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Review�

In order to predict outcomes for games without (pure strategy) Nash equilibria, we need to extend such concepts as strategies and equilibria

Randomization of strategies
Mixed Strategy Nash equilibria

The need for randomizing moves in a game usually arises when one player prefers a coincidence of actions, while his rival prefers to avoid it

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Review�

Mixed Strategy specifies that an actual move is chosen randomly from the set of pure strategies with some specific probabilities

A mixed strategy for Player 1 is a probability distribution (r,1-r), where r is the probability of playing Head, and 1-r is the probability of playing Tail
The mixed strategy (1,0) is the pure strategy Head, and the mixed strategy (0,1) is the pure strategy Tail

Tail

Head

Tail

Head

-1 , 1	1 ,-1
1 ,-1	-1 , 1

Player 1

Player 2

1-r

1-q

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Review�

How to find mixed strategy equilibrium:

Find the best response for player 1, given player 2’s mixed strategy
Find the best response for player 2, given player 1’s mixed strategy
Use the best responses to determine mixed strategy Nash equilibria

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Battle of the Sexes: Nash equilibrium�

There are two Nash equilibria in the Battle of the Saxes:

(Mall, Mall) is a Nash equilibrium
(Game, Game) is a Nash equilibrium

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

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Battle of the Sexes: Mixed strategy equilibrium�

Suppose Steve and Rebecca are playing a mixed strategy:

Steve randomizes according to probability distribution (q,1-q)
Rebecca randomizes according to (r,1-r)

The expected payoff for Rebecca:

Eu₁((r,1-r),(q,1-q)) = 2×r×q + 1×(1-r)×(1-q) = r×(3q-1) + 1×(1-q)

The expected payoff for Steve:

Eu₂((r,1-r),(q,1-q)) = 1×r×q + 2×(1-r)×(1-q) = q×(3r-2) + 2×(1-r)

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

1-r

1-q

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Battle of the Sexes: Mixed strategy equilibrium�

The expected payoff for Rebecca:

Eu₁((r,1-r),(q,1-q)) = 2×r×q + 1×(1-r)×(1-q) = r×(3q-1) + 1×(1-q)

Rebecca’s best response B₁((q,1-q)) to Steve’s mixed strategy (q,1-q) is the following:

For q<1/3, the best response is to play r=0 (Game)
For q>1/3, the best response is to play r=1 (Mall)
For q=1/3, indifferent 0≤r≤1

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

1-r

1-q

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Battle of the Sexes: Mixed strategy equilibrium�

The expected payoff for Steve:

Eu₂((r,1-r),(q,1-q)) = 1×r×q + 2×(1-r)×(1-q) = q×(3r-2) + 2×(1-r)

Steve’s best response B₂((r,1-r)) to Rebecca’s mixed strategy (r,1-r) is the following:

For r<2/3, the best response is to play q=0 (Game)
For r>2/3, the best response is to play q=1 (Mall)
For r=2/3, indifferent 0≤q≤1

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

1-r

1-q

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Battle of the Sexes: Mixed strategy equilibrium�

Rebecca’s best response:

For q<1/3, play r=0; For q>1/3, play r=1; For q=1/3, play 0≤r≤1

Steve’s best response:

For r<2/3, play q=0; For r>2/3, play q=1; For r=2/3, play 0≤q≤1

1/3

2/3

Rebecca’s best response

Steve’s best response

Mixed Strategy Equilibrium

Nash Equilibrium

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Battle of the Sexes: Mixed strategy equilibrium�

There are two pure strategy Nash equilibria in the “Battle of the Sexes”:

((1,0), (1,0))
((0,1), (0,1))

There is also one Mixed Strategy Equilibrium:

((0.67,0.33), (0.33,0.67))

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

0.67

0.33

0.67

0.33

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Experiment #4: Results (2020 CWRU)�

Equilibrium

Behavior

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

0.67

0.33

0.67

0.33

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

0.68

0.32

0.69

0.31

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Experiment #4: Results (2020 CWRU)�

Equilibrium

Behavior

		Steve
		Mall	Game
Rebecca	Mall	2 , 2	0 , 0
Rebecca	Game	0 , 0	1 , 1

0.33

0.67

0.33

		Steve
		Mall	Game
Rebecca	Mall	2 , 2	0 , 0
Rebecca	Game	0 , 0	1 , 1

0.99

0.01

0.00

1.00

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Why mixed strategy may not work?�

People are not good at randomizing

Most people (including students) are unable to generate random outcomes without the aid of a random generator
Professional soccer players (i) randomize much better than students making serially independent choices and (ii) they equate their strategies' payoffs to the equilibrium ones (Palacios‐Huerta and Volij, 2008)

People make mistakes

Small amount of noise and payoff asymmetries generate inconsistent behavior (Goeree and Holt, 2001)

People are risk-averse

While mixed strategy assumes risk-neutrality, most people are risk averse (Goeree et al., 2003)

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2×2 Games: Expected payoff�

Assume that player 1 plays (r,1-r) and player 2 plays (q,1-q)

Player 1’s expected payoff is

Eu₁((r,1-r),(q,1-q)) = r×q×u₁(s₁₁,s₂₁) + r×(1-q)×u₁(s₁₁,s₂₂) + + (1-r)×q×u₁(s₁₂,s₂₁) + (1-r)×(1-q)×u₁(s₁₂,s₂₂)

Player 2’s expected payoff is

Eu₂((r,1-r),(q,1-q)) = q×r×u₂(s₁₁,s₂₁) + q×(1-r)×u₂(s₁₂,s₂₁) + + (1-q)×r×u₂(s₁₁,s₂₂) + (1-q)×(1-r)×u₂(s₁₂,s₂₂)

		Player 2
		s₂₁	s₂₂
Player 1	s₁₁	u₁(s₁₁,s₂₁), u₂(s₁₁,s₂₁)	u₁(s₁₁,s₂₂), u₂(s₁₁,s₂₂)
Player 1	s₁₂	u₁(s₁₂,s₂₁), u₂(s₁₂,s₂₁)	u₁(s₁₂,s₂₂), u₂(s₁₂,s₂₂)

1-r

1-q

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2×2 Games: Mixed-Strategy Equilibrium�

A pair of mixed strategies ((r*,1-r*), (q*,1-q*)) is a Mixed Strategy Equilibrium if (r*,1-r*) is a best response to (q*,1-q*), and (q*,1-q*) is a best response to (r*,1-r*)

That is,

Eu₁((r*,1-r*),(q*,1-q*)) ≥ Eu₁((r,1-r),(q*,1-q*)), for all 0≤r≤1
Eu₂((r*,1-r*),(q*,1-q*)) ≥ Eu₂((r*,1-r*),(q,1-q)), for all 0≤q≤1

		Player 2
		s₂₁	s₂₂
Player 1	s₁₁	u₁(s₁₁,s₂₁), u₂(s₁₁,s₂₁)	u₁(s₁₁,s₂₂), u₂(s₁₁,s₂₂)
Player 1	s₁₂	u₁(s₁₂,s₂₁), u₂(s₁₂,s₂₁)	u₁(s₁₂,s₂₂), u₂(s₁₂,s₂₂)

1-r

1-q

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Theorem 1�

THEOREM 1: A pair of mixed strategies ((r*,1-r*), (q*,1-q*)) is a Mixed Strategy Nash equilibrium if and only if� Eu₁((r*,1-r*), (q*,1-q*)) ≥ Eu₁(s₁₁, (q*,1-q*))� Eu₁((r*,1-r*), (q*,1-q*)) ≥ Eu₁(s₁₂, (q*,1-q*)) � Eu₂((r*,1-r*), (q*,1-q*)) ≥ Eu₂((r*,1-r*), s₂₁)� Eu₂((r*,1-r*), (q*,1-q*)) ≥ Eu₂((r*,1-r*), s₂₂)

That is, no player can do better by playing a pure strategy

		Player 2
		s₂₁	s₂₂
Player 1	s₁₁	u₁(s₁₁,s₂₁), u₂(s₁₁,s₂₁)	u₁(s₁₁,s₂₂), u₂(s₁₁,s₂₂)
Player 1	s₁₂	u₁(s₁₂,s₂₁), u₂(s₁₂,s₂₁)	u₁(s₁₂,s₂₂), u₂(s₁₂,s₂₂)

1-r

1-q

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Theorem 1: Illustration�

Use THEOREM 1 to check whether ((2/3,1/3), (1/3,2/3)) is a Mixed Strategy Equilibrium

Rebecca’s expected payoff of playing “Mall” is Eu₁(Mall,(1/3,2/3)) = 2×1/3 = 2/3
Rebecca’s expected payoff of playing “Game” is Eu₁(Game,(1/3,2/3)) = 1×2/3 = 2/3
Rebecca’s expected payoff of mixed strategy is Eu₁((2/3,1/3), (1/3,2/3)) = 2×2/3×1/3 + 1×1/3×2/3 = 2/3

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

2/3

1/3

2/3

1/3

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Theorem 1: Illustration�

Use THEOREM 1 to check whether ((2/3,1/3), (1/3,2/3)) is a Mixed Strategy Equilibrium

Steve’s expected payoff of playing “Mall” is Eu₂((2/3,1/3),Mall) = 1×2/3 = 2/3
Steve’s expected payoff of playing “Game” is Eu₂((2/3,1/3),Game) = 2×1/3 = 2/3
Steve’s expected payoff of mixed strategy is Eu₂((2/3,1/3),(1/3,2/3)) = 1×2/3×1/3 + 2×1/3×2/3 = 2/3

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

2/3

1/3

2/3

1/3

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Theorem 1: Illustration�

Rebecca:

Eu₁((2/3,1/3), (1/3,2/3)) ≥ Eu₁(Mall,(1/3,2/3))

Eu₁((2/3,1/3), (1/3,2/3)) ≥ Eu₁(Game,(1/3,2/3))

Steve:

Eu₂((2/3,1/3),(1/3,2/3)) ≥ Eu₂((2/3,1/3),Mall)

Eu₂((2/3,1/3),(1/3,2/3)) ≥ Eu₂((2/3,1/3),Game)

Hence, ((2/3,1/3), (1/3,2/3)) is a Mixed Strategy Equilibrium by THEOREM 1

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

2/3

1/3

2/3

1/3

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Theorem 2�

THEOREM 2: Let ((r*,1-r*),(q*,1-q*)) be a pair of mixed strategies, where 0<r*<1, 0<q*<1. Then ((r*,1-r*), (q*,1-q*)) is a Mixed Strategy Equilibrium if and only if � Eu₁(s₁₁, (q*,1-q*)) = Eu₁(s₁₂, (q*,1-q*))� Eu₂((r*,1-r*), s₂₁) = Eu₂((r*,1-r*), s₂₂)

That is, each player is indifferent between her two pure strategies

		Player 2
		s₂₁	s₂₂
Player 1	s₁₁	u₁(s₁₁,s₂₁), u₂(s₁₁,s₂₁)	u₁(s₁₁,s₂₂), u₂(s₁₁,s₂₂)
Player 1	s₁₂	u₁(s₁₂,s₂₁), u₂(s₁₂,s₂₁)	u₁(s₁₂,s₂₂), u₂(s₁₂,s₂₂)

1-r

1-q

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Theorem 2: Illustration�

Use THEOREM 2 to find a Mixed Strategy Equilibrium

At the equilibrium Rebecca should be indifferent between going to Mall or Game Eu₁(Mall,(q,1-q)) = Eu₁(Game,(q,1-q))
This implies that 2×q + 0×(1-q) = 0×q + 1×(1-q)
Hence, q* = 1/3 and 1-q* = 2/3

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

1-r

1-q

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Theorem 2: Illustration�

Use THEOREM 2 to find a Mixed Strategy Equilibrium

At the equilibrium Steve should be indifferent between going to Mall or Game Eu₂((r,1-r),Mall) = Eu₂((r,1-r),Game)
This implies that 1×r + 0×(1-r) = 0×r + 1×(1-r)
Hence, r* = 1/3 and 1-r* = 2/3

		Steve
		Mall	Game
Rebecca	Mall	2 , 1	0 , 0
Rebecca	Game	0 , 0	1 , 2

1-r

1-q

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Theorem 2: Application�

Using THEOREM 2 is a fast way to find a Mixed Strategy Equilibrium in a 2×2 games

However, THEOREM 2 is not intuitive:

At the Mixed Strategy Equilibrium, you randomize in order to make your opponent indifferent
Row randomizes to make column indifferent
Column randomizes to make row indifferent
By doing so, each player is best responding to each other

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Theorem 2: Asymmetric Matching Pennies�

There are no pure strategy Nash equilibria
At the Mixed Strategy Equilibrium:

Player 1 is indifferent between H and T: Eu₁(Head,(q,1-q)) = Eu₁(Tail,(q,1-q)) implies q* = 0.5
Player 2 is indifferent between H and T: Eu₂((r,1-r),Head) = Eu₂((r,1-r),Tail) implies r*= 0.17
The equilibrium is: ((0.17,0.83), (0.5,0.5))

Tail

Head

Tail

Head

-1 , 4	1 ,-1
1 ,-1	-1 , 0

Player 1

Player 2

1-r

1-q

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Theorem 2: Tennis�

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Theorem 2: Tennis�

THEOREM 2 predicts that each type of a serve (left or right) should yield the same payoff to the receiver in the Mixed Strategy Equilibrium

Empirical findings (Walker and Wooders, 2001):

The receiver’s probability of winning the point is the same disregarding whether the ball was served to the left or to the right
However, serves are negatively correlated (i.e., too much randomization)

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Employee Monitoring�

Employee can Work or Shirk

Salary $100K if not caught shirking
Salary $0 if caught shirking
Cost of effort is $50K

Managers can Monitor or Not Monitor

Profit if employee works is $200K
Profit if employee shirks is $0
Cost of monitoring is $10K

		Manager
		Monitor	Not Monitor
Employee	Work	50 , 90	50 , 100
Employee	Shirk	0 , -10	100 , -100

1-r

1-q

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Employee Monitoring�

What is the pure strategy Nash Equilibrium for this game?

There are none!

But the Existence theorem (Nash, 1951) guarantees the existence of a Mixed Strategy Equilibrium

		Manager
		Monitor	Not Monitor
Employee	Work	50 , 90	50 , 100
Employee	Shirk	0 , -10	100 , -100

1-r

1-q

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Employee Monitoring�

Employee’s best response:

Shirk (r=0) if q<0.5
Work (r=1) if q>0.5
Any mixed strategy (0≤r≤1) if q=0.5

Manager’s best response:

Monitor (q=1) if r<0.9
Not Monitor (q=0) if r>0.9
Any mixed strategy (0≤q≤1) if r=0.9

		Manager
		Monitor	Not Monitor
Employee	Work	50 , 90	50 , 100
Employee	Shirk	0 , -10	100 , -100

1-r

1-q

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Battle of the Sexes: Mixed strategy equilibrium�

Therefore, ((0.9,0.1),(0.5,0.5)) is a mixed strategy equilibrium

Application: this game illustrates why sometimes workers shirk (10% of the time) and managers do not always monitor them (50%)

0.5

0.9

Employee’s best response

Manager’s best response

Mixed Strategy Equilibrium

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How about more complicated games?�

Next Time!

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Thank you!

Roman Sheremeta, Ph.D.

Professor, Weatherhead School of Management

Case Western Reserve University

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References�

Watson, J. (2013). Strategy: An Introduction to Game Theory (3rd Edition). Publisher: W. W. Norton & Company. (Chapters 4 & 11)
Palacios‐Huerta, I., & Volij, O. (2008). Experientia docet: Professionals play minimax in laboratory experiments. Econometrica, 76(1), 71-115.
Goeree, J.K., & Holt, C.A. (2001). Ten little treasures of game theory and ten intuitive contradictions. American Economic Review, 1402-1422.
Goeree, J.K., Holt, C.A., & Palfrey, T.R. (2003). Risk averse behavior in generalized matching pennies games. Games and Economic Behavior, 45(1), 97-113.