Haggling that ends at the largest product

One cuts and the other chooses divides a cake between two people in two moves, and it works because each move is final. Most divisions are not made that way. A buyer and a seller, a union and an employer, two heirs over an estate: one names a split, the other refuses and names another, and the haggling goes on until somebody accepts. Nothing in the rules says when it ends or where. The only pressure is that every refusal costs time, and time costs both sides something.

John Nash proposed in 1950 that a reasonable bargain should maximise the product of what the two sides gain, and justified it with axioms — conditions any fair rule ought to meet — rather than with a model of how anybody actually haggles. The same product turns up as a rule for dividing goods, and there too it arrives as a principle. Ariel Rubinstein showed in 1982 that it is also what haggling produces. In a game of offers and counter-offers with nothing but the cost of delay to discipline it, there is exactly one outcome that never relies on an empty threat, it is reached at once, and as the delay between offers shrinks it converges to Nash’s product.

Offers, refusals and the cost of waiting #

The game is short to state. Player 1 proposes a split of a sum of one, giving herself $x$. Player 2 accepts, and the game ends, or rejects and after one round of delay proposes a split of his own. Player 1 accepts or rejects that, and so on, with no limit on the number of rounds. A share received after $t$ rounds of delay is worth $\delta^t$ times as much as the same share received now, where the delay factor $\delta$ is a number between nought and one: close to one if offers come fast or the players are patient, close to nought if every round wastes a great deal.

The difficulty is that almost anything can happen in such a game if threats are taken at face value. Player 1 can announce that she will refuse every offer below ninety per cent, and if player 2 believes her he should give her ninety per cent at once. But such a threat is empty if carrying it out would hurt her: when an offer of eighty arrives, refusing it means waiting a round for something she might not get. Reinhard Selten’s notion of a subgame-perfect equilibrium rules empty threats out by demanding that every player’s plan be optimal from every position the game can reach, including positions the plan says will never happen. Rubinstein’s theorem is about equilibria of that kind.

Counting back from a deadline #

With a deadline the game can be solved by working backwards, which Ingolf Ståhl did in 1972. Suppose the game must end after $T$ rounds, and whoever proposes in the last round can demand everything, since the other player gets nothing by refusing. In the round before, the proposer must offer at least $\delta$ times that to make refusal pointless, and can keep $1 - \delta$. In the round before that, she must leave $\delta(1 - \delta)$ and keeps $1 - \delta + \delta^2$. And so on back to the first round.

The first proposer’s share is a partial sum of an alternating geometric series, $1 - \delta + \delta^2 - \cdots \pm \delta^{T-1}$, which is $(1 - (-\delta)^T)/(1 + \delta)$. A geometric series is the sum that fits in one square, and this one zig-zags as the deadline moves by one round, because a deadline falling on the other player’s turn hands him the final say. The swings shrink by a factor $\delta$ each round, and the sequence closes in on $1/(1 + \delta)$.

The logic is the same as in the subtraction game where the winner is decided by who meets the first long run: with a definite end, every position is solved from the end backwards, and the first move inherits everything that follows. The surprise is what happens when there is no end.

No deadline, one answer #

Without a deadline there is no last round to count back from, and the obvious worry is that the game has many equilibria, or none. Rubinstein proved it has exactly one. When both players have delay factor $\delta$, the first proposer asks for $1/(1 + \delta)$, the other accepts immediately, and no other subgame-perfect behaviour exists.

The cleanest proof, due to Avner Shaked and John Sutton, takes the largest and smallest amounts the first proposer could get in any equilibrium, $M$ and $m$. Whatever the second player gets when it is his turn to propose is at most $M$ as well, by symmetry. So the first proposer need never offer him more than $\delta M$, since he cannot expect more by refusing, and she keeps at least $1 - \delta M$; giving $m \ge 1 - \delta M$. By the mirror argument, she can never keep more than $1 - \delta m$, giving $M \le 1 - \delta m$. Together the two inequalities force $M = m = 1/(1 + \delta)$. The infinite game is squeezed into one point by the fact that it looks the same from every round.

The shape of the answer deserves a second look. The first proposer’s advantage over an even split is $1/(1 + \delta) - \tfrac12 = (1 - \delta)/(2(1 + \delta))$, and it is exactly what refusing her costs the other player: if he refuses, he gets to propose, but only after a round’s delay, so his alternative is worth $\delta$ times the proposer’s share. The first mover’s bonus is the price of waiting, not a reward for power, and it vanishes as offers can be made faster. With offers a second apart and a day’s delay costing a few per cent, $\delta$ is within a whisker of one and the split is even to many decimal places.

The product as a limit #

Equal shares of a sum of money is the answer when both players value money at face value. When their values differ the game still has one equilibrium, but its shares are no longer a simple formula. Write $x$ for the share player 1 asks for and $y$ for the share player 2 offers her. Each proposer offers exactly enough to make the other indifferent between accepting now and proposing next round, so

$$u_2(1 - x) = \delta\, u_2(1 - y), \qquad u_1(y) = \delta\, u_1(x).$$

These two conditions pin down $x$ and $y$, and the figure at the top solves them for a player who values a share $s$ at $\sqrt s$ against one who values it at $s$. As $\delta$ approaches one, $x$ and $y$ squeeze together, and the point they squeeze onto is where the product $u_1 u_2$ is largest along the frontier of possible splits. Taking logarithms of the two conditions and subtracting shows why: in the limit they say that the relative loss to one player from moving the split slightly equals the relative gain to the other, which is the condition for the product to be at its peak.

Nash’s 1950 solution was the split that maximises $u_1 u_2$, derived from four axioms: that the answer should not depend on the units in which either player measures value, that it should be efficient, that it should treat symmetric players symmetrically, and that removing options nobody chose should not change it. Axioms of this kind pin down a sharing rule, but they do not say why anybody haggling would end up there. Rubinstein’s game does. It was the first complete case of what Ken Binmore, Rubinstein and Asher Wolinsky called in 1986 the Nash programme: supporting a cooperative solution, defined by axioms, with a non-cooperative game whose equilibrium produces it.

Other cooperative answers to a division come from other axioms. The nucleolus, which makes the loudest objection as quiet as possible, and the Kalai–Smorodinsky solution, which gives each player the same fraction of the most they could hope for, both pick different points on the same frontier, and each can be supported by some bargaining game of its own. What sets Nash’s product apart is that the simplest game — offers, refusals, delay — picks it out with no further design.

One equilibrium, where patience usually brings many #

The uniqueness is more surprising than it first looks, because patience in a repeated game does the opposite. When two players meet again and again and value the future highly, almost any pattern of play can be sustained by the threat of punishment later — the folk theorem — and patience multiplies the equilibria rather than removing them. In Rubinstein’s game the same patience leaves exactly one.

The difference is what is at stake each round. In a repeated game the players receive something every period, so the future is a stream of payoffs that a threat can withhold, and a sufficiently patient player will put up with a lot to keep it flowing. In bargaining the sum is divided once, and until it is divided nobody receives anything; the only thing a threat can withhold is time, and time is equally costly to the one making the threat. A player who refuses an offer to punish the other is punishing herself by the same factor $\delta$. That symmetry is what the Shaked–Sutton squeeze uses, and it leaves no room for the self-enforcing punishments that make repeated games so permissive.

Patience as bargaining power #

The two players need not be equally patient. If each round takes time $\Delta$ and the players discount time at rates $r_1$ and $r_2$, their delay factors are $e^{-r_1 \Delta}$ and $e^{-r_2 \Delta}$, and the first proposer’s share is $(1 - \delta_2)/(1 - \delta_1\delta_2)$.

As offers come faster the first mover’s bonus disappears, and the split tends to $r_2/(r_1 + r_2)$: each player’s share is in proportion to the other’s impatience. A player who discounts twice as fast gets a third of the sum. That is again a Nash product, but a weighted one, maximising $u_1^{1/r_1} u_2^{1/r_2}$ with each player’s patience as the weight on its own value. The axioms Nash used included symmetry, and dropping it gives exactly this family; Rubinstein’s game says what the weights mean.

The practical reading is old wisdom with a number attached. In a strike, the side that can hold out longer — a union with a large strike fund, an employer with a large inventory — has a lower rate of impatience and a larger share, and not by a vague amount: in this model the shares are in the ratio of the other side’s costs of waiting.

Caution costs a share #

Values also differ in shape. A player who values a share $s$ at $s^\alpha$ with $\alpha < 1$ is cautious about risk: a sure half is worth more to her than an even chance of everything or nothing, because the curve $s^\alpha$ bends down.

In the limit of fast offers her share is $\alpha/(1 + \alpha)$: a third when $\alpha = \tfrac12$, half when she is neutral about risk, more than half when she is a gambler. The game never mentions risk — there is no lottery in it, only delay — but delay acts on her through the same curve. A cautious player loses more, in her own terms, from the round of waiting that a refusal costs, and so the threat of refusal frightens her more and she concedes more to avoid it. If the game is changed so that each refusal risks a chance that the negotiation breaks down altogether, rather than a delay, the same limit appears, which is Binmore, Rubinstein and Wolinsky’s other result: delay and breakdown are interchangeable as the pressure that drives the bargain.

A way out that matters only if it beats the deal #

Many bargains are struck in the shadow of an alternative. A worker can take another job; a buyer can go to another shop. The usual intuition is that such an outside option raises a player’s share — that the sensible split gives each side its option and then divides what is left over between them. The equilibrium says something sharper.

An option worth less than what bargaining would give him anyway changes nothing at all. Threatening to take it is empty, because carrying out the threat would leave him worse off than carrying on, and Selten’s condition throws the threat away. An option worth more than his bargaining share is paid exactly — he gets the option’s value and not a penny above it, since anything more would be given away for nothing. The rule is $\max(\delta/(1 + \delta), w)$, and the split-the-difference intuition is wrong everywhere except where the option is worth the whole sum. Binmore, Shaked and Sutton tested this in a laboratory in 1989 and found that bargainers behaved far more like the equilibrium than like the intuition — a rare case of a game-theoretic prediction that sounds counter-intuitive and survives contact with real people.

What the model leaves out #

The model’s assumptions are strong, and each one can be seen failing in practice. Both players know everything: the size of the sum, each other’s values and each other’s patience. Real negotiations involve private information — a seller does not know how much the buyer would pay — and then delay becomes a way of signalling strength, and the equilibrium can involve strikes and long holdouts, which the complete-information game never produces. A union that strikes is, in the model, either telling the employer something or making a mistake.

Laboratory bargainers also care about fairness in ways the model ignores. In the one-shot ultimatum game, where a single offer is made and accepted or rejected with no further rounds, the equilibrium says the proposer should offer almost nothing and the responder should accept. Werner Güth, Rolf Schmittberger and Bernd Schwarze found in 1982 that proposers typically offer between a third and a half, and responders often reject offers below about a quarter — giving up money to punish a split they consider unfair. Rubinstein’s game is more forgiving to the theory than the ultimatum game, because the threat of a counter-offer makes even splits a prediction rather than an anomaly, but the gap shows that the delay factor is not the only thing on a bargainer’s mind.

And the figures solve the game rather than demonstrate it. Each curve is computed from the equilibrium conditions, which are a theorem’s conclusion; nothing here shows that the equilibrium is unique, or that players reach it. Uniqueness is Rubinstein’s proof, sketched above, and reaching it is an empirical question.

Still open: haggling when nobody knows the other’s price #

When both sides have private information — a seller who knows her cost and a buyer who knows his value, neither knowing the other’s — Roger Myerson and Mark Satterthwaite proved in 1983 that no bargaining procedure at all can guarantee that a sale happens whenever it should, while giving both sides a reason to take part and to tell the truth. Some efficient deals are always lost. Alternating offers with private information typically has many equilibria, with different amounts of delay and different splits, and the refinements proposed to choose among them disagree.

The theorem belongs to the same theory of mechanisms as the result that very different auctions earn a seller the same on average: both ask what a procedure can achieve when each participant knows something the others do not, and both answer by counting what each must be paid to reveal it.

Which outcome haggling under two-sided uncertainty actually predicts is not settled. It is known what cannot be achieved, and many particular equilibria are understood; a theorem as clean as Rubinstein’s — one outcome, characterised by a simple formula, converging to a recognisable rule as delays shrink — does not exist for that case, and the question of which procedure loses the fewest good deals is still studied, including by the designers of the online platforms on which many such bargains now take place. The question is the bargaining counterpart of choosing a rent that is both fair and hard to game: once each party knows something the rule does not, the rule’s guarantees and its incentives pull apart.

An axiom recovered as a limit #

Bargaining by alternating offers, with nothing but the cost of delay to end it, has exactly one subgame-perfect outcome, and as offers come faster it converges to the split that maximises the product of the two players’ values. The first mover’s advantage is the price of a round’s delay; patience is power in exact proportion; caution about risk costs a share; and an outside option matters only if it beats the deal, and then is paid exactly. Nash wrote the product down as the consequence of reasonable axioms. Rubinstein’s game derives it as the consequence of two people who could each walk away from any offer, but who lose a little every time they do.