Solution
A Shrinking Pie in Six Rounds
Show the problem again
Players 1 and 2 bargain over a pile of coins. In round 1 the pile has 100 coins and player 1 proposes how to split it. If player 2 rejects, the pile shrinks by 15 coins (to 85) and player 2 proposes a split of the new pile. The roles alternate and the pile shrinks by 15 coins after every rejection, so in round 6 there are 25 coins and player 2 makes the final proposal. If the round-6 offer is rejected, both players get nothing. Coins are indivisible, each player wants as many coins as possible, and a player who is indifferent between accepting and rejecting accepts. In the subgame-perfect equilibrium, how many coins does player 1 keep in round 1?
Worked solution
The answer is 45. Work backward. In round 6, player 2 proposes and a responder accepts anything (rejecting yields 0), so the proposer keeps all 25. In round 5 (pile 40, player 1 proposes), the responder can get 25 by rejecting, so the proposer offers 25 and keeps 15. In round 4 (pile 55, player 2 proposes), player 1 can get 15 by rejecting, so the proposer keeps 40. In round 3 (pile 70), the responder gets 40 by rejecting, so the proposer keeps 30. In round 2 (pile 85), the responder gets 30 by rejecting, so the proposer keeps 55. In round 1 (pile 100), player 2 can obtain 55 by rejecting, so player 1 offers 55 and keeps 45.
Source: Finite-horizon alternating-offer bargaining in the style of Stahl (1972) and Rubinstein, with a pie that shrinks each round. Statement written for AxiomIQ.