Topic: Epic Cursing (Read 9764 times)

jotheonah · « **on:** December 18, 2011, 05:42:53 pm »

The card that give out the most curses in a single play is (probably) not the one you think.

In a four-player game, a single play of this card can give out 9 curses.

What is it?

How is this accomplished?

Can anyone top that?

def · « **Reply #1 on:** December 18, 2011, 05:49:35 pm »

Ambassador a Ill-Gotten Gains?

AJD · « **Reply #2 on:** December 18, 2011, 05:54:11 pm »

The cheap answer is that a single play of King's Court can hand out nine curses by, you know, using it on a Witch. But I like def's answer better.

michaeljb · « **Reply #3 on:** December 18, 2011, 06:13:57 pm »

def beat me to it.

jotheonah · « **Reply #4 on:** December 18, 2011, 06:34:43 pm »

def's was the answer I had in mind.

I just neglected to mention that 3 of the curses go to the person who played the card.

Qvist · « **Reply #5 on:** December 19, 2011, 05:00:49 am »

Great puzzle. But you get 3 curses by yourself. Counter question:
"What single card gives out the most curses not affecting yourself and how is this accomplished?"
(King's Court a Witch doesn't count, because these are two cards)

I get 2 curses in a 2-player game and 9 curses in a 4-player game too

Davio · « **Reply #6 on:** December 19, 2011, 05:53:16 am »

Quote from: AJD on December 18, 2011, 05:54:11 pm

The cheap answer is that a single play of King's Court can hand out nine curses by, you know, using it on a Witch. But I like def's answer better.

Cheap answer: KC a Mint, reveal IGG 3 times

Some Good Answers in decreasing order of likelihood
1st answer: Play Thief, every other player reveals an IGG, you gain 3 IGGs, dealing out 9 Curses, none of which to yourself
2nd answer: Play Saboteur, every other player reveals a $7+-Coster and chooses to replace it with IGG. You get 3 Curses, the others get 2.
3rd answer: Play Swindler, every other player reveals a $5-Coster which you replace with IGG, you get 3 Curses, every opponent gets 2.

jotheonah · « **Reply #7 on:** December 19, 2011, 10:07:42 am »

Along the same lines as Davio's, Jester could produce those results. What I had in mind was a play that was guaranteed to give out X curses, not dependent on luck or opponents decisions.

Thisisnotasmile · « **Reply #8 on:** December 19, 2011, 10:18:38 am »

No play is guaranteed to give out curses. All attacks can be blocked by Moat or Lighthouse and IGG can be blocked by Trader. No other cards can give out Curses (I think).

Davio · « **Reply #9 on:** December 19, 2011, 10:21:25 am »

Now that this has been solved, a funny little follow-up question could be: What is the least amount of cards needed to empty out the Curse pile? The distribution of Curses doesn't matter.

You can solve this for 2p, 3p and 4p and your opponents don't have Lighthouses, Moats or Traders.
You may set up any or all decks to make this happen and use perfect shuffling.

To start things off, you could of course play Village,KC-KC-Sea Hag-Sea Hag-Sea Hag, Sea Hag to give your opponent 10 Curses in 2p, or deal out 30 Curses in 4p. This requires 7 cards total.

Can this next one be beat? KC-KC-Bridge-Bridge-Bridge, buy out the IGG pile, 5 cards.

@Tinas: Masquerade, of course, but that's cheating and if you TR it, you end up with what you gave away most of the time.

theory · « **Reply #10 on:** December 19, 2011, 10:23:09 am »

KC-Bridge, play another Bridge, play Talisman. You have $5 and 5 buys, IGG's cost $1, buy 5 and get 5 for free.

jotheonah · « **Reply #11 on:** December 19, 2011, 10:40:33 am »

You need a +action to play that Bridge, so you end up back at 5 cards.

And TINAS, you are right of course. I'm still getting the hang of how very specific one must make one's language in order to talk about Dominion accurately.

Let's say a single-card play that is guaranteed to initiate 9 curse-gaining events, regardless of whether those events will be stopped or superseded by reactions or durations.

Also, suddenly I'm realizing that Ambassador, IGG, Trader would be really sweet combo in 4-player, giving out 6 curses and gaining you 3 silvers. Thought not worth the effort to set up just for that obviously.

Thisisnotasmile · « **Reply #12 on:** December 19, 2011, 10:45:59 am »

Quote from: jotheonah on December 19, 2011, 10:40:33 am

Also, suddenly I'm realizing that Ambassador, IGG, Trader would be really sweet combo in 4-player, giving out 6 curses and gaining you 3 silvers. Thought not worth the effort to set up just for that obviously.

Theoretically yes, but if Ambassador and IGG are both available and your opponents haven't stocked up on Traders too they probably shouldn't be playing Dominion.

Lekkit · « **Reply #13 on:** December 19, 2011, 10:59:26 am »

KC->Any +buy card. 2 Talismans. Buy 4 Curses get 8 ones for free.

Qvist · « **Reply #14 on:** December 19, 2011, 11:37:18 am »

Davios first answer I had in mind too.
So, has anybody an idea to deal out at least 2 curses in a 2-player match with one card (without KC or TR)?

Thisisnotasmile · « **Reply #15 on:** December 19, 2011, 11:42:07 am »

Golem?

Or Bridge/Highway/Princess + Talisman -> Buy IGG It uses more than one card but there's only one card involved in the cursing effect. Well actually that's not exactly true either... I dunno.

Edit: Buy a card from a pile with two Embargo Tokens or if you have to deal the Curses to your opponent do it while Possessing them.

def · « **Reply #16 on:** December 19, 2011, 01:52:10 pm »

Already thought of Embargo earlier:

KC KC Embargo + buy card -
Embargo, say, Copper and buy 4 of them, gaining 12 Curses with 4 cards. Or 24 with 5 cards (with 2 + buy cards)

Qvist · « **Reply #17 on:** December 19, 2011, 03:26:28 pm »

Didn't thought of Golem. But it belongs into the same category as TR/KC. But I think we can count that, because it's really playing only one card.
With playing one card I mean, you can achieve that at the start of you turn without playing another card before. So TR/Witch doesn't count, but Golem would fit. But all other tries involve buying. It's not buying, you deal out 2 curses by playing one card at the start of your turn.

Anon79 · « **Reply #18 on:** December 20, 2011, 12:31:02 am »

Quote from: def on December 19, 2011, 01:52:10 pm

Embargo, say, Copper (...)

Real men Embargo the Curses.

Lekkit · « **Reply #19 on:** December 20, 2011, 03:25:17 am »

I think it's impossible to deal out 2 curses with one card to your opponent in a 2-player game. There is however a LOT of ways to do it if you can take them yourself.

Qvist · « **Reply #20 on:** December 20, 2011, 03:49:34 am »

Quote from: Lekkit on December 20, 2011, 03:25:17 am

I think it's impossible to deal out 2 curses with one card to your opponent in a 2-player game. There is however a LOT of ways to do it if you can take them yourself.

It's possible. I already made it.
Hint: Interaction with IGG

def · « **Reply #21 on:** December 20, 2011, 03:52:47 am »

Remake 2 4$-cards into 2 Ill-Gotten Gains.

Qvist · « **Reply #22 on:** December 20, 2011, 04:01:26 am »

Exactly. That's the straight-forward solution.
Or like I've done it:

Remake 2 Ill-Gotten-Gains into 2 Border Villages and gain 2 Ill-Gotten-Gains.

See here.

Lekkit · « **Reply #23 on:** December 20, 2011, 05:03:36 am »

Then I guess I misunderstood you as I'd say that it requires 3 cards. Remake and the cards to remake.

Qvist · « **Reply #24 on:** December 20, 2011, 05:21:56 am »

Sorry I'm not a native speaker.
I tried to say that you deal out curses by only playing one card, so only one card is in play when those curses are dealt out (so no KC/TR and basically no Golem too)
I did not meant to say only one card is required. It's obvious that this is not possible.

Davios first answer met the requirements for 9 curses in a 4-player game and def guessed right for 2 curses in a 2-player game.

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Author Topic: Epic Cursing (Read 9764 times)