dunbruha wrote:I am going over the sample adventure, and I am wondering how many trolls are suggested in the encounter F-A. The paragraph above says "This creature is a troll...", but then says "...the trolls will stomp out of the forest...". The monster stats say "Trolls", but have no mention of a number. How many trolls?
I would think looking at it that it depends on the size of the player character party. Maybe if all the player characters are used, all 4 trolls shown on the table should be used. But if only a few PCs are being used scale it to 1 or 2? Just a thought.
The StormPatriarch is correct. To find out the number of opponents find the total party CS, then divide by the creature's CS.
On page 65 I calulate the CS of a PC, although the Reflex MRV (4) was added to the CS as a mistake (not that it would really have that big of an effect...). Only the max MRV of the highest branch of each potential harm category should have been added to the total. But anyway, correcting this...
For example, Nathal the elven wizard has his highest ranked Ability Branches as follows:
Close combat: Melee D4 > Sword D8 > Long sword D6 (18)
Unarmed D4 > Pugilism D4 +1 (bonus due to cestus) (9)
Missile combat: Ranged D12 > Bow D6 > Short Bow D4 (22)
> Thrown weapons D4 (16)
Arcane combat: Arcanum D10 > Mystic D8 > Wizardry D6 (24)
> Primordial 12 (24)
Total Challenge is sum of highest branches = (18+22+24) 64.
Then, check out the sidebar on page 64 of the Core Rules:
Challenge score and designing encounters] If Nathal had a brother with the exact same statistics (-4 to account for my accidental inclusion of Reflexes), their group CS would be 128. To determine a suitable challenge for them, divide the group CS by the CS of the opposing monsters, or use a combination of opponents with different CS ratings, subtracting one at a time from the total Group score. Always round down.
For example, our two elves above might face a pack of goblins of the fodder variety. If these goblins had a CS of eight (+4 for ranged which was unlisted but assumed) (TY-fodder, TR D4-close, D4-unarmed, HP 4, RS 2, BR D4), then an easy fight would involve 16 or 17 opponents. This may seem like a lot, but remember that fodders are weak and automatically lose initiative to the PCs in every battle phase.
Standard creatures would have no less than a CS of 12 (TR 2D4 Melee, 1D4 Unarmed (use highest of two close combat scores / Ranged D4, Arcane D0, HP 8, RS 4, BR D4), so the two elves could face off against ten or eleven such opponents; a tougher fight. Exceptional creatures have no less than a CS of 16 (TR 3D4-close, HP 12, RS 6, BR D4), so the two elves could face off against eight such opponents; less opponents but tougher still. Alternatively, the two elves could face off against five of the weakest Extraordinary (80 points) creatures, 2 Standard foes (24), and three Fodders. Full-fledged enemies may or may not pose a greater threat, as some may have abilities that are formidable but not combat orientated.