Multiple attackers / defenders

From: Richard Develyn <Richard.Develyn_at_...>
Date: Fri, 10 Mar 2000 09:45:25 -0000

Have I understood this correctly:

Against, say, 3 attackers:

My first defense would be normal
My second at -3
My third at -6

If I was making 3 attacks, however, all three attacks would be at -6.

Is that right?

Now followers can reduce these penalties, however I'm not sure that this rule is a good idea. Let me give an example:

Jrrrz the great troll has skill 10w2 with a Pole Axe ('tickler', with an extra +5 edge - who said there were no +1 swords in Glorantha:-)) and wears lead plate mail also enchanted to +5. His skill is therefore 10w2 ^ 10 (I think I've got this right - edgewise). He has 50 APs and is, I'm sure you'll all agree, a bit of a dude.

He has three followers, all trollkin, stark-bollock naked, wielding sticks. They have a skill of 6 and 6 AP.

If Jrrrz and his trollkin were treated as individuals, the encounter would have:

a) One 'attack' at 10w2^10
b) One 'defense' at 10w2^10
c) Three 'attacks' at 6
d) Three 'defenses' at 6

If Jrrrz used his trollkin as followers, then the encounter would present you with:
a) Four 'attacks' at 10w2^10
b) Four 'defenses' at 10w2^10

Since Jrrrz would be allowed to multiply attack and defend an extra three times each without penalty because of his followers.

This doesn't really seem right. I'm tempted to get rid of the follower-negates-multi-penalty rule. Then you would have:

  1. Four 'attacks' at 1w2^10
  2. One 'defense' at 10w2^10
  3. One 'defense' at 7w2^10
  4. One 'defense' at 4w2^10
  5. One defense at 1w2^10

Which is what he's entitled to anyway - but the trollkin do give him their APs.

Have I got this right? And what do you think?


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