Attacking with 2 Wild Beastmasters
Asked by Slurpeee 8 years ago
Does attacking with 2 Wild Beastmasters bounce their effects between each other and make an infinite amount of damage?
So when you say "whenever this card attacks" it's pretty much the same thing as creatures with the same name?
December 28, 2015 5:29 p.m.
Raging_Squiggle says... #3
Yes, when I mean "this card" I literally mean the card that the effect is printed on.
If you have let's say.. 4 Stuffy Dolls. You activate the tap ability of one of them. This act does Not happen 4 times just because you have 4 cards named Stuffy Doll. You would have to activate all 4 of the abilities to get 4 instances of the ability.
Whenever a card mentions itself in its text, it can be replaced with "this card" to avoid such confusion.
December 28, 2015 6:23 p.m.
Another question, how do you tally up the damage summed up if I swing with a 3/3 Wild Beastmaster, a 5/5 Wild Beastmaster and a 1/1 creature.
December 29, 2015 12:29 a.m.
Epochalyptik says... #5
You control both Wild Beastmasters' abilities, so you choose the order in which they are stacked. If you want to do the most damage, you should put the 3/3's triggered ability onto the stack first, then put the 5/5's above it. This way, the 5/5's ability will give your 3/3 +5/+5, which will then carry over to the (former) 3/3's ability. So creatures you control would get +5/+5 and then +8/+8 for a total of +13/+13.
You'd end up with a 16/16, an 18/18, and a 14/14.
December 29, 2015 1 a.m.
Raging_Squiggle says... #6
He'd actually end up with a 8/8 Wild Beastmaster (the 3/3 with +5/+5), a 13/13 Wild Beastmaster (the 5/5 with +8/+8), and a 14/14 creature (the 1/1 with +13/+13).
The ability from Wild Beastmaster specify each other creature aside from this card will get the buffs.
December 29, 2015 1:21 a.m.
Wouldn't their orginal attack totals be what is sent to the other creatures first, then they get their own buffs?
For example:
3/3 Wild Beastmaster would get +5/+5
5/5 Wild Beastmaster would get +3/+3
1/1 creature would only get a +3/+3 and +5/+5
So in total, attacking with two 8/8 Wild Beastmasters and a 9/9 creature?
December 29, 2015 4:31 a.m.
EpicFreddi says... #9
Slurpeee no, because when the 5/5 Beastmaster would get the buff from the (original) 3/3 Beastmaster it's already 8/8, so it would get +8/+8.
3/3 Wild Beastmaster would get +5/+5; 5/5 Wild Beastmaster would get +8/+8; 1/1 creature would get +5/+5 and +8/+8.
December 29, 2015 4:38 a.m.
It somewhat doesn't make sense to me, the 5/5 is sending out two different attack totals to other creatures (first +5/+5, then +8/+8), his ability triggers twice? Why doesn't the other trigger twice also?
December 29, 2015 4:52 a.m.
EpicFreddi says... Accepted answer #12
No ability triggers twice.
The 3/3 triggers first, then the 5/5. So the stack would look like this:
- 5/5 Ability
- 3/3 Ability
The 5/5 ability checks the power after resolving and gives +5/+5 to the other creatures. After this one resolved, the 2nd one will start resolving. The ability checks and sees: Okay, the beastmaster is now 8/8 (3/3 before the other triggered and +5/+5 from the ability). So the ability gives additional +8/+8 to other creatures.
So the 5/5 Beastmaster gets +8/+8.
The 3/3 Beastmaster gets +5/+5.
And the token (originally 1/1) gets +5/+5 and +8/+8 (from both beastmasters).
So we end up with a 8/8, 13/13 and 14/14.
December 29, 2015 5:06 a.m.
So I've read your comment a couple times and tried to figure out what was happening, I think I've got it and agree with you.
First the 5/5 is sending +5/+5 to the 1/1 (making it 6/6) and to the 3/3 (making it 8/8). Then the 3/3 (which is now 8/8) sends back +8/+8 to the 5/5 (making it 12/12) and 6/6 (the previously 1/1) which makes it 14/14.
What we're left with:
8/8 Wild Beastmaster
12/12 Wild Beastmaster
14/14 Creature
December 29, 2015 5:22 a.m.
EpicFreddi says... #15
The 12/12 Beastmaster should be 13/13 because 5+8=13 but yes, you got it right. :p
December 29, 2015 5:24 a.m.
Raging_Squiggle says... #17
Slurpee yes you have it. Almost. 8+5 is 13 though not 12. ;)
You will have an 8/8 Wild Beastmaster, a 13/13 Wild Beastmaster, and a 14/14 creature.
December 29, 2015 5:26 a.m.
EpicFreddi says... #18
On a quick side note: Last week i was playing my Ezuri, Claw of Progress EDH, Rite of Replicationed my Beastmaster with kicker and I had 14 Experience counters. Have fun calculating that.
15/15 Triggers, gives everything +15/+15.
The 16/16 triggers, +16/+16 to everything.
The 32/32 triggers, +32/32 to everything,
The 64/64 triggers, +64/64 to everything.
The 128/128 triggers, +128/+128 to everything.
The 256/256 triggers, +256/+256 to everything.
And last but not least: The 512 triggers, +512/+512 to everything.
We'd end up with everything being 1000+. Fun times.
Raging_Squiggle says... #1
No, whenever a trigger happens, it happens once for each source for each time the conditions are met for it to trigger. If you have 2 Wild Beastmasters that are both declared as attackers, each card's ability will trigger. You can stack them in whichever order you want (If one of them has a buff from something else, to get the most value), but it won't go infinite.
Something to note: Notice how Wild Beastmaster says "whenever Wild Beastmaster attacks..." That can also be written as "whenever this card attacks..." So, if you attack with 2 of them, it won't trigger infinite times that way either.
December 28, 2015 5:02 a.m.