Adjusting EL for larger parties [Archive] - Wizards Community

Adjusting EL for larger parties [Archive] - Wizards Community

Post/Author/DateTime Post
MrPowell82
07-23-07, 11:28 AM I am about to begin DMing a campaign that will have 6 PC's. What is a good rule of thumb as to how many creatures and of what CR would make a good EL?

Everything in the DMG refers to a party of 4 PC's...

For example would an EL 2 encounter for a party of 4 2nd level PC's be the same as a EL 4 encounter for a party of 6 2nd level PC's?

Also, what about mixed groups?

Thanks for your guys' help!
Ebonarc
07-23-07, 11:34 AM Increase the EL by one.

You have a 50% increase in players. To counteract this, you need a 50% increase in challenge. Two monsters of equal CR are an EL two higher than their CR. Adding a third increases it to an EL three higher than the CR. Therefore, if you add in monsters that constitute an EL 2 lower than average party level, you increase the EL by one and keep the challenge to the same for 6 PCs as it would be for 4.

Note that this does not work for ELs 2 and lower.
Gryllmyre
07-23-07, 11:34 AM I suppose you could calculate the EL for the party itself (4 5th level PC's = EL 9, 6 5th level PC's = EL 10) and subtract 4 to get the typical EL they would face.
Leilond
07-23-07, 12:04 PM I always suggest to add up monster and do not make them stronger

Example
Four 4th level character can beat a CR 4 monster (EL 4)
If you double the pary members (8 characters) and you double the power of the monster (CR 8, EL 8) you going to surely experiment more than one character death, if not a TPK
It is surely better to add monster in "percentage".
4 against one is like 8 against two.
MrPowell82
07-23-07, 10:22 PM So let me make sure I have this right...

My party is 6 lvl 2 PC's

So the appropriate EL for that party should be 3

So I could throw two CR 2 monsters at them

or 3 CR 1 monsters

or 1 CR 3

Is that right?

Also, Gryllmire said:
"(4 5th level PC's = EL 9, 6 5th level PC's = EL 10)"

based on how you guys explained it, shouldn't 6 5th level PC's be an EL 11? (each one CR=5 plus 6 for total characters=11)

I am a bit confused

Keep the responses coming!!
Ebonarc
07-24-07, 10:57 AM My party is 6 lvl 2 PC's

So the appropriate EL for that party should be 3Correct.So I could throw two CR 2 monsters at them1 CR 1 and 1 CR 2.3 CR 1 monsters

or 1 CR 3Correct and Correct.

Also, Gryllmire said:
"(4 5th level PC's = EL 9, 6 5th level PC's = EL 10)"

based on how you guys explained it, shouldn't 6 5th level PC's be an EL 11? (each one CR=5 plus 6 for total characters=11)
Divided into pairs, you get 3 EL 7s. Add together two of the EL 7s and you get an EL 9. Adding together the EL 9 and the EL 7 (using the mixed pair rule) you get an EL 10)

Post/Author/DateTime	Post
MrPowell82 07-23-07, 11:28 AM	I am about to begin DMing a campaign that will have 6 PC's. What is a good rule of thumb as to how many creatures and of what CR would make a good EL? Everything in the DMG refers to a party of 4 PC's... For example would an EL 2 encounter for a party of 4 2nd level PC's be the same as a EL 4 encounter for a party of 6 2nd level PC's? Also, what about mixed groups? Thanks for your guys' help!
Ebonarc 07-23-07, 11:34 AM	Increase the EL by one. You have a 50% increase in players. To counteract this, you need a 50% increase in challenge. Two monsters of equal CR are an EL two higher than their CR. Adding a third increases it to an EL three higher than the CR. Therefore, if you add in monsters that constitute an EL 2 lower than average party level, you increase the EL by one and keep the challenge to the same for 6 PCs as it would be for 4. Note that this does not work for ELs 2 and lower.
Gryllmyre 07-23-07, 11:34 AM	I suppose you could calculate the EL for the party itself (4 5th level PC's = EL 9, 6 5th level PC's = EL 10) and subtract 4 to get the typical EL they would face.
Leilond 07-23-07, 12:04 PM	I always suggest to add up monster and do not make them stronger Example Four 4th level character can beat a CR 4 monster (EL 4) If you double the pary members (8 characters) and you double the power of the monster (CR 8, EL 8) you going to surely experiment more than one character death, if not a TPK It is surely better to add monster in "percentage". 4 against one is like 8 against two.
MrPowell82 07-23-07, 10:22 PM	So let me make sure I have this right... My party is 6 lvl 2 PC's So the appropriate EL for that party should be 3 So I could throw two CR 2 monsters at them or 3 CR 1 monsters or 1 CR 3 Is that right? Also, Gryllmire said: "(4 5th level PC's = EL 9, 6 5th level PC's = EL 10)" based on how you guys explained it, shouldn't 6 5th level PC's be an EL 11? (each one CR=5 plus 6 for total characters=11) I am a bit confused Keep the responses coming!!
Ebonarc 07-24-07, 10:57 AM	My party is 6 lvl 2 PC's So the appropriate EL for that party should be 3Correct.So I could throw two CR 2 monsters at them1 CR 1 and 1 CR 2.3 CR 1 monsters or 1 CR 3Correct and Correct. Also, Gryllmire said: "(4 5th level PC's = EL 9, 6 5th level PC's = EL 10)" based on how you guys explained it, shouldn't 6 5th level PC's be an EL 11? (each one CR=5 plus 6 for total characters=11) Divided into pairs, you get 3 EL 7s. Add together two of the EL 7s and you get an EL 9. Adding together the EL 9 and the EL 7 (using the mixed pair rule) you get an EL 10)