Uncommon situation with the Girlfriend card

[Jay]

I find an answer:

Originally Posted by MM

for Illusion, you can play it on either: If played on the Mate, that will replace the Mate with a different Monster; play it on the original Monster, and you'll be facing that new Monster and its Mate.

so
Plutonium Dragon + Ancient + Mate + Enraged = Two Dragon, 35 combat strength each (total 70)

I play Illusion with Lame Goblin, and...

if I play it on Mate, i'm facing Lame Goblin and Ancient Enraged Plutonium Dragon
if I play it on Plutonium Dragon, i'm facing Lame Goblin + Lame Goblin Mate
Right?
----

you get 2 levels from the Dead Broke, Ancient Plutonium Dragon and 6 Treasures left behind by a parrot that used to be an Ancient Plutonium Dragon's Mate.

yes. agreed. it was fewer wrong situation(

another situation

1. Plutonium Dragon + Mate
2. Someone plays Ancient
2. I'm, as a Wizard, discard my hand to Charm Plutonium Dragon
3. Someone plays Baby, Drunk, Miniature, Bedridden and Drunk ("-5 to monster, draw 1 fewer Treasure" each)
4. I kill the Ancient Baby Drunk Miniature Bedridden Drunk Plutonium Dragon

I get 2 level and 1 Treasure from killed Dragon and (cause treasures would be based on the final Monster) only 1 treasure from charmed Dragon.
Right?

[Clipper]

Originally Posted by Jay

I find an answer:



so
Plutonium Dragon + Ancient + Mate + Enraged = Two Dragon, 35 combat strength each (total 70)

I play Illusion with Lame Goblin, and...

if I play it on Mate, i'm facing Lame Goblin and Ancient Enraged Plutonium Dragon
if I play it on Plutonium Dragon, i'm facing Lame Goblin + Lame Goblin Mate
Right?
----

Yep.

another situation

1. Plutonium Dragon + Mate
2. Someone plays Ancient
2. I'm, as a Wizard, discard my hand to Charm Plutonium Dragon
3. Someone plays Baby, Drunk, Miniature, Bedridden and Drunk ("-5 to monster, draw 1 fewer Treasure" each)
4. I kill the Ancient Baby Drunk Miniature Bedridden Drunk Plutonium Dragon

I get 2 level and 1 Treasure from killed Dragon and (cause treasures would be based on the final Monster) only 1 treasure from charmed Dragon.
Right?

Just to be clear, I assume you're charming the Mate in your second "2" step. And yes, according to the ruling MunchkinMan made in this thread, you would end up with 2 levels and 2 Treasures total.

[MunchkinMan]

Originally Posted by Jay

I find an answer:



so
Plutonium Dragon + Ancient + Mate + Enraged = Two Dragon, 35 combat strength each (total 70)

I play Illusion with Lame Goblin, and...

if I play it on Mate, i'm facing Lame Goblin and Ancient Enraged Plutonium Dragon
if I play it on Plutonium Dragon, i'm facing Lame Goblin + Lame Goblin Mate
Right?

Correct.

another situation

1. Plutonium Dragon + Mate
2. Someone plays Ancient
2. I'm, as a Wizard, discard my hand to Charm Plutonium Dragon
3. Someone plays Baby, Drunk, Miniature, Bedridden and Drunk ("-5 to monster, draw 1 fewer Treasure" each)
4. I kill the Ancient Baby Drunk Miniature Bedridden Drunk Plutonium Dragon

I get 2 level and 1 Treasure from killed Dragon and (cause treasures would be based on the final Monster) only 1 treasure from charmed Dragon.
Right?

Correct, since that seems to match up with what I said.

[Jay]

1. Plutonium Dragon + Mate
2. Someone plays Ancient
3. I'm, as a Wizard, discard my hand to Charm Plutonium Dragon
4. Someone plays Baby, Drunk, Miniature, Bedridden and Drunk ("-5 to monster, draw 1 fewer Treasure" each)
5. I kill the Ancient Baby Drunk Miniature Bedridden Drunk Plutonium Dragon

or maybe:

if I Charm Mate, I get 6 treasure from charmed Dragon.
if I Charm original Plutonium Dragon, I get only 1 treasure from charmed Dragon.

[Clipper]

Originally Posted by Jay

or maybe:

if I Charm Mate, I get 6 treasure from charmed Dragon.
if I Charm original Plutonium Dragon, I get only 1 treasure from charmed Dragon.

No. That's one of the things the FAQ is there for. You have to choose the Mate as the monster you Charm and you will get two Treasures, one from each Dragon, after all the Dehancers are applied.

Oh, and because I just noticed it, you can't play two Drunk enhancers to the same monster, but that's probably just a typo on your part.

[Flemroid]

Originally Posted by Clipper

No. That's one of the things the FAQ is there for. You have to choose the Mate as the monster you Charm and you will get two Treasures, one from each Dragon, after all the Dehancers are applied

I couldn't find where in the FAQ it said that if you are charming a monster or its mate that you must choose the mate?

[Clipper]

Originally Posted by Flemroid

I couldn't find where in the FAQ it said that if you are charming a monster or its mate that you must choose the mate?

Charming the other monster would leave an unpaired Mate in play which makes no sense. In any case, I've also seen it ruled on the forums that way. It took me a little while to find it again, but here's the MM post backing it up (click the little arrow to get to the thread):

Originally Posted by MunchkinMan

To put it most simply: The Mate creates a separate Monster that duplicates one Monster card and the Monster Enhancers played on it, and nothing else. The Mate card represents that Monster, and since it is a separate Monster, you must Run Away from it separately, and if you're going to use a card or ability to remove a Monster and you want to get rid of one of the "twinned" Monsters, get rid of the Mate.

[Flemroid]

Gotcha. Makes sense, I figured the mate would just represent the last monster that was in play before it was removed.

That would then make it possible to remove the original monster, leave the mate and then wander in your own level 1 monster or something and have it mate that instead.

[Andrew Hackard]

Originally Posted by Flemroid

Gotcha. Makes sense, I figured the mate would just represent the last monster that was in play before it was removed.

That would then make it possible to remove the original monster, leave the mate and then wander in your own level 1 monster or something and have it mate that instead.

No, it wouldn't -- the Mate stays the same even if the original monster leaves. But it would mean that you had to remember what the Mate represented, and since it's a clone of the monster it Mated, it's easier just to ditch the Mate and keep the other.

[Flemroid]

Originally Posted by Andrew Hackard

No, it wouldn't -- the Mate stays the same even if the original monster leaves. But it would mean that you had to remember what the Mate represented, and since it's a clone of the monster it Mated, it's easier just to ditch the Mate and keep the other.

Ok, sorry I'm just a little confused because earlier MM agreed with Jay on:

Originally Posted by Jay

Plutonium Dragon + Ancient + Mate + Enraged = Two Dragon, 35 combat strength each (total 70)

I play Illusion with Lame Goblin, and...

if I play it on Mate, i'm facing Lame Goblin and Ancient Enraged Plutonium Dragon
if I play it on Plutonium Dragon, i'm facing Lame Goblin + Lame Goblin Mate

So in this instance the Mate would be changing when the original monster leaves?
So if the monster leaves it stays the same, but if the card lets you replace the monster, for example Illusion, then the Mate would change?