AI-Lab
Cooperative Game Theory

Coalitions

Cooperative games — the setup

In a cooperative game, a group of players can form any coalition they like. Each possible coalition has some value it can guarantee on its own — money it can earn, costs it can avoid, votes it can pass. The question this lab is about: once everyone has joined the grand coalition, how should they divide the spoils?

This isn't a voting question — the players have already agreed to cooperate. The question is how to divide the joint surplus their cooperation creates, fairly enough that no sub-group walks away.

To split the spoils fairly we need to answer a sharper question first: how much did each player actually contribute to the group? "Contribution" is slippery in teamwork. A player's solo value v({i}) ignores synergy; her share of v(N)/n ignores skill. The right notion is the marginal contribution — how much value i adds when she joins a coalition S: v(S ∪ {i}) − v(S). But that number depends on which coalition S she joins, so a single answer needs us to average it across all the coalitions she could have walked into. That average — the Shapley value — is what this lab builds up.

Why care? Three motivating scenarios

Before the formulas, three stories where the same question keeps showing up: a group together produces value — how much of it does each member really deserve? The members can be politicians, fighters, or pixels in an image.

Scenario 1
Coalitions in politics
A parliament has 120 seats. No party wins a majority alone, so several parties join a governing coalition. They pass a budget worth, say, v(N) = 61 ministries / favors / portfolios to split among themselves.

Now imagine a small 4-seat party sitting inside a 62-seat coalition. On paper its share looks like 4/62 ≈ 6%. But suppose this little party threatens to walk out: the coalition falls from 62 to 58 seats, below the 61-seat majority — the government collapses, a rival bloc forms, and everyone else loses all 61 portfolios. The big parties have to keep this 4-seat party happy, so its real contribution to the group is far more than 4 seats' worth.
Scenario 2
Warriors & the lone magician
You're building a raiding party. Soldiers alone can fight: v({S, S}) = 20. A magician alone is squishy — she can squeeze off one fireball before getting cut down: v({M}) = 5. But put the magician behind the soldiers and her spells obliterate the enemy: v({S, S, M}) = 100.

On her own the magician barely scratches the surface (5), yet joining the soldiers she adds 80 more (100 − 20). Her solo value massively understates her real contribution to the group. How much of the loot does she deserve?
Scenario 3
Explainable AI (SHAP)
A model reads a review and outputs P(positive) = 0.92. A vision model looks at an X-ray and outputs P(pneumonia) = 0.78. Which words pushed the sentiment toward "positive"? Which pixels of the X-ray flagged the pneumonia? Doctors, regulators and customers all need to know which input features actually drove the prediction.

Treat each word (or pixel patch) as a player. The "coalition" is the subset of features the model is allowed to see; v(S) is the probability it outputs. We want to credit each feature for its share of the prediction — how much that single word, or that single pixel patch, moved the model's output.
A cooperative game
players: N = {1, 2, …, n} coalitions: SN // any subset, 2n of them characteristic v: 2N → ℝ function: v(S) = value S can guarantee on its own v(∅) = 0 by convention
The fundamental questions
φi
Fair share
How much should each player get?
A division (φ1, …, φn) of v(N). The Shapley value is one canonical answer.
Core
Will anyone leave?
Stable allocations
Splits where no coalition does better by walking away. Sometimes the core is empty — sometimes huge.
v(N)
Grand coalition
All players together
When v(N) ≥ Σ v({i}) the grand coalition is profitable. Usually we assume it forms.
v(S∪{i})−v(S)
Marginal contribution
What player i adds
The extra value i brings to coalition S. The atomic unit of cooperative game theory.
The lattice of coalitions (n = 3)
{1,2,3} GRAND COALITION (N) {1,2} {1,3} {2,3} {1} {2} {3}

For 3 players there are 23 = 8 coalitions. For 10 players, 1024. For the UN Security Council (15 members), 32 768. A full game specifies v(S) for every one of them.

Basic definitions — what kind of game is it?

A quick glossary of the most useful properties a cooperative game can have, plus the standard "kinds of player" that show up everywhere in the lab — null, dummy, veto, symmetric. Each card gives the formal condition and a one-line meaning. Use it as a reference while you work through the later tabs.

Properties a game can have — structural conditions on v(·)
Monotone— bigger is never worse
v(S) ≤ v(T) whenever S ⊆ T
Adding members can only keep or increase the coalition's value. Weaker than superadditive (a monotone game can still have v(S∪T) < v(S)+v(T)).
Efficient— nothing left on the table
Σᵢ xᵢ = v(N) all of v(N) is distributed
An allocation x is efficient if the players' shares sum exactly to the grand-coalition value — no surplus held back, nothing invented out of thin air. Shapley is efficient by construction (Σ φᵢ = v(N)); this is the first of the four Shapley axioms covered in tab 03.
Additive— games stack linearly
φ(v + w) = φ(v) + φ(w) for any two games v, w on the same players
If the players play two games on the same day — splitting the morning revenue v and the afternoon revenue w — running the solution rule on each separately and adding the shares gives the same answer as running it once on the combined game v + w. The Shapley value satisfies this; it's the fourth Shapley axiom from tab 03.
Kinds of game — named families that recur
Superadditive game— merging never hurts
v(S ∪ T) ≥ v(S) + v(T) for all disjoint S, T
Two non-overlapping coalitions are at least as valuable together as apart. Implies the grand coalition N is at least as good as any partition — so usually the players want to form it.
Convex game— supermodular, increasing returns
v(S ∪ T) ≥ v(S) + v(T) − v(S ∩ T) for all S, T ⊂ N
A game is convex (a.k.a. supermodular) when overlapping coalitions stack better than they fall apart — equivalent to: a player's marginal contribution grows as the coalition she joins grows. Strictly stronger than superadditive. For convex games, the Shapley value always lies in the Core — fair and stable (see tab 05).
Essential game— cooperation pays
v(N) > Σᵢ v({i})
The grand coalition produces strictly more than the sum of solo values. If a game isn't essential, there's nothing to bargain over — everyone might as well go alone.
Simple game— voting / win-lose
v(S) ∈ {0, 1} monotone, v(N) = 1
Coalitions are either winning (1) or losing (0). This is the world of weighted voting — UN Security Council, EU Council, shareholder votes. Shapley becomes a power index — try the "Shareholder Voting" game in the Examples tab (tab 08).
Symmetric game— anonymous
v(S) depends only on |S|
Names don't matter — only how many players are in the coalition. By symmetry, every player gets the same Shapley value v(N)/n.

Hierarchy of niceness:  convexsuperadditiveessential. Monotonicity is independent. Most "real" games people study (voting, runway cost-sharing, ML feature attribution) are at least superadditive — which is why Shapley is so widely applicable.

Kinds of player — how individual players relate to v(·)
Null player— contributes nothing, ever
v(S ∪ {i}) = v(S) for every S ⊆ N \ {i}
Player i changes no coalition's value. The null-player axiom (tab 03) forces φᵢ = 0 — fair, because she literally adds nothing.
Dummy player— always worth exactly v({i})
v(S ∪ {i}) = v(S) + v({i}) for every S ⊆ N \ {i}
Player i always contributes her solo value, no synergy. Shapley pays her exactly v({i}). A null player is the special case where v({i}) = 0.
Veto player— nothing happens without her
v(S) = 0 for every S ⊆ N with i ∉ S (simple game)
Player i is indispensable: every winning coalition must include her. The structural opposite of a null player. In the UN Security Council each P5 member is a veto player — and Shapley shows the five P5 share ≈ 98% of the power between them.
Symmetric players (i ~ j)— interchangeable
v(S ∪ {i}) = v(S ∪ {j}) for every S not containing i or j
Players i and j contribute the same to every coalition that excludes them both. The symmetry axiom (tab 03) forces φᵢ = φⱼ — equal contribution, equal pay.
The Shapley value — two equivalent formulas

Player i's Shapley value φi is the average marginal contribution of i over all possible coalitions she could join.

"All possible coalitions" can be counted two ways — over subsets or over orderings — and the formulas look different but give the same number.

Form 1 · subset weighted sum
φi(v) = ΣSN\{i} |S|! · (n − |S| − 1)! n! · v(S ∪ {i}) − v(S) )
weight on subset S marginal contribution of i to S
Form 2 · average over orderings
φi(v) = 1 n! · Σπ ∈ Π(N) v(Pπ,i ∪ {i}) − v(Pπ,i) ]
average across all n! orderings marginal contribution of i when she arrives
Π(N) = the set of all n! permutations of the players · Pπ,i = the players who arrive before i in ordering π
"Imagine the players walk in one at a time in a random order. Each player gets credit for the value she adds when she walks in. Sum that credit over every possible order, then divide by n!."
Why the weights in Form 1?

It looks ugly. |S|! · (n−|S|−1)! / n! ?? Where does that come from?

It's just counting orderings. For player i to "arrive after exactly the players in S":

  • |S|! ways to order the players in S (they arrive first, in any sequence)
  • (n − |S| − 1)! ways to order the remaining n − |S| − 1 players (they arrive after i)
  • divided by n! total orderings → the probability that i arrives right after exactly S

So Form 1 and Form 2 say the same thing. Form 2 (orderings) is the intuitive one. Form 1 (subsets) is what you implement in code.

Players n
Focal player i
Pick a subset S ⊆ N\{i}

Click through every subset on the left — for every choice of S, the matching-ordering count is exactly |S|! · (n − |S| − 1)!. Add those counts across all 2n−1 subsets and you recover all n! orderings. That's why dividing each subset's contribution by n! gives a clean probability — the Form 1 weight.

A 3-player walk-through

A tiny 3-player game. Every coalition's value v(S) is laid out below — only the pair {1,2} and the grand coalition produce anything.

v({1})0
v({2})0
v({3})0
v({1,2})80
v({1,3})0
v({2,3})0
v({1,2,3})90

Players 1 and 2 are productive together (80); player 3 only bumps the grand coalition by 10. Walk through all 3! = 6 orderings of the players. For each ordering π, write down each player's marginal contribution — the bump in v when she walks in. Then average across the 6 rows:

👆 Click any MC cell above to see exactly how that value was computed.

Reading a row. Take 2 → 1 → 3: player 2 walks in first, so MC₂ = v({2}) − v(∅) = 0. Player 1 walks in next, so MC₁ = v({1,2}) − v({2}) = 80 − 0 = 80. Player 3 walks in last, so MC₃ = v({1,2,3}) − v({1,2}) = 90 − 80 = 10. Each row sums to v(N) = 90 (every player gets credited exactly once when they arrive).

The columns. Player 1 collects 80 in two orderings (when she arrives second after 2 alone) and 90 in two orderings (when she arrives last after either {2,3} or {3,2}) → total 260, average 260 / 6 = 43.33. Player 2 is symmetric → also 43.33. Player 3 only ever brings 10 (and only in two of the six orderings where she arrives last after {1,2} or {2,1}) → total 20, average 3.33.

Sum of averages: 43.33 + 43.33 + 3.33 = 90 = v(N). The Shapley shares add to the grand coalition's value — that's efficiency, our first axiom.

The four axioms — and a stunning theorem

What does it mean for a sharing rule to be "fair"? Shapley (1953) wrote down four properties that any reasonable rule should satisfy, then proved that exactly one rule satisfies all four — his.

Σ φ = v(N)
Efficiency
No value lost or invented
Whatever the grand coalition is worth, it is fully distributed — nothing left over, nothing pulled from thin air.
i ≡ j ⇒ φij
Symmetry
Equal contributors → equal pay
If swapping i and j in every coalition leaves v unchanged, they must receive the same payoff.
φnull=0
Null player
No contribution → no payoff
If v(S ∪ {i}) = v(S) for every S, then i contributes nothing — and gets nothing.
φ(v+w)=φ(v)+φ(w)
Additivity
Two games stack linearly
Run two games on the same players? Each player's total Shapley value is the sum from each. Also called linearity.
Shapley's theorem
Shapley (1953)
Theorem. Among all functions φ from games to allocation vectors, exactly one satisfies efficiency + symmetry + null player + additivity:
φi(v) = ΣSN\{i} |S|! · (n − |S| − 1)! n! · v(S ∪ {i}) − v(S) )
The four axioms uniquely nail down the formula.

This is a beautiful axiomatic result: pick the axioms, get the formula for free. Variants of this proof technique recur all over economics — Arrow's impossibility theorem (Voting lab), Nash bargaining, etc.

Sanity check — re-run the 3-player game
Recall the tiny game. Only the productive pair {1,2} and the grand coalition have non-zero value:
v(∅) = 0 v({1}) = 0 v({2}) = 0 v({3}) = 0 v({1,2}) = 80 v({1,3}) = 0 v({2,3}) = 0 v({1,2,3}) = 90
Shapley value: φ₁ = 43.33 φ₂ = 43.33 φ₃ = 3.33
Efficiency ✓ holds
Σi φᵢ = v(N)
All Shapley shares should add up to the grand-coalition value — nothing left over, nothing invented. Plug in the three numbers:
φ₁ + φ₂ + φ₃ = 43.33 + 43.33 + 3.33 = 90 = v(N) ✓
Symmetry ✓ holds
if v(S∪{i}) = v(S∪{j}) for every S not containing i, j ⇒ φᵢ = φⱼ
Swap players 1 and 2 in every coalition. Each v(S) stays the same (the game treats them as interchangeable productive players). Symmetry then forces equal Shapley shares:
φ₁ = φ₂ = 43.33 ✓
Null player ✓ holds (player 3 isn't null)
i is null ⇔ v(S ∪ {i}) = v(S) for every S ⊆ N\{i} ⇒ φᵢ = 0
The axiom only forces φᵢ = 0 for players who add literally nothing to every coalition. Test player 3 against her 4 subsets:
v({3}) − v(∅) = 0 · v({1,3}) − v({1}) = 0 · v({2,3}) − v({2}) = 0 · v({1,2,3}) − v({1,2}) = 90 − 80 = 10 ≠ 0
Player 3 does contribute to the grand coalition (she lifts it from 80 to 90), so the "for every S" condition fails. She is not a null player — and the axiom is satisfied because its premise doesn't hold, leaving φ₃ free to be positive.
Premise fails ⇒ axiom holds vacuously. φ₃ = 3.33 > 0 ✓ — exactly her share of that single non-zero contribution.
Additivity ✓ holds
φ(v + w) = φ(v) + φ(w) for any two games v, w on the same N
Decompose this game into two simpler pieces:
vA(S) = 80 · 𝟙[S ⊇ {1,2}]   (the productive-pair piece)
vB(S) = 10 · 𝟙[S = N]   (the grand-coalition bonus)
Then v = vA + vB. Compute Shapley on each:
φ(vA) = (40, 40, 0)   φ(vB) = (3.33, 3.33, 3.33)
φ(vA) + φ(vB) = (43.33, 43.33, 3.33) = φ(v) ✓
Shapley Playground — watch the orderings, build your own game

Pick a small game. We'll walk through every permutation of the players, accumulate each player's marginal contributions, and watch the Shapley values converge.

Game
Ordering0/6
All orderings
Shapley values (running average)

The Core — will anyone walk away?

Shapley answers "what's fair?". The Core answers a different question: "what's stable?". A payoff vector x is in the core if no coalition can do strictly better by defecting.

The Core
Core(v) = { x ∈ ℝn : Σi xi = v(N) (efficient) Σi ∈ S xiv(S) for every SN (no coalition is unhappy) }
Equivalent definition · in words
The core is the set of payoff allocations x ∈ ℝN satisfying:
1. Efficiency: Σi ∈ N xi = v(N)
2. Coalitional rationality: Σi ∈ C xi ≥ v(C)  for all subsets (coalitions) C ⊆ N
Two names for the same two conditions. Efficiency says the grand coalition's value is fully shared out — same axiom as in tab 01. Coalitional rationality says every sub-coalition C earns at least what it could earn on its own; otherwise C would walk away and the grand coalition would collapse.
Worked example · Shapley is fair, but not in the Core

The clearest way to feel why Core ≠ Shapley is to find a game where the Shapley split is something some coalition strictly prefers to escape from. Here it is, in 3 players.

v({A})0
v({B})0
v({C})0
v({A,B})80
v({A,C})20
v({B,C})20
v({A,B,C})85

Compute Shapley. A and B are symmetric, C contributes a little (the grand coalition lifts the {A,B} value of 80 by 5, up to 85). Running the Form-1 sum (which you can do in the Shapley Playground with the Custom preset) gives:

φA = 38.33 φB = 38.33 φC = 8.33 Σ = 85 = v(N) ✓ efficient
Check every coalition · who would defect?
{A}:xA = 38.33  ≥  v({A}) = 0✓ stays
{B}:xB = 38.33  ≥  v({B}) = 0✓ stays
{C}:xC = 8.33  ≥  v({C}) = 0✓ stays
{A, B}: xA + xB = 76.67  <  v({A,B}) = 80 ✗ defects!
{A, C}:xA + xC = 46.67  ≥  v({A,C}) = 20✓ stays
{B, C}:xB + xC = 46.67  ≥  v({B,C}) = 20✓ stays
{A, B, C}:Σ xi = 85 = v(N)✓ efficient
Verdict. A and B can produce 80 on their own — yet the Shapley split hands them only 76.67 together. They have a strict incentive to walk away from C, take the 80, and split it however they prefer. (Yes, they leave 5 units of v(N) − v({A,B}) on the table — but that 5 was C's "synergy with the grand coalition", and A+B don't need her share to come out ahead.) So the Shapley vector (38.33, 38.33, 8.33) is fair but not stable — it is not in the Core.
A concrete defection · A and B both strictly prefer (41, 39)
Suppose A and B abandon C and split the 80 they can produce alone as (xA, xB) = (41, 39). Compare side-by-side with the Shapley payoff:
Player Shapley Defection (A,B) Δ
A 38.33 41 +2.67 ↑
B 38.33 39 +0.67 ↑
C 8.33 0 −8.33 ↓
Σ 85 80 −5
Both A and B end up strictly better off. A goes from 38.33 to 41 (+2.67), B goes from 38.33 to 39 (+0.67). C is the loser — she drops from 8.33 to 0 — but she has no vote in this: the coalition {A, B} can produce 80 on its own and doesn't need her permission to walk away. The defection leaves 5 units of grand-coalition surplus (85 − 80) on the table, but that's fine for A and B: they still come out ahead. Any split (xA, xB) with xA + xB = 80 and both xA > 38.33, xB > 38.33 makes both defectors happier.

The Core for this game is the set of (xA, xB, xC) with Σ = 85, every x ≥ 0, xA+xB ≥ 80, xA+xC ≥ 20, xB+xC ≥ 20. Solving: xA+xB ≥ 80 together with xA+xB = 85 − xC forces xC ≤ 5. So C's share in any stable allocation is at most 5 — strictly less than her Shapley value of 8.33. That single inequality is what kicks Shapley out of the Core: she's being paid more than the {A,B} coalition is willing to leave on the table.

Core ↔ Nash equilibrium · a coalitional analogue

The Core is to cooperative game theory what Nash equilibrium is to non-cooperative game theory. In both:

  • Nash: no single player can do strictly better by changing her strategy, given everyone else's.
  • Core: no coalition (any sub-group, including all singletons) can do strictly better by leaving with its guaranteed value v(C).

The Core is strictly stronger than Nash: it closes deviations under arbitrary group sizes, not just individuals. Restricted to singletons, "xi ≥ v({i})" is exactly the individual rationality condition — the cooperative version of "no unilateral profitable deviation". Add the requirement that no 2-player, 3-player, …, (n−1)-player group can defect either, and you arrive at the Core.

In the worked example above, the Shapley vector passes every singleton check (Nash-style individual rationality) but fails the {A,B} pair check — a deviation only a coalition can spot. That gap is exactly why the Core was introduced as a separate stability concept.

Empty Core · Unique Core

The Core is a set of allocations, and that set can take two very different special shapes:

Empty Core
Literally no allocation in ℝⁿ satisfies every coalition. There is no Core point at all. The grand coalition cannot hold together: at every proposed split, some coalition has a strict incentive to walk away. The game is too contested for cooperation to settle.
Unique Core
Exactly one allocation satisfies every constraint. The math leaves no negotiation room: the answer is forced. Unique Cores are rare and only happen when all the coalition constraints are tight simultaneously.
Example · Empty Core— the 3-player majority game
v(S) = 1 if |S| ≥ 2, else 0 v({i}) = 0, v({i,j}) = 1, v(N) = 1
Any two of the three players can produce 1 on their own; alone they're worth 0. Core constraints require x1+x2 ≥ 1, x1+x3 ≥ 1, x2+x3 ≥ 1. Adding the three: 2(x1+x2+x3) ≥ 3, so Σ xi ≥ 1.5. But efficiency says Σ xi = v(N) = 1 < 1.5 — a flat contradiction. No allocation exists in the Core. Whichever pair forms, the third player can offer the other two more than they're getting and break the deal.
Example · Unique Core— symmetric pairs, tight grand coalition
v({i}) = 0,   v({i,j}) = 50 for every pair,   v(N) = 75
Adding the three pair constraints: 2 Σ xi ≥ 150, so Σ xi ≥ 75. Efficiency says Σ = 75 exactly — so all three pair constraints must be tight equalities. Solving xA+xB = xA+xC = xB+xC = 50 forces xA = xB = xC = 25. The Core is the single point (25, 25, 25) — and Shapley, by symmetry, lands right on it. Fair and stable, and uniquely so.
Theorem · the core of a simple game
In a simple game, the core is empty iff there is no veto player. If there are veto players, the core is exactly the set of payoff vectors in which every non-veto player gets 0.
Why: in a simple game v(S) ∈ {0,1}. A non-veto player i is in some winning coalition that doesn't need her — so the rest can guarantee 1 without her. Any core payoff vector must give her xi = 0 (else N\{i} could split <1 and improve). With no veto player, the constraints force every xi = 0, contradicting Σ xi = v(N) = 1 — so the core is empty.
Example: in the UN Security Council (tab 08 → Examples), the five P5 are the only veto players. The core consists of every split of "1 unit of power" among the P5 with the 10 non-permanent members getting nothing — which matches the Shapley value's ≈ 98% / 2% verdict almost exactly.
Two theorems · convex games
Definition · convex game
A game G = (N, v) is convex if for all S, T ⊂ N, v(S ∪ T) ≥ v(S) + v(T) − v(S ∩ T)
Theorem 1
Every convex game has a non-empty core.
Theorem 2 (Shapley, 1971)
In every convex game, the Shapley value lies in the core.
Convexity is the "supermodular" / increasing-marginal-contributions condition you already met in tab 01. Together these theorems are the cleanest result in the lab: when the game is convex, the fair answer (Shapley) and the set of stable answers (Core) coincide — and the Shapley value is one of them. You get both at once. The Convex (big core) example in the visualiser below illustrates this.
Build your own 3-player game

Drag the sliders to set every v(S) yourself. The triangle shades in the Core (all stable splits), drops a gold dot for the Shapley value, and lists every coalition constraint — green if no one defects, red if someone walks away.

The triangle is the 2-simplex of efficient allocations (x₁+x₂+x₃ = v(N)). Each vertex is "one player gets everything"; the centre is equal split.

Presets:

Set v(S)

v({A})10
v({B})10
v({C})10
v({A,B})30
v({A,C})30
v({B,C})30
v(N)60
Click anywhere inside the triangle to inspect that allocation — you'll see xA, xB, xC and whether the point sits in the Core.

Live diagnosis

Coalition rationality

Things to try:   (1) start from Balanced, raise v({A,B}) past 40 — watch the Core shrink and Shapley fall outside. (2) start from Empty Core, raise v(N) until the shaded region reappears. (3) use Convex and verify that Shapley sits cleanly inside the Core no matter how you wiggle the singletons.

Examples · six real-world games

Pick a game. Edit the inputs. Watch Shapley update in real time. The first four are bargaining / cost-sharing problems that show up in real politics and economics; the fifth is a live explainable-AI demo — every word of a movie review is a player, and Shapley tells you which words drove the model's prediction.

Army Builder
7 units · guards
UN Security Council
15 players · veto
Airport Runway
cost share
🏛
Parliament Coalition
parliament · 120 seats
🗳
Shareholder Voting
4 owners · quota 51%
Explainable AI
SHAP · sentiment

The setup. Build a battle squad from a roster of 7 units. Each unit deals damage over a limited survival time; the squad's value is total damage dealt before everyone falls. The catch: a wizard alone dies almost instantly (1 second), but with 3 soldiers guarding him he survives 7 seconds — and his 10 dps suddenly matters. Healers are zero-dps alone but extend everyone else's survival.

The characteristic function is v(S) = Σᵢ∈S dpsᵢ × survᵢ(S), where survival depends on how many soldiers and healers are in S. Shapley distributes the credit fairly across every unit — including the guards who never landed a hit but kept the wizard alive.

Try:
Adjust the squad — use [−] / [+] to add or remove units of each type (each type capped at 4, total capped at 8)

What to look for. The wizard's solo value is just 10 — but his Shapley value (the fair credit averaged across every team) is much higher, because in the roughly half of orderings where ≥1 soldier already arrived, he contributes 30–70. Each soldier looks individually modest (15 dmg solo) but gets a big Shapley bump for enabling the squishy ranks. The healer, who deals zero damage, still earns positive Shapley — she lengthens everyone else's survival. This is the complementarity premium of cooperative game theory in action.