A Shrinking Pie in Six Rounds

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?

• nash equilibrium

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.