how does Chord of Calling work with x cost creatures like Mistcutter Hydra?
Asked by PlagueRats 10 years ago
If you pay 10 for Chord of Calling can you fetch a Mistcutter Hydra and put it on the battlefield with six +1/+1 counters on it?
Also with Mistcutter Hydra, would it's "can't be countered" ability have any useful effect? I assume Chord of Calling could be countered as usual and would have to resolve before I search my library or disclose the card I intend to fetch.
In addition to Rhadamanthus's answer, I will also note that Chord of Calling doesn't allow you to cast the creature without paying it's casting cost. The found creature is merely put onto the battlefield. That means you don't/can't pay any additional costs (for examples, Joraga Warcaller's multikicker or Wren's Run Vanquisher's additional cost)
Rhadamanthus says... Accepted answer #1
You only choose the value of X in the cost of a spell (or ability) when you're casting (or activating) it. Any other time, X defaults to 0. Chord of Calling puts a creature onto the battlefield without casting the creature as a spell, so any X is 0.
It's also important to note that when an effect tells you to cast something "without paying its mana cost", then the only legal choice for X is 0.
Only spells/abilities on the stack can be countered, so Mistcutter Hydra's "can't be countered" ability only matters when it's being cast.
January 12, 2015 8:02 p.m.