Question

Consider the below 2x2 normal form game.

		Player 2
		C	D
Player 1	A	5,2	1,3
	B	4,3	2,1

      Assume that Player 1 plays A with p probablity and B with (1-p) probability as Player 2 plays C with q probability and D with (1-q). Given that players are rational, which of the following is true?

	a.	If p = 1/2, then Player 2 Plays C.
	b.	If p =2/3, then Player 2 Plays D.
	c.	If q =1/4, then Player 1 Plays A.
	d.	If q = 1/2, then Player 1 Plays B.
	e.	All of the above.
	f.	Only (a) and (d).
	g.	None of the above.

Answer 1

Assume that Player 1 plays A with p probability and B with (1-p) probability as Player 2 plays C with q probability and D with (1-q)

Now for player 1 to be indifferent between playing A and B, we must have:

E(A) = E(B)

=> 5q + 1(1-q) = 4q + 2(1-q)

=> 5q + 1 - q = 4q + 2 - 2q

=> 2q = 1 => q = 1/2 and if q < 1/2, E(A) < E(B)

Now for player 2 to be indifferent between playing C and D, we must have:

E(C) = E(D)

2p + 3(1-p) = 3p + 1(1-p)

=> 2p + 3 - 3p = 3p + 1 -p

=> 3p = 2 => p = 2/3 and if p < 2/3, E(C) > E(D)

Thus, the answer is

a.

If p = 1/2, then Player 2 Plays C

All oher options give wrong claims about the startegy fo players and are hence incorrect