PC Survival in Star Wars

[rgard]

Psst! He's playing Saga edition.

Oops!

 

[Prince Atom]

Challenges?

A single CL 2 Clone Trooper (non-heroic 6th) is intended to be a challenge to a group of 4 2nd level heroes. Two of them together is actually a CL 3 encounter, and is more of a major encounter/boss battle.

And yet my copy of the Saga Edition sayeth:

For example, a 1st-level hero should find a CL1 stormtrooper challenging. By extension, four CL 1 stormtroopers should prove challenging to four 1st-level heroes.

So it seems to me that a CL 2 clone trooper, at 6th level nonheroic, should be a pushover for a group of 4 2nd-level heroes. You'd need 4 such clone troopers to be a challenge. Do you know something I don't?

And I'm not clear on how you arrived at the CL 3 calculation in your original post.

Seriously, I haven't even run one session under Saga Edition rules. I'm not trying to snark at anyone here; am I missing something?

TWK

 

[Nebulous]

From the errata:

p. 247 – Building an Encounter
Change the second paragraph to say the following:
"A challenging encounter is one the heroes should overcome with minor to moderate damage to themselves and some depletion of their resources. A single obstacle, threat, or situation of Challenge Level n is challenging for a group of 4 of similar level. For example, a group of 1st-level heroes should find a CL 1 stormtrooper challenging. A single enemy is a difficult encounter for a character of a level equal to the enemy's CL."
Under Combining Different CLs, add the following sentence at the end of the first sentence: "The combined CL for the encounter is either this result or the highest single CL + 2, whichever is more."
Add the following sentence to the end of the paragraph: "Most encounters should not include a single enemy whose CL is more than 3 levels higher than the average party level."
Also, the last line should say, "For each additional hero," not "Four each additional hero."

 

[Prince Atom]

Ah, okay. My book is in error. Thanks a bunch!

But I'm still not clear on calculating the total CL for multiple opponents who are all the same CL. Could someone explain that to me in small words?

Thanks!

TWK

 

[3d6]

OK. Lets say you have five stormtroopers, each of which are CL 1.

There are two methods for calculating group CL.

First, add up the total CL of the group (in this case, 1 X 5), and divide by 3. In this case, that results in a CL of 1 (5/3, rounding down).

The second method is to add two to the highest individual CL of the group. In this case, the highest individual CL is 1, so the result is 3.

Then, you use whichever method gives a higher CL. Therefore, the group CL of five CL 1 stormtroopers is CL 3.

 

[Prince Atom]

I thought that was the process for determining overall CL for encounters with threats of different CLs. Certainly all the examples on p. 247 are of opponents with different CLs.

But I guess you gotta use the methodology you've got.

TWK