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The Dummy Paradox of the Bargaining Set pp. 443-446 $100.00
Authors:  Bezalel Peleg and Peter Sudholter
Abstract:
We consider the weighted majority game (N, v0) which has the tuple (3; 1, 1, 1, 1, 1, 0)
as a representation (see (3)). The maximum payoff to the dummy (the last player) in the
bargaining set of (N, v0) is shown to be 2/7 (see Remark 2). If we now increase v0(N) by
δ, 0 < δ < 2/3, then the maximum payoff to the last player in the new game, in which
this player is no longer a dummy and contributes δ to N, is smaller than 2/7 and strictly
decreasing in δ (see Lemma 1).
We recall some definitions and introduce relevant notations. A (cooperative TU) game
is a pair (N, v) such that ∅ = N is finite and v : 2N → R, v(∅) = 0. For any game (N, v) let
I(N, v) = {x ∈ RN | x(N) = v(N) and xi ≥ v({i}) for all i ∈ N}
denote the set of imputations. (We use x(S) = i∈S xi for every S ⊆ N.) Let (N, v) be a
game, x ∈ I(N, v), and k, l ∈ N, k = l. Let
Tkl = {S ⊆ N \ {l} | k ∈ S}.
An objection of k against l at x is a pair (P, y) satisfying
P ∈ Tkl, y(P) = v(P), and yi > xi for all i ∈ P. (1)
We say that k can object against l via P, if there exists y such that (P, y) is an objection
of k against l. Hence k can object against l via P, if and only if P ∈ Tkl and e(P, x, v) > 0,
where e(S, x, v) = v(S) − x(S) is the excess of S at x for S ⊆ N.
A counter objection to an objection (P, y) of k against l is a pair (Q, z) satisfying
Q ∈ Tlk, z(Q) = v(Q), zi ≥ yi for all i ∈ Q ∩ P and zj ≥ xj for all j ∈ Q \ P. 


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The Dummy Paradox of the Bargaining Set pp. 443-446