Are you trying to do Assumptions -> ForAll[x, score[x] >= 0] ? I don't think ForAll works as an assumption. It's used inside Reduce or Resolve.
Many thanks - that's exactly what I'm trying to do. With an arbitrary x elements, how can I declare the assumption that score[x]>0 for every element?
Aha - cracked it with:
probresult[g_, factor_] := Simplify[PDF[PoissonDistribution[factor], score[g]], Assumptions -> score[g] >= 0]
Very many thanks!
