A random assignment of n indivisible objects to n agents is specified by its assignment matrix and implemented by drawing a deterministic assignment from a Birkhoff-von Neumann decomposition. Kawase et al. observed that the choice of decomposition matters for fairness: a matrix that is envy-free in the sense of stochastic dominance (SD-EF) can be decomposed so that some agent envies another with probability close to 1. They call a decomposition Dec-EF if every agent envies every other agent with probability at most 1/2, proved that every SD-EF matrix admits a Dec-EF decomposition when n <= 3 or when there are at most two distinct preferences, and left the general case open. We settle the first open case: every SD-EF matrix with four agents admits a Dec-EF decomposition. The worst case over the SD-EF polytope of a profile is attained at a vertex, and our computer-aided proof enumerates all 26,927 vertices for the 762 profiles up to symmetry in exact arithmetic and certifies each by a rational decomposition. The same method settles five agents with at most four distinct preferences and the probabilistic serial rule for all five-agent profiles, and adversarial search up to seven agents finds no counterexample. For general n, an envy-budget identity shows that 1/2 is the best possible threshold. We prove that every SD-EF matrix with at most two distinct rows admits a Dec-EF decomposition, and that the maximum-entropy decomposition is Dec-EF whenever all agents but two share a preference; the latter proof rests on a new monotonicity lemma for weighted least-squares rankings. In general, natural decompositions fail: greedy Birkhoff-von Neumann can come arbitrarily close to envy probability (n-1)/n, and maximum entropy fails at n = 4 when all preferences differ. Deciding whether an arbitrary random assignment, not necessarily SD-EF, admits a Dec-EF decomposition is strongly NP-complete.

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