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Effective Encounters for a Party of 7

What I've done in the past is figure the EL of the party if they were encountered by a group of adventurers. Remember that EL goes up by two when you double the number of creatures.

One sixth level PC would be an EL of 6
Two sixth level PC's would be an EL of 8
Four sixth level PC's would be an EL of 10
Eight sixth level PC's would be an EL of 12

So 7 sixth level PC's would have an EL of 11 or so (11.75 to be precise). I would say that, for this party, an EL 11 or 12 encounter would be a fair fight (I wouldn't be suprised to see either side destroyed).

Anyway... just my two cents.
 
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Ok, same question and still looking for an answer...

How does a party of 7 PC's at 6th level compare to 4 PC party?

8th level party?
9th?
10?

More?
 


Well, the easiest way would be to use the tables as given and simply increase the numbers of all opponents by 75% (round fractions as you see fit), and increase the level of single opponents by +1 to +2, when the increase in numbers doesn't make sense (i.e. for unique opponents or boss-type creatures).

You could also go by the EL+2 shift, the Dog Faced God proposed, i.e. take appropriate EL for a group of four 6th level PCs and add 2 to that EL to base the encounter on that number.

Bye
Thanee
 

I hope this helps you.

Effective encounters are those that challenge the players. An attack by a swarm of bees would be just as deadly. Remember that bees die after stinging. Now send a swarm and get them poisoned.

Traps. Falling down a pit that closes. then maybe water flowing in. Drowning is such a dangerous thing to happen.

It isn't always about monsters.


AS for 7- Level 6 characters. If you are designing a combat without the RTTTOEE, then you would need to do this. The normal cr of a monster is for 4 pcs. SO a cr6 would be easy for this group, but 2 would make them use more resources. Remember the CR is not designed to let them use up all their resources, just 20%. So what happens if it isn't 1 cr6 creature but a whole bunch of creatures that equal ECL 6?

10 level 1 characters might not sound like a hard fight. Now, make them hide and attack from cover. hmm harder fight isn't it? Now add in a couple of psy war or psions, using their mind attacks to stun. 1 stunned character is 1 less that will attack them. What happens if the pcs encounter 10 psion kobolds? with a possibility of all of them being stunned and tied up very fast. Once stunned and captured, they would believe that the kobolds are a much higher level than them, or more specialized. Not so, it's just how they are played out.

Email me, I'll send you a html file that I downloaded from a page that no longer exist called Encounter Calculator. It'll give you an idea and after testing it a few times, the html file will come in handy when you make encounters.

It tells you if the pcs will find it challenging, very hard, deadly, and obscene. hehe.

email: kyramus@yahoo.com
 

I'd start with 7 "units", each with CR equal to average party level -4. Then start "combining" them. Two units combined have CR equal to average party level -2. When you're satisfied with the number and spread of the units, you can replace them with creatures whose CR equals the CR of the "units." This can give you any combination of multiple weak creatures to one or two very powerful creatures.

Of course, this makes random encounters very annoying, but you could just use two random encounters, each with EL equal to the party level (this will avoid getting a single nasty monster).

Using one creature is too risky. If your party has spellcasters or rogues, that creature is dead. If the creature is a brute, on the other hand, it may very well slaughter the party. If the creature is a spellcaster that uses proper tactics (eg Fly, Improved Invisibility, then Fireball, Fireball, Fireball, etc) then the amount of damage put out may severely hurt the party.
 

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