Bargaining by taking turns

The previous lesson picked a split by asking what properties a good answer should have, and it described nobody making an offer, refusing one, or storming out.

One offer, take it or leave it

Start with the shortest possible negotiation. Party A proposes a division of £100 and party B either accepts, in which case it happens, or rejects, in which case both get nothing. This is the ultimatum game, and backward induction settles it in a line.

At the final node B compares the offer with zero. Any positive amount beats nothing, so B accepts anything above zero. Anticipating that, A offers the smallest positive amount and keeps the rest. If money is divisible into pennies the unique subgame perfect equilibrium is A keeping £99.99, and if it is perfectly divisible the equilibrium is A keeping everything with B indifferent and accepting.

The result is not an artefact of the tiny horizon; it is the whole logic of commitment from the earlier lesson, arrived at for free. A has the power to make a take-it-or-leave-it offer, which is a commitment device the rules handed over, and the entire surplus follows the commitment. What makes the prediction interesting is that it is so easy to test, and it fails badly, which the second half of this lesson takes up.

Letting the other side answer back

Give B a right of reply. If B rejects A's offer, a period passes and B makes a counter-offer, which A may accept or reject; if A rejects, both get nothing. Delay is costly: a pound agreed one period later is worth $\delta$ pounds now, with $\delta$ the discount factor of the repetition lesson.

Solve backwards. In the second period B is making a take-it-or-leave-it offer, so B takes the whole pie, worth $\delta$ to B in first-period terms. So in the first period B will accept anything worth at least $\delta$, and A offers exactly that, keeping $1-\delta$. With $\delta =0.9$, A keeps just 10 per cent. Having the last word is worth almost everything when players are patient.

Now add a third period in which A proposes again. In period three A takes everything, so in period two B must leave A a share worth $\delta$, keeping $1-\delta =0.1$. In period one A must leave B a share worth $\delta (1-\delta )=0.09$, keeping 0.91.

The shares as the horizon grows, still at $\delta =0.9$, run 1, 0.1, 0.91, 0.181, 0.8371, 0.2466, 0.7781, and so on. They oscillate, because whoever holds the last offer takes everything and each extra period flips who that is, and they converge, because the swings shrink by a factor of $\delta$ each time. The limit is 0.5263.

Example. With $\delta =0.8$ and three periods of alternating offers, what does the first proposer keep?

In period three the first proposer takes everything. So in period two the second party must leave them $\delta =0.8$, keeping 0.2. In period one the first proposer must leave the other $\delta (0.2)=0.16$, so keeps 0.84.

Now you. Same $\delta =0.8$, but with four periods. What does the first proposer keep, and why has the answer moved so far?

Never having the last word

Ariel Rubinstein's 1982 paper removed the last word entirely. Offers alternate indefinitely: A proposes, B accepts or counters, A accepts or counters, forever, with each period of delay shrinking the pie by the factor $\delta$. There is now no final node, so backward induction has nowhere to start.

Use stationarity instead. Every subgame that begins with a given player proposing looks exactly like every other, so if a player's equilibrium share when proposing is $x$, it is $x$ in every such subgame. Then the responder's choice is between accepting what they are offered now and rejecting to become the proposer one period later, which is worth $\delta x$ to them. The proposer offers exactly $\delta x$, no more, and keeps the rest:

At $\delta =0.9$ the proposer keeps 0.5263 and the responder 0.4737, matching the limit of the finite sequence. At $\delta =0.5$ the proposer keeps $2/3$. As $\delta$ rises to 1 the split approaches even, and the proposer's advantage, $(1-\delta )/(1+\delta )$ of the pie, is exactly the cost of one period's delay.

Rubinstein's real achievement was uniqueness rather than the formula. Stationarity was assumed above; his proof does not assume it. Let $M$ be the highest and $m$ the lowest share a proposer receives in any subgame perfect equilibrium. A responder can always secure $\delta m$ by rejecting, so no proposer keeps more than $1-\delta m$, giving $M\le 1-\delta m$. A responder will never turn down more than $\delta M$, so every proposer can secure at least $1-\delta M$, giving $m\ge 1-\delta M$. The two inequalities force $M=m=1/(1+\delta )$. The equilibrium is unique among all subgame perfect equilibria, which is a much rarer outcome than the previous lessons might suggest.

Two features of the solution deserve attention because they are testable. Agreement is immediate: the first offer is accepted, and no delay is ever observed on the equilibrium path. And the shrinking of the pie is never actually paid; it works entirely through the threat of paying it, in the same way that the commitment lesson's capacity investment worked through a comparison that was never reached.

Example. Two firms bargain over £100,000 of joint surplus, alternating offers monthly, and each values money a month later at 0.9 of its present worth. What is the equilibrium division?

The proposer keeps $1/(1+0.9)=0.5263$, so £52,632, and the responder takes £47,368. The proposer's advantage is £5,264, which is 10 per cent of the £52,632 the responder would be waiting a month for, and that is the entire content of "first-mover advantage" here.

Now you. The same firms bargain annually rather than monthly, so that a year's delay leaves them valuing money at 0.5 of its present worth. What is the division now?

Impatience decides

Nothing forces the two parties to be equally patient, and once they are not, the formula says something with real content.

Let party 1 discount at ${\delta }_1$ and party 2 at ${\delta }_2$. Write $x$ for party 1's share when party 1 proposes and $y$ for party 2's share when party 2 proposes. Party 1, proposing, must leave party 2 the discounted value of party 2's own proposer share, so $x=1-{\delta }_{2}y$. Symmetrically $y=1-{\delta }_{1}x$. Substituting the second into the first gives $x=1-{\delta }_2+{\delta }_1{\delta }_{2}x$, so

With ${\delta }_1=0.9$ and ${\delta }_2=0.8$, party 1 receives $0.2/0.28=0.7143$. Patience is worth 71 per cent of the pie against an opponent who loses a fifth of the value with every round of delay. Swap the two and party 1's share falls to 0.3571.

The comparative static is the useful part. Your share rises with your own patience and falls with your opponent's, and the mechanism is entirely about who can afford to wait: an impatient party is one for whom every rejection is expensive, and the other side prices that. This is what a strike fund buys, what a firm's inventory buys, and what a mortgage payment due next week takes away.

Example. A union discounts at ${\delta }_1=0.95$ per week and the firm at ${\delta }_2=0.7$ per week, because the firm's plant is idle and the union has a strike fund. The union proposes first over a £2 million surplus. What does the union get?

$x=(1-0.7)/(1-0.665)=0.3/0.335=0.8955$, so £1.79 million. Almost the entire surplus goes to the side that can wait, and the firm's weekly loss of 30 per cent of the value is what pays for it.

Now you. Suppose the firm builds up stock so that its weekly discount factor rises from 0.7 to 0.9, and the union's stays at 0.95. What is the union's share now?

Meeting the axioms in the limit

Two apparently unrelated answers to the same question now sit side by side, and they turn out to be the same answer.

Let the interval between offers be $\Delta$ and let each party have a continuous discount rate $r_i$, so ${\delta }_i=e^{-r_i\Delta }$. As $\Delta$ shrinks, $1-{\delta }_i\approx r_i\Delta$, and substituting into the formula gives

The share depends only on the ratio of impatience, and the first-mover advantage vanishes, because the advantage was worth one period's delay and the period has become negligible. Numerically, with $r_1=5$ per cent and $r_2=15$ per cent, the discrete formula gives 0.768 at $\Delta =1$, 0.752 at $\Delta =0.1$ and 0.7502 at $\Delta =0.01$, converging on the predicted 0.75.

That limit is exactly the asymmetric Nash bargaining solution of the previous lesson, with bargaining weights $r_2/(r_1+r_2)$ and $r_1/(r_1+r_2)$. Ken Binmore, Ariel Rubinstein and Asher Wolinsky set this out in 1986, and the moral is the one Nash had hoped for: an axiomatic answer earns its keep when a procedural model of the haggling delivers it. When it is the disagreement point rather than impatience that drives the outcome, a different procedural model converges on the plain Nash solution with the outside options in the disagreement point, so the axioms are not one thing but a family whose members depend on what actually threatens the negotiation.

What people actually do

The ultimatum game is the easiest prediction in this course to test, and it is comprehensively wrong.

Werner Güth, Rolf Schmittberger and Bernd Schwarze ran it in 1982. The theory says offer the smallest positive amount and have it accepted. What happens, across hundreds of replications since, is that the modal offer is 40 to 50 per cent of the pie, mean offers cluster around 40 per cent, and offers below about 20 per cent are rejected roughly half the time. Rejection means choosing nothing over something, and it persists when the stakes are raised: studies paying two or three months' income in Indonesia and in India still find low offers refused.

The cross-cultural evidence sharpens rather than dissolves the puzzle. Joseph Henrich and colleagues ran the game in fifteen small-scale societies in 2001 and found the variation between societies far larger than anything within industrialised samples. The Machiguenga of Peru, who live in largely independent households, offered about 26 per cent on average and almost never rejected. The Lamalera whale hunters of Indonesia, whose livelihood requires large cooperative crews, offered about 58 per cent. Among the Au and Gnau of Papua New Guinea, where accepting a gift creates an obligation, offers above half were rejected as often as offers below it. The prediction fails in a different direction in each place, which is evidence that the payoffs in the laboratory are not the payoffs in the participants' heads.

Alternating offers fares no better. Jack Ochs and Alvin Roth's 1989 experiments found first offers far from the subgame perfect prediction, and, more damagingly, found that most rejections were followed by a disadvantageous counter-offer: the rejecting party countered with a demand worth less to themselves, after discounting, than what they had just turned down. No adjustment to the discount factor rescues that, since it is not a mistake about arithmetic but a refusal to be treated a certain way.

What the model still buys

The natural conclusion, that bargaining theory is useless, is too quick. Three things survive.

The comparative statics survive. Whatever the level of offers, the direction of the effects is right: the more patient side does better, better outside options improve terms, and the ability to make the final offer is worth something. These predictions hold in the experiments even where the point predictions do not.

The mechanism survives. Impatience is a real cost of disagreement, and the model correctly identifies that the cost of failing to agree, rather than any notion of deservingness, is what moves the split. Anyone preparing for a negotiation who spends their effort on their alternative rather than on their argument is applying it correctly.

And the failure is informative. The model predicts immediate agreement, and real disputes involve strikes, lockouts, delayed settlements and litigation that both sides know will be settled eventually. The best explanation for delay is the assumption this course has not yet dropped: that each side knows what the other's payoffs are. If a party's patience or reservation value is private, refusing an offer becomes a way of signalling that you are the tough type, and delay stops being a mistake and becomes information transmitted at a price.

Where this leaves us

Two routes to the same split have now been travelled, one axiomatic and one procedural, and they converge as the delay between offers goes to zero. Both assume that each party knows the other's preferences exactly.

That assumption has been in force since the first lesson, and it is false in most of the situations the subject is applied to. A bidder does not know what a painting is worth to the person beside them, an insurer does not know how careful a customer is, and a negotiator does not know how badly the other side needs a deal. The next lesson gives each player a private type and a belief about everyone else's, which is the last major piece of machinery in the course, and applies it to the setting where the theory has done most practical good: auctions.