Question

There are two people dividing 5 yuan by biding simultaneously. If the sum of their bids equals 5, the bid is the amount of money they get. If the sum of their bids is
below or above 5, then no one gets anything.

Suppose all their bids are integers, draw this game in normal form and solve for the Nash equilibria.

Answer 1

If Sum of bids of Player 1 and Player 2 = 5 then they will get the amount of bid and  If the sum of their bids is
below or above 5, then no one gets anything.

Hence we have the following Normal form of game.

Player 2

		0	1	2	3	4	5
	0	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,5)
	1	(0,0)	(0,0)	(0,0)	(0,0)	(1,4)	(0,0)
Player 1	2	(0,0)	(0,0)	(0,0)	(2,3)	(0,0)	(0,0)
	3	(0,0)	(0,0)	(3,2)	(0,0)	(0,0)	(0,0)
	4	(0,0)	(4,1)	(0,0)	(0,0)	(0,0)	(0,0)
	5	(5,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)

If Player 1 chooses 0 then Player 2 will choose 5 and if Player 2 chooses 5 then Player 1 can choose 0 and hence (0,5) is a nash equilibrium.

If Player 1 chooses 1 then Player 2 will choose 4 and if Player 2 chooses 4 then Player 1 will choose 1 and hence (1,4) is a nash equilibrium.

If Player 1 chooses 2 then Player 2 will choose 3 and if Player 2 chooses 3 then Player 1 will choose 2 and hence (2,3) is a nash equilibrium.

If Player 1 chooses 3 then Player 2 will choose 2 and if Player 2 chooses 3 then Player 1 will choose 2 and hence (3,2) is a nash equilibrium.

If Player 1 chooses 4 then Player 2 will choose 1 and if Player 2 chooses 1 then Player 1 will choose 4 and hence (4,1) is a nash equilibrium.

If Player 1 chooses 5 then Player 2 will choose 0 and if Player 2 chooses 0 then Player 1 will choose 5 and hence (1,4) is a nash equilibrium.

Hence, Nash equilibrias are strategy :

(0,5) , (1,4) , (2,3) , (3,2) , (4,1) and (5,0)