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#### NoMoreFun

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##### Non transitive challenges - 2nd Edition
« on: September 15, 2018, 09:37:21 pm »
+5

I posted 2 challenges in 2014 to try and find non transitive strategies. With Landmarks it may be easier, and the winning strategies the first time around involved Masquerade pins, so I'm posting them again.

Challenge 1: Find 3 strategies (A, B, C), all of which beat Big Money, that have a Rock/Paper/Scissors relationship (i.e. A beats B, B beats C, C beats A).

There will be 3 win loss ratios between the strategies. The winner will be the triad with the highest win loss ratio in the triads closest matchup (so a triad with 99/1, 99/1, 51/49 is not as good as a triad with 70/30, 70/30, 65/35)

Challenge 2: Find 3 strategies, A, B and C such that in a 3 player game, when it comes to the percentage of wins

Strategy A>Strategy B>Big Money
AND
Strategy B>Strategy C>Strategy A
(Or B>A>C, or C>B>A)
Strategy C also must beat 2 Big Money bots in 3 player (so it's a real strategy that can win games).

The winner is the strategy that demonstrates the most decisive shift in the balance; the highest possible score is 200 (which would be a 100:0 changing to a 0:100).

Common Rules
• 2nd edition cards only: No 1st edition only cards from Base and Intrigue, and cards that function differently (like Masquerade) function as they do in the 2nd edition
• Every Card/Event/Landmark referenced in all 3 strategies must fit into a single kingdom (10 cards and maximum 2 Events/Landmarks with exceptions like Young Witch)
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#### faust

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##### Re: Non transitive challenges - 2nd Edition
« Reply #1 on: September 24, 2018, 07:24:01 am »
+2

I feel that there should be some rule that disqualifies a triple (A,B,C) if there is a strategy D that beats all of A, B, C on the same board.
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#### ahyangyi

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##### Re: Non transitive challenges - 2nd Edition
« Reply #2 on: January 17, 2022, 07:45:45 am »
+1

Yep, I would assume that all dominated strategies would be iteratively removed before we find the intransitivity in the remaining strategies.

https://en.wikipedia.org/wiki/Strategic_dominance

And yes, thread necromancy but we don't have new posts in this board. And... I'm kinda thinking about writing my own simulator to design a few tailored kingdoms. Needed that thing to get people into Dominion. And intransitivity is obviously a nice trait for this kind of tailored kingdoms.

People look at Dominion and say "well this game's power lies in the randomization but each of the actual game is very shallow" or "this game is mostly multiplayer solitaire", neither of which is very true, but it's good to have a prepared example where you have to pay attention to what your opponent is doing and counteract accordingly and the best strategy is still not clear after playing the same kingdom three times.
« Last Edit: January 17, 2022, 07:49:34 am by ahyangyi »
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#### ephesos

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##### Re: Non transitive challenges - 2nd Edition
« Reply #3 on: January 20, 2022, 02:21:07 am »
+1

Oh geez, it has been over 7 years since that post. Good times.

I'm tempted to follow the same line as before: strategy A is robust to attacks, but slow to end the game, strategy B is all about disruptive attacks, and strategy C is a slow combo that's weak to attacks, but wins if uncontested. Maybe something like Pursue Patron Travelling Fair, with a slow trasher like Counterfeit that can't get rid of Curses. Once you build up some Coffers, you can start to multiply them by revealing Patrons with Pursue. Add in Mission to double the growth rate so that it has a better chance of beating A, and maybe City Gate for a little more consistency (topdecking a Patron from your hand).
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#### ahyangyi

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##### Re: Non transitive challenges - 2nd Edition
« Reply #4 on: February 10, 2022, 09:19:39 am »
+1

I'm thinking about how "apart" the strategies are. Sometimes we can have A,B,C on paper, but in reality, one of them can be easily transformed to another as they observe their opponents' purchases. For example, A beat B, B beat C, C beat A, but A can change halfway into B when they see their opponents do C. Then the "hybrid A" might still just dominate the game.

Having combos seems like a good starting point, since they usually have a higher sinking cost than other strategies, and switching is hard. Not sure about the Pursue Patron Travelling Fair/Mission combo though, Patron sounds like too compatible with other strategies.
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#### kieranmillar

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##### Re: Non transitive challenges - 2nd Edition
« Reply #5 on: February 11, 2022, 02:24:47 am »
0

FYI Patron has been erratad so Pursue / Patron no longer works.
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#### ephesos

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##### Re: Non transitive challenges - 2nd Edition
« Reply #6 on: February 11, 2022, 07:36:18 pm »
0

FYI Patron has been erratad so Pursue / Patron no longer works.

RIP, that's 2/2 on my combos I guess. Maybe something else that's infinite with Capitalism Mandarin, gaining Mandarin to put lots of Action-Treasures back on deck. Bit hard to do it without buying anything, but you can use Kiln + Way of the Horse to gain a Mandarin while putting another back to replace it, and Storyteller for draw. The main question is how much can Attacks really disrupt the deck, since Mandarin topdecks everything and thus ignores junk. But the setup takes a while, so attacks might disrupt that.
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#### Awaclus

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##### Re: Non transitive challenges - 2nd Edition
« Reply #7 on: July 01, 2022, 12:00:38 am »
+6

Here's a solution to Challenge 1, featuring some Copper strategies.

Kingdom is Banquet, Tower and Masterpiece.

A — a Masterpiece/Tower strategy:

Buy Masterpiece if available \$ > 6 and count in supply (Silver) > 1
Buy Duchy if count in supply (Copper) = 0
Buy Gold if count in supply (Copper) > 0
Buy Masterpiece if available \$ >= 5 and count in supply (Silver) = 2
Buy Estate if count in supply (Copper) = 0

B — simple Banquet/Tower strategy that simply empties the Copper pile and then buys green:

Buy Duchy if count in supply (Copper) = 0
Buy Estate if count in supply (Copper) = 0

C — a Banquet/Tower strategy that has been modified to counter A by using Banquets to gain Masterpieces, and also to get rekt hard by B (it still loses to B without the intentional crippling, but only like 55-45):

Buy Duchy if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Estate if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Banquet if count in supply (Masterpiece) > 0
Buy Banquet if count in deck (Copper) > count in all opponent's decks (Copper)

A beats B 98.93-0.83 and beats BMU 88.41-11.09
B beats C 100.0-0.0 and beats BMU 99.46-0.48
C beats A 96.74-2.82 and beats BMU 99.97-0.03.

(using Geronimoo's accurate simulation)
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#### Awaclus

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##### Re: Non transitive challenges - 2nd Edition
« Reply #8 on: July 01, 2022, 12:12:32 am »
+3

You could trivially get 100.0-0 to all three by doing something silly, like a Smithy/BM deck only ever buying anything if the Wharf pile has 10 cards, a Wharf/BM deck only ever buying anything if the Gear pile has 10 cards and a Gear/BM deck completing the cycle. But then it would also be trivial to come up with a strategy that beats all three by buying one of each BM terminal. In my solution, there should theoretically exist a strategy that beats all three (and I don't think a solution exists where that is not the case), but it shouldn't be as easy to find.

While simulating things, I did also notice that the simple Banquet/Tower strategy beats the regular optimized Cultist/BM 100.0-0, so if you can come up with a strategy that beats Banquet/Tower and loses hard to Cultist/BM, there could potentially be a better solution there.
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#### Holger

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##### Re: Non transitive challenges - 2nd Edition
« Reply #9 on: July 01, 2022, 09:30:35 am »
+2

Here's a solution to Challenge 1, featuring some Copper strategies.

Kingdom is Banquet, Tower and Masterpiece.

A — a Masterpiece/Tower strategy:

Buy Masterpiece if available \$ > 6 and count in supply (Silver) > 1
Buy Duchy if count in supply (Copper) = 0
Buy Gold if count in supply (Copper) > 0
Buy Masterpiece if available \$ >= 5 and count in supply (Silver) = 2
Buy Estate if count in supply (Copper) = 0

B — simple Banquet/Tower strategy that simply empties the Copper pile and then buys green:

Buy Duchy if count in supply (Copper) = 0
Buy Estate if count in supply (Copper) = 0

C — a Banquet/Tower strategy that has been modified to counter A by using Banquets to gain Masterpieces, and also to get rekt hard by B (it still loses to B without the intentional crippling, but only like 55-45):

Buy Duchy if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Estate if count in supply (Copper) = 0 and count in deck (Copper) > count in all opponent's decks (Copper)
Buy Banquet if count in supply (Masterpiece) > 0
Buy Banquet if count in deck (Copper) > count in all opponent's decks (Copper)

A beats B 98.93-0.83 and beats BMU 88.41-11.09
B beats C 100.0-0.0 and beats BMU 99.46-0.48
C beats A 96.74-2.82 and beats BMU 99.97-0.03.

(using Geronimoo's accurate simulation)

Nice find! To clarify: I assume Banquet's gain preferences are the same as the buy preferences, so B's Banquets always gains 3 Coppers and C's Banquets always gain Masterpiece unless the pile is empty? And does the "intentional crippling" refer to the "if count ..." conditions?

FWIW, I would consider intentional crippling against the spirit of the challenge. If you allowed it, you could solve the challenge 100-0 in any kingdom that contains 3 cards a, b, c such that BM+either of a, b, c beats pure BM:
Player A plays BM+a, except they stop doing anything if card c is gained by the opponent.
Player B plays BM+b, except they stop doing anything if card a is gained by the opponent.
Player C plays BM+c, except they stop doing anything if card b is gained by the opponent.

Quote
In my solution, there should theoretically exist a strategy that beats all three (and I don't think a solution exists where that is not the case), but it shouldn't be as easy to find.

Any finite game has a (probabilistic) "optimal strategy" for each player by Nash theory. Dominion has only finitely many (relevantly different) gamestates in any kingdom without tokens or infinite loops (like the kingdom you gave), so in such a game there should indeed always be a strategy that beats all three.
« Last Edit: July 01, 2022, 09:32:44 am by Holger »
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#### Awaclus

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##### Re: Non transitive challenges - 2nd Edition
« Reply #10 on: July 01, 2022, 11:07:11 am »
+1

Nice find! To clarify: I assume Banquet's gain preferences are the same as the buy preferences, so B's Banquets always gains 3 Coppers and C's Banquets always gain Masterpiece unless the pile is empty? And does the "intentional crippling" refer to the "if count ..." conditions?

Yes and yes, except for the same conditions that exist in B; those are there to make it better.

FWIW, I would consider intentional crippling against the spirit of the challenge. If you allowed it, you could solve the challenge 100-0 in any kingdom that contains 3 cards a, b, c such that BM+either of a, b, c beats pure BM:
Player A plays BM+a, except they stop doing anything if card c is gained by the opponent.
Player B plays BM+b, except they stop doing anything if card a is gained by the opponent.
Player C plays BM+c, except they stop doing anything if card b is gained by the opponent.

Quote
In my solution, there should theoretically exist a strategy that beats all three (and I don't think a solution exists where that is not the case), but it shouldn't be as easy to find.

Any finite game has a (probabilistic) "optimal strategy" for each player by Nash theory. Dominion has only finitely many (relevantly different) gamestates in any kingdom without tokens or infinite loops (like the kingdom you gave), so in such a game there should indeed always be a strategy that beats all three.

It probably is against the spirit of the challenge, but it is a bit of a spectrum. In principle, it is not possible to find an A>B>C>A relationship without some of the strategies doing something suboptimal, and there is no fundamental difference between an A>B>C>A relationship that occurs "naturally" and one that occurs because some of the strategies are doing something obviously stupid, besides the obviousness of the stupidity.

I think there is also a relevant difference between a trio where all strategies are crippled against a specific opponent and a trio where only one is, especially if it's basically only possible to trigger the criteria by doing something reasonably similar to that opponent such that you would expect any strategy that triggers it to have similar winrates too (the difference being that it is still difficult to find the strategy that beats all three).
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#### Holger

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##### Re: Non transitive challenges - 2nd Edition
« Reply #11 on: July 01, 2022, 12:09:49 pm »
+2

It'd be hard to formalize this spirit indeed. IMO a trio of strategies shouldn't qualify if someone finds another strategy D that beats (or ties) all three.

In your example, I think you get such a strategy D by removing the crippling from strategy C, and only buying Masterpiece when there is less than 10 copies of it in the supply.

Then D would tie with B, being the identical strategy when the opponent doesn't gain any Masterpieces.
D beats C due to the latter's crippling (and I think it would tie with uncrippled C, as Banqueted Masterpieces give the same \$ and VP as Coppers).
And D probably still beats A, as contesting the Masterpieces only after A has bought the first one should still be enough to prevent A from emptying the Silver pile fast enough.
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