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Stormthorn
2009-05-14, 12:10 PM
Ok, im sure this is DMing 101 but i dont have that hanbook so i need to ask. How does CR increase as you add multiple creatures? I dont think two CR2 are a CR4 enounter but at the same time it would make the fight more difficult.

Douglas
2009-05-14, 12:17 PM
Technically it doesn't do anything to CR. The term you're looking for is Encounter Level (EL).

Two creatures of CR X have an EL of X+2.
One creature of CR X and another of CR X-2 have an EL of X+1.

For calculating EL, you can treat a group of monsters as a single monster of CR equal to the group's EL.

XP is still assigned according to each individual monster's CR, however.

Stormthorn
2009-05-14, 01:11 PM
Ok. Cool. Im not sure how you got those numbers tho. A CR5 and a CR5 are a CR7 but a CR 5 and a CR3 are a CR6. Well...that rules out averaging the numbers to get the CR. And its not adding them...

Douglas
2009-05-14, 01:30 PM
It's an approximation of an exponential progression. +2 CR equates to doubling the number of monsters.

It's in the DMG somewhere near the XP rewards table, I think.

Sstoopidtallkid
2009-05-14, 01:31 PM
Ok. Cool. Im not sure how you got those numbers tho. A CR5 and a CR5 are a CR7 but a CR 5 and a CR3 are a CR6. Well...that rules out averaging the numbers to get the CR. And its not adding them...I think it involves large amounts of liquor and a dartboard.

raitalin
2009-05-14, 01:37 PM
I'm pretty sure you take the average of all the relevant CRs and then add one for each CR beyond the first, so:

5 CR1=1+1+1+1+1=5/5=1+4=EL 5

2 CR3 4 CR2 4 CR1=3+3+2+2+2+2+1+1+1+1=20/10=2+9=EL 11

Granted, that's really just a starting point; the environment, preparation and party vs. encounter matchup changes it to pretty much whatever the DM feels is appropriate.

Douglas
2009-05-14, 01:44 PM
I'm pretty sure you take the average of all the relevant CRs and then add one for each CR beyond the first, so:

5 CR1=1+1+1+1+1=5/5=1+4=EL 5

2 CR3 4 CR2 4 CR1=3+3+2+2+2+2+1+1+1+1=20/10=2+9=EL 11
No, you don't. If you actually check the DMG you will find that my answer is exactly correct by RAW.

5 CR1 = 2 EL3 + 1 CR1 = 1 EL5 + 1 CR1 = approximately EL 5.

2 CR3 + 4 CR2 + 4CR1 = 1 EL5 + 2 EL4 + 2 EL3
1 EL5 + 2 EL4 + 2 EL3 = 1 EL5 + 1 EL6 + 1 EL 5 = 2 EL5 + 1 EL6
2 EL5 + 1 EL6 = 1 EL7 + 1 EL6 = between EL 8 and 9.

raitalin
2009-05-14, 01:59 PM
No, you don't. If you actually check the DMG you will find that my answer is exactly correct by RAW.

5 CR1 = 2 EL3 + 1 CR1 = 1 EL5 + 1 CR1 = approximately EL 5.

2 CR3 + 4 CR2 + 4CR1 = 1 EL5 + 2 EL4 + 2 EL3
1 EL5 + 2 EL4 + 2 EL3 = 1 EL5 + 1 EL6 + 1 EL 5 = 2 EL5 + 1 EL6
2 EL5 + 1 EL6 = 1 EL7 + 1 EL6 = between EL 8 and 9.

Huh, ya know, I've been using the formula for d20 modern in all my games (which is where I got this) as I just assumed it was the same. It does make sense that ELs would come out a little higher in d20M, as recovery is so much more difficult.

Another_Poet
2009-05-14, 02:46 PM

http://www.penpaperpixel.org/tools/d20encountercalculator.htm

Stormthorn
2009-05-15, 12:02 PM
No, you don't. If you actually check the DMG you will find that my answer is exactly correct by RAW.

5 CR1 = 2 EL3 + 1 CR1 = 1 EL5 + 1 CR1 = approximately EL 5.

2 CR3 + 4 CR2 + 4CR1 = 1 EL5 + 2 EL4 + 2 EL3
1 EL5 + 2 EL4 + 2 EL3 = 1 EL5 + 1 EL6 + 1 EL 5 = 2 EL5 + 1 EL6
2 EL5 + 1 EL6 = 1 EL7 + 1 EL6 = between EL 8 and 9.

This is why it would be nice if i had the DMG. I cant find a DM localy but i can find a few people willing to play, which means if i want to play then i have to DM. But my only resource is the SRD.

Tsotha-lanti
2009-05-15, 12:24 PM
Ok. Cool. Im not sure how you got those numbers tho. A CR5 and a CR5 are a CR7 but a CR 5 and a CR3 are a CR6. Well...that rules out averaging the numbers to get the CR. And its not adding them...

Every time you double the threat, the EL (not CR!) rises by 2.

You'll notice this is how the XP table works, to a point. Double the threat is double the XP and +2 CR. (That is, every time you go up two CR on the table, the XP value doubles; mostly.)

It's really simple mathematics, and not even remotely random.

And dude, why not just buy the DMG? Or have everyone who wants to play chip in to buy one.

Stormthorn
2009-05-16, 11:34 PM
And dude, why not just buy the DMG? Or have everyone who wants to play chip in to buy one.

I just got a job yesterday?

AslanCross
2009-05-17, 05:51 PM
I just use the encounter calculator, but I always keep in mind that two monsters of the same CR gives us EL=(Monster CR+2).