Solution to the Pie Division Bargaining Problem

This problem describes a classic alternating-offer bargaining game, which can be analyzed using game theory. The core assumption is that both Johnny and John are rational players, meaning they will always act to maximize their own utility (the amount of pie they receive, adjusted for its "attractiveness").

1. Initial Setup and Definitions

• Players:
  • Johnny (Player A), who proposes first.
  • John (Player B).
• Game Structure: An alternating-offer bargaining game over a finite or infinite horizon. The total value of the pie at the start is normalized to 1.
• Discount Factors: The "attractiveness" lost each hour is a time cost. We can represent this with a discount factor for each player, which is the value that remains after one hour of waiting.
  • Johnny's discount factor: δA=1−100X. This is the value Player A assigns today to receiving one unit of the pie in the next round.
  • John's discount factor: δB=1−100Y.
  • A player is more "patient" if their discount factor is higher (i.e., they lose less value per hour).
• Core Principle: A rational player will accept an offer if it is at least as good as the value they expect to get by rejecting the offer and waiting for the next round. Delay is costly for both players, so it is in their mutual interest to reach an agreement as soon as possible.

Part A: Lola Returns in 6 Hours (Finite Horizon)

In a game with a known, finite end, the solution can be found using backward induction. We start from the last possible action and work our way back to the beginning.

The Reasoning (Step-by-Step):

• Terminal Outcome (After Round 6): If no agreement is reached by the 6th hour, Lola eats the pie. The payoff for both players is (0, 0).
• Round 6 (B proposes): This is the last chance. If B's offer is rejected by A, the payoff will be (0, 0). Therefore, B can offer A an infinitesimally small slice of the pie, ϵ, and keep the rest, 1−ϵ. Player A, being rational, will accept this offer because ϵ>0. In the limit, the payoff for this round is (0 for A, 1 for B).
• Round 5 (A proposes): Player A knows what will happen in Round 6 if his offer is rejected. If they move to Round 6, B will receive a payoff of 1. However, because one hour passes, the value of that future payoff to B today is discounted to δB×1=δB. To make B accept the offer in Round 5, A must offer B a share worth at least δB. A has no incentive to offer more.
  • A's proposal: (1−δB,δB). B will accept, as receiving δB now is at least as good as receiving 1 in the next (devalued) round.
• Round 4 (B proposes): Player B knows that if his offer is rejected, Player A will receive a payoff of 1−δB in Round 5. The present value of that payoff to A today is δA×(1−δB). Therefore, B must offer A a share of at least this value to ensure acceptance.
  • B's proposal: (δA(1−δB),1−δA(1−δB)). A will accept.

This logic continues, rolling back to the very first round. The share a player demands is what remains after giving the other player just enough to make them indifferent between accepting now and rejecting to wait for their next turn.

Conclusion for Part A:

The friends will not wait. A rational player will always accept a deal that is better than the discounted value of the next-best alternative. Johnny (Player A), knowing the entire sequence of events, will make an offer in the very first round that John (Player B) will immediately accept. The division will be:

• Johnny's (A) Share: SA=1−δB(1−δA(1−δB(1−δA(1−δB)))))
• John's (B) Share: SB=δB(1−δA(1−δB(1−δA(1−δB))))