Mixed Strategy Equilibrium
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Zero-Sum Game
• Matching Pennies
Player 2 Heads
Player 1
Heads
Tails
1
-1
Tails
-1
1
-1
1
1
-1
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No Nash Equilibrium?
• The distinguishing feature of zero-sum game
is that each player would like to outguess the
other since there is no “win-win” situation.
• Example: Porker (bluff or not), Battle (by land
or by sea), Tennis (left or right to serve)
• When each player would always like
outguess the other(s), there is no Nash
equilibrium.
 Contradicting Nash’s theorem??
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Mixed Strategy
• A mixed strategy for a player is a probability
distribution over some (or all) of her
strategies.
• The strategy we have studied so far, i.e.,
taking some action for sure, is called a purestrategy.
• When the outcome of the game is uncertain,
we assume that each player maximizes
expected value of her payoff.
 Expected utility theory (von Neumann and
Morgenstern, 1944)
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Matching Pennies Again
• Introducing mixed strategies
Player 2 Heads (p)
Player 1
Heads
(q)
Tails
(1-q)
Tails (1-p)
1
-1
-1
1
-1
1
1
-1
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How to Find Equilibrium?
• If a player takes both “Heads” and
“Tails” with positive probability, she
must be indifferent between these two
pure strategies, i.e., the expected
payoff derived by choosing Heads must
be equal to that by choosing Tails.
• -p+(1-p)=p-(1-p), hence p=0.5.
• q-(1-q)=-q+(1-q), hence q=0.5.
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How to Verify
Equilibrium?
• Note that if p=0.5, Player 1 does not have a
strict incentive to change her strategy from
q=0.5.
• Similarly, Player 2 does not have a strict
incentive to change his strategy from p=0.5,
if q=0.5.
 Therefore, p=q=0.5 constitutes a mixedstrategy equilibrium.
Q: Is this equilibrium reasonable/stable?
A: Yes (each player ends up randomizing two
strategies equally if the rival is smart
enough).
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Modified Matching
Pennies
• Suppose the payoffs in the up-left cell
changes as the following:
Player 2 Heads (p)
Player 1
Heads
(q)
Tails
(1-q)
Tails (1-p)
2
-2
-1
1
-1
1
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1
-1
Indifference Property
• Under mixed-strategy NE, Player 1
must be indifferent between choosing
H and T:
 -2p+(1-p)=p-(1-p), hence p=0.4.
• Similarly, Player 2 must be indifferent
between choosing H and T:
 2q-(1-q)=-q+(1-q), hence q=0.4.
• You can easily verify that (p,q)=(0.4,0.4)
indeed becomes a mixed-strategy NE.
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Existence of NE
Theorem (Nash, 1950)
• If a game has finite number of players and
actions, then there exists at least one Nash
equilibrium, possibly involving mixed
strategies.
 The proof uses the Kakutani’s fixed-point
theorem.
 Best response mappings satisfy the
condition of the fixed-point theorem, and
hence have a fixed point, which is equivalent
to a NE!
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Infinite Games
• The finite game assumption is not necessary
but gives a sufficient condition for the
existence of NE.
• There are many games that do not satisfy the
conditions of the Theorem but nonetheless
have one or more NE. (e.g., Bertrand model,
Cournot model)
Q: Are there any games which do not even
have a mixed-strategy equilibrium?
A: Yes, e.g., Integer game.
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Reporting a Crime
• A crime is observed by a group of n people.
Each person would like the police to be
informed but prefers that someone else make
the phone call.
• Players: The n people.
• Strategies: “Call” or “Don’t call”
• Payoffs:
0 if no one calls.
v if someone else calls but she does not.
v-c if the player calls.
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Pure-Strategy Equilibrium
• This game has n pure-strategy NE, in each of
which exactly one person calls. (if that
person switches to not calling, her payoff
falls from v-c to 0; if any other person
switches to calling, his payoff falls from v to
v-c.)
• If the members of the group differ in some
respect, then these asymmetric equilibria
may be compelling as steady states.
• For example, the social norm in which the
oldest person in the group makes the phone
call is stable.
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Mixed-Strategy
Equilibrium
• There is a symmetric mixed strategy
equilibrium in which each person calls with
positive probability p<1.
• In any such equilibrium, each person’s
expected payoff to calling is equal to her
expected payoff to not calling.
 v-c=v(1-Pr{no one else calls})
 c/v=Pr{no one else calls}
• Notice that the probability that no one else
calls is independent of n, but constant!
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Implication
Pr{no one calls}
=Pr{i does not call}xPr{no one else calls}
=(1-p)c/v
• You can show that 1-p is increasing in n, i.e.,
the probability that person i does not call
increases as n becomes larger.
• This implies that the probability that no one
calls is also increasing in n.
 The larger the group, the less likely the
police are informed of the crime!
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