7

Ravager of the Fells (the flip-side of Huntmaster of the Fells) says,

Whenever this creature transforms into Ravager of the Fells, it deals 2 damage to target opponent and 2 damage to up to one target creature that player controls.

What happens if I cannot target my opponent (e.g. due to Witchbane Orb giving her hexproof)? Are her creatures protected as well?

2 Answers 2

11

You must be able to declare legal targets for all targets of a spell or ability. Since Witchbane Orb prevents you from targeting your opponent, you cannot choose their creature to target either. Therefore CR603.3d forces you to remove it from the stack.

It should be noted that if your opponent gains Hexproof after you choose valid targets, the spell will still resolve because the creature is still a valid target (it knows that it is controlled by the same player that you targeted, even if it cannot target that opponent)

114.1. Some spells and abilities require their controller to choose one or more targets for them.The targets are object(s), player(s), and/or zone(s) the spell or ability will affect. These targets are declared as part of the process of putting the spell or ability on the stack. The targets can't be changed except by another spell or ability that explicitly says it can do so.

114.1d A triggered ability is targeted if it identifies something it will affect by using the phrase "target [something],"where the "something" is a phrase that describes an object, player, or zone. The target(s) are chosen as the ability is put on the stack;see rule 603.3d.

603.3d The remainder of the process for putting a triggered ability on the stack is identical to the process for casting a spell listed in rules 601.2c–d. If a choice is required when the triggered ability goes on the stack but no legal choices can be made for it, or if a rule or a continuous effect otherwise makes the ability illegal, the ability is simply removed from the stack.

1
  • SO, if the opponent gains Hexproof, the damage to opponent will fizzle, but the creature damage is good? Commented May 31, 2018 at 20:05
1

I don't think it's possible to have a "target creature that player controls" if you can't locate a "that player" - by having a legal "target opponent". If you can't find an opponent to target, I don't see how much sense can be made of the second part at all.

My instinct on this one, therefore, is that the second target must be dependent on the first at the point of target selection. If the player gains hexproof between that time and the point of resolution, it won't do anything to make the target creature illegal, as ikegami has rightly clarified in the comments!

6
  • 1
    The way the rules fix this dilemma is that the dependency is verified afterwards. Once you have made all selections they are simultaneously verified. Additionally no targets at all are chosen from the games perspective unless they are all selected.
    – Guvante
    Commented Apr 15, 2013 at 6:10
  • 1
    @thesunneversets, You are mistaken. It is possible to obey "target creature that player controls" if the player has hexproof. It is the creature being targeted, not the player. For example, if the targetted player were to flash in Witchbane Orb after the triggered ability in question were added to the stack, the ability would not be countered since it still has a legal target.
    – ikegami
    Commented Apr 15, 2013 at 15:11
  • Another example: "Choose a player. Deal 2 damage to each of two target creatures that player controls." could choose a player with hexproof and target his creatures.
    – ikegami
    Commented Apr 15, 2013 at 15:13
  • Only continuous effects have dependencies, and there are no continuous effects in the OP's scenerario.
    – ikegami
    Commented Apr 15, 2013 at 15:15
  • Oh yes, I agree totally that flashing in Witchbane Orb would not make the creature an illegal target. But I'm right that you can't target it in the first place without a targetable opponent at the time, yes? Thanks for the useful clarification. Commented Apr 15, 2013 at 17:02

You must log in to answer this question.

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