Just so I know I'm doing things right...
Using Bandit templates in an encounter state that each bandit is rated CR 1/2.
So, going by the CR table, it's meant to be CR+5 for 6 enemies at once, so as each CR is actually 1/2, I decided the CR would be CR3 as there are 6 enemies at CR 1/2 each.
Is this right or would the party get more exp than the 800 a CR3 encounter gives?
Thanks! :)
Yar!
6 CR 1/2 creatures would give the equivalent of a single CR 4 creature, thus it is a CR 4 encounter. Each one individually would give 200 exp, 6 of them fought individually at different times of the day would give a total of 1200, just as a single CR 4 encounter would. One could argue that it would be a greater challenge due to action economy now favoring the enemies, and count it as a CR 5 encounter (which would fit the math of the CR Equivalencies table better, as the fractional CR's are simply steps in a sequence, not actual fractions. They could have named then colors instead of fractions, and it would still be the same... the CR+# simply moves it up that many steps in the sequence, making CR 1/2+5 = 5. CR 1/6+5 would = 2, etc).
~P
EDIT: looking at those tables more, the CR Equivalencies part works well for CRs 1 and higher. CR's lower than 1 grant exp based on a fraction of 1 = to their listed fraction rounded to the nearest 5. This is a fraction of the EXP given, NOT a fraction of the enemies encountered. Using the CR Equivalencies formula for the factional CR's doesn't work very well, as the math doesn't line up. I'd sooner just add up all the values individually and give that amount (in all other cases, it works out to exactly the same value anyways).
EDIT 2: You'll also note that the entire section on designing encounters appears to assume that you're determining the CR before you are determining the creatures that make it up.
Step 1 - Determine APL: ...
Step 2 - Determine CR: ...
Step 3 - Build the Encounter: ...
If you reverse steps 2 and 3, you can end up with strange encounters that do not fit nicely with the exp by encounter CR table. For example: 2 CR 2 creatures with 1 CR 1/6 creature and 3 CR 1/8 minions gives a total exp of 1415 xp. A CR 4 encounter should give only 1200, but the next step up (CR 5) gives 1600 exp. What CR is the encounter?! Well guess what: it doesn't fit with the Encounter CR formula, because steps 2 and 3 got switched. Yeah, it's a more natural and organic encounter, but it is not how that section of the CRB was designed or intended* to be used. It appears to assume you're building encounters for the party based on their APL and an appropriate challenge for them, as opposed to what may or may not actually exist in a given location.
* I'm no developer, so my assumption of intent is just that: my assumption.
~P
Ah, so I should actually ramp it up to CR4 or 5?
Which one would you do personally?
They took a level 3 and level 2 rogue out (these guys were mean and worked in pairs) with ease. was surprised, so just threw in more basic men to make it seem like a decent encounter.
If they didn't roll so high, the two rogues would have been dealing +7 Earthbreaker - 2d6 + 10, 2d6 sneak and Short Sword x2 1d6 + 4 + 1d6 sneak x2.
Would have been nasty!
Maybe I should reward them as a CR5 then for dealing with things so effectively, though afterwards the party turned on each other as they saw a Cleric who claimed to ally with them, attacking the bandits, then healing one that was left because he was an indifferent medic, so turned on him.
I don't think I should penalise them for that though. I chucked them in the cells of Sandpoint for that.
Edit: So yea, CR3 then? Because 6 halves are 3? Or just put it to 4 going with the CR + 4 thing?
Hmm, well it's weird because I have to work out the CR based on the encounter. I find it easier than working out the CR first, then adding creatures as I first need to know what CR the amount of creatures I make include would make. :/
Yar.
Yeah, added a second edit to my previous post. Persoanlly, I'd give them individual exp for each creature and not bother with "Encounter CR" at all. So in this case, 6 CR 1/2 creatures would give 200 exp each, for a total of 1200 exp if they are all defeated.
Of course, atm I'm doing a "no exp" game, where they level up by reaching various plot points. But when I do do exp, I do it all on an individual creature basic, instead of "Encounter Cr".
But it would technically be CR 4, as 6 CR 1/2's = 1200 = CR 4.
~P
Yar.
Hmm, well it's weird because I have to work out the CR based on the encounter. I find it easier than working out the CR first, then adding creatures as I first need to know what CR the amount of creatures I make include would make. :/
Why do you have to? Just for ease? I'd honestly not try to find the exact encounter CR, because you will often not fit into one nicely doing it that way. You can approximate it, based on the amount of EXP the creatures would give you, but it will only be an approximation.
(and seriously, the math is more accurate if you figure it out by exp than by CR, as the fractional CR's are fractions of the EXP given from CR 1, and do not work as fractions of the CR if the CR ever goes higher than 1).
~P
Ah, yea I think that's probably a much easier way of working it out then. For multiple creatures that aren't listed as campaign encounters, I'll just scrap CR entirely and sum up the exp from each creature individually. :)
Thanks very much, me hearty!
Yarrrr!
Yar!
They took a level 3 and level 2 rogue out (these guys were mean and worked in pairs) with ease. was surprised, so just threw in more basic men to make it seem like a decent encounter.
If they didn't roll so high, the two rogues would have been dealing +7 Earthbreaker - 2d6 + 10, 2d6 sneak and Short Sword x2 1d6 + 4 + 1d6 sneak x2.
Would have been nasty!
That sounds more like good tactics combined with sheer dumb luck. Sometimes that happens, sometimes they roll bad and the enemy rolls good and they get stomped. CR cannot perfectly account for the randomness of a d20, it can only bring you close. Try not to let one lucky fight skew everything else.
But I will also say to remember that a CR equal to the APL is supposed to be - on average - an average fight (which in my experience is usually actually fairly easy). Sometimes luck is on your side and it's too easy, sometimes luck is against you and it's harder than theory-crafting would suggest, but on average it is what it is.
...
And you're welcome! Glad to help, and best in your gaming endeavors!
~P
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Award XP based on individual creatures defeated, not the overall CR of the encounter (usually the two will align, but not always).
When designing encounters, it is usually best to have a total number of enemies approximately equal to the number of PCs plus their "pets" (familiars, animal companions, mounts, etc.).
If the PCs are having many encounters per day, they should each be a total CR right around APL. This is most easily handled by giving yourself an XP budget for each encounter based on the CR equal to the APL. Spend these XP on several lower-CR creatures until the total is right around your budget.
If the PCs are facing only one or two encounters per day (such as in a wilderness campaign), you can afford to give them tougher encounters, usually CR = APL + 2 is about right. You still want to try to design such encounters to have multiple creatures, however.
Ah, yea the group in question was a group of 4 level 2's, but I brought in a level 4 cleric (3 cleric, 1 fighter) to aid them and they turned on him, though I did give him to a player to use, with just instructions on how to use him as he was sheerly a gmpc I made originally who would randomly aid the group.
My GM gave the dice to one of the players in the game I play in last week to roll a random encounter and threw 12 worgs at our 4 level 4's and one druid wolf companion. One of the players were away and as a result, because it was a CR9 encounter technically as they're all CR2 each...my guy and the fighter went down together. Was terrible as I really didn't want to lose that character. :(
Ah, so I should actually ramp it up to CR4 or 5?
Which one would you do personally?
They took a level 3 and level 2 rogue out (these guys were mean and worked in pairs) with ease. was surprised, so just threw in more basic men to make it seem like a decent encounter.
If they didn't roll so high, the two rogues would have been dealing +7 Earthbreaker - 2d6 + 10, 2d6 sneak and Short Sword x2 1d6 + 4 + 1d6 sneak x2.
Would have been nasty!
As NPCs with NPC gear, the L3 Rogue is CR 2 and the L2 Rogue is CR 1; as an encounter, they combine to a EL 3 encounter. Your APL 2 group should, by the CR system, be just a little challenged for it.
Keep in mind that the CR system assumes that the creatures are encountered in environments that are suited to them -- maybe not optimized for them, but suited. An ambush monster (such as Rogues often are) isn't its normal CR if it isn't in an environment that it has a decent change of attacking the party from surprise.
A bit off-topic ... are those numbers for damage right? A L3 Rogue with +10 damage seems high, and a L2 Rogue having equal damage (+4 at that) with both TWF weapons seems a bit unusual (but possible) as well.
You can also work out the CR by XP. It will get you the same result.
Step 3—Build the Encounter: Determine the total XP award for the encounter by looking it up by its CR on Table: Experience Point Awards. This gives you an “XP budget” for the encounter. Every creature, trap, and hazard is worth an amount of XP determined by its CR, as noted on Table: Experience Point Awards. To build your encounter, simply add creatures, traps, and hazards whose combined XP does not exceed the total XP budget for your encounter. It's easiest to add the highest CR challenges to the encounter first, filling out the remaining total with lesser challenges.
are those numbers for damage right? A L3 Rogue with +10 damage seems high, and a L2 Rogue having equal damage (+4 at that) with both TWF weapons seems a bit unusual (but possible) as well.
Yep, they are.
The level 3 Rogue had an Earthbreaker, +4 Str, was two-handing, using power attack and was going to sneak before he failed miserably.
That equals +6 to hit because of 5 str, 2 bab, -1 Power Attack and 2d6 (earthbreaker) + 10 (due to dirty fighter trait which they both had) + 2d6 sneak in flanking.
I looked into it beforehand and weapons don't seem to effect whether or not you can sneak attack, so yea, he would have been nasty if he could hit, especially as they were going to copy each other!
That's 4d6 + 10 from him alone in one attack!
34 damage....hmm.
The Rogue wasn't bad either. 1d6 sneak, but 1d6 + 4 with each weapon, also due to dirty fighter. 3d6 + 8, so 26 max damage from him and they were going to flank the same opponent each, but were surprised by two barbarians storming in, a witch lowering the hammer guys AC and a magic missile from the magus.
I don't think they got a single attack off. Well, not a successful one at least. Hammer guy was ready for them, but the other guy failed his perception check (he had +7!!!).
I was disappointed as I thought I'd have fun with them. :(