Three players 1,2, and 3, are deciding on how to divide a cake worth $1, using t

Question

Three players 1,2, and 3, are deciding on how to divide a cake worth $1, using the following procedure: Player 1 first divides the cake into three portions: x,y,and z such that x+y+z=1, x,y,z greater than or equal to 0. Player 2 then picks which of the three portions to consume. Next, player 3 picks the picks the one from the remaining two portions, followed by player 1 consuming the last one left.

a) Identify the subgame perfect equilibrium of this game. How will player 1 divide the cake?

b) Suppose that player 1 picks his portion before player 3 does (with the rest of the procedure unchanged). Identify the sublime perfect equilibrium of this game. How will player 1 divide the cake?

Explanation / Answer

a) If player2 chooses first she may choose any of the three as the unique values of x,y,z are not specified.

If she chooses x then, player 3 is left with y and z. Similarly if she chooses y then, player 1 gets z and vice versa. Similar logic applies for the rest. As player 1 is left with no other chice she won't gamble and hence x=y=z=1/3

b) In this case also she gets the second choice. Player2 will pick up the most if she chooses first.Hence, player1 will not risk to take less as more is better. Hence, here also x=y=z=1/3

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