Dominion Strategy Forum
Dominion => Puzzles and Challenges => Topic started by: Voltgloss on January 10, 2012, 02:04:30 pm
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The setup:
- You are playing second in a two-player match.
- Fool's Gold is on the board. You can choose the other cards present to fill out the kingdom.
- Your opponent is going to buy Fool's Gold, and nothing but Fool's Gold, until the pile is empty.
- Your opponent is not going to willingly sabotage himself (e.g., hand you a Fool's Gold if you play Masquerade, or discard treasure instead of Estates if you play Militia).
- You have perfect shuffle luck.
The challenge:
- Have six Fool's Golds in your deck as soon as the Fool's Gold pile empties.
I see at least one way to accomplish this. Perhaps there are more?
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Use a Thief.
Open Thief/FG, on turn 3 and 5 steal a FG from your opponent. On turn 6 they'll buy the last FG, but you'll have 6 and they'll have 4.
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That's certainly one way, Jorbles. Though it wasn't the way I originally had in mind. Can you find another method that doesn't use Thief?
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Open IGG/FG.
On turns 3 and 6 he draws the curse with 3 estates, so he can only buy 4 total FGs in the first 6 turns.
On turn 3 you buy nomad camp which you play on turns 4 and 5 to get double FG turns, and you buy the last FG on turn 6.
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Hm. Does this work?
P1 will buy the fifth FG on his turn 5; therefore, we have until the end of our turn 4 to buy six FG.
Turns 1, 2: Open Hamlet-Margrave.
Turn 3: Draw Hamlet,Margrave, 3xCopper. Play Hamlet, draw an estate and discard it for +1 buy. Play Margrave, drawing 3x Copper.This gives 6 copper in-hand, 3 buys, with Copper, 2xEstate left in the draw pile. Buy 3 FG.
Turn 4: Draw a new hand of Copper, 2xEstate, Hamlet, Margrave. Play Hamlet, drawing a Fool's Gold and discarding an Estate for +1 Buy. Play Margrave, drawing 2x Fool's Gold. You again have 3 buys and far more than enough coin to buy the remaining three FG.
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I knew there had to be a way to use Hamlet. It's the only $2 with a +Buy. ftl's solution should also work with Council Room subbed in for Margrave.
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It's the only $2 with a +Buy.
You forgot Pawn. Which should be able to replace Hamlet in the solution. Herbalist also has +buy, but wouldn't work for this.
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It's the only $2 with a +Buy.
You forgot Pawn. Which should be able to replace Hamlet in the solution.
Indeed, Pawn can work as well, but only if you also substitute Council Room for Margrave as Jorbles mentioned. Otherwise, you don't cycle enough cards to get both of your opening buys in hand again by turn 4.
I also love how HiveMindEmulator's solution relies on using perfect shuffle luck as a weapon to bog down Player 1's buying speed. Ingenious.
Amusingly, my original idea is different from all of the solutions offered so far! My solution:
Turn 1: Draw Copper x4, Estate. Buy Nomad Camp and topdeck it.
Turn 2: Draw Nomad Camp, Copper x3, Estate. Play Nomad Camp; buy Wharf.
Turn 3: Draw Estate, reshuffle, draw Copper x3, Wharf. Play Wharf, drawing Copper and Estate. Buy Talisman.
Turn 4: Draw Nomad Camp, Estate, Copper x3. Last turn's Wharf then triggers a reshuffle and draws Talisman and either a Copper or an Estate (doesn't matter). Play Nomad Camp, then Talisman with your Coppers. You now have $6 or $7, three buys, and Talisman in play, allowing you to pick up all six remaining Fool's Golds at once.
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It's the only $2 with a +Buy.
You forgot Pawn. Which should be able to replace Hamlet in the solution. Herbalist also has +buy, but wouldn't work for this.
Doh, you're right.
I'm surprised at the level of complexity Voltgloss had in mind when he came up with this puzzle originally, but I do like how many different interactions there are that allow this to be solved.
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Doesn't Mint/Fool's Gold accomplish this pretty simply?
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No, you can only get 5 FGs by the end of turn 4 that way.
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Doesn't Mint/Fool's Gold accomplish this pretty simply?
Turn 1: Buy Mint
Turn 2: Buy FG
Turn 3: Hand: Mint, FG, Copper, Estate, Estate - Mint->FG, buy FG (3 in deck)
Turn 4: Hand: Estate, Copper, Mint, FG, FG - Mint->FG, buy FG (5 in deck)
Before you get your turn 5, opponent buys his 5th, the last, FG.