6
\$\begingroup\$

This is D&D 5e. We were attacking a baddie. The wizard cast charm monster and charmed it.

It has an Intelligence score of 4. Is it smart enough to know that the other party members are, in fact, companions of the wizard?

The significance of whether the charmed baddie knows what companions are is in the interpretation of the companions phrase of the spell.

\$\endgroup\$
12
\$\begingroup\$

The spell's effects relate to "you or your companions", regardless of whether the target realizes they're your companions

The charm monster spell description says (emphasis mine):

You attempt to charm a creature you can see within range. It must make a Wisdom saving throw, and it does so with advantage if you or your companions are fighting it. If it fails the saving throw, it is charmed by you until the spell ends or until you or your companions do anything harmful to it. The charmed creature is friendly to you. When the spell ends, the creature knows it was charmed by you.

(The charm person spell has mostly identical wording, aside from only working on humanoids, and otherwise works the same way.)

The spell doesn't say these things depend on whether the charmed creature realizes they're your companions. As such, this means these effects - advantage on the save if companions are fighting it, and the charmed condition ending if your companions do anything harmful to it - apply simply based on whether they're actually your companions, regardless of whether the targeted creature knows they are.

The spell doesn't provide an in-universe explanation for how the spell determines that the creatures are your companions, so you may have to substitute your own reasoning to explain it in-game. However, out of game, that is simply how the spell works; it seems reasonable to interpret "your companions" as referring to the other player characters (and any possible allied NPCs) if the caster is a player character - assuming the PCs are all working together as a party, of course.

\$\endgroup\$
14
\$\begingroup\$

4 Intelligence is plenty to recognize allies and enemies

An ordinary wolf has an intelligence score of 3, and it can recognize allies and enemies just fine. It even has an ability (Pack Tactics) that requires it to know who its allies and enemies are. Even further, the Feeblemind spell reduces a creature's intelligence to 1, and even then the creature can still identify its allies. Furthermore, 4 Intelligence is implied to be the lower limit of sentience and self-awareness, as evidenced by spells like Detect Thoughts, which works only on creatures with 4 or higher Intelligence. So the monster with 4 intelligence may be very dumb, but it is smarter than a wolf, and it is smart enough to be considered sentient. So it ought to be able to identify allies and enemies.

Admittedly, this monster may not be intelligent enough to understand the concept of multiple factions, but if that's the case, then by its limited understanding, all creatures hostile to it are allied with each other against it. So no matter how you slice it, if your allies attack the charmed monster, the charm spell will end. (The same would likely be true of a wolf.)

\$\endgroup\$
  • 4
    \$\begingroup\$ It may also be worth noting that the spell says the charm effect ends when "you or your companions do anything harmful to it"; the spell doesn't say it depends on whether the charmed creature realizes they're your companions. \$\endgroup\$ – V2Blast Jan 19 at 2:33
  • \$\begingroup\$ I think this is actually even more telling than the actual answer. I had not read the spell that way, but it seems like exactly the right interpretation. Whether someone is or is not a "companion" is not up to any quality of the the charmed person or monster, or up to the caster, but is up to the magic itself (ie, the DM) as to what is or is not a companion. \$\endgroup\$ – Jack Jan 19 at 3:00
  • \$\begingroup\$ @V2Blast I think that's worthy of its own answer, honestly. \$\endgroup\$ – Ryan C. Thompson Jan 19 at 3:25
  • \$\begingroup\$ @RyanC.Thompson: Fair enough! Wasn't sure if it was enough to stand on its own. \$\endgroup\$ – V2Blast Jan 19 at 3:36

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.