The order everybody arrives in

Three partners work together and earn nine. Any two of them, working without the third, would earn something as well: the first two would earn six, the first and third four, the second and third two. Each alone earns nothing. What is each partner owed?

The question has no answer until owed is given a meaning, which is the shape of every question in this field: a rule for choosing has to be stated before it can be argued about. What makes this one remarkable is that four conditions, none of which anybody would refuse, leave exactly one rule standing.

What a player adds #

Start with the wrong answer, because it is instructive. Split it equally: three each, which is the rule a cake gets when nobody’s contribution is distinguishable. That respects the symmetry of the situation and ignores everything that makes the partners different — the first is in two profitable pairs and the third in one, and equal shares do not notice.

Start again. Suppose the partners arrive one at a time and each is paid what the group’s earnings go up by when they walk in. If the first arrives first, they add nothing, because one partner alone earns nothing; if they arrive last, they add the whole nine minus whatever the other two were earning on their own, which is seven. The number depends entirely on the order, so no single order is a defensible rule.

Average over all six orders, then. That is the number in the figure, and there is nothing arbitrary left: the arrival order was the only choice, and it has been integrated away.

The result is four, three and two. The first partner gets more than an equal share and the third less, and the amounts are not guesses — they are what each adds, on average, to whoever got there first.

The glove game is the sharper test of whether the rule is doing anything. Every player is essential in the sense that the group earns nothing without a right glove and nothing without at least one left; and the answer is nevertheless lopsided, because in four of the six orders the right glove arrives after at least one left and contributes the whole value.

The four conditions #

The reason to trust the average is not that it feels fair. It is that four separately reasonable requirements have exactly one solution between them.

It adds up. The shares total what the whole group earns; nothing is left over and nothing is invented.

Alike is treated alike. If two players contribute the same amount to every group not containing them, they get the same share. This is a demand about the game, not about the players: it says the rule may not consult anything except the numbers.

No pay for no work. A player who adds nothing to any group gets nothing.

Two jobs at once are the sum of two jobs. If a group runs two independent ventures, the share from the combination is the sum of the shares from each. This is the condition that is easy to state and hard to see the force of, and it is the one that pins the answer down.

The figure runs the first three against three sharing rules and finds the failures rather than quoting them. Splitting equally breaks no pay for no work the moment a game has a player who adds nothing; paying each what they earn alone and splitting the rest equally breaks it in exactly the same place.

The three conditions in the figure do not by themselves single out one rule — several rules pass all three on these games — and the fourth is what closes the gap. Shapley’s theorem, from 1953, is that the four together admit exactly one rule, and it is the average over orders.

Why the proof is a decomposition #

The argument for uniqueness is worth a paragraph because it is the additivity condition doing all the work, and it is the sort of argument that is invisible unless somebody points at it.

Consider the simplest possible games: pick a group, and say the value is one if a coalition contains that whole group and zero otherwise. Call these the unanimity games. On such a game, the first three conditions already force the answer — everybody outside the chosen group is a dummy and gets nothing, everybody inside is interchangeable and so gets an equal share, and the shares add to one. So each member of the group gets one over its size, and there is no freedom at all.

Now the key fact: every game is a combination of unanimity games, uniquely, with coefficients that can be positive or negative. They form a basis. So additivity extends the forced answer on the basis to a forced answer everywhere, and uniqueness follows.

That is why the fourth condition is the load-bearing one. The first three pin the rule down on a spanning set; the fourth says the rule is linear; and a linear map is determined by what it does to a basis. The whole theorem is that observation applied to a space of dimension $2^n - 1$.

The formula, and why it looks worse than it is #

The closed form is the one that appears in textbooks and it is genuinely off-putting:

$$\varphi_i = \sum_{S \not\ni i} \frac{|S|! \, (n - |S| - 1)!}{n!} \big( v(S \cup \{i\}) - v(S) \big).$$

Read as an average over orders it is not off-putting at all. The bracket is what player $i$ adds to the group $S$. The fraction in front is the proportion of orders in which exactly the members of $S$ arrive before $i$: there are $|S|!$ ways to arrange them beforehand, $(n - |S| - 1)!$ ways to arrange the rest afterwards, and $n!$ orders in total.

So the sum is not over orders but over groups, with each group weighted by how many orders produce it — which turns a sum of $n!$ terms into a sum of $2^{n-1}$ terms and is the only reason the value is computable for a dozen players rather than six.

The figure lists the orders instead, because a reader has no reason to believe the weights until the thing being weighted is visible. The two computations are the same number and the figure performs the slow one.

Where the number sits relative to what a group could demand #

There is a second question about a division, and it is not the same question. Instead of asking what each player deserves, ask whether any group would walk away.

For the three partners the two answers agree: four, three and two is a split no pair can improve on by leaving. That is a pleasant coincidence rather than a theorem, and the next essay is about how often it fails.

In the glove game they part company sharply. The only split no coalition can beat is the one giving the whole value to the holder of the right glove, because the right-hand holder can pair with either left and threaten to. The Shapley value gives that player two thirds and the left-hand holders a sixth each — a division nobody could enforce and which most people would call fairer.

That gap is the honest content of the pair. One rule answers “what is owed”; the other answers “what can be held”. They are different questions and they have different answers, and a field that ran the two words together would have nothing to say about the glove game at all.

A worked disagreement #

It is worth putting two of the rules side by side on one game and reading off exactly where they part, because the axiom table reports verdicts and not reasons.

Take the partnership: nine between three, with the pairs worth six, four and two. Splitting equally gives three, three, three. The average over orders gives four, three, two.

The second partner gets the same share under both rules, and that is not an accident. Swap the first and third partners and the game is not the same — the pairs are worth six and four rather than four and six — so the two are not interchangeable, and the middle share coinciding is arithmetic rather than symmetry.

Where the rules differ is the pair worth six against the pair worth two. Under equal shares those two numbers change nothing at all: the rule reads only the total. Under the average over orders, the first partner is in the six-pair and the third is not, and the gap between their shares — two units — is exactly the gap the pair values create.

So the disagreement is not about fairness in the abstract. It is about whether a rule is allowed to read the values of the coalitions that never form. The equal split refuses to; the average over orders insists on it; and the second condition, alike is treated alike, is what forces the insistence, because two players who differ in the coalitions they help must be allowed to differ in what they are paid.

Where it fails, and what it needs #

Every coalition’s worth has to be a number, and stated. A game here is a value for each of the $2^n$ subsets, and providing that is often the entire difficulty; in a real setting nobody knows what a group of four would have earned without the fifth, because that group never existed.

The count grows fast. There are $n!$ orders, and while the closed formula avoids listing them, it still sums over all $2^n$ coalitions. Above a couple of dozen players the exact value stops being computable and is estimated by sampling orders, which reintroduces the randomness the averaging was meant to remove.

Additivity is the condition to argue with. Two ventures run by the same group are rarely independent, and if they interact then the combined game is not the sum and the condition does not apply. Every objection to the Shapley value that has survived is an objection to this condition, and every alternative rule in the literature relaxes it.

And a value is not an enforcement. The rule says what a player is owed; it does not say what a player can get. In the glove game those differ by a wide margin, and no theorem here says which one a group of people should adopt.

Nor is the value a price. It has the units of the game’s value and it responds to the numbers in the game, which makes it tempting to read as a market outcome. It is not: no theorem here says a market would produce it, and where the field does price things — a constraint’s shadow value in a linear program — the number arises from an optimisation rather than from a list of axioms, and the two constructions have nothing in common but the word.

Where it came from #

Lloyd Shapley wrote the paper in 1953, as a graduate student, in a volume of contributions to the theory of games — the field then a decade old and dominated by von Neumann and Morgenstern’s two-person zero-sum theory. The move away from that was the whole point: a game where players cooperate and the question is how to divide the proceeds is a different mathematical object from a game where one player’s gain is another’s loss, and it needed its own theory.

The axiomatic method is what makes the paper still readable. Rather than proposing a formula and defending it, Shapley proposed conditions and derived the formula, which turns a matter of taste into a theorem — anybody who dislikes the answer must say which condition to drop. That style is now standard across the field, and this site’s essays on Arrow’s conditions and on stable matchings are two more instances of it; the difference is that those two produce impossibilities and this one produces a unique answer.

What the pictures cannot show #

Every game here has three players, because six orders fit on a page and twenty-four do not. Nothing about the argument depends on three, and nothing about the figures demonstrates that.

The values in the games are whole numbers chosen to make the arithmetic legible, and the exactness of the shares — four, three, two — is a consequence of that choice. In general the Shapley value is a fraction with $n!$ in the denominator, and the figures would then be tables of fractions.

And the axiom table checks three conditions rather than four. Additivity is a statement about pairs of games and cannot be tested on a single one; testing it needs a family, and the figure would then be a table of games rather than a table of rules. The essay states it, the theorem needs it, and the picture does not check it — which is the one place on this page where a claim is quoted rather than performed.

The ladder from here #

Below: one cuts and the other chooses, the smallest fair-division rule there is, and the value from both sides, where a two-player game’s worth is pinned down by an equality rather than by axioms. Sideways: a lottery over whole assignments, which is the same move of averaging over a set of extreme outcomes, and four conditions and no rule, where the axiomatic method produces an impossibility instead. Above: the core, the nucleolus, weighted values, and the games where the value cannot be computed exactly.

What is worth carrying away #

A quantity that depends on an arbitrary choice can sometimes be repaired by averaging over every version of the choice. What a player contributes depends on when they arrive; average over all arrival orders and the dependence is gone, and what is left is a well-defined number.

The reason to trust it is not the averaging, which is a construction, but the characterisation, which is a theorem. Four conditions and one answer is a much stronger statement than a formula with a plausible derivation, because it converts every future objection into a specific one: name the condition.