Murderous Redcap+Falkenrath Torturer+ Mikaeus, The Unhallowed= Infinite Damage?
Asked by julesleos 12 years ago
So here's the scenario:
Mikaeus, the Unhallowed is in play, giving all of your non-human creatures undying.
Falkenrath Torturer is in play, allowing you to sacrifice creatures at will.
You cast Murderous Redcap and he comes in, dealing 2 damage to your opponent.
You activate Falkenrath Torturer 's ability and do sacrifice Murderous Redcap . He goes to the graveyard, both persist and undying trigger in APNAP order. Since you control both, you decide which triggers first (doesn't matter). You choose Undying, and Murderous Redcap comes back into play with a +1/+1 counter on it and deals 3 damage directly to your opponent. You sac him again with Falkenrath Torturer and this time only persist triggers. He comes back in with a -1/-1 counter on him and deals 1 damage directly to your opponent. Rinse, lather, repeat.
Am I correct in assuming that this combo would work? Has anyone experienced this first hand?
I don't see a problem with that combo except that you forgot to add the bonus from Mikaeus to damage calculations and so Murderous Redcap would deal 4 and 2 damage on alternate sacrifices.
May 23, 2013 3:12 a.m.
@Dallie Murderous Redcap wouldn't enter with both counters since only undying of persist can trigger on him at a time. When it first dies both trigger, but only the first one to resolve will see it in graveyard and return it to play thus giving it a counter. After that they will trigger alternating every sacrifice since there is a counter on it.
May 23, 2013 3:14 a.m.
Draugo: Nice spot on his static buff. That would indeed up the damage, but it would always deal 3 damage. Counters still annihilate each other, so he would enter as a 3/3.
May 23, 2013 3:15 a.m.
Draugo: My earlier post was before I saw what you had posted. You're right, I totally gapped that. You are correct.
May 23, 2013 3:16 a.m.
@Dallie no problem, it was easy to guess that you were posting that response at the same time as me :)
Dallie says... #1
Actually, this is exploitable
121.3. If a permanent has both a +1/+1 counter and a -1/-1 counter on it, N +1/+1 and N -1/-1 counters are removed from it as a state-based action, where N is the smaller of the number of +1/+1 and -1/-1 counters on it...
Meaning that as the Murderous Redcap enters, his +1/+1 counter from Undying and his -1/-1 counter from Persist will annihilate each other; much like matter and antimatter do.
Thus he would enter as a 2/2 with both counters on it. Then his ETB triggers, and when it resolves, the game would see a 2/2, because the counters annihilated each other when SBA's were checked, entering, thus having Murderous Redcap deal 2 damage.
May 23, 2013 3:11 a.m.