Mixed Strategy Nash Equilibrium

04. Mixed Strategy Nash Equilibrium

Ryan Fang1

1ryf1@psu.edu

Decision Making and Strategy in EconomicsEcon 402 (Spring 2015)

Reading

Osborne: Chapter 4

1 Mixed Strategies

2 Mixed Strategy Nash Equilibrium

3 Best Responses

4 Dominance

5 Finding All Mixed Strategy Nash Equilibria

Matching Pennies Revisited

• Matching pennies:

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• No pure strategy Nash equilibria.

• But how would you play the game?

A Different Interpretation

• Think about a penalty kick in a soccer game.

• Player 1 is the goalie and has to choose the direction toattempt the save, left or right.

• Player 2 is the shooter and has to choose the direction tokick the ball, left or right.

• How should the shooter and the goalie play the game?

Mixed Strategies

• Players can randomize their choice of actions.

• A mixed strategy of a player specifies the probabilities withwhich she is going to choose each of her actions.

• Formally, a mixed strategy, αi , of player i is a probabilitydistribution over her set of actions Ai .

• We write αi ∈ ∆Ai , where ∆Ai denotes the set of allprobability distributions over Ai .

• For every a ∈ Ai , αi (a) is the probability with which a isplayed.

Probability Distributions

• Given a finite set Ω = ω1, ω2, . . . , ωN, a probabilitydistribution over Ω is simply a n-vector, i.e., a ordered list ofn numbers, (p1, p2, . . . , pN) such that 0 ≤ pn ≤ 1 for alln = 1, 2, . . . ,N, and

∑Nn=1 pn = 1.

• ∆Ω is the set of all such n-vectors.

Example

• In the Matching Pennies, player i ’s set of actions are H andT .

• A mixed strategy of player i , αi , can specify that i chooses Hwith probability 1

2 and T with probability 12 .

• Thus, αi (H) = 12 and αi (T ) = 1

Example

• Another mixed strategy of player i , α′i , can be defined as:α′i (H) = 1

3 and α′i (T ) = 23 .

• Alternatively, we can write αi =(

)and α′i =

where the order of the probabilities are understood.

• Moreover, player i can also assign probability 1 to action H.(1, 0) is also a probability distribution over Ai .

• In this case, we say that i plays a pure strategy H. Thus, apure strategy is just a special mixed strategy.

Example

• As before, we can define a mixed strategy profile as a list ofmixed strategies chosen by the players.

• For example, we can have: (α1, α2) =((

Thus, player 1 plays H and T with probability 12 while player

2 plays H with probability 13 and T with probability 2

Expected Payoff

• Recall that, as part of a game, we have to specify theplayers’ preferences over all possible action profiles.

• So, how do we define the players’ preferences over mixedstrategy profiles?

• We can define players’ preferences over “lotteries”.

• A (mixed) strategy profile induces a lottery over the actionprofiles.

Expected Payoff

• If the players play (α1, α2) =((

)), and if they

randomize over their actions independently (which is alwaysassumed to be the case in our course), then:

• with probability 12 ×

13 they will end up playing (H,H);

23 they will end up playing (H,T );

13 they will end up playing (T ,H); and,

23 they will end up playing (T ,T ).

Expected Payoff

• Thus, the strategy profile (α1, α2) =((

))induces a lottery, which offers (the outcome induced by)(H,H) with probability 1

6 , (H,T ) with probability 13 , (T ,H)

with probability 16 , and (T ,T ) with probability 1

• A famous theorem by mathematician John Von Neumannand economist Oskar Morgenstern allows us to representplayers’ preferences over lotteries by their expected payoffs.

• That is, we take the (Bernoulli) payoffs to player i in theevents (H,H), (H,T ), etc., and calculate the expected valueof these payoffs based on the probabilities that these eventswill occur.

Example

• If the players’ (Bernoulli) payoffs are given as in the followingtable:

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Then, if the players choose (α1, α2) =((

player 1’s expected payoff is going to be1× 1

6 + (−1)× 13 + (−1)× 1

6 + 1× 13 = 0.

• Similarly, player 2’s expected payoff is(−1)× 1

6 + 1× 13 + 1× 1

6 + (−1)× 13 = 0.

Example

• BoS:

•Bach Stravinsky

Bach 2,1 0,0Stravinsky 0,0 1,2

• If the players choose (α1, α2) =((

)), player 1’s

expected payoff is going to be 2× 16 + 1× 1

3 = 23 .

• On the other hand, player 2’s expected payoff is1× 1

6 + 2× 13 = 5

Expected Payoff

• In general, given α1 ∈ ∆A1 and α2 ∈ ∆A2, where both A1and A2 are finite, player i ’s expected payoff is:

Ui (α1, α2) =∑

a1∈A1

∑a2∈A2

α1 (a1) · α2 (a2) · ui (a1, a2)

1 Mixed Strategies

3 Best Responses

4 Dominance

Expected Payoff Maximization

• We say that a player’s preference over lotteries is representedby her expected payoffs if she prefers one lottery to anotherif and only if her expected payoff from the first lottery ishigher than that from the second lottery.

• Recall that we always assume that the players are rational,which means they always choose the best option available tothem. Thus, when choosing between lotteries (or mixedstrategies), they always choose those that maximize theirexpected payoffs.

Mixed Strategy Nash Equilibrium

• As before, we can think of Nash equilibrium in mixedstrategies as a (stochastic) steady state, where no player hasany incentive to deviate.

• Formally, a strategy profile α∗ is a Nash equilibrium in mixedstrategies (or mixed strategy Nash equilibrium) if and only if,for each player i , we have:

Ui(α∗i , α

∗−i)≥ Ui

(αi , α

∗−i)

for all αi ∈ ∆Ai

• Again, αi can be a pure strategy for every i . In which case,α∗ is called a pure strategy Nash equilibrium. Thus, a PSNEis just a special MSNE.

Example

• Let’s return to Matching Pennies:

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

))a mixed strategy Nash

equilibrium?

• No. Why?

• Because by deviating to (0, 1), player 1 can strictly increaseher expected payoff.

• Check: U1

((12 ,

))= 0, and

((0, 1) ,

))= (−1)× 1

3 + 1× 23 = 1

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• No. Why?

• Check: U1

((12 ,

))= 0, and

((0, 1) ,

))= (−1)× 1

3 + 1× 23 = 1

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• No. Why?

• Check: U1

((12 ,

))= 0, and

((0, 1) ,

))= (−1)× 1

3 + 1× 23 = 1

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• No. Why?

• Check: U1

((12 ,

))= 0, and

((0, 1) ,

))= (−1)× 1

3 + 1× 23 = 1

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• Yes. Why?

• Because, given that her opponent plays(

), a player’s

payoff is always 0 no matter what mixed strategy shechooses.

• Later on, we will show that this is in fact the unique mixedstrategy Nash equilibrium.

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• Yes. Why?

), a player’s

Example

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• Yes. Why?

), a player’s

Existence of NE

• Recall that we established previously that there are no purestrategy Nash equilibria of the Matching Pennies game.

• Now, by allowing the players play randomized strategies, wehave existence of Nash equilibria.

• In fact, Nash proved that so long as the players’ action setsare finite, there always is at least one Nash equilibrium inmixed strategies.

More Examples

• The Prisoners’ Dilemma:

•Quiet Fink

Quiet 2,2 0,3Fink 3,0 1,1

• The unique Nash equilibrium in mixed strategies is whereboth players choose F with probability 1.

• In particular, there are no Nash equilibria in which the playersstrictly mix (i.e., play every action with strictly positiveprobabilities).

More Examples

• The Coordination Game:

•Up Down

Up 2,2 0,0Down 0,0 1,1

• A mixed strategy Nash equilibrium is where both players playαi =

• In this equilibrium, both players’ expected payoffs are equalto 2

• Recall that (U,U) and (D,D) are also Nash equilibria. Inthese equilibria, both players’ expected payoffs are 2 and 1,which are strictly higher.

More Examples

• BoS:

•Bach Stravinsky

• A mixed strategy Nash equilibrium is where player 1 plays(23 ,

)and player 2 plays

). In equilibrium, both

players’ expected payoffs are equal to 23 .

1 Mixed Strategies

3 Best Responses

4 Dominance

Expected Payoff

• Suppose that player 1 plays the mixed strategy (p, 1− p) andplayer 2 plays the mixed strategy (q, 1− q).

•C (prob. q) D (prob. 1-q)

A (prob. p) a,b c,dB (prob. 1-p) e,f g,h

Expected Payoff

• We can calculate the probability for each of the four possibleaction profiles to be played.

A (prob. p) a,b (pq) c,d (p (1− q))B (prob. 1-p) e,f ((1− p)q) g,h ((1− p) (1− q))

Expected Payoff

• U1 ((p, 1− p) , (q, 1− q)) =

a (pq) + c [p (1− q)] + e [(1− p)q] + g [(1− p) (1− q)]

• U2 ((p, 1− p) , (q, 1− q)) =

b (pq) + d [p (1− q)] + f [(1− p)q] + h [(1− p) (1− q)]

Example

• Suppose that player 1 plays the mixed strategy (p, 1− p) andplayer 2 plays the mixed strategy (q, 1− q).

•C (prob. 1

3) D (prob. 23)

A (prob. 12) 2,2 0,3

B (prob. 12) 3,0 1,1

Example

• We can calculate the probability for each of the four possibleaction profiles to be played.

C (prob. 13) D (prob. 2

A (prob. 14) 2,2

(14 ×

13 = 1

(14 ×

23 = 1

)B (prob. 3

4) 3,0(

13 = 1

(34 ×

23 = 1

Example

C (prob. 13) D (prob. 2

A (prob. 14) 2,2

(14 ×

13 = 1

(14 ×

23 = 1

)B (prob. 3

4) 3,0(

13 = 1

(34 ×

23 = 1

)• U1 = 2× 1

12 + 0× 16 + 3× 1

4 + 1× 12

• U2 = 2× 112 + 3× 1

6 + 0× 14 + 1× 1

Mixed Strategy Nash Equilibrium

• As before, we can define Nash equilibrium in mixed strategiesusing best responses.

• Formally, a strategy profile α∗ is a Nash equilibrium in mixedstrategies (or mixed strategy Nash equilibrium) if and only ifα∗i is a best response to α∗−i for every i .

•Head

)Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• No. Because by deviating to (0, 1), player 1 can strictlyincrease her expected payoff from 0 to 1

3 . So,(

)is not a

best response to(

•Head

)Head (p) 1,-1 -1,1

Tail (1− p) -1,1 1,-1• In fact, let p be the probability that player 1 chooses H. Herexpected payoff given her opponent’s mixed strategy is:

1(13p)

+ (−1)

(1− p)

13− 2

which is strictly decreasing in p! Thus, her payoff ismaximized by setting p = 0. In other words, her unique bestresponse is to play T with probability 1.

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Is (α1, α2) =((

equilibrium?

• Yes. Because, given that her opponent plays(

player’s payoff is always 0 no matter what mixed strategy shechooses.

• In other words, given that her opponent plays(

), any

mixed strategy of a player is a best response.

Best Responses

• It turns out that it is always the case that a player’s bestresponse to her opponent’s strategies is either a single purestrategy or a continuum of mixed strategies.

• To see why, let’s go back to the expected payoff calculations.

Expected Payoff

• U1 ((p, 1− p) , (q, 1− q)) =

a (pq) + c [p (1− q)] + e [(1− p)q] + g [(1− p) (1− q)]

• U2 ((p, 1− p) , (q, 1− q)) =

b (pq) + d [p (1− q)] + f [(1− p)q] + h [(1− p) (1− q)]

Expected Payoff

• U1 ((p, 1− p) , (q, 1− q)) =

a (pq) + c [p (1− q)] + e [(1− p)q] + g [(1− p) (1− q)]

• We can rewrite this as: U1 ((p, 1− p) , (q, 1− q)) =

p [a (q) + c (1− q)] + (1− p) [eq + g(1− q)]

Expected Payoff

p [a (q) + c (1− q)] + (1− p) [eq + g(1− q)]

• But [a (q) + c (1− q)] is simply player 1’s expected payoff ifshe chooses the pure strategy A given player 2’s mixedstrategy (q, 1− q).

• Similarly [eq + g(1− q)] is her expected payoff if shechooses the pure strategy B.

Expected Payoff

• Thus, writing U1 (A, (q, 1− q)) = a (q) + c (1− q) andU1 (B, (q, 1− q)) = eq + g(1− q), we have:

U1 ((p, 1− p) , (q, 1− q))

= pU1 (A, (q, 1− q)) + (1− p) U1 (B, (q, 1− q))

• In other words, player 1’s expected payoff from playing(p, 1− p) is simply a weighted average of her expectedpayoffs from playing pure strategies A and B.

Expected Payoff

U1 ((p, 1− p) , (q, 1− q))

= pU1 (A, (q, 1− q)) + (1− p) U1 (B, (q, 1− q))

• How do you maximize a weighted average?

• By putting more weight on values that are higher you canincrease a weighted average.

• So, if there is a value that is higher than all the other ones,you should put all the weights on that value.

Expected Payoff

U1 ((p, 1− p) , (q, 1− q))

= pU1 (A, (q, 1− q)) + (1− p) U1 (B, (q, 1− q))

• How do you maximize a weighted average?

• By putting more weight on values that are higher you canincrease a weighted average.

• So, if there is a value that is higher than all the other ones,you should put all the weights on that value.

Best Responses

U1 ((p, 1− p) , (q, 1− q))

= pU1 (A, (q, 1− q)) + (1− p) U1 (B, (q, 1− q))

• Thus, when U1 (A, (q, 1− q)) > U1 (B, (q, 1− q)),U1 ((p, 1− p) , (q, 1− q)) is maximized by setting p = 1.That is, player 1’s unique best response is the pure strategyA.

• Similarly, when U1 (A, (q, 1− q)) < U1 (B, (q, 1− q)), player1’s unique best response is the pure strategy B.

Best Responses

U1 ((p, 1− p) , (q, 1− q))

= pU1 (A, (q, 1− q)) + (1− p) U1 (B, (q, 1− q))

• However, when U1 (A, (q, 1− q)) = U1 (B, (q, 1− q)),U1 ((p, 1− p) , (q, 1− q)) = U1 (A, (q, 1− q)) regardless ofthe value of p. Therefore, any mixed strategy is a bestresponse!

• The same logic applies to player 2.

• Let’s return to the matching pennies game.

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Suppose that player 2 plays the mixed strategy (q, 1− q).What is U1 (H, (q, 1− q)) and U1 (T , (q, 1− q))?

• U1 (H, (q, 1− q)) = 1× q + (−1)× (1− q) = 2q − 1

• U1 (T , (q, 1− q)) = (−1)× q + 1× (1− q) = 1− 2q

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• U1 (H, (q, 1− q)) = 1× q + (−1)× (1− q) = 2q − 1

• U1 (T , (q, 1− q)) = (−1)× q + 1× (1− q) = 1− 2q

• Thus, when q < 12 , we have

U1 (H, (q, 1− q)) < U1 (T , (q, 1− q)) and player 1’s uniquebest response is the pure strategy T .

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• U1 (H, (q, 1− q)) = 1× q + (−1)× (1− q) = 2q − 1

• U1 (T , (q, 1− q)) = (−1)× q + 1× (1− q) = 1− 2q

• When q > 12 , we have U1 (H, (q, 1− q)) > U1 (T , (q, 1− q))

and player 1’s unique best response is the pure strategy H.

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• U1 (H, (q, 1− q)) = 1× q + (−1)× (1− q) = 2q − 1

• U1 (T , (q, 1− q)) = (−1)× q + 1× (1− q) = 1− 2q

• Finally, when q = 12 , we have

U1 (H, (q, 1− q)) = U1 (T , (q, 1− q)) and any mixedstrategy of player 1’s is a best response.

• We can summarize this information in a graph: seeblackboard.

• (The graph is on P.112 of Osborne)

• We can determine player 2’s best responses in a similarfashion.

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• Suppose that player 1 plays the mixed strategy (p, 1− p).What is U2 ((p, 1− p) ,H) and U2 ((p, 1− p) ,T )?

• U2 ((p, 1− p) ,H) = (−1)× p + 1× (1− p) = 1− 2p

• U2 ((p, 1− p) ,T ) = 1× p + (−1)× (1− p) = 2p − 1

•Head Tail

Head 1,-1 -1,1Tail -1,1 1,-1

• U2 ((p, 1− p) ,H) = (−1)× p + 1× (1− p) = 1− 2p

• U2 ((p, 1− p) ,T ) = 1× p + (−1)× (1− p) = 2p − 1

• Thus, player 2’s best response is H if p < 12 , T if p > 1

2 , andany mixed strategy if p = 1

• We can draw the same graph for player 2’s best responsefunction.

• By graphing the best response functions of the two players,we can find the unique point where they intersect,((

• This is our Nash equilibrium.

• BoS:

•Bach Stravinsky

• Again, suppose that player 2 now plays (q, 1− q). What isplayer 1’s best response?

• U1 (B, (q, 1− q)) = 2q, U1 (S , (q, 1− q)) = 1− q

• When q < 13 , U1 (B, (q, 1− q)) < U1 (S , (q, 1− q)),

BR1 = S

• When q > 13 , U1 (B, (q, 1− q)) > U1 (S , (q, 1− q)),

BR1 = B

• When q = 13 , any mixed strategy is a best response.

• BoS:

•Bach Stravinsky

• Again, suppose that player 2 now plays (q, 1− q). What isplayer 1’s best response?

• U1 (B, (q, 1− q)) = 2q, U1 (S , (q, 1− q)) = 1− q

• When q < 13 , U1 (B, (q, 1− q)) < U1 (S , (q, 1− q)),

BR1 = S

• When q > 13 , U1 (B, (q, 1− q)) > U1 (S , (q, 1− q)),

BR1 = B

• When q = 13 , any mixed strategy is a best response.

• BoS:

•Bach Stravinsky

• Now suppose that player 1 plays (p, 1− p). What is player2’s best response?

• U2 ((p, 1− p) ,B) = p, U2 ((p, 1− p) ,S) = 2 (1− p)

• When p < 23 , U2 ((p, 1− p) ,B) < U2 ((p, 1− p) ,S),

BR2 = S

• When p > 23 , U2 ((p, 1− p) ,B) > U2 ((p, 1− p) ,S),

BR2 = B

• When p = 23 , any mixed strategy is a best response.

• BoS:

•Bach Stravinsky

• Now suppose that player 1 plays (p, 1− p). What is player2’s best response?

• U2 ((p, 1− p) ,B) = p, U2 ((p, 1− p) ,S) = 2 (1− p)

• When p < 23 , U2 ((p, 1− p) ,B) < U2 ((p, 1− p) ,S),

BR2 = S

• When p > 23 , U2 ((p, 1− p) ,B) > U2 ((p, 1− p) ,S),

BR2 = B

• When p = 23 , any mixed strategy is a best response.

• We can again summarize this information in a graph:blackboard.

• (The graph can be found on P.113 of Osborne)

• We can also graph player 2’s best response function.

• We now find three points at which the two best responsefunctions intersect: ((1, 0) , (1, 0)), ((0, 1) , (0, 1)), and((

• These are our Nash equilibria.

1 Mixed Strategies

3 Best Responses

4 Dominance

Dominance Allowing for Mixed Strategies

• Recall that the analysis of a game can be simplified by firsteliminating all strictly dominated actions.

• We argued before that a strictly dominated action can neverbe played in a Nash equilibrium.

• Recall the definition for a strictly dominated action we hadearlier:

• The action ai is strictly dominated, if there exists a′i , suchthat:

• ui(a′i , a−i

)> ui (ai , a−i ) for all a−i ∈ ×j 6=iAj .

• According to this definition, a′i strictly dominates ai if it givesa strictly higher payoff to player i regardless of what purestrategy profile the opponents play.

• To what extend does this definition need to be adapted totake into account of mixed strategies?

• A natural extension should probably look like this:

• The action ai is strictly dominated, if there exists αi ∈ ∆Aisuch that:

• Ui (αi , α−i ) > Ui (ai , α−i ) for all α−i ∈ ×j 6=i ∆Aj .

• That is, we allow for mixed strategies when:

• considering potential dominant strategies• considering the opponents’ strategy profiles

• But is this strictly necessary?

• The answer is: not really.

• What we need is the following:

• The action ai is strictly dominated if there exists a mixedstrategy αi such that:

• Ui (αi , a−i ) > Ui (ai , a−i ) for all a−i ∈ ×j 6=iAj .

• Thus, all we need in addition to our old definition is to allowfor mixes strategies when we consider potential dominantstrategies. Why?

Example

• Prisoners’ Dilemma:

•Quiet Fink

Quiet 2,2 0,3Fink 3,0 1,1

• u1 (F ,Q) = 3 > 2 = u1 (Q,Q);u1 (F ,F ) = 1 > 0 = u1 (Q,F ).

• Thus, F strictly dominates Q for player 1 according to theold definition.

• However, can you find any mixed strategy α2 of player 2 suchthat U1 (F , α2) fails to be strictly higher than U1 (Q, α2)?

Example

• Prisoners’ Dilemma:

•Quiet Fink

Quiet 2,2 0,3Fink 3,0 1,1

• u1 (F ,Q) = 3 > 2 = u1 (Q,Q);u1 (F ,F ) = 1 > 0 = u1 (Q,F ).

• Thus, F strictly dominates Q for player 1 according to theold definition.

• However, can you find any mixed strategy α2 of player 2 suchthat U1 (F , α2) fails to be strictly higher than U1 (Q, α2)?

• No. Why?

Expected Payoff

• U1 ((p, 1− p) , (q, 1− q)) =

a (pq) + c [p (1− q)] + e [(1− p)q] + g [(1− p) (1− q)]

• U2 ((p, 1− p) , (q, 1− q)) =

b (pq) + d [p (1− q)] + f [(1− p)q] + h [(1− p) (1− q)]

Expected Payoff

• U1 ((p, 1− p) , (q, 1− q)) =

a (pq) + c [p (1− q)] + e [(1− p)q] + g [(1− p) (1− q)]

q [a (p) + e (1− p)] + (1− q) [cp + g(1− p)]

Expected Payoff

q [a (p) + e (1− p)] + (1− q) [cp + g(1− p)]

• [a (p) + e (1− p)] is player 1’s expected payoff if she choosesthe mixed strategy (p, 1− p) and player 2 plays the purestrategy C .

• Similarly [cp + g(1− p)] is her expected payoff if player 2chooses the pure strategy D.

Expected Payoff

• Thus, writing U1 ((p, 1− p) ,C) = a (p) + e (1− p) andU1 ((p, 1− p) ,D) = cp + g(1− p), we have:

U1 ((p, 1− p) , (q, 1− q))

= qU1 ((p, 1− p) ,C) + (1− q) U1 ((p, 1− p) ,D)

• In other words, player 1’s expected payoff when player 2chooses (q, 1− q) is simply a weighted average of herexpected payoffs when 2 plays pure strategies C and D.

Strict Dominance

• Now suppose that x1 > x2 and y1 > y2, then, for anyq ∈ [0, 1], we must have:

[qx1 + (1− q) y1]− [qx2 + (1− q) y2]

= q (x1 − x2) + (1− q) (y1 − y2)

• Thus, if a strategy (mixed or pure) yields strictly higherpayoffs to a player than another strategy regardless of heropponents’ pure strategy profiles, then it must be the casethat, the former yields strictly higher payoffs than the latterregardless of her opponents’ mixed strategy profiles.

Example

• Consider the following payoff function of player 1:

L RT 3, 0,M 1, 2,B 0, 5,

• Are there any dominated strategies for player 1?

Example

L RT 3, 0,M 1, 2,B 0, 5,

• T is NOT strictly dominated by M or B.• M is NOT strictly dominated by T or B.• B is NOT strictly dominated by T or M.

Example

L RT 3, 0,M 1, 2,B 0, 5,

• Can M be played in any mixed strategy Nash equilibrium?

Example

L RT 3, 0,M 1, 2,B 0, 5,

• Can M be played in any mixed strategy Nash equilibrium? No.• M is strictly dominated by the mixed strategy α =

( 12 , 0,

Example

L RT 3, 0,M 1, 2,B 0, 5,

• M is strictly dominated by the mixed strategy α =( 1

2 , 0,12

• U1 (α,L) = 12 × 3 + 1

2 × 0 = 1.5, U1 (M,L) = 1; and,• U1 (α,R) = 1

2 × 0 + 12 × 5 = 2.5, U1 (M,R) = 2.

Example

• M is strictly dominated by the mixed strategy α =(

12 , 0,

• U1 (α,L) = 12 × 3 + 1

2 × 0 = 1.5, U1 (M,L) = 1; and,• U1 (α,R) = 1

2 × 0 + 12 × 5 = 2.5, U1 (M,R) = 2.

• By the argument we gave above, α yields a strictly higherpayoff to player 1 than M does, regardless what mixedstrategy profile her opponents play. Therefore, M can neverbe a best response to any mixed strategy profile of 1’sopponents, and, hence, can never be part of a Nashequilibrium.

Dominance by Mixed Strategies

• Thus, even though an action is not strictly dominated by anyother actions, it is still possible for it to be strictlydominated by a mixed strategy.

• When we allow for mixed strategies, there are more chancesfor an action to be dominated.

Dominated Mixed Strategies

• Any mixed strategy that assigns positive probability to astrictly dominated action is strictly dominated. Why?

• Therefore, a strictly dominated action can never be part ofany mixed strategies used in a Nash equilibrium.

• We will not lose any mixed strategy Nash equilibrium if weeliminated all strictly dominated actions.

1 Mixed Strategies

3 Best Responses

4 Dominance

Finding all Nash Equilibria

• Find all mixed strategy Nash equilibria of the following game:

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

An Algorithm

• For each player i , choose a subset Si of her set of actions, Ai .

• Check whether there exists a mixed strategy Nashequilibrium α∗ such that each strategy α∗i assigns positiveprobability to all members of Si and only members of Si .(That is, Si is the support of α∗i .)

• Repeat the analysis for every collection of subsets of theplayers’ sets of actions.

Characterizing Mixed Strategy Nash Equilibria

• A mixed strategy profile α∗ is a MSNE if and only if, for eachplayer i :

• the expected payoff, given α∗−i , associated with every actionthat is chosen with positive probability by α∗i is the same; and,

• the expected payoff, given α∗−i , associated with every actionthat is chosen with zero probability by α∗i is no higher thanthe expected payoff to any action that is chosen with positiveprobability.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• First, find all pure strategy Nash equilibria (i.e., first look atSi ’s that are singleton sets).

• No PSNE.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Now, are there any MSNE where player 1 randomizes over Tand M and player 2 randomizes over L and R?

• Let (p, 1− p) be the probabilities with which 1 chooses Tand M, and (q, 1− q) be the probabilities with which 2chooses L and R.

• Can there be any pair (p, q) that is a MSNE?

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Let (p, 1− p) be the probabilities with which 1 chooses Tand M, and (q, 1− q) be the probabilities with which 2chooses L and R.

• Can there be any pair (p, q) that is a MSNE?• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Tand M.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Tand M.

• That is, 2p + 3 (1− p) = 3p + 2 (1− p) and2q + 2 (1− q) = 0q + 3 (1− q)

• Solving the set of equations, we find that (p, q) =( 1

)satisfy the conditions.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• When q = 13 , player 1’s expected payoff from playing both

Tand M is 2, and her expected payoff from playing B is 1.So, we have a MSNE!

• The MSNE is(( 1

2 ,12 , 0),( 1

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Now, are there any MSNE where player 1 randomizes over Mand B and player 2 randomizes over L and R?

• Let (p, 1− p) be the probabilities with which 1 chooses Mand B, and (q, 1− q) be the probabilities with which 2chooses L and R.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Let (p, 1− p) be the probabilities with which 1 chooses Mand B, and (q, 1− q) be the probabilities with which 2chooses L and R.

• Can there be any pair (p, q) that is a MSNE?• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Mand B.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Mand B.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• When q = 12 , player 1’s expected payoff from playing both M

and B is 1.5, and her expected payoff from playing T is 2.So, this is not a MSNE after all.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Now, are there any MSNE where player 1 randomizes over Tand B and player 2 randomizes over L and R?

• Let (p, 1− p) be the probabilities with which 1 chooses Tand B, and (q, 1− q) be the probabilities with which 2chooses L and R.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Let (p, 1− p) be the probabilities with which 1 chooses Tand B, and (q, 1− q) be the probabilities with which 2chooses L and R.

• Can there be any pair (p, q) that is a MSNE?• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Tand B.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• If (p, q) is a NE, we must have that: given p, 2 is indifferentbetween L and R, and, given q, 1 is indifferent between Mand B.

• But there cannot be any p ∈ [0, 1] that can solve the firstequation! Because, when player 1 never chooses M, R strictlydominates L for player 2!

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• But there cannot be any p ∈ [0, 1] that can solve the firstequation! Because, when player 1 never chooses M, R strictlydominates L for player 2!

• Thus, we cannot have a MSNE in which 1 randomizesbetween T and B and 2 randomizes between L and R.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Finally, are there any MSNE where player 1 randomizes overT , M, and B and player 2 randomizes over L and R?

• Let (p, q, 1− p − q) be the probabilities with which 1 choosesT , M, and B, and (r , 1− r) be the probabilities with which 2chooses L and R.

• Can there be any vector (p, q, r) that is a MSNE?

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• Let (p, q, 1− p − q) be the probabilities with which 1 choosesT , M, and B, and (r , 1− r) be the probabilities with which 2chooses L and R.

• Can there be any vector (p, q, r) that is a MSNE?• If (p, q, r) is a NE, we must have that: given p and q, 2 isindifferent between L and R, and, given r , 1 is indifferentbetween T , M, and B.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• If (p, q, r) is a NE, we must have that: given p and q, 2 isindifferent between L and R, and, given r , 1 is indifferentbetween T , M, and B.

• That is, 2p + 3q + 1 (1− p − q) = 3p + 2q + 2 (1− p − q)and 2r + 2 (1− r) = 0r + 3 (1− r) = 3r + 0 (1− r)

• We can’t find a vector (p, q, r) such that these equations areall satisfied. Therefore, there cannot be any MSNE here.

Example

L RT 2,2 2,3M 0,3 3,2B 3,1 0,2

• In the end, we conclude that the unique MSNE of this gameis(( 1

2 ,12 , 0),( 1

Example

L RT 4,2 2,3M 2,3 2,2B 1,1 3,2

Mixed Strategy Nash Equilibrium

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