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For three distinct players $i,j,k$ in a normalized simple cooperative game, enumerate the predecessor coalitions in the Shapley value formula:Subtracting the formula with $i$ and $j$ exchanged proves the displayed identity. Every coalition value is zero or one, so $2({\varphi }_i-{\varphi }_j)$ is an integer. This rules out, for example, coordinates differing by $1/6$, although each may separately satisfy the Shapley value denominator in a simple game. The difference identity itself holds for arbitrary three-player transferable utility games; the integer restriction uses the zero-one values.

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 $1/6$, by the Shapley value denominator in a simple game. But $6(5/12)=5/2$ and $6(3/12)=3/2$ are not integers. The vector sums to one, but fails this necessary denominator condition.