Let's say that I determined the best course of action is using the Help action to give my Rogue buddy an advantage on his attack instead of attacking by myself. Would that attack then gain the sneak attack bonus since he's at attacking at advantage?

  • 3
    \$\begingroup\$ Possible duplicate of Can a Familiar aid a Rogue's Sneak Attack? \$\endgroup\$
    – Luke
    Nov 22, 2018 at 4:35
  • \$\begingroup\$ This is a weird case... on the one hand, it seems like a distinct question being slightly more broad, but on the other, the answers clearly overlap \$\endgroup\$
    – Ifusaso
    Nov 22, 2018 at 5:57
  • \$\begingroup\$ @Axoren The real question is, would someone looking for an answer to THIS question click on the other question. I'd say no. \$\endgroup\$ Nov 22, 2018 at 7:06
  • 1
    \$\begingroup\$ If anything, the other question is a duplicate of this one, as it's a more specific example of this more general question. \$\endgroup\$
    – V2Blast
    Nov 22, 2018 at 7:43

1 Answer 1


Sneak Attack says:

Once per turn, you can deal an extra 1d6 damage to one creature you hit with an attack if you have advantage on the attack roll.

The Help action says:

If your ally attacks the target before your next turn, the first attack roll is made with advantage.

Sneak Attack can be used if the attack roll is made with advantage (or an enemy of the target to be with 5 feet); the Help action gives that advantage, as long as the rogue is the first ally to attack the target (and the advantage isn't canceled out by disadvantage). It's that simple.

  • \$\begingroup\$ "as long as the rogue is the first ally to attack the target" - why is that? The wording seems to suggest you are aiding a specific ally, not just the next person attacking that target... \$\endgroup\$
    – Chris
    Nov 22, 2018 at 9:47
  • \$\begingroup\$ Ah. rpg.stackexchange.com/questions/128393/… seems to answer my question. Strange that the wording is seemingly different in different places. \$\endgroup\$
    – Chris
    Nov 22, 2018 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .