Complex EL calculations

[Galfridus]

Has anyone out there derived any guidelines for calculating ELs in non-standard situations (that is, when the monsters aren't all the same CR or a "matched pair")? I have an encounter with CRs 6, 4, 4, and 3, and I don't know how to estimate the EL.

 

[kreynolds]

Go to The Game Hub and register with them. Then go to the DM tools tab and check out the encounter calculator. It's really slick but their site appears to be having some trouble right now.

 

[Petrosian]

You can use the mixed pair and pair to mush them together...

6 4 4 3 .... does not exactly work out....

close is... take the 4 and 3 as a mixed pair...

a 4 and 2 would be a 5...
a 4 and a 4 would be a 6

so a 4 and a 3 is BETWEEN 5 and 6.

lets say 5, give them the benefit of the round.

so now we have

6 5 (the combined 4 and 3) and 4.

4 and 6 is a mixed pair 7.

that leaves 7 (combined 6 and 4) and 5 (combined 4 and 3.)

7 and 5 make a mixed pair 8...

so we have an EL 8 encounter.

 

[Sir Hawkeye]

Mathematically speaking:

Creatures with CRs of X,Y, and Z

(sqrt(2))^X
+(sqrt(2))^Y
+(sqrt(2))^Z
-----------------
(sqrt(2))^N

Where N is the approximate ECL. This works for any number of creatures. For 3,4,4,6:

2*sqrt(2)
+4 [which is sqrt(2)^4)
+4
+8
---------
about 19, which is closest to 16 (from the powers of sqrt(2)), which is sqrt(2)^8, making the total ECL slightly higher than 8.

 

[Galfridus]

Sir Hawkeye,

I like your formula! Is it something you derived from analysis of the charts, or did it come from somewhere else?

 

[Sir Hawkeye]

Basically from the charts and the language surrounding it. "Double # creatures = CR+2" indicates a geometric progression in sqrt(2) so I just used that with the CR as power, and It worked with the examples in the book.

And they said 3e would be simpler