Rolling Thunder and the Heroic keyword.
Asked by MrQuagsire 9 years ago
When I cast Rolling Thunder, can I target my own creatures and deal 0 damage to them? The Gatherer claims that Aurelia's Fury (and seemingly many other X spells that deal to any number of creatures) do not work that way: "1/24/2013 You announce the value of X and how the damage will be divided as part of casting Aurelia's Fury. Each chosen target must receive at least 1 damage." The rulings for Rolling Thunder don't mention. I don't understand the logic behind the Aurelia's Fury ruling, and it seems like what I'm trying to do works, but I figured I'd ask.
Thank you for your time!
GoblinsInc says... Accepted answer #2
Oh, and this may help.
601.2d. If the spell requires the player to divide or distribute an effect (such as damage or counters) among one or more targets, the player announces the division. Each of these targets must receive at least one of whatever is being divided.
November 30, 2014 7:25 p.m.
MrQuagsire says... #3
@GoblinsInc unfortunate that, guess I'm stuck with only four Cauldron Haze effects. thank you!
November 30, 2014 7:33 p.m.
Rolling Thunder is still great. a lot of heroic creatures get counters effectively negating the point of damage dealt.
December 1, 2014 3:15 p.m.
MrQuagsire says... #5
kanofudo but it still requires 1 mana per target, so for my purposes I might as well use Launch the Fleet.
December 1, 2014 4 p.m.
Could Rolling Thunder by any chance hit planeswalkers? It says to deal damage to creatures and/or players, could you redirect hitting the player and dealing damage to the planeswalker reducing its loyalty counters?
GoblinsInc says... #1
from gatherer
10/4/2004 If X is greater than zero, you can't choose zero targets. You must choose between 1 and X targets. If X is zero, you can't choose any targets.
November 30, 2014 7:23 p.m.