Question

1. In class, relevant to gerrymandering, I talked about the game of Divide and Choose. There’s a pile of goodies. Little Annie divides the big pile into two piles. Then Little Bobby chooses which of the two piles he wants, and Annie gets the other. The advantage of the method is that both children can see to it that they don’t envy what the other child gets. Annie can divide the big pile into two piles of equal value to her, and Bobby can choose the one he likes better, and neither can complain.

If the cake is worth 100 total to both, and the icing is worth 0 to Annie and 50 to Bobby, then Annie can do this:

Pile 1: all the icing, and 26% of the cake      Value to Annie 26. <----------

Value to Bobbie 76

Pile 2: no icing,and 74% of the cake             Value to Annie 74

Value to Bobbie 74. <------------ (this arrow is supposed to be connected to the above arrow)

Bobby will choose pile 1 because 76 is better than 74 (arrows). Bobby will get 76, and she’ll get pile 2 and get 74. (They’ll get the outcomes I put in bold.)

What would each get if Annie put half the icing and half the cake in each pile? Is that outcome Pareto efficient?

2. I’ll talk more about voting soon, but here’s something you can answer now. Let’s assume that there are exactly four candidates B, C, R and T in a primary election, and there are exactly four citizens voting on them. The citizens rank the four candidates starting from their favorite on the left to their least favorite on the right. So Citizen 1 prefers T to R to C to B.

Citizen 1.   T, R, C, B            

Citizen 2.   B, T, C, R

Citizen 3.   B, T, R, C

Citizen 4.   C, T, R, B

Which candidates are Pareto efficient for these four citizens? If a candidate is not Pareto efficient say why. (Hint: as usual, to decide whether a candidate is PE, look for another candidate that everyone prefers (“everyone”!).)

Answer 1

1)

As per the given situation, if Annie put half the icing and half the cake in each pile, following are the values for Annie and Bobbie;

Here, Pile I would be equal to Pile II, as half the icing and half the cake is in each pile.

For Annie (Pile I = Pile II) –> [½ Icing + ½ cake] -> 0 (from Icing) + 50 (from Cake) = Value to Annie 50

For Bobbie, Again, (Pile I = Pile II) –> [½ Icing + ½ cake] -> 25 (from Icing) + 50 (from Cake) = Value to Bobbie 75

This outcome is not Pareto Optimal. As we can calculate that marginal rate of substitution between icing and cake for Annie is zero (MUi/MUc = 0/100) because her marginal utility from icing is zero. Similarly, for Bobbie it is ½ (MUi/MUc = 50/100), he is ready to exchange 1 unit of icing for 2 units of cake. Hence Pareto efficiency can be obtained by equating values for both.

Following distribution will lead to equal values than what is obtained above.

As Annie as no value for Icing, it better that she put all icing in one pile and divide cake as such total value of both are equal. Hence,

Pile 1 -> [0 icing + 75 % of cake] -> Value to Annie 75 (0 icing + 75 cake) and Value to Bobbie 75 (0 icing + 75 cake)

Pile 2 -> [all icing + 25% of cake] -> Value to Annie 25 (0 icing + 25 cake) and Value to Bobbie 75 (50 icing + 25 cake)

As Bobbie is indifferent between two piles and choosing pile 2 lead to equal value to both and a Pareto efficient distribution.

____________________________________________________________________________________________

2)

Let’s examine all candidates set of preferences by four citizens –

Candidate B – Citizen 1 -> TRC, means for citizen 1, B is the last preference, he will first choose candidate T, then candidate R and thirdly candidate C.

Similarly, Citizen 2 -> B

                Citizen 3 -> B

                Citizen 4 -> CTR

Candidate B is not a Pareto efficient as he is last preference for two citizens.

….

Next, Candidate C – Citizen 1 -> TR

                                Citizen 2 -> BT

                                Citizen 3 -> BTR

                                Citizen 4 -> C

Candidate C is also not a Pareto efficient as he is last preference for citizen 3 and second last for citizen 1 and 2.

……

Next, Candidate R – Citizen 1 -> T

                                Citizen 2 -> BTC

                                Citizen 3 -> BT

                                Citizen 4 -> CT

Candidate R is also not a Pareto efficient as he is last preference for citizen 2 and second last for citizen 3 and 4.

….

Next, Candidate T – Citizen 1 -> T

                                Citizen 2 -> B

                                Citizen 3 -> B

                                Citizen 4 -> C

Candidate R is a Pareto efficient as he is the first preference of citizen 1 and second for the remaining citizens. He has the best set of preference among all candidates.