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Thread: Stupid rules question about CR
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2009-05-14, 12:10 PM (ISO 8601)
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Stupid rules question about CR
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.
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2009-05-14, 12:17 PM (ISO 8601)
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Re: Stupid rules question about CR
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.Like 4X (aka Civilization-like) gaming? Know programming? Interested in game development? Take a look.
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2009-05-14, 01:11 PM (ISO 8601)
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Re: Stupid rules question about CR
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...
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2009-05-14, 01:30 PM (ISO 8601)
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Re: Stupid rules question about CR
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.Last edited by Douglas; 2009-05-14 at 01:31 PM.
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2009-05-14, 01:31 PM (ISO 8601)
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Re: Stupid rules question about CR
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FAQ is not RAW!Avatar by the incredible CrimsonAngel.
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I play primarily 3.5 D&D. Most of my advice will be based off of this. If my advice doesn't apply, specify a version in your post.
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2009-05-14, 01:37 PM (ISO 8601)
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Re: Stupid rules question about CR
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.
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2009-05-14, 01:44 PM (ISO 8601)
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Re: Stupid rules question about CR
No, you don't. If you actually check the DMG you will find that my answer is exactly correct by RAW.
Your first example:
5 CR1 = 2 EL3 + 1 CR1 = 1 EL5 + 1 CR1 = approximately EL 5.
Your second example:
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.Last edited by Douglas; 2009-05-14 at 01:51 PM.
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2009-05-14, 01:59 PM (ISO 8601)
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Re: Stupid rules question about CR
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2009-05-14, 02:46 PM (ISO 8601)
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Re: Stupid rules question about CR
The CR calculator answers all your questions and more:
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2009-05-15, 12:02 PM (ISO 8601)
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2009-05-15, 12:24 PM (ISO 8601)
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Re: Stupid rules question about CR
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.
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2009-05-16, 11:34 PM (ISO 8601)
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2009-05-17, 05:51 PM (ISO 8601)
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Re: Stupid rules question about CR
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).
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