3
\$\begingroup\$

Is it true that not even the person/creature/thing doing the initial action that starts a combat is able to use a full-round action?

According to all sources I can find, the initial attack already is part of the surprise round, so you're only entitled to a move action or standard action; but that leads to an (in my opinion) absurd scenario, so I just wanted to be sure whether there is something I'm missing:

Let's say sorcerer A is hiding in the bushes, 1HP left, waiting to finish sorcerer B (1HP left) off with magic missiles before B does the same to A. Then he suddenly spots sorcerer B - but sadly, he's some feet out of range. Now sorcerer A wants to use metamagic to double his magic missiles range and fire that on sorcerer B ... but now his magic missile takes a full-round action to cast, so sorcerer A cannot surprise sorcerer B with it anymore.

Is this interpretation right RAW-wise or am I missing something?

\$\endgroup\$
3
  • \$\begingroup\$ Are you looking for an answer that solely confirms your understanding? Or is there more to this question? That is, an answer can certainly just say Yep and provide a link to the surprise rules, but that seems like it would be… unsatisfactory? Unhelpful? For instance, it's possible a question like How is game balance affected by a house rule saying that creatures can take a full-round action during the surprise round? would yield more robust answers. \$\endgroup\$ Jul 20, 2018 at 12:51
  • \$\begingroup\$ You're probably right; my phrasing is very narrow here. But as I already got a fully valid answer, I'll maybe ask a new question later about how to houserule the absurdity away without breaking the game. \$\endgroup\$
    – Blutkoete
    Jul 20, 2018 at 14:45
  • \$\begingroup\$ Just so I can confirm the issue: What exactly is the absurdity? I mean, so that his spell can hit Sorcerer B from a greater distance, Sorcerer A has opted to change how he normally casts his spell so that it takes longer to cast—therefore sacrificing the advantage he'd normally enjoy—, and Sorcerer A is aware of the potential consequences. I guess I'd kind of like a definition of absurd: Is that chain of events unfair? Unrealistic? Some other adjective? (Honestly, I'm not trying to be a rules apologist here but understand the question!) \$\endgroup\$ Jul 20, 2018 at 15:59

2 Answers 2

3
\$\begingroup\$

That is correct. The best this sorcerer can do is start a full-round action. At least, barring something like time stop.

Generally, the game is designed to avoid situations where someone doesn’t even get to roll.1 In this situation, the game wants to give B a chance to roll initiative, so they have a chance to perhaps respond or escape.

Note that A could choose to sneak closer to B before initiating the attack, so that a standard action is sufficient. That gives B the chance to roll Spot and/or Listen, though. Still, you can do even better by hiding behind something—with total cover—so that you can 5-foot-step out as a free action and then shoot magic missile at B. That really gives B no opportunity to respond, but it requires a specific setup and great timing.

  1. Though the game doesn't really do a good job of this, as the rest of the answer demonstrates ways around it. And at the high end of optimization, going first is almost always the same as winning a fight. And things like nerveskitter and celerity mean a spellcaster can go first every time, even when they shouldn’t.
\$\endgroup\$
0
2
\$\begingroup\$

What the game does not specify is what happens if after a surprise round, you still have unaware combatants. For instance, in your example, what if As starting a full round action did not alert B? Ultimately, this is not covered in the rules, so it's up to the DM. The way I rule it is, if at the end of a surprise round, you still have unaware combatants, another surprise round ensues, but that's purely a house rule at this point.

I think KRyan's suggestion that sneaking closer wouldn't spoil the surprise supports this very idea - you would have to use a move action to sneak closer.

\$\endgroup\$

You must log in to answer this question.

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