Regeneration + Deathtouch + Double Blocking

Asked by DruneGrey 9 years ago

Say that I have a Ruthless Ripper (face-up), enchanted with Molting Snakeskin and also a Brave the Sands active. On my opponents turn he attacks with, let's just say, 2x Kjeldoran Frostbeast. I declare Ruthless Ripper blocking both and activate the Regenerate feature of Molting Snakeskin.

What is the sequence of events assuming nothing else is played?
Does my creature deal 3 damage to each simultaneously activating deathtouch, receive simultaneous damage from each of those creatures and then have regeneration activate saving my creature?
I guess this boils down to, when double blocking is it "as if" I have two creatures that share a toughness pool with damage calculating at the same time?

megawurmple says... #1

Before combat damage is dealt, you active the ability of Molting Snakeskin so your Ruthless Ripper is now a 3/1 with deathtouch and a regeneration shield. In this situation, you will choose how the Ruthless Ripper distributes its damage among the creatures it is blocking (in this case 2 to one and 1 to the other) and combat damage is dealt all at once. Both Kjeldoran Frostbeast will die due to deathtouch. Because all the combat damage is dealt at once, the single regeneration shield will protect your Ripper from all creatures it is blocking and it will live.

February 4, 2015 5:31 p.m.

Seraphicate says... #2

In addition to that, the ability from Kjeldoran Frostbeast triggers, and you may, once again, choose to regenerate your Ruthless Ripper for .

February 4, 2015 5:34 p.m.

Rhadamanthus says... Accepted answer #3

Assigning damage when blocking 2 attackers with one blocker (only possible if you do have something like Brave the Sands, etc.) works like assigning damage when one attacker is blocked by 2 blockers. You choose a damage assignment order for the attackers and then assign the blocker's total power between them, having to assign at least lethal damage to the first attacker in order before assigning any to the second. Since Ruthless Ripper has deathtouch, 1 damage counts as lethal.

So a legal damage assignment between the Frostbeasts in your example if they're in X-Y order is either 3-0, 2-1, or 1-2. They'll assign and deal a total of 4 damage to the Ripper. At end of combat, both of the Frostbeasts' delayed triggered abilities will go off (even though they're currently dead) and Ripper will get destroyed unless you regenerate it again for each of them.

February 4, 2015 5:34 p.m.

megawurmple says... #4

Bugger, completely forgot to read Kjeldoran Frostbeast's triggered ability. As stated by the others, you'll have to regenerate again for each instance of the triggered ability, meaning that you'll have to activate it a total of 3 times if you want your Ruthless Ripper to make it as far as the second main phase.

February 4, 2015 6:04 p.m.

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