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Author Topic: Can Dingan Win?  (Read 818 times)

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Deadlock39

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Can Dingan Win?
« on: August 09, 2016, 10:40:11 am »
+2

Inspired by this post... also by Dingan.  I watched this a long time ago and it felt like there was a win, but I haven't had the energy to try working it out.

At 11:05 of this game against Wandering Winder
https://www.youtube.com/watch?v=FyM9IShvz1M&t=665

At the start of this turn, Dingan had in his deck:
4 Coppersmith
1 Hermit
3 Storeroom
5 Madman
5 Pawn
7 Copper

Prior to that point in the turn, he played 1 Madman, then 1 Storeroom (0 coins), then 1 Madman.
He plays a Storeroom at this point, but I think that may be a mistake. He cycles no cards and plays a Madman, so I think he could have played the Madman first and drawn his deck at that point.

The question is, can you generate 46 coins and 8 buys from this position to win the game?  How about 44 coins and 7 buys for a tie? (WW went first).

My gut says at least the tie was possible, but I haven't worked it out.  Can anyone find a line to do it?

« Last Edit: August 09, 2016, 02:56:54 pm by Deadlock39 »
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faust

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Re: Can Dingan Win?
« Reply #1 on: August 09, 2016, 02:57:31 pm »
+1

Well, his deck is 25 cards. Play Madman, have 8 cards in hand. One is a Storeroom and one is a Madman. Playing the Madman gets you to 14 cards in hand (of 23 remaining), and 3 actions. Then you:

Play Storeroom (now 13 cards in hand)
  cycle for more Madmen
  discard a card for +1$ (now 12 cards in hand)
Play Madman (now 22 cards in hand, of 23 remaining, and 3 actions)
Play Storeroom (21 cards in hand)
  cycle for more Madmen and the last Storeroom
  discard 9 cards for +9$ (12 cards in hand)
Play Madman to draw deck of 21 cards
Play last Storeroom (20 cards in hand)
  discard 9 cards for +9$
Play last Madman to draw deck.

Now you have 19$, 4 buys and 3 Actions. Play 4 Pawns for +1 action, +1 buy, 1 for +1action, +1$ and 3 Coppersmiths. Then playing 7 Coppers brings you to $48.

So there is no guarantee. But you could discard one more card witht the last two Storeroom plays (which leaves your hand 1 card short of the deck the first tiem and 2 cards short the second time, and if the 2 cards you didn't draw are a Hermit and a Coppersmith, then you're good (which may be manageable by clever cycling with the Storerooms).

EDIT: Edited to correct a mistake.
« Last Edit: August 10, 2016, 08:49:02 am by faust »
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

Mic Qsenoch

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Re: Can Dingan Win?
« Reply #2 on: August 09, 2016, 03:12:44 pm »
+6

Yes, there was a win. As you point out we know he already had another Madman in hand when he played the 2nd Storeroom.

He should play the 3rd Madman before any more Storeroom to finish drawing the deck. This gives 21 Cards in hand (whole deck at this point). He can discard up to 14 cards and still redraw the whole deck with the last two Madmen (though only 11 is needed for the win). So play Storeroom for $14, redraw with 2 Madmen, play all Pawns for action/buy, play all 4 Coppersmiths.

5*7 + 14 = $49 money
2 Storeroom + 5 Pawns + Starting = 8 buys.
« Last Edit: August 09, 2016, 03:17:23 pm by Mic Qsenoch »
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Deadlock39

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Re: Can Dingan Win?
« Reply #3 on: August 09, 2016, 03:20:21 pm »
0

Well, his deck is 25 cards. Play Madman, have 8 cards in hand. One is a Storeroom and one is a Madman. Playing the Madman gets you to 14 cards in hand (of 23 remaining), and 3 actions. Then you:

Play Storeroom (now 13 cards in hand)
  cycle for more Madmen
  discard a card for +1$ (now 12 cards in hand)
Play Madman (now 22 cards in hand, of 23 remaining, and 3 actions)
Play Storeroom (21 cards in hand)
  cycle for more Madmen and the last Storeroom
  discard 9 cards for +9$ (12 cards in hand)
Play Madman to draw deck of 21 cards
Play last Storeroom (20 cards in hand)
  discard 9 cards for +9$
Play last Madman to draw deck.

Now you have 19$, 4 buys and 3 Actions. Play 3 Pawns for +1 action, +1 buy, 2 for +1action, +1$ and 3 Coppersmiths. Then playing 7 Coppers brings you to $42.

So there is no guarantee. But you could discard one more card witht the last two Storeroom plays (which leaves your hand 1 card short of the deck the first tiem and 2 cards short the second time, and if the 2 cards you didn't draw are a Hermit and a Coppersmith, then you're good (which may be manageable by clever cycling with the Storerooms).

This looks pretty similar to what I worked out after I posted it.

...but I think you calculated the value of Copper after 3 Coppersmiths incorrectly?

Deadlock39

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Re: Can Dingan Win?
« Reply #4 on: August 09, 2016, 03:26:21 pm »
0

Yes, there was a win. As you point out we know he already had another Madman in hand when he played the 2nd Storeroom.

He should play the 3rd Madman before any more Storeroom to finish drawing the deck. This gives 21 Cards in hand (whole deck at this point). He can discard up to 14 cards and still redraw the whole deck with the last two Madmen (though only 11 is needed for the win). So play Storeroom for $14, redraw with 2 Madmen, play all Pawns for action/buy, play all 4 Coppersmiths.

5*7 + 14 = $49 money
2 Storeroom + 5 Pawns + Starting = 8 buys.

Nice.

The solution I worked out after posting, was slightly worse, very similar to faust's

Play the 3rd Madman to draw everything.
21 cards in hand: Play Storeroom, discard 9 for coins
11 cards in hand 9 in the discard: Play Madman 4 to draw everything
19 cards in hand: Play Storeroom, discard 8 for coins
10 cards in hadn 8 in the discard: Play Madman 5 to draw everything
Play 4 Pawns for buys, 1 for coin, 3 Coppersmith
$17 from Storeroom, $1 from Pawn, $28 from Copper/Coppersmith = 46 coins.
Starting buy + 3 from Storeroom + 4 from Pawn = 8 buys.

Mic Qsenoch

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Re: Can Dingan Win?
« Reply #5 on: August 09, 2016, 03:33:42 pm »
0

Well, his deck is 25 cards. Play Madman, have 8 cards in hand. One is a Storeroom and one is a Madman. Playing the Madman gets you to 14 cards in hand (of 23 remaining), and 3 actions. Then you:

Play Storeroom (now 13 cards in hand)
  cycle for more Madmen
  discard a card for +1$ (now 12 cards in hand)
Play Madman (now 22 cards in hand, of 23 remaining, and 3 actions)
Play Storeroom (21 cards in hand)
  cycle for more Madmen and the last Storeroom
  discard 9 cards for +9$ (12 cards in hand)
Play Madman to draw deck of 21 cards
Play last Storeroom (20 cards in hand)
  discard 9 cards for +9$
Play last Madman to draw deck.

Now you have 19$, 4 buys and 3 Actions. Play 3 Pawns for +1 action, +1 buy, 2 for +1action, +1$ and 3 Coppersmiths. Then playing 7 Coppers brings you to $42.

So there is no guarantee. But you could discard one more card witht the last two Storeroom plays (which leaves your hand 1 card short of the deck the first tiem and 2 cards short the second time, and if the 2 cards you didn't draw are a Hermit and a Coppersmith, then you're good (which may be manageable by clever cycling with the Storerooms).

This looks pretty similar to what I worked out after I posted it.

...but I think you calculated the value of Copper after 3 Coppersmiths incorrectly?

Yeah, I didn't check all the Madman drawing stuff, but assuming that's okay then faust's method can give $48 and 8 buys, as he missed $7 from Coppersmith, but he'll need to swap one of the Pawns to buys.
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Dingan

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Re: Can Dingan Win?
« Reply #6 on: August 09, 2016, 03:34:51 pm »
+3

\_(ツ)_/

What a trip down memory lane!  I actually kind of remember this game.  Was thinking of it when I posted here:
http://forum.dominionstrategy.com/index.php?topic=12757.msg487181#msg487181
I remember I was thinking really hard and doing the math and what not, but then I got the red warning message that tells you it's about to forfeit the game, so I panicked and just kind of did whatever.  Or something.  Not sure.  Good solution, MQ.
« Last Edit: August 09, 2016, 03:39:04 pm by Dingan »
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faust

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Re: Can Dingan Win?
« Reply #7 on: August 10, 2016, 08:47:44 am »
0

Right, I only counted the Coppers as being worth $3 (because 3 Coppersmiths). That was kinda stupid :D
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.
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