 Solution

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.