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Magic (The Gathering) => Rules => Topic started by: Particle on August 22, 2013, 12:44:43 AM

Title: Garruk relentless
Post by: Particle on August 22, 2013, 12:44:43 AM
Got a question about {garruk relentless}. When he has zero loyalty counters, does a trigger go on the stack for him to transform but does nothing since state based moves him to graveyard? Or no trigger? Seems like wording should say when garruk has one or two loyalty counters, not two or less.
Title: Re: Garruk relentless
Post by: Gorzo on August 22, 2013, 06:34:52 AM
Quote from: CbStrad on August 22, 2013, 01:53:42 AM
He flips, then dies because of having no counters

Short version: He won't flip first. He'll be moved to the graveyard before the transform trigger resolves.

Long version: when {Garruk Relentless} is dropped to 0 (or less) loyalty, at the next state-check two things are seen. 1) Garruk's loyalty is 2 or less. The triggered ability to transform goes on the stack, and simultaneously 2) Garruk has 0 loyalty - he is moved to the graveyard, and this state-based action does not use the stack.
So once those two things happen at the same time, you're left with Garruk in your graveyard, and a dead, useless trigger on the stack that just fizzles away (and is most often simply ignored in gameplay, though it does occur/attempt to resolve and fail)
Title: Re: Garruk relentless
Post by: Particle on August 22, 2013, 12:27:46 PM
So should the wording read when garruk has one or two loyalty counters instead of two or less to avoid useless triggers?
Title: Re: Garruk relentless
Post by: Keyeto on August 22, 2013, 02:03:02 PM
Quote from: Particle on August 22, 2013, 12:27:46 PM
So should the wording read when garruk has one or two loyalty counters instead of two or less to avoid useless triggers?
Essentially, yes. However, for anything that cares about triggers going off (nothing comes to mind), it's good to note that the trigger does still occur, even if it fails to resolve.