Solution
Bargaining with Costly Delay
Show the problem again
Players 1 and 2 bargain over one dollar. In round 1 player 1 proposes a split, which player 2 accepts or rejects. After a rejection, player 2 proposes in round 2, then player 1 in round 3, and so on forever, alternating, with no deadline. If agreement on a split $(x_1,x_2)$ is reached in round $t$, player $i$ gets payoff $x_i - c_i\,(t-1)$, where the per-round waiting costs are $c_1 = 3/10$ and $c_2 = 1/10$. Never agreeing is worse than any agreement for both, so play always ends. A player who is indifferent accepts. Look for the subgame-perfect equilibrium. What share of the dollar does player 1 obtain? Give a reduced fraction.
Worked solution
The answer is 1/10. Let $a$ be the share a proposer keeps when player 1 is proposing, and $b$ the share when player 2 is proposing (shares of the dollar; the waiting cost is subtracted separately). A responder who rejects becomes the next proposer, and that is worth his future share minus one more round of his own waiting cost, so the proposer must offer him $\max(0,\text{future share}-c)$. Hence $a = 1-\max(0,\,b-c_2)$ and $b = 1-\max(0,\,a-c_1)$. Try $a=c_2=1/10$: then $a-c_1 = -1/5 < 0$, so $b=1$, and $a = 1-(1-1/10) = 1/10$, which is consistent. Player 2 gets $9/10$ when player 1 proposes, and player 1 is shut out when player 2 proposes. Intuitively, player 1 waits at a cost of $0.3$ per round, more than the whole share he could hope to win back, while player 2 waits at only $0.1$: the cheaper waiter dominates, and the other can only extract the difference in waiting costs. A backward-induction computation over horizons of 30 to 101 rounds gives $1/10$ for both parities of the horizon.
Source: Rubinstein (1982), the model with fixed per-round bargaining costs rather than discounting. Statement written for AxiomIQ.