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A house sale stalls over the last few thousand dollars. Every week the deal drags on, both sides lose: the buyer pays rent, the seller pays a second mortgage. The clock, more than any argument, decides the price.

Strategic bargaining

The axiomatic answer said what a fair split looks like. Rubinstein asked how bargaining actually plays out when offers alternate and delay is costly. The result is sharp: a single subgame-perfect split, agreed immediately, set by who loses less from waiting.

From earlier lessons

Two quick ones before we start.

A subgame-perfect equilibrium requires optimal play...

In Nash bargaining, a better outside option gives you...

Predict, then play

You and a bot alternate offers to split a pie of 1. Each round of delay shrinks its value by $\delta$ . You move first. Propose how much to keep; the bot responds by playing the equilibrium.

Why did that happen?

Take a guess, then reveal the reason

Why settle at once instead of haggling for rounds? Because both sides can see how it ends. If you lowball, the responder rejects, becomes the proposer next round, and claims the proposer’s share of a slightly smaller pie. Knowing that, you offer them exactly what waiting is worth to them, no more, and they accept immediately. Delay only shrinks the pie, so no one chooses it in equilibrium.

The math

Let the proposer’s equilibrium share be $x$ . By symmetry of the stationary game, if the responder rejects, they become the proposer and claim $x$ of a pie discounted to $\delta$ , worth $\delta x$ to them now. So the proposer must leave them at least $\delta x$ :

x = 1 - \delta x \quad\Longrightarrow\quad x = \frac{1}{1+\delta}.

At $\delta = 0.9$ the proposer keeps $\tfrac{1}{1.9} \approx 0.526$ : barely more than half. The advantage of moving first vanishes as players get patient.

Definition, Rubinstein bargaining

With alternating offers and a common discount factor $\delta \in (0,1)$ , the infinite-horizon game has a unique subgame-perfect equilibrium: the first proposer offers $(\tfrac{1}{1+\delta}, \tfrac{\delta}{1+\delta})$ and it is accepted at once. With different discount factors the proposer’s share is $\tfrac{1-\delta_2}{1-\delta_1\delta_2}$ : the more patient side gets more.

As $\delta \to 1$ the split approaches $(\tfrac12, \tfrac12)$ , matching the Nash bargaining solution with equal footing. Two different questions, the axiomatic and the strategic, converge on the same answer. That link is the Nash program.

Work it out

Solve for the split, then vary patience

The proposer leaves the responder their continuation value. If the responder rejects, they propose next round and keep $x$ of a pie worth $\delta$ , so their continuation is $\delta x$ . The proposer keeps the rest: $x = 1 - \delta x$ , giving $x = 1/(1+\delta)$ .

At $\delta = 0.5$ the proposer keeps $1/1.5 \approx 0.667$ ; at $\delta = 0.9$ , about $0.526$ ; at $\delta \to 1$ , exactly $0.5$ . Patience compresses the first-mover advantage.

With unequal patience, the proposer (player 1) keeps $\tfrac{1 - \delta_2}{1 - \delta_1 \delta_2}$ . If player 1 is very patient ( $\delta_1$ near 1) and player 2 is not, player 1 takes almost the whole pie. Relative patience is the source of bargaining power.

Key idea

Alternating-offer bargaining ends in the first round at a split set by patience: the proposer keeps 1/(1+delta). The one who loses less from waiting wins the larger share.

Check yourself

Check yourselfapply

Both players share a discount factor delta = 0.9. What is the proposer's equilibrium share, and when is the deal struck?

Check yourselfanalyse

As both players become perfectly patient (delta approaches 1), what happens to the split?

Check yourselfbrings back: subgame perfection

Why is the outcome agreement in the very first round, rather than a long sequence of rejected offers?

Sources

Rubinstein, A. (1982). Perfect equilibrium in a bargaining model. Econometrica, 50(1), 97-109.