Question

Q3 Consider the game below between Tony and Bil Each has two strategies. Tony's payoffs are given first. The game has a unique Nash equilibrium. Bill Drink Pass Safe-3, 3 3, -3 Risky 3, -33, 3 (1). What is the mixed strategy Nash equilibrium? What is Tony's expected payoff? What is Bill's expected payoff? (20 points) (2). Recently, the rules of the game have changed. If Tony selects Safe and Bill selects Pass, the payoffs become (6, 6) instead of (3, 3). All other payoffs remain the same. What is the new mixed strategy Nash equilibrium? (20 points)

PLZ TYPE THE ANSWER NOT HAND WRITEING CUZ CANT READ IT

Answer 1

1) let Tony play safe (S) with probability P

& Play Risky with probability (1-p)

Bill play Drink with probability q,

Play pass with probability (1-q)

So EU_{​​​​​T} = -3pq+ 3p(1-q)+3q(1-p)-3(1-p)*(1-q)

= -3pq + 3p -3pq +3q -3pq-3(1-p-q+pq)

= -9pq+3p+3q-3+3p+3q-3pq

= -12pq+6p+6q -3

Now differentiate EU of Tony with respect to P

Then -12q+6 = 0

q* = .5

similarly for Bill

EU_{​​​​​B} = 3pq -3p(1-q)-3q(1-p)+3(1-p)(1-q)

= 3pq -3p+3pq-3q+3pq+3-3p-3q+3pq

= 12pq-6p-6q+3

Now differentiate it with respect to q & put equal to zero

so 12p -6= 0

p* = .5

so mixed strategy NE : (P,q) = (.5,.5)

now expected payoff:

EU_{​​​​​T} = -12*.25 +6-3 = 0

Similarly for Bill, EU = 0

B)

Tony/ bill	Drink	Pass
Safe	(-3,3)	(6,-6)
Risky	(3,-3)	(-3,3)

Now EU_{​​​​​T} = -3pq +6p(1-q)+3q(1-p)-3(1-p)(1-q)

= -3pq +6p-6pq +3q -3pq-3+3p+3q-3pq

= -15pq +9p +6q-3

Differentiate it wrt P & put equal to zero.

-15q +9 = 0

So q* = 9/15 = .6

for bill

EU_{​​​​​B} = -3pq-6p(1-q)-3q(1-p)+3(1-p)(1-q)

= -3pq -6p +6pq -3q +3pq +3-3p-3q +3pq

= 9pq-9p -6q +3

Differentiate it with respect to q & put equal to zero.

9p -6 = 0

p* = 6/9 = 2/3

mixed strategy NE :( P,q) = (2/3, .6)