1
\$\begingroup\$

Situation: I'm currently playing in CtL 2E at the moment and working out a Huntsman designed to challenge the local combat monster, a magic swordsman. The hunter wears cold iron plate mail,* which offers 4/2 armor, and he also has the Unbreakable Dread Power (p. 256). The power reads as-follows:

"The hobgoblin is nearly indestructible. Any attack that does not score an exceptional success inflicts only a single point of bashing damage. Exceptional successes inflict damage as normal. Attacks incorporating iron or another of the goblin’s frailties bypass this power."

On page 148 of the Chronicles of Darkness 2E rulebook, the exact Dread Power is listed for Horrors, though with the description altered slightly to be more generic. I listed it in the tags for that purpose, as this is an important question for Horrors, too.

  • (No, I don't hold a grudge against him. It was his idea, actually. He wanted a problem that can't be solved with brute force.)
\$\endgroup\$
2
\$\begingroup\$

It seems to me that the combination of the two means that, unless an exceptional success is scored, the Huntsman will reduce the attack to 1 Bashing damage via Unbreakable and then processes the damage via armor, completely nullifying an attack that wasn't an exceptional success or his bane. I believe this because, reviewing the rules for Armor on page 187 of CtL 2E and 94 of CoD 2E, it says to subtract your general armor from the damage taken.

The way Unbreakable works is that, when you receive damage from an attack and certain conditions haven't been met, you swap out the damage taken for 1 Bashing. After Unbreakable has activated, THEN you take the new damage, and when you take damage from an attack, you check your armor score.

Therefore, you need a minimum of three successes to have a chance of hurting this Huntsman with an attack by default.

\$\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.