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One seller has a good that two buyers want, but only one can have it. The buyers would love to split the surplus with the seller. They cannot, and the reason is simple: whatever they are offered, the seller can always deal with the other buyer instead.

Coalitional games and the core

So far players acted alone. Cooperative game theory asks what groups can achieve by working together, and how the joint value gets divided. A characteristic function $v(S)$ assigns each coalition $S$ the total value it can create on its own. The core keeps allocations no coalition can improve on by breaking away.

From earlier lessons

Two quick ones before we start.

In bargaining, your disagreement payoff is what you get if...

The value a coalition can guarantee itself is given by its...

Predict, then play

The seller and either buyer can create value 100 together; the two buyers alone make nothing. Split the 100 among the three and see which coalitions object.

Why did that happen?

Take a guess, then reveal the reason

The two buyers are substitutes: the seller needs only one of them. So the buyers compete, and competition drives their share to zero. Any allocation that hands a buyer a positive amount can be undercut, because the seller and the excluded buyer can form a coalition worth 100 and split it in a way both prefer. Stability requires that no such blocking coalition exists, and here that forces the entire surplus to the scarce seller.

The math

Definition, coalitional game and the core

A coalitional (transferable-utility) game gives each coalition $S$ a value $v(S)$ , the total its members can guarantee together. An allocation $x$ divides the grand-coalition value $v(N)$ among the players. The core is the set of allocations that are efficient ( $\sum_i x_i = v(N)$ ) and unblocked: for every coalition $S$ , $\sum_{i \in S} x_i \ge v(S)$ . If some coalition could get more on its own, it blocks, and the allocation is not in the core.

Here $v(\{s\}) = v(\{b_1\}) = v(\{b_2\}) = v(\{b_1,b_2\}) = 0$ , $v(\{s,b_1\}) = v(\{s,b_2\}) = 100$ , and $v(N) = 100$ . A core allocation $(x_s, x_1, x_2)$ must satisfy $x_s + x_1 \ge 100$ (else the seller and buyer 1 block) and $x_s + x_2 \ge 100$ . Together with efficiency, $x_s + x_1 + x_2 = 100$ , the first inequality gives $x_2 \le 0$ . But the singleton coalition $\{b_2\}$ can guarantee $0$ , so core membership also requires $x_2 \ge 0$ . Thus $x_2=0$ ; symmetrically $x_1=0$ ; hence $x_s=100$ . The core is the single point $(100, 0, 0)$ .

The core can also be empty (the three-player majority game has no unblocked allocation), which motivates other solution concepts in the next lessons.

Work it out

Pin down the core

Write the blocking conditions. The coalition of the seller and buyer 1 can secure $100$ , so any stable split must give them at least that: $x_s + x_1 \ge 100$ . Symmetrically $x_s + x_2 \ge 100$ . Efficiency requires $x_s + x_1 + x_2 = 100$ .

From $x_s + x_1 \ge 100$ and the total being $100$ , we get $x_2 \le 0$ . The singleton coalition $\{b_2\}$ is worth $0$ , so core membership also gives $x_2 \ge 0$ . Hence $x_2 = 0$ . The same argument gives $x_1 = 0$ . Therefore $x_s = 100$ : the unique core allocation is $(100, 0, 0)$ . No matter how the buyers coordinate, competition between them leaves the seller with everything.

Key idea

The core keeps only allocations no coalition can beat by going it alone. When one side is scarce, competition on the other side collapses its share to nothing, and the core can even be empty.

Check yourself

Check yourselfapply

Why does the allocation (80, 10, 10) fail to be in the core?

Check yourselfanalyse

Why does the entire surplus go to the seller?

Check yourselfbrings back: nash equilibrium

How does core stability differ from Nash-equilibrium stability?

Sources

Gillies, D. B. (1959). Solutions to general non-zero-sum games. In Contributions to the Theory of Games IV. Princeton University Press.
von Neumann, J., and Morgenstern, O. (1944). Theory of Games and Economic Behavior. Princeton University Press.