Mixed Strategy Equilibrium in Game Theory

Q1)
a)
2
1
Let player 2, mixed strategy be given by p = prob(L)
3p = p + ε(1-p)
3p = p + ε - εp
2p = ε - εp
2p = ε(1-p)
ε = 2 p
1− p
b) Let p be probability of player 1 choosing U
ε ∈[0 , 1] for each p ∈[0 , 1]
Player 2 would play D insteady U
(q = 0) if 0 < ε(1 – p)
 P < 1
P < 1 => ε = 0
P = 1 => ε∈(0 , 1)
c) Player 1
p = 1
ε∈(0 , 1)
q = 1
q = 1, then ε = ½ in order to be true that p = 1. Hence BNE are restricted to
p =1, ε∈( 1
2 , 1) and q = 1
this will support the 3rd pure Nash equilibrium (UU, L)
d) BNE of which player 2randomize U and D
L R
U 1, 3 0, 1
D 0, 0 2, ε

Player 2 plays L
Player 2 plays L in steady of R
R(p =1) if ½ [3q + 0(1-q)] > ½ [1q + ε(1-q)]
3/2q > ½ [q + ε - εq]
3/2q – 1/2q > ½ ε(1 –q)
q > ½ ε(1-q)
this can be summarized as
q > ½ ∈ ( 1−q )=¿ p=1
q = 1 => ε ∈ [ 0 ,1 ]
Q2
20.14 Find at least one mixed – strategy Bayes – Nash equilibrium in the Bertrand pricing game
First mixed strategy
Suppose the husband plays F with probability 1. A type 1 wife’s expected payoff from playing F
is therefore 1
Playing O is 3(1, -1). The wife plays a mixed strategy only if she were indifferent between F and
O, that is, if λ -3(1-λ) or λ = ¾
Second mixed strategy
Type 2 wife would get the same payoffs from playing F as type 1 wife gets from playing O,
similarly same payoffs would play O as a type 1 wife gets from playing F
Third mixed strategy
The husband would be indifferent only when; 3μ1ρ + 3μ2(1 -ρ ¿ = (1 - μ1)ρ+1−μ2)(1- ρ ¿,
therefore F and O would be best responses, therefore, λ = ¾ would be best response
Fourth mixed strategy
If mixed strategies (m1, m2) would satisfy ρ ( 4 μ 1−1 ) =(1−ρ)(1−4 μ 2), the husband would
have best mixed strategy response of λ = ¾
Fifth mixed strategy
When the husband would play λ = ¾ then mixed strategy pair (m1, m2) would be best response
for the wives

20.15
Regardless of r value, there must be a mixed strategy Bayes Nash equilibrium when;
λ = ¾, μ1, μ2 = (1/4, ¼)
Any addition mixed strategy Bayes Nash equilibrium would characterized by λ = ¾ and (m1,
m2) should satisfy ρ ( 4 μ 1−1 ) =(1−ρ)¿)
20.16
Two pure strategies Bayes Nash equilibrium would exist whenρ> 3
4 . When the husband F and
wives play (F, O) in the first case. The second case is when the husband would play O and wives
would play (O, F). One pure strategy Bayes Nash equilibrium would exist when ¼ ≤ ρ< 3
4
20.17
If ρ< 1
4 , then r =0, where the husband knows that her wife is type 2. There would be no pure
strategy Nash equilibrium. Since the husband wants to go to a same event but the wife wants to
go to a different event.
20.18
Probability of of husband playing F = 1
The expected playoff of wife type 1 playing F = 1
Type 1 wife playing O the playoff = 3(1 -1)
Mixed strategy that would be played by the wife if she would be indifferent between F and O
would be when λ-3(1-λ) or when λ = ¾
Type 2 would get the same payoff as she play F while type 1 would be playing O, similarly when
type 2 would play O she would get the same playoff while type 1 wife plays F
20.19