Sample Adventure: how many trolls?

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dunbruha
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Sample Adventure: how many trolls?

Post by dunbruha »

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?
StormPatriarch
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Re: Sample Adventure: how many trolls?

Post by StormPatriarch »

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. :)
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kragar00
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Re: Sample Adventure: how many trolls?

Post by kragar00 »

Dan suggested using 1 troll per PC... If you have a bunch of fighters in the party, this may be fine..
In play, I thought that drew things out a little longer than I preferred (primarily because they have 176 hit points)... I prefer a 2:3 ratio of trolls to PCs, so if you have 3 PCs, have them face 2 trolls... You can adjust the number based on the combative power of the party...
Conversely, you can drop the troll hit points to about 100 and have 1 troll per PC... In practice, either of these 2 methods works pretty well (or at least it did when I ran the sample adventure)...
dancross
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Re: Sample Adventure: how many trolls?

Post by dancross »

kragar00 wrote:Dan suggested using 1 troll per PC... If you have a bunch of fighters in the party, this may be fine..
In play, I thought that drew things out a little longer than I preferred (primarily because they have 176 hit points)... I prefer a 2:3 ratio of trolls to PCs, so if you have 3 PCs, have them face 2 trolls... You can adjust the number based on the combative power of the party...
Conversely, you can drop the troll hit points to about 100 and have 1 troll per PC... In practice, either of these 2 methods works pretty well (or at least it did when I ran the sample adventure)...
1 troll per PC is too tough if the idea is to stay and fight to the death...the troll's CS score being 77. If every pregen in the quickstart were used they should be able to take a few trolls down and/or escape, if not take them all on, but it's a tough fight.

Usually I'll find the largest maximum Potential-Harm in Melee & Unarmed (both close combat...use the higher number), Ranged, and Arcane, add up those three categories to find a character's CS. So if Fighter PC guy has Melee D8, Unarmed D6, Ranged D4, and Arcane D6, his CS is only 18 (8+4+6, disregarding the lower of the two max P-Harm scores in close combat). If there were four equal fighters like that it would give them a combined "party CS" of 72. That party should take on one troll, and it would take them probably five to eight rounds to take down that single monster.

The pregens in the quickstart are more powerful than that, so throwing two or three trolls at them should work. However, the number of trolls in the quickstart should be strong indicator that not all encounters are meant to end in total desctruction of every creature. Maybe it's my "old skool" thinking, but sometimes I encourage the PCs to fight to escape...
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Re: Sample Adventure: how many trolls?

Post by dancross »

StormPatriarch wrote:
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)
Reflexes: D4
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.
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Re: Sample Adventure: how many trolls?

Post by dancross »

By the way, I was dissatisifed with calculating CS for simple monsters less than full-fledged, so Randy and I came up with a revision that is much faster. I think I've posted it elsewhere, but here it is again. This would change the numbers in the core text a bit, but not by too much.

Challenge Score allows the GM to balance the number of monsters encountered with the heroes’ total potential to inflict harm. The system as presented in the core rules works, but is unnecessarily complex. We suggest you use this similar method instead.

Make the BASE CS score = base HP (before multipliers area applied). At the base level, HP is proportional to damage done (for most monsters).

Any multipliers to hit points (HP) or resilience (RS) scores add to the creature’s overall Challenge Score. Add all of the multipliers together and consult the table below, which updates and improves upon the CS tables given in the core rules on page 66, combining them into one easy reference:

x2 to HP/RS* = +25% to CS
x3 to HP/RS* = +50% to CS
x4 to HP/RS* = +75% to CS
x5 to HP/RS* = +100% to CS
x6 to HP/RS = + 125 % to CS
x7 to HP/RS = + 150 %
x8 to HP/RS = + 175 %
x9 to HP/RS = + 200 %
x10 to HP/RS = + 225 %
x11 to HP/RS = + 250 %
x12 to HP/RS = + 275 %
x13 to HP/RS = + 300 %
x14 to HP/RS = + 325 %
x15 to HP/RS (max) +350 %

If a simple critter had 10 base hitpoints, and you made it "tough" with a x10 multiplier to HP it would have a CS of 10 (his base) + 10 x 2.25 (added from your chart) for a total CS of 33. If the critter gained a x10 to HP and a x5 to RS, then it would have a x15 combined, increasing CS by 350%, for a final CS of 42.5 (10 + 10 x 3.5% ) or just 42.
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dunbruha
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Re: Sample Adventure: how many trolls?

Post by dunbruha »

Thanks for all the help. I am planning to run the sample adventure with my group tomorrow night!
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Re: Sample Adventure: how many trolls?

Post by dancross »

dunbruha wrote:Thanks for all the help. I am planning to run the sample adventure with my group tomorrow night!
Awesome, let us know how it goes!
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