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Let's say a rogue and an ally are adjacent to an invisible enemy. The rogue would have disadvantage on the attack roll because the target is invisible, but at the same time an ally is adjacent to the target.

How does sneak attack work in this situation?

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2 Answers 2

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Sneak attack says (PHB p.96):

You don't need advantage on the attack roll if another enemy of the target is within 5 feet of it, that enemy isn't incapacitated, and you don't have disadvantage on the attack roll. So, as described you cannot sneak attack.

However, advantage and disadvantage says (PHB p.173):

If circumstances cause a roll to have both advantage and disadvantage, you are considered to have neither of them, and you roll one d20. This is true even if multiple circumstances impose disadvantage and only one grants advantage or vice versa. In such a situation, you have neither advantage nor disadvantage.

So if the rogue had advantage to cancel the disadvantage (say, be being hidden) and had an ally adjacent the foe (since the disadvantage negates the advantage and vice versa) then they can sneak attack.

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Sneak Attack specifically states:

You don't need advantage on the attack roll if another enemy of the target is within 5 feet of it, that enemy isn't incapacitated, and you don't have disadvantage on the attack roll. (PHB p.96, emphasis mine)

If you have disadvantage on the attack roll then you can't Sneak Attack.

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  • \$\begingroup\$ Corrolary question : If the rogue is hidden or is himself invisible, he would normally have advantage on his attack roll. Against an invisible opponent, advantage and disadvantage would negate each other, meaning the roll is performed without modifiers. Does this mean an invisible rogue can sneak attack an invisible opponent? \$\endgroup\$
    – Dungarth
    Commented Dec 22, 2016 at 3:17
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    \$\begingroup\$ @Dungarth I think that is worthy of its own question \$\endgroup\$
    – Dale M
    Commented Dec 22, 2016 at 3:21
  • \$\begingroup\$ @DaleM - I asked, and apparently the answer is yes, according to RAW. \$\endgroup\$
    – Dungarth
    Commented Dec 22, 2016 at 5:19

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