What I ment was, if you attack, say, 6 targets with a long burst, 1 bullet each, do they take a -5 dodge penalty each for receiving a long burst? The rules say you take the -5 if the attacker did a long burst. It says that's because of the amount of bullets flying at you (so RAI is that it's dependant on that), but it doesn't actually make it dependant on that in the actual rules, and Multiple Attacks is silent on it as well. So RAW all 6 targets would receive the -5. Which is silly.
Page 180, the red box titled
Not Enough Bullets answers this.
For example, Wombat attempts to fire Full Auto (Complex
Action) but his Ares Alpha only has 7 rounds left. 10 bullets
needed minus 7 bullets left is a 3 bullet shortage and a 3 point
reduction in the –9 defense modifier, making it a –6 on defense.
Full Deck fires a Long Burst that empties the last 5 rounds
from his Uzi V. Since he is 1 round short he only imposes a –4
defense penalty to his opponent.
The defense penalty is reduced by 1 for every bullet short of the "required" amount going at a target. a 6 round burst imposes a -5 Defense penalty. If you only have 4 bullets left, you are 2 bullets short, so the defense penalty is -3. If you are splitting it so that 1 bullet flies at each of 6 targets, then each target is 5 bullets short of the "required" amount, so the -5 becomes -0 for each of them. Another way to read this would be that the Defense Penalty suffered by a target is equal to one less than the total number of bullets being fired at them.
At this point, reducing your dice pool among so many targets, while they all get to roll full defense pools, just makes it very likely that you will; Miss, Glitch, or Critically Glitch.