Yes. Use the simple cooperative game in which a coalition wins exactly when it contains at least two players. In every permutation, precisely the second arrival changes a losing coalition into a winning one. Each player is second in two of the six permutations, so each Shapley value is . This constructs the required vector while verifying that singleton coalitions lose and the grand coalition wins.
No. In a simple cooperative game, a player's marginal contribution along any permutation is zero or one. With three players there are six equally likely permutations, so every Shapley value is an integer multiple of , by the Shapley value denominator in a simple game. But and are not integers. The vector sums to one, but fails this necessary denominator condition.
Yes. Give players weights , respectively, and let the quota be three. This weighted voting game is a simple cooperative game whose minimal winning coalitions are and . In orders , player 3 supplies the winning marginal contribution. In order player 1 does, and in order player 2 does. Averaging the six marginal contribution vectors gives exactly as the vector of Shapley values.
No. The denominator test from part ii is satisfied here, so a stronger obstruction is needed. Put and . In a three-player simple cooperative game, the Shapley value formula gives, for distinct ,Subtract the corresponding expression for :Each coalition value is zero or one, so this difference must be an integer multiple of . This is the obstruction from pairwise Shapley differences in a three-player simple game. The proposed first two coordinates differ by , which is not an integer multiple of . Therefore the vector cannot occur, even though each coordinate separately passes the denominator test.
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