all (p2=0). Now solve as a 2x2 and
find p1=prob(DL)=0.25 and prob(Lob)=0.75.
How is it done? (not in the book):
(DL for Navr) 50p1+70(1-p1) = 80p1+60(1-p1) (CC for Navr)
-20p1+70 = 20p1+60
10 = 40p1 so p1=1/4.
Evert’s maximin (65) is the same as Navratilova’s minimax. Usually, this method works
(3 lines that depict a “curve” as figure 7.9)
b. Exceptional cases
If the payoffs are “weird” we can get special configurations. If both payoffs for Evert
from Lob are 70, the Lob line is horizontal. Both kinks: q=1/3 and q=5/7. Then, not a
unique q for Navratilova.
If the 3 lines intersect at one point: Evert is willing to mix among all three pure strategies
(if she wants) when Navratilova chooses q=0.6 (in the example of the book; not graphed
in PPT). Indeterminate equilibrium.
c. Domination by a mixed strategy
What if new payoffs are like fig 7.10? See fig 7.11 for the graph. Lob lies everywhere
below the “upper envelope”. For q<0.6 Evert would pick DL and CC if q>0.6 (never
Lob, even when Lob is not dominated).
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Guillermo Sabbioni
Lob is never a BR. Lob is in fact dominated by a (right) mixture of DL and CC. Mix line:
average of the heights of DL and CC (weighted using p and 1-p). The line rotates through
the intersection point.
Right mix: p=0.25 (you take a look how to calculate it: page 206). Recall the elimination
of the “never a BR” set of strategies? In 2 player games, it is the same as “dominated by a
mixed strategy”.
Expanding to mixed strategies: We can do iterated elimination of strategies (pure or
mixed) that are dominated by other strategies (pure or mixed). Remaining set: set of
rationalizable strategies.
5. Mixing when both players have 3 strategies
We now need 2 variable algebra (each mixed strategy is defined by 2 probabilities; for
the third one… use 1 minus the other two probs).
a. Full mixture
See fig 7.12. It is a zero-sum game. No equilibrium in pure strategies. Kicker mixes with
probs: pL, pR and 1-pL-pR (for center). Not possible to graph. Against each of the
Goalie’s pure strategies:
If Left by goalie: 45pL+85pC+95pR
If Center: 90pL+0pC+95pR
If Right: 90pL+85pC+60pR
Kicker will choose p’s such that goalie is indiff between her 3 pure strategies. Then, the 3
expressions have to be equal. Solving a 2x2 system:
pL=0.355 and pR=0.457 so pC=0.188
Same for the Goalie yields the equilibrium q’s.
b. Some strategies unused.
Lower payoffs from Center to the Kicker: see fig 7.13. Set up the problem to find q such
that the Kicker is indiff between his 3 pure strategies: you will find qC<0!!!! Impossible
to find a q such that the Kicker chooses Center (that is the meaning of qC<0). Then qC
(for Goalie) has to be zero.
As a consequence the Kicker does not use Center either. Some intuition: notice that
Center is worse than before (70 vs 80) for Kicker. Solve the 2x2 excluding Center for
both: pL=0.4375 and qL=0.375 for Kicker expected payoff=73.13.
Verify that nobody wants to bring in her Center strategy: when Goalie is playing this qmix, the Kicker would get 70 if playing Center (lower than 73.13 from his p-mix so don’t
play Center). If Goalie plays Center, the Kicker gets 92.8 with his p-mix (larger than
73.13 so Goalie should not incorporate Center to his q-mix either).
Finally: redefinition of the opponent’s indifference principle: each player’s equilibrium
mix is such that the opponent is indifferent among all the strategies used in her
equilibrium mix.
6. Evidence on mixing
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Guillermo Sabbioni
Read yourselves (there is some evidence, but not conclusive).
7. How to use mixed strategies in practice
Read yourselves. It seems people alternate “too much”. True randomization is very hard.
SUGGESTED EXERCISES: 2, 3, 4, 6.
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