Question

Problem 1: A Bargaining Game

Two players, PI and PII, have to divide 23 M&M's: 12 regular ones (R's) and 11 peanut-coated ones (P's). PI likes both types equally but PII likes only the P's (she will have no use for any R's she gets). Thus, u1(r,p) = r + p and u2(r,p) = p. Moreover, both players' preferences are known to both and the bargaining will be restricted to the allocations of M&M's, e.g., no monetary side payments or other modifications involved.

Part 1a: PI Divides and PII Chooses
PI divides the M&M's into two piles in any way he chooses. Then PII selects one of the piles, leaving the other one to PI. Use the backwards induction method to compute the outcome of the game.

Part 1b: PII Divides and PI Chooses
The same as in Problem 1, but with the order reversed.


Problem 2

Suppose that PI's preferences are the same and still known as in Problem 1 (it is common knowledge to everybody that he likes both types of M&M’s equally), but he is not sure about PII's preferences. He thinks that there is a 2/3 chance that she likes only P's and a 1/3 chance that, like himself, she likes both. (She knows which of the two types she likes, and he knows that she knows her own taste. But he does not know what her taste is.)

i. How would this change his strategy in Problem 1a?
ii. In this situation, should he prefer that she do the dividing? Give a very brief reason for your conclusion.

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Part 1a: PI Divides and PII Chooses

PI divides the M&M's into two piles in any way he chooses. Then PII selects one of the piles, leaving the other one to PI. Use the backwards induction method to compute the outcome of the game.

Solution:

Let ri and pi be the number of regular and peanut-coated M&M’s in pile i for i=1,2. Then given two piles by PI, PII will select the pile with the largest number of peanut-coated M&M’s. Without loss of generality, let p1>p2. Then PII will select pile p1 and his utility will be p1 while PI will have a utility of r2+p2. Notice that p2∈{1,2,3,4,5} and PI wants to maximize his utility. Hence, PI will assign p2=5 and assign r2=12. This will force p1=6 and r1=0.

This is a Nash Equilibrium, since either player cannot gain from deviating from their respective positions. Thus PI will pile as follows: pile 1 consists of 6 peanut-coated M&M’s while pile 2 consists of 5 peanut-coated M&M’s and 12 regular ones. The utilities for this equilibrium would be 17 and 6 for PI and PII, respectively....

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