Question

d e f

a 5,3 3,5 8,5

b 1,2 0,2 9,3

c 6,3 2,4 8,9

The game above has a Nash Equilibrium in which Player 1 plays strategy_ and Player 2 plays strategy _ with probability at least _ (Please, do not use fractions, if your answer is 2/5 use 0.4)

Answer 1

Player 1 has 3 choices a, b and c

Player 2 has three choices d, e and f

We have given payoffs

		Player 2
		d	e	f
Player 1	a	5,3	3,5	8,5
	b	1,2	0,2	9,3
	c	6,3	2,4	8,9

Look at player 1 first,

Compare a and b, 5>1, 3>0 but 8<9 so no dominant strategy.

Compare b and c, 1<6, 0<2 but 9>8 again no dominant strategy

Again compare a and c, 5<6, 3>2 and 8=8 so no dominant strategy

Now look for Player 2,

For player 2, d is a dominated strategy as compare to e and f, it has the lowest payoffs

Compare d and f, f had high payoff than d

We can delete d and we have

		Player 2
		e	f
Player 1	a	3,5	8,5
	b	0,2	9,3
	c	2,4	8,9

Between a and b,

Player 1 is playing a with probability p and b with probability (1-p) when player 2 chooses e

His payoff from this strategy,

3p + (1-p)*0 = 3p > 2 since p is between 0 and 1

When player 2 plays f and player 1 plays a with probability p and b with probability (1-p)

His payoff from this strategy,

8p + 9(1-p) = 9 – p > 8 since p is between 0 and 1

So player 1 will not play c as c gives him either 2 (when player 2 plays e) or 8 (when player 2 plays f)

So c could be eliminated

		Player 2
		e	f
Player 1	a	3,5	8,5
	b	0,2	9,3

When player 1 plays a, best strategy for player 2 is to choose either e or f

When player 1 plays b, best strategy for player 2 is to choose f

When player 2 plays e, best strategy for player 1 is to choose a

When player 2 plays f, best strategy for player 2 is to choose b

So there is no pure nash equilibrium

We check for mixed strategy

Player 1: a with probability p and b with prob (1-p)

Player 2: e with probability q and f with prob (1-q)

			q	1- q
			Player 2
prob			e	f
p	Player 1	a	3,5	8,5
1-p		b	0,2	9,3

So expected payoff will be

E(e) = 5p + 2(1-p)= 2 + 5p

E (f) = 5p +3(1-p) = 3 + 2p

E(e) = E (f)

2 + 5p = 3 + 2p

p = 0.33