12
\$\begingroup\$

I want to create a "boss and minions" encounter for my low-to-mid-level party, and I love the idea of legendary and lair actions as a way to give the encounter an appropriately "epic" feel. Problem is, I can't seem to find any material in the DMG that directly addresses how these abilities affect the CR of the creature in question.

The DMG (p. 278) has this to say:

When calculating a monster's damage output, also account for special off-turn damage-dealing features, such as auras, reactions, legendary actions, or lair actions.

Does this mean that I perform the "average damage across three rounds" calculation, assuming that legendary actions get used as frequently as possible on a party of four adventurers, to help determine the Offensive CR of the creature? What about if I have a not-specifically-damaging feature (such as a creature that attempts to charm another creature as a legendary action)?

Are there official rules to handle situations like these, or does this fall into "estimate then playtest" territory?

\$\endgroup\$
7
\$\begingroup\$

Definitely calculate the average damage of the effect to add to CR.

Charm is harder to calculate, so I will link to a good example of monster creation for 5th edition here.

Off the top of my head, a charm effect like the spell would give a DC 15 saving throw to avoid it. I will assume average character can pass this 60% of the time. Assuming a 4 person party will output 40 damage a round, each extra point of AC on a creature will remove 1/20 of that damage = 2 damage.

Assuming a character will fail a DC 15 wisdom saving throw 30% of the time, this equates to a party damage reduction of 0.3 * 10 damage per round = 3 damage per round. Given the damage to AC equivalency of 2 damage to 1 AC we get the Charm ability equal to an extra 1.5 AC

My assumptions on average damage per round and average DC 15 failure are wrong, most likely, but this illustrates the calculation required at least.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.