I think I see what you are saying. If you have an unmatched system (50 ohm input sources, and 200 ohm input impedance amplifier that has a 1 dB NF in a 200 ohm system) then you are throwing input power away with a big reflection loss.
|ρ|=(200-50)/(200+50) = 0.6
So transmission "thru" this mismatch is |t|=[1-(0.6^2)]^0.5 =0.8
20 LOG 0.8 =-1.9 dB
So with the mismatch, your effective NF should be 1+1.9=2.9 db
With a 1 dB loss saw that corrects the mismatch exactly, you should have a NF=1+1=2 dB
Looks to me like the SAW wins.