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Topics - jonaskoelker

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Here's a loop:
This nets you 1 coffer per iteration. Add Travelling Fair or a +buy token from Teacher to buy the whole kingdom for a horse. Putting a +action token from Teacher on Merchant Guild saves you from needing a village.

Merchant Guild is convenient and unique in that it effectively reduces the cost of Cavalry below zero. You can use other cost reducers instead (e.g. Highway) if you net money via some other channel (e.g. a +$1 token). You can also use Villa instead of Cavalry, but then you need to come up with a buy somehow. You can't do both substitutions unless there's something I've missed.

If you play the following kingdom, there's a slow but guaranteed way of going off:

Here's how to go off (slowly, and probably without many surprises):
  • Get to Merchant Guild plus Soldier
  • Do the Teacher business
  • Get to 5xMerchant Guild
  • Gain the Horse pile
  • Draw your deck and go off
To get to Merchant Guild plus Soldier: either open Merchant Guild + Peasant and promote the Peasant, or open Peasant + Peasant, promote the Peasant to Soldier, pass until you reshuffle, then play Peasant using Way of the Horse the first time you draw it. That divides the $9 of your deck across exactly two hands; buy Merchant Guild on the hand that hits $5. (You play Peasant either on the first or second hand of your shuffle, or the first hand of the next shuffle if it misses.)

To do the Teacher business: promote your Soldier to Teacher (passing while you wait to draw a Traveller), then play and call Teacher repeatedly to place the tokens, then leave Teacher on the mat. (You may optionally put +$1 on the Peasant pile, if you think it speeds up the next step.)

To get to 5xMerchant Guild: buy Merchant Guild whenever you can. Play Merchant Guild whenever you can. If you can't buy Merchant Guild on the turns where you play Merchant Guild, instead buy a Peasant (gaining a coffer). Play the Peasant using Way of the Horse whenever your draw it. (You can interleave this step with the previous; if you prioritize the Teacher terminals the previous step is guaranteed to eventually complete.)

To gain the Horse pile: with 5xMerchant Guild plus your starting cards, you're guaranteed to hit $4 eventually (you can't spread 3xEstate across three hands such that there's at least two estates in each, and your non-estates are each worth $1). Buy a Cavalry. Play it every time you draw it, until you've gained the Horse pile. You can speed this up by buying more than one Cavalry (so you can play them more often) and by buying arbitrary actions (e.g. Peasant) which serve as Horses (by the Way).

To draw your deck: hold your horses until you reshuffle. Then play a Horse whenever you have one (including Cavalry using Way of the Horse). Pass if you haven't drawn your entire deck. Eventually (in at most four turns) you will. In detail: you have 15 stop cards, so playing 11xHorse will get you to a hand size greater than the number of stop cards; at that point you will draw your deck. That means you can burn 20 horses and still have enough (by "burn" I mean play a Horse on a turn where you don't draw your deck). But you can burn at most 10 horses: every Horse you burn gets you past a stop card; eventually you will have drawn past all of them, or rather enough.

Formally, after discarding but before drawing your hand in the clean-up phase, the number of horses in your deck exceeds the number of stop cards in your hand by at least 11:
  • when you first reshuffle it's 31 vs. 15 (and 31-15 = 16 > 11)
  • after discarding once it's 31-X vs. 15-5-X where X is the number of burned horses, and 31-X - (15-5-X) = 31 - 10 = 21 > 11
  • after discarding twice it's 31-X-Y - (15-5-X-5-Y) = 31 - 5 = 26 > 11
  • after discarding three times it's 31-X-Y-Z - (15-5-X-5-Y-5-Z) = 31 - 0 = 31 > 11
  • eventually there will be zero stop cards left in your deck, along with 11 or more horses
  • (this analysis generalizes to any positive initial hand size, though for a hand size of 1 you'll need 15 horses)
Technically you'll only need to play 10xHorse (with a hand size of 5): that draws all your stop cards. During the first loop iteration you'll draw two more horses off of Cavalry's on-gain. Play Horse until you draw the Cavalry you gained, then resume the loop.

If my math is wrong and you need way more horses than I think: gain an unbounded number of coffers the slow way, by repeatedly buying (and later returning) Peasant. Then buy 9 cards from every pile of actions cards, and put the entire Peasant line into your deck (30 cards). Then gain the Horse pile. That ought to be enough.

I tried designing a kingdom with Rebuild and 9 other Dark Ages cards that would be interesting. This is my first revision:

(Shelters and Provinces, no landscapes)

Would you consider this kingdom interesting to play (and also: fun)? If an interesting Rebuild kingdom (100% Dark Ages) is something that exists in your world, could this be made more interesting, and if so how?

My design process is not particularly sophisticated: pick Rebuild, then pick all the cards that interact with how a normal Rebuild game plays out.

Trashing non-Rebuild cards increases your Rebuild density and trashing your money density up to $1 increases your Rebuild and Duchy gain rate. Junking un-trashes your opponent. Scavenger and Pillage do card (de-)selection, Market Square and Rogue interact with the trash(ing). You know, "+$2, gain a Duchy" is great. Feodum gives Overgrown Estate two paths along which it can be promoted into Duchy, and Squire lets you gain the expensive attacks without Rebuild or Duchy being part of the opportunity cost. Death Cart might give you Province money, and if 2xMarauder pile the ruins it's nice to gain. Squire/Feodum with Rogue support might let Rebuild play a non-mirror for once. Is Rebuild/Feodum a thing? ???

The missus and I had fun playing the kingdom and I found it interesting, so mission success, mostly. I have been underwhelmed by Marauder and Death Cart; everything else played roughly how I expected. Maybe I'm underestimating Marauder, though? Forager is happy to trash the ruins, but making $1 off of two cards makes it hard(er) to hit $5.

The card selection effect of Scavenger is very overt. Sage and Wandering Minstrel are more indirect in their Rebuild-finding effects, and maybe more interesting but perhaps also clumsy enough that they won't see play? Sage will find a lot of non-Rebuild cards and Wandering Minstrel will help your terminals collide. Maybe making them work is an interesting challenge? I think not, but it seems worth experimenting with.

I think Poor House might be a worthwhile Death Cart replacement in v2 of the kingdom. It'll help you hit $5 with support (Silver, Forager, Squire), but at the cost of terminal space; it might be easy to pick up off a spare Forager buy.

Anyways, I've rambled enough. Would you find the kingdom interesting and fun to play?

Game Reports / City Quarter, Witch, Lurker, slow trashing with benefits
« on: September 02, 2018, 10:43:07 am »
I just played the following kingdom:

Code: [Select]
City Quarter, Lurker, Pearl Diver, Swindler, Island, Poacher, Sacrifice, Salvager, Merchant Ship, Witch, Banquet, Triumphal Arch
I drew a random kingdom, then replaced Ghost Ship with Witch; I think my gf might have injured me if I had Ghost Ship'd her CQs into oblivion every turn ;)

So, this kingdom. Drawing cards is good and CQ is key in this. It's pricey and weak early. It can be gained with Lurker, but that's only non-risky if you're already drawing your deck and Lurker is not amazing early. It really hates non-actions, e.g. Curse and treasures. Those can be trashed with Salvager and/or Sacrifice, but both of those are slow. There are cute tricks you can do with Lurker; e.g. Salvager/Witch or Sacrifice/Island, then regain the trashed card. Or you can steal actions with Swindler/Lurker. There's virtual money in Swindler and Merchant Ship, but the duration effect of the latter causes a lowers action density compared to non-duration actions. Then again, it also increases your CQ density. A minor interaction is that Witch is great at emptying one pile fairly early, which turns Poacher into Oasis which is not super strong. Also, Lurker is great at end-game pile control.

Getting to the point where you're drawing deck is a drawn-out process. You're getting curses dumped on you, your trashing is terminal, the villages are expensive and the draw is weak until you have a high action density. In both my games, by the time I was drawing deck I had to green on a tight schedule—the missus really likes Island.

Variants and Fan Cards / "Demon", the Druid of Hexes
« on: April 09, 2018, 10:11:39 am »
The basic idea is this:

Action — Attack
[Some moderate bonus, I'm thinking +]
Each other player receives [some of the set-aside hexes]

Setup: Set aside the top n Hexes face up.

This looks similar to Tormentor. Cool, if you were re-designing Nocturne don't include them both. I don't care that much that it's similar. Demon makes it feasible for the same Hex to be dealt out multiple times in succession, which is different, and when Demon is stacked, it might change which Hex is strongest; Bad Omens followed by Famine seems like a pretty mild attack.

If you set aside the top 1 Hexes and it's Delusion, and the on-play causes each other player to take at least one of the set-aside Hex, I think it's ridiculously powerful. Also, a stackable Demon that always gives out Greed is BANANAS. So it should set aside at least two Hexes, and it can't give out all of them. Which ones should it give out? How should that be determined?

If it's simply "the attacker chooses some subset of a fixed size", we're back at perma-Delusion, so the victim has to make some choices. If both attacker and victim makes choices it might become slow. Let's try out the simplest version of this idea:

Action — Attack
Each other player receives a set-aside Hex of their choice (leaving it there).

Setup: Set aside the top 2 Hexes face up.

If the two hexes are Plague and Poverty, this is almost always a Torturer with a +$2 on the first play, but it stacks differently—and a lot of the power of Torturer is the stacking. In fact, any time Poverty is set aside, Demon is weaker than Militia: either it's a $5 Militia—okay, more expensive is not strictly worse, edge cases everywhere—or the other Hex is weaker than the Militia attack in which case it's even weaker. Provided your opponent chooses correctly. Hm. Put that in the "problems" column.

If one of the hexes is Misery and the other is a very strong one, Demon might simply become a terminal Harem. Oh wait, there's Twice Miserable, it might become a terminal Silver worth 4 VP for the first copy (if you get to play it twice) and 0 for each subsequent one. That... sounds fine? I'm not the most expert Dominion player so I might mis-evaluate things, but it seems like it might be slightly bland mandatory buy due to the on-play being worth 4 VP. One of them is definitely better than a Duchy. That is, if Misery is set aside, for which the probability 1 in 6.

Probably the strongest pair is Delusion/Envy—that's bound to hurt an engine one way or the other. Luckily, Demon provides some unenviable virtual money. Self-synergy? ;) — anyways, this combinations seems really strong, but it screws with everyone evenly and is thus balanced (lol), and in any case this pair only comes up 1 out of every 666 66 Demon games.

Let's think up one that involves attacker choice as well. Two modes I can think of are pie rule and vetoing.

Pie Rule Demon
Action — Attack
Choose a set-aside Hex. Each other player receives that Hex or the two others, their choice (leaving them there).

Setup: Set aside the top 3 Hexes face up.

Veto Demon
Action — Attack
Choose a set-aside Hex. Each other player receives a different set-aside Hex of their choice (leaving it there).

Setup: Set aside the top 3 Hexes face up.

If they only set aside 2 Hexes, the pie rule becomes victim chooses (which we already did) and the veto becomes attacker chooses (which is bad). So at least 3 Hexes, and that's probably plenty.

With Veto Demon, the attacker vetoes the (situationally) weakest Hex, the victim vetoes the strongest and takes the middle one. With Pie Rule Demon, the attacker chooses as even a split as possible, which if intuition serves is always done by picking the most powerful. So with Pie Rule, the victim takes the middle and weakest Hex. There might be complications, e.g. if the victim takes Bad Omens followed by Locusts, it's close to just being a Locusts, but Pie Rule on the face of it looks stronger than Veto. Also, I'm not quite sure how to compare middle-of-3 vs. weakest-of-2.

However, if it's fairly obvious to both attacker and victim which is most powerful on both the first, second and nth play of Demon, once you think about it enough, the more complex choice rules which include attacker choice might slow the game down for very little benefit, even if picking their poison is fun. So, based on thinking about this and never playing with it, the first design looks best, though the other two should be tried out and probably dropped if they aren't noticeably better.

The basic idea is this:

Action — Reserve
Cost: ?
[some on-play ability]

When another player plays an Attack card, you may call this. If you do, you are unaffected by that Attack.

Regarding the cost and on-play, I'm thinking something like

Action — Reserve
+1 action
+2 cards
Discard 2 cards.
Put this on your Tavern mat.

When another player plays an Attack card, you may call this. If you do, you are unaffected by that Attack.

Where Lighthouse gives you The Field's Gift and Moat nets you The Sea's Gift, this gives you The Wind's Gift. They all cost , and the Boons are balanced so this is balanced, right? ;) — more seriously, Dungeoning seems like a reasonable minor bonus that isn't present on any canon attack-blocker.

The most interesting aspect of Shield that leaps out at me is that it interacts very differently against attacks that are spammed vs. attacks that are played once per turn—archetypal examples are Minion and Mountebank. It's probably fine against Mountebank and not so much against Minion, unless you pick up a large pile of shields.

In deck-drawing engines, it plays different from Lighthouse: you need two Lighthouses to block all attacks, but one Shield per attack you want to block. It also becomes weaker the more players there are: you need one Shield per attack-per-turn per player. The strength of Shield is of course that you can play it any time before the attack happens and be protected, rather than just the turn before or after the attack is played. Also, it has self-synergy: each Shield helps you cycle to your other Shields :)

The on-play effect provides some modest after-the-fact mitigation of topdecking and junking attacks, but it's probably not something you want to keep in hand after being hit by Militia, unless the sifting is key to kicking off. Against Relic, Minion and other random-smaller-hand attacks (such as... I'm drawing a blank), the sifting might be good though. That is, the on-play ability of Shield has a Watchtower-like quality in that it interacts differently with different types of attacks.

The basic idea is this:

Cost: ?

When you buy this, you may overpay for it. For each you overpaid, take a coin token.

Assuming its on-play is just money, at which <price, value> points would this be non-broken?

The no-strictly-better rule suggests that it should cost at least if it provides , at least if it gives and at least if it gives .

At or greater, I wonder when you'd ever buy it.

At for , I have a hard time seeing myself taking it over Gold. Maybe when I'm playing big money and hit and I'm confident I won't shuffle again and a Duchy is no good, which is... nearly never?
At for , it feels underwhelming compared to Gold. If you only play it once and you pay for it, you get more money than a Gold would've gotten you, plus the flexibility of coin tokens.
At for , it's a $5 Silver with a coin token. Compared to other $5 Silvers, it seems... reasonable? You'll probably take it for some times, and it'll probably be quite good for in some situations.

At for it competes with Masterpiece. Most of the time, I think you'd rather have Silvers in deck, but it's probably worth picking up one. It seems not too different from for
At or for it seems like it could easily be quite good—it lets you pile up a lot of coin tokens on your penultimate shuffle, which lets you hit Province much more reliably on your last shuffle.

All of this assumes you're playing something moneyish that buys Provinces, preferably one each turn. On most boards, that's not the strongest strategy available. I have a hard time seeing myself buying Savings in an engine: if I can spend all my money I'd rather pick up engine pieces. If I can't, why not? If I'm choked on buys, a large stockpile of coin tokens isn't likely to do me any good.

One exception I can see: if I have a Surplus of buys and I would rather threaten quad province that buy double province, saving up money on a 1-for-1 basis with a minor deduction seems good. One might worry that it amplifies first player advantage in an engine mirror: if both players save up to octuple province turns, it's no good being the second player to have enough money and buys for eight provinces.

One obvious card to compare Savings to is Capital: both of them shift your money across time. Capital gives you benefit when you play it in exchange for a later payment. Savings requires payment first and gives benefits later, but that's in relation to when you buy rather than play it, so in some sense you get the benefits earlier than with Capital.

My first guess would that it would be okay but nothing to write home about at for . It'd be about the same at for , but I should definitely also try out for and for just to find out how bananas broken it is.

I recently played on this kingdom:

Code: [Select]
Bazaar, Ghost Ship, Merchant Ship, Tactician, Treasury, Embargo, Native Village, Pearl Diver,
 Treasure Map, Navigator

The missus and I (typically) play 3 games on each kingdom, this not being an exception. In the first two games, I built a DoubleTac deck with roughly this composition:

2 Tactician
2 Merchant Ship
1 Ghost Ship
4 Navigator
4 Bazaar
2 Native Village
1 junk card to discard to Tactician

Your starting deck, your provinces and maybe that one Silver you got early on should be on your Native Village mat.  Maybe you leave one in your deck to have a Tactician discard, unless you picked up a Pearl Diver on a spare buy, with an exception to the exception if you start double-provincing before you're completely thin.

It takes a while to get going: without the draw of Tactician you can't reliably steer your Native Village thinning, and early on you don't have enough money to play DoubleTac profitably.  Once you're up, though, you'll reliably double Province and play Ghost Ship, even if you get Ghost Ship played on you.

And when I say reliably, I mean super reliably: starting your turn with Navigator and having several more to play means you can not only draw your deck, you can draw it in almost any order you like, and you can mat your provinces with perfect precision.  Navigator is considered a weak card and I agree, but playing it on this board felt really good.  And besides, with DoubleTac you need virtual money which Navigator delivers.  Also, you're going to hit $4 a few times, and what else are you going to buy?

In the third game, I used Tactician to buy and then connect two Treasure Maps, then played SingleTac/money with two tacticians.  It felt fairly fast but not super dominant against what my gf was doing (which was moneyish).

What would you do on this board?


Math ahead! In this article I try to explain some mathematical concepts and how they relate to Dominion, in particular with respect to infinite loops.  With this mathematical toolkit you should be able to understand why infinite loops take the shape they do, and be better able to predict which cards do and do not enable new loops. The strategic relevance of the presented concepts in your day-to-day play is... unclear.  If this article belongs in some other (sub)forum, let Theory know.

Edit: If you are unfamiliar with the infinite loops, I wrote a survey of them in a post in this thread, based on trivialknot's suggestion. You may want to read this first.

I'm interested in feedback: could I organize it better, is it too mathy, have I expressed the math clearly enough, is there anything you think could be improved? Was it useful to you? Can you think of applications of this approach I haven't mentioned?

A theory of finite processes

If I sell all my Pokemon cards and don't buy any new ones, I will eventually have none left, i.e. this is a finite process.  The number of cards I own is an upper bound on how many more sales I have to make; observe that this number always decreases, but never below 0.

A set of numbers A is called well-ordered if, for any set non-empty set B which is a subset of A, there is an element of B which is smaller than all the others.  The natural numbers is an example of a well-ordered set. The set of integers is not: the set of negative numbers is non-empty, but there is no smallest negative number, since -1 > -2 > -3 > ... and so on, infinitely.

In a well-ordered set, there cannot be an infinite decrease sequence: the set of all numbers in the sequence is a non-empty set and by well-ordering must have a smallest element, let's say the nth in the sequence. But since the sequence is decreasing, the (n+1)th element is less than the nth, violating the assumption that the nth element was the smallest.

So, if we can map some process onto a decreasing sequence of natural numbers, or of elements of some other well-ordered set, we can conclude that the process cannot be infinite.

This works great in the case of selling pokemon cards: the number of cards I own is a decreasing sequence.  To apply this to Dominion, we will need something more than a single number.

If A and B are two well-ordered set, their cartesian product—the set of pairs (x, y) with x in A and y in B—is also a well-ordered set, if we use the lexicographic ordering; that is, we take (x1, y1) < (x2, y2) to be true iff either x1 < x2 or x1 = x2 and y1 < y2.  To see that A×B is well-ordered, note that since A is well-ordered, in any non-empty set {(x1, y1), (x2, y2), ...} there is some smallest x-value, x*.  Among the pairs {(x*, y_a), (x*, y_b), ...} there is some smallest y-value, y*.  The pair (x*, y*) is smaller than all the other pairs. To see this, take some other pair (x', y'). Either x* < x', and so (x*, y*) < (x', y') by the lexicographic ordering, or else x* = x' and y* < y'.

In particular, pairs of natural numbers form a well-ordered set.  If I repeatedly sell some Pokemon cards, sell some Magic cards and/or trade some of my Pokemon cards for some Magic cards, I will eventually have neither Pokemon nor Magic cards left.  The pair (p, m) always goes down whenever I complete a transaction, where p is the number of pokemon cards I own and p:Pokemon::m:Magic. Note that the number of Magic cards I own may increase, but only if p simultaneously decreases.  The (p, m) pairs form a decreasing sequence (p1, m1) > (p2, m2) > ..., so the process is finite. (It may help to observe that if p stays constant, m eventually decreases to 0, at which point p must decrease by at least 1—so p also eventually reaches 0.)

If we have three well-ordered sets, A, B and C, then B×C is well-ordered, and so is A×(B×C). In general, a cartesian product A1×(A2×(A3×...)) of any finite number of well-ordered sets is itself well-ordered. If I make a list of card games, e.g. Pokemon, Magic, Yu-Gi-Oh, etc., such that I will sell cards from any game, or exchange cards from games earlier in the list for cards from games later in the list (but not the other direction), eventually I get rid of all my cards.  So long as the number of cards from the first n-1 games remains constant, the number of cards from the last game must decrease to 0, so the number of cards from game n-2 must decrease by at least one; repeat this, and so eventualy the number of cards from n-3, n-4, ..., 1 must all decrease to 0.

For convenience, let's use (x1, x2, ..., xn) as a shorthand for (x1, (x2, (..., (x_(n-1), xn))).

Note that in the triple (p, m, y) where ...::y:YuGiOh, I can trade some Pokemon cards for Yu-Gi-Oh cards while keeping the number of Magic cards constant. In general, the tuple can contain numbers which either decrease or stay the same at each step of the process, provided the tuple-as-a-whole decrease.

The last theorem: if some sequence of tuples (x1, x2, ..., xn) > (y1, y2, ..., yn) > ... forms a decreasing sequence, you can append anything after the tuples, and you can prepend anything that never increases (but maybe sometimes doesn't decrease, i.e. stays the same for one or more steps) and the sequence will still be decreasing. Consider (x0, x1, ..., xn, x_(n+1), ..., xm) and (y0, y1, ..., yn, y_(n+1), ..., ym) with x0 >= y0. If x0 > y0, then (x0, ..., xm) > (y0, ..., ym). Otherwise, (x0, ..., xm) > (y0, ..., xm) iff (x1, ..., xm) > (y1, ..., ym) by definition of the lexicographic comparison. Repeat this argument for x1, x2, etc., until either we conclude that (x0, ..., xm) > (y0, ..., xm) or we get to xn and yn. By assumption, (x1, ..., xn) > (y1, ..., yn); given that xi=yi for i=1...n-1, it must be the case that xn > yn (by definition of the lexicographic comparison).

These decreasing sequences in well-ordered sets are sometimes called termination functions. (Or rather, the functions f that map from states of the process to some well-ordered set, such that f(x) > f(y) if state y can occur after state x, are called termination functions.)

Applications to Dominion

How does this relate to Dominion? Given that there are infinite loops in Dominion, it's impossible to prove that there aren't any. In particular, we can't use the above methodology to make such a proof. Or rather, if we do the proof will be incorrect.  That's what I will do.  The point of doing this is that it's instructive to try and fail: when trying to do so, you will think of the quantitative aspects of Dominion which but for a few exceptions are always decreasing. Knowing the almost-monotone subprocesses will make the exceptions stand out, and being aware of the currently known exceptions will help you fiddle with loops.

Throning a Warehouse sets up four effects to be resolved: draw 3, discard 3, draw 3, discard 3. I assume without proof that no matter what sequence of plays you make, the not-yet-resolved effects will eventually resolve, no matter how many reactions you play.  I think the rules are such that if you may reveal Moat, you may reveal it any number of times.  Given that almost all reactions can either do something a finite number of times or are idempotent and thus have trivial fixpoint/limit behavior, one could make some rules which forbid degenerate infinite plays, and under those rules my assumption would actually be true. But the way everyone actually plays (except maybe a few jerks who run out of willing opponents?), I think that it is true.

Of course, the number of not-yet-resolved effects is not a termination function on its own: it's 0 at the start of your action phase and goes to 4 if you Throne a Warehouse. But this requires you to put a card into play. So let's try that as a termination function: (number of cards not currently in play, number of yet unresolved effects).

Playing a reserve card puts it on your tavern mat rather than in play, so this doesn't decrease the termination function. Also, calling a reserve card puts it in play, which increases the termination function. To remedy this, let's try the following: (number of cards neither in play nor on your tavern mat, cards not in play, pending effects).  This works for both non-reserve actions, playing a reserve and calling a reserve.

This is the core termination function which "proves" that you can only play a finite number of actions each turn.  Except: if you play a self-trasher, such a Mining Village or Pixie, the number of cards not in play increases when you trash it, proving that it is in fact not a termination function. Likewise, if you gain a Mandarin or buy a Bonfire you can take cards out of play.

With Bonfire you face the challenge of having to get cards out of the trash, so we could use (cards not in the trash, not in play/tavern, not in play, pending effects) as a termination function, except (a) we already know that Bonfire violates this, and more importantly (b) the number of cards not in the trash can increase, if we play Lurker, Rogue or Graverobber.

Likewise, any quantitative feature of Dominion game states that (almost) always stay the same or decrease could be prepended.  The interesting ones I know are the following:
  • If we split the buy phase in two, the turn structure is (Action, Treasures, Purchase, Night, Clean-up); the number of steps to do is decreasing, but for Villa.  Note that Horn of Plenty can gain Villa during the treasures sub-phase.
  • Card only leave the supply, never enter it. Exceptions are Ambassador (it only returns cards on net in 2-player games), and a few self-returners such as Spoils, Madman and Wish.
  • Cards never leave the trash, except for Lurker, Rogue and Graverobber.
  • Cards never leave your opponents' decks, but for Masquerade, Thief, Noble Brigand and trashing attacks.
  • Buying a card except with Black Market on net costs you a buy, except for Forum
  • When you buy a card, a card leaves the supply, except if you have Trader and the Silvers are out.
  • When you turn a card in the trash face down with Necromancer, it isn't turned face up again until the turn ends or the card leaves the trash.
Note in particular that Necromancer isn't on any list of exceptions. When Nocturne came out, some people speculated that Necromancer might be useful for constructing infinite loops. I think it won't: to play some card an arbitrary number of times, you must already be able to take it out of the trash and put it back the same number of times; so Necromancer can perhaps feed off an already existing loop (and maybe improve their capabilities), but not create any new ones. To find new loops or evaluate whether a new card might enable a new loop, find some almost weakly decreasing quantity which the new card increases.

How loops relates to pins

I have skipped over the fact that the clean-up phase takes cards out of play. If you can set up a pin such that your opponent(s) can't do anything on their turn(s), or can only do things which won't prevent you from doing what you want, you can set up an infinite process which spans multiple turns and uses the clean-up step to take cards out of play.

For example, with the kingdom {Scrying Pool, Scheme, Miser, Monument, Pirate Ship, Minion, Rabble, Relic, Storyteller, King's Court}, you can use multiple KC'd Rabbles to make your opponents topdeck three blanks, then play Monument for 1 VP, play Relic with Storyteller then play Minion to make them discard their hand and draw a hand of 3 blanks. Miser, Scheme and Scrying Pool are there to ensure perfect reliability: use 5 Scrying Pools to draw your deck (which has no Copper in it thanks to Miser), play 5 Schemes, do the above, then topdeck 5 Scrying Pools for next turn.  Pirate Ship prevents your opponents' decks from growing too big for your Rabbles to topdeck three blanks.  All they can do is buy Copper or Curse so they can't end the game, as you need no more than 9 copies of any card and can probably make do with less. (But not in practice, most likely, since it takes too long to set up.)

The limits of this approach

This approach doesn't say much about what I call finite loops. There may be some set of k cards such that playing one of each in the right order will gain and draw one more copy of each plus a victory card of your choice.  Such a process is finite since it runs out of copies to gain (assuming you don't put any back), but by assumption it empties the Province or Colony pile into your deck, which is often a strong move.

Is there something special about Dominion? Could we also apply this type of analysis to e.g. M:tG? Yes and no. In M:tG, within the span of a single turn it's more common for permanents to tap than untap, for cards to leave than enter your deck and for cards to enter than leave your graveyard, but the exceptions to these are all very numerous; I attribute this in part to M:tG having many more cards and in part to a difference in design philosophy. What distinguishes Dominion is that the main almost-correct termination function is very restrictive and has very few exceptions, and the other almost-never-increasing quantities likewise have very few exceptions. You get more mileage out of this approach in Dominion than in e.g. M:tG.

Appendix: the example pin kingdom, in pictures

Edit: Actually, you might need to add Young Witch (with Scheme as the Bane) to make sure your opponents have enough junk in their deck; otherwise your Minion might make them discard some of their junk, and your Rabbles will fail to topdeck enough junk to lock them down.

Puzzles and Challenges / Win with an all-blue deck
« on: March 18, 2018, 05:21:42 pm »
Challenge: win the game using only blue cards.

  • The kingdom should be full-random-possible (0-2 events/landmarks, only Colony if 1+ Prosperity pile, etc.)
  • You may only play cards if they have type Reaction
  • You may only gain cards if they have type Reaction*
  • When the game ends, all cards in your deck must have type Reaction
  • The game must end in a victory for you

* To be exact: check the contents of your deck whenever the implicit stack is empty, i.e. whenever you can't do (or are done doing) reactions and there are no still-unresolved effects. If the cards have been added between the present and most recent such point in time don't currently have type Reaction, you lose the challenge.

For example, if you buy a Cache, you lose the challenge after having gained two Coppers, but no earlier, and then only if one or more of the three cards are in your deck.

Variant 2:
  • The kingdom should be full-random-possible (0-2 events/landmarks, only Colony if 1+ Prosperity pile, etc.)
  • Apart from one exception of your choice, you may only play cards if they have type Reaction
  • When the game ends, all cards in your deck proper (not including mats, etc.) must have type Reaction
  • Colony must be in the kingdom, and at the end of the game all colonies must count towards your score (i.e. either be in your deck or on a mat, or set aside, or some such shenanigans)
  • At the end of the game, there must be at least one Hovel in your deck

Games are played against one opponent who does nothing. Solutions should work even with worst-case shuffle luck.

As of the time of writing, the blue cards are the following:

Code: [Select]
Moat, Secret Chamber, Diplomat, Watchtower, Horse Traders, Fool's Gold, Tunnel, Trader, Beggar, Market Square, Hovel, Caravan Guard, Faithful Hound
Note: in the first variant you're not allowed to play copper. I think naming 'copper' as the exception in variant 2 makes it unsolvable, but I don't have a proof.

This challenge is due to Adam Horton. He never fully spelled out a set of rules. Variant 2 is my pedantification of what he said informally, and the first puzzle is inspired by variant 2. I posted one of my solutions close to where he posted the challenge, so to avoid spoilers I will defer posting a link for a little while.

Puzzles and Challenges / Trash all the things!!
« on: January 02, 2018, 06:16:20 pm »
Objective: put all the cards in the trash as fast as possible.

Pick a kingdom of your choice, satisfying the recommended rules. Solo play, the shuffles always favor you, 0 <= #Events + #Landmarks <= 2, Colony and Platinum are allowed if you have at least one randomizer card from Prosperity and required if all randomizers are from Prosperity; similarly for shelters.

I'm not sure that Black Market helps you; if you use it, some people like solutions that work online—bonus points for making them happy.

Hard mode: 2+ players, your opponents do nothing, you have trash all of their cards as well.

Rules reminder: if you trigger the game-end condition and buy Donate on the same turn, you never get to do the Donate thing. (That's a Dominion rule, not a this-particular-challenge rule.)

Some cards are described as swingy.

User rrenaud once compiled a sorted list, based on how much entropy was left in the win-when-in-kingdom rates once rating was factored out, and concluded that Goons, Ambassador and Tournament are high-skill cards, whereas Embassy is a high-randomness card.

What makes Embassy high-randomness is the fact that there is an Embassy-centered strategy (Embassy BM) which is fairly strong and easy to play correctly.

What makes Tournament high-skill, I think, is that it requires good judgment and game sense to know when to go for it, how best to go for it, and which prizes to pick when you connect it with a Province. Somehow, [joke]rrenaud's ironclad mathematical proof—proof, I tell you![/joke]—that Tournament is high-skill seems not to have changed the common perception of Tournament as swingy.

I think I have a different definition of swingy which classifies Tournament as swingy. It allows for a 2x2 matrix of swingy yes/no and high-skill yes/no; we could for example use rrenaud's measure to indicate whether a card is high-skill.

Whenever you make a play, you can compare you chances of winning (assuming optimal play) in the game state before and after making that play.

Weak cards are cards that reliably only improve your win chances by a small amount (or not at all, or even harm you). Strong cards are cards that usually do something better than weak cards for your win chances.

Swingy cards are cards that are usually ok-ish but sometimes have an unusually large impact on your win chance when played. That is, your win chance after is significantly better than your win chance before. Playing it causes a big swing.

I equivocated between "making a play" and "playing a card". I only used the latter phrase for convenience—Cultist can be highly swingy, in particular if exactly one player opens 5/2 (but also if it misses shuffles, or the first ruins hit vs. miss shuffles). In the case of opening 5/2, it's the dealing of the opening hands that's the swingy play. That play is made by RNGsus, so really the definition should read "the move greatly changes the win percentages of the players", without using 'you[r]'.

And considering this further, having Province in hand when you play Tournament (and being first to do so) is often largely attributable to shuffle luck. When the shuffle has been made in a way that favors you, actually making the swingy play is just playing optimally—so really, I think swingy is something like "some of the moves made by RNGsus have the capacity to greatly alter the win probabilities of the players".

The degree to which a card is swingy is then the capacity of RNGsus to impact win chances, averaged across all the kingdoms in which the cards occur. The inherent swinginess of Tournament is attributed a little bit to e.g. Smithy, because they can occur on the same kingdom, but Embassy's randomness is also attributed a little bit to Smithy by the same logic. I claim without evidence that each card's inherent swinginess will be attributed overwhelmingly to itself, and each card's measured swinginess will be its inherent swinginess, adjusted a little bit in the direction of the average inherent swinginess across all cards, but the ranking by measured swinginess and the ranking by inherent swinginess will be the same.

A small pedantic footnote: instead of win probabilities I should probably use "expected number of match points" (loss=0 and win,draw=1/#winners), such that the players' win probabilities expected number of match points sum to 1.

I skipped over some rigor in a few places, but I think my definition roughly approximates what people mean by swingy. I would love to hear whether people agree or disagree, and especially if you think you can improve this definition. I think of it as a 'first draft' :)

Edit: today I relearned that the internet is a poor medium for carrying tone of voice, which is often crucial to indicate when you're joking. To address this, I have inserted [joke] tags in the appropriate places.

Solo Challenges / Copper Rush: Drain the Copper pile as fast as possible
« on: December 18, 2017, 05:36:38 pm »
Here's the challenge: empty the Copper pile into your deck as fast as possible.

Normal mode: you have to gain a Copper n times, where n is the number of Copper in the pile at the start of the game, and the pile must be empty.

Hard mode: all the Copper must be in your deck (deck includes hand and discard pile).

You pick the kingdom, including Shelters and Colonies (if you have at least one Dark Ages/Prosperity card), and 0-2 events/landmarks. Assume perfect shuffle luck.

Bonus points if you don't use an infinite combo, i.e. if there exists some n such that in the turn where you satisfy the ordinary win condition, you couldn't satisfy the same condition if the Copper pile started with n Coppers.

Example solutions:
  • Buy Copper every turn. This takes 53 turns. (Hard mode)
  • Banquet for a Copper every turn. This takes 18 turns. (Hard mode)
  • Turn 1 Alms for a Villa, buy King's Court off 5xCopper, Villa, Baker token. Turn 2 buy Beggar. Turn 3 through 8: King's Court a Beggar, gaining 9 Copper and buying Donate, trashing everything except King's Court and Beggar. (Normal mode)

I recently played on this lovely kingdom (IRL, and won): Patrician, Settlers, Farmers' Market, Engineer, Sacrifice, Temple, Archive, Charm, Crown, City Quarter, Advance, Triumphal Arch.

Here's my own analysis of the board: you can build a trashed down engine that draws deck with City Quarter and payloads Crowned Farmers' Markets, sprinkled with one or two treasures if need be. If you run out the Farmers' Markets, or at an opportune moment, dig into the Emporium pile. Extremely late in the game, your +buy can come from Charm, but probably you 3-pile before that. With Engineer to gain cheap actions and Advance to turn them into Emporium and Crown, piles can run out in no time.

There are no attacks that disrupt the engine. All the alt-VP on the board favors the engine: Crown plus Farmers' Market gives you better control over it than any money strategy can ever have; thanks to Advance you can justify picking up Temple for the points; and Triumphal Arch only (really) rewards engines. So you're playing (more or less exactly) that engine.

Given that you can draw deck and there are no tricks with Forum discarding your Provinces and Archive setting them aside for temporary thinning, Archive seems not worth going for.

Charm is either a Silver with a +buy, or it gains; but what to gain? Buy Patrician, gain Settlers, meh? Buy Crown gain... more Charm? But you want actions rather than treasures to make City Quarter good. Buy Temple, gain Sacrifice, after you've already added a trasher to your deck? I think Charm is only good in your deck if its the only way you have of getting enough +buy.

Given that there's plenty of virtual money payload, to make City Quarter good you're going to want actions more than treasures; but if you have more Crowns than Farmers' Markets, adding in a single Gold, maybe two, can be okay.

Temple trashes more cards per play than Sacrifice. I think that makes it better—but after the Estates are out, Sacrifice provides more economy which impacts building. An building is exactly the part I'm completely uncertain about.

  • How do you put this engine together?
  • Do you get one or two trashers? Or even more? Which one(s)?
  • Do you just mass Farmers' Market until you add CQ, or do you mix Farmers' Market with cheap actions (Patrician, Settlers), advancing them into Crown? By massing Farmers' Market (e.g. by trashing Engineer) you can win the split and secure a greater multiple of 4 VP, if you have the time to cash them all in, but if done early it overterminals your deck quite a bit.
  • On your $2 turns, do you take Patrician, Settlers or Engineer? I think Settlers is good early, while you still have Copper, whereas Patrician will eventually be better than a blank cantrip when you have lots of Crowns and Emporiums; but by then you probably should be drawing deck with City Quarter.
  • When do you get City Quarter? When your action density crosses 1/2? Earlier, later?
  • Have I missed anything?

On there were some Nocturne cards that were not yet transcribed. I transcribed them. I caught two mistranscriptions by myself, one only after saving it. Given my fallibility, some proofreading would be nice.

The diff is at

The transcribed cards are:
  • Bard
  • Changeling
  • Cobbler
  • Conclave
  • Den of Sin
  • Guardian
  • Monastery
  • Night Watchman
  • Sacred Grove
  • Secret Cave
  • Tormentor
  • Tracker
  • Tragic Hero

Pick a kingdom of your choice. Make a sequence of plays, after which you know 0 or more of the cards on top of your deck. What's the longest sequence of known cards you can set up?

Example solution:

The kingdom is Baker, Vault, Native Village, Ratcatcher, Duchess.
Open Vault/Ratcatcher, and buy Native Village and Duchess later.
Trash all Estates and mat the Ratcatcher.
Put all but 5 cards on your NV mat.
On the megaturn, unmat all your cards with NV; discard 7xCopper to Vault; play Duchess, shuffle, look at a Copper and put it back.

Now your deck has seven known cards on top, all of which are Copper.

Here's the simplest infinite money loop I'm aware of:

  • 2xOverlord in hand.
  • 1xOverlord in your discard pile (and nothing else).
  • 0 cards in your deck.
  • You can play Overlord as King's Court, Mining Village and Lurker.
Play one Overlord as King's Court, then:
  • Play Overlord as Mining Village (drawing Overlord), trashing it for $2.
  • Play the same Overlord as Lurker, gaining itself back from the trash.
  • Play the same Overlord as King's Court.
Repeat the loop with the Overlord you drew, drawing the first Overlord back.

With 3xOverlord in hand and an empty discard pile, one iteration of the loop—i.e. Overlord as King's Court, then Overlord as Minining Village, Lurker and King's Court—gets you to the right game state, so really the setup is one tripled action, one Overlord in hand and one Overlord in your discard pile.

To bring King's Court into Overlord range, use Ferry or 2x(Overlord-as-)Highway. To have 0 cards in deck, use your favorite trasher, or Bonfire/Hireling/Lurker, or Native Village, or whatever else you like. You can payload other self-trashers than Mining Village, but be mindful that you need to draw the other Overlord. Teacher tokens can make a lot of things possible here. To make good use of $∞, add Travelling Fair to the board.

One weakness of this loop is that it's extremely restrictive; other loops let you play an arbitrary payload card of your choice infinitely often. The loop I'm presenting here needs to spend two out of three Overlord plays on Lurker and King's Court, so the first of the three has to be a self-trasher (or else all your cards eventually end up in play and the process stops).

Some loops without this restriction:

Kingdom: Overlord, Lurker, Raze, Watchtower, Crown, Mandarin

Kingdom, by cost:
2: Lurker
2p: Scrying Pool
3: Ambassador, Masquerade
4: Cutpurse, Thief, Pirate Ship, Villa
7: King's Court
Events/Landmarks: Bonfire, Travelling Fair

There are more loops in the same thread.

There's an old and very bad thread on the board which I don't want to Necromancer. In it, eHalcyon makes some claims about probabilities. Here I do the math to see if those claims are accurate.

If you had five 5 coppers and 5 estates, there is just as much chance of you drawing ccccc as there is of you drawing ccece, even if one "looks" more random to the human mind. 

If we put little stickers on the cards labeled "1" through "10", there would be 10! different orderings. But since we don't distinguish between the 5 estates nor the 5 coppers, there are only 10! / (5!*5!) = 252 = 2*2*3*3*7 orderings.

How many of those orderings start with "ccccc"? As many as there are (non-distinguished) ways of ordering the remaining cards, which is just 1 (that ordering being "eeeee").

The number of orderings that start with "ccece" equals the number of orderings which end with some reordering of "cceee". There are [5 choose 2] = 10 such orderings.

Hence, you're 10 times as likely to draw "ccece" than you are to draw "ccccc". You are even more likely to draw "3 coppers and 2 estates in any order" as you are to draw "ccece" in that specific order, because there are more orders (and all orderings are equally likely).

[D]rawing a hand of all actions, then a hand of all treasures followed by a hand of all greens seems really non-random to us.  But with perfect randomness, there is just as much chance of drawing these clumps as there is of getting three hands each with mixes of all three.

Restated, if you draw three batches each of five balls from a jar with five green, five white and five yellow balls, not putting the balls back between draws, you're equally likely to draw three monochromatic batches as you are to draw three rainbow batches (all colors represented).

The total number of orderings is 15! / (5!*5!*5!) = 756756.

There are six orderings of monochromatic batches: GWY, GYW, WGY, WYG, YGW, YWG. (This is shorthand for ggggg+wwwww+yyyyy etc.); 6 out of 756756 is a bit less than 1 in 100 000.

A rainbow batch has gwyXZ in some order, where X and Z are both members of {g, w, y}. X and Z can either be equal or different. This means that a rainbow batch either has three of one color and one of each of the rest, or two pairs and one loner.

Suppose the first batch is rainbow, with a color occurring thrice; let's say it's green, so it's gggwy in some order. There are 2g+4w+4y left. If all batches are rainbow, the remaining two batches are either [gwyyy+gwwwy], [gwwyy+gwwyy] or [gwwwy+gwyyy].

Note that gggwy has 5! / (3!*1!*1!) = 20 orderings, and gwwyy has 5! / (1! * 2! * 2!) = 30 orderings.

So there are 20 * (20*20 + 30*30 + 20*20) = 34000 orderings that start with gggwy. But picking green was arbitrary; there are just as many orderings that start with wwwgy or yyygw. (Having picked green, we shouldn't count the swap of white and yellow as separate, because that would double-count the gwyyy+gwwwy and gwwwy+gwyyy follow-ups. It would also double-count gwwyy+gwwyy, because it isn't different from gyyww+gyyww.)

So that's 3*34000 = 102000 orderings that start with a rainbow batch with one color repeated thrice.

If instead it starts with two pairs, let's say gwwyy in some order, the rainbow follow-ups are [gwwyy+gggwy], [ggwwy+ggwyy], [ggwyy+ggwwy] and [gggwy+gwwyy], where all hands can be reordered.

That's 30 * (30*20 + 30*30 + 30*30 + 20*30) = 90000 orderings. But this again is with green arbitrarily picked to be the starting loner, counting those with a white or yellow loner first (and the other two colors paired) gives us 3*90000 = 270000 orderings.

In total, that's 102 000 + 270 000 = 372 000 orderings out of 756 756, or approximately half.

So rainbow batches are about 1 in 2 compared to monochromatic batches which are about 1 in 100 000, so rainbow batches are about 50 000 times as common.


I think eHalcyon's quoted claims are incorrect. Here are some similar claims which I think are correct:
  • The ordering ccccc eeeee is just a likely as the ordering ccece eceec.
  • The ordering ggggg wwwww yyyyy is just as likely as the ordering wggyg wgwyw ywyyg.
I agree that monochromatic hands feel 'less random', whatever the hell that means, to most people (including me). I think eHalcyon's point is accurate, but slightly misstated, which in turns means that this whole thread is a bunch of pedantic nitpicking.

Am I doing f.ds'ing right? ;)

As the title says, the Poor House challenge is this: play poor house in a kingdom of your design, such as to maximize [the amount of money you have before playing Poor House] minus [the amount of money you have after playing Poor House]. That is, maximize the net loss.

This is not quite the same as maximizing the amount of treasures you reveal: if you have $0, play Poor House and reveal 5 treasures, you have $0 afterwards as well, for a net loss of $0 rather than 1. If you had $1+ and made the same play, you would lose $1. To lose money, you must first have that money (modulo the +$4 from Poor House).

I'm highly confident that the worst possible score is -5: if your +$1 token is on Poor House and you reveal no treasures, you gain $5, which is the same as losing -$5.

I'm even more confident that the best score is greater than 0 ;)

Here are some variables you might want to tweak when picking a kingdom:
  • The number of players in the game (2-6).
  • Whether or not to use Colonies.
  • Whether or not to use Shelters.
  • Whether to have 0-2 or a greater number of events plus landmarks.
  • The size of the Black Market deck (from 0 up to the number of randomizers not in the kingdom).
Pick whichever set of variables you like the best. Colonies without Prosperity randomizers is cool too, likewise for Shelters and Dark Ages. If your kingdom is within the limitations of recommended play and/or what the online implementation allows, you win double the bragging rights ;)

Your opponents play as follows: they never play any cards. They never buy anything. When one of them has to make a decision not covered by these rules, they all resign and the game ends.

Recently I had the following 2p kingdom come up. How would you play it? Catapult/Rocks, Chariot Race, Gladiator/Fortune, Castles, Temple, Archive, Legionary, Wild Hunt, Overlord, Royal Blacksmith — Palace, Ritual.

No villages, the only +buy is Fortune, the only non-terminals are Archive and Chariot Race. That makes the board rather slow.

Here are some strategies:
  • BMU: buy the most expensive of [Province, Gold, Silver] that you can afford until you run out the provinces (tactically bump up Palace points near the endgame, and/or buy Duchy, maybe). Has a hard time with Legionary (I guess); maybe Archive can help a little? But this deck gets thick very quickly (whatever "quickly" means on this board), so you won't have Archive often, or many of them simultaneously, unless you delay ramping up Palace points.
  • BM+Archive+Legionary: trash a bit, get some Archives, get a Legionary and a few Golds, attack a lot, buy Provinces and maybe sprinkle in some Rituals. Perhaps Overlord plays here?
  • Ritual golden(ish) deck: trash down, get some Archives, get a Fortune, repeatedly trash Curse, trash a Gold for VP, buy a replacement Gold. Transition into Provinces when the Curses are out.
  • Jam your deck full of Chariot Races and expensive cards (Gold and Province seems good), some Archives maybe, milk CR for VP. Sounds not terribly awful, but in my experience with Chariot Race it has been... lackluster.
  • Play one Wild Hunt per turn, gain 40 VP and an estate around turn 45 ??? sounds terrible, quite easy to contest, at which point... question mark?
I think Castles can fit into BM/Archive/Legionary, maybe in Ritual/golden if you have the time once the Curses run out. I haven't tried putting them into BMU, but I think it likes Province better; BMU's VP/turn rate drops off pretty sharply once it hits 7 Gold/Silver/Copper triples. But in general they seem like an expensive and slow VP supplement, probably not worth going for unless you put effort into getting the two scaling ones ($3 Humble and $10 King's).

I lean towards thinking BM/Archive/Legionary is strongest. The draw is limited, so the discard attack is strong, it can green reasonably soon and use Archive as temporary pseudo-thinning of green cards, so it should be possible to build to a stage where in can semi-almost-reliably attack and buy Province for a small handful of turns. My very lovely opponent never bought Legionary (we played ~5 games on this board) which I think helped me a lot; she swore a bit when I played mine ;D

What do you guys think? What's best here? Which cards never play?


I have $6 and two buys; when should I get Gold vs. Cache vs. 2xSilver? I'm playing BM on a board with no terminal draw; should I go for a cantrip copper over Silver? I'm playing a wonky board with nothing going on; is Hermit or Ironworks a better BM enabler?

Here I'll do some money density math to give partial answers to such questions.

Comparing purchases, the generic math

Let m be the total money in your deck and c be the number of cards, not counting cantrips (because they draw a replacement card when you play them). Then your money density, d, is m/c.

If you add a bundle of c1 cards which in total produce m1 money, your new money density d_1 is (m+m1)/(c+c1). If you add another bundle of c2 cards which produce m2 money, your new money density d_2 is (m+m2)/(c+c2).

The first bundle is better, in the sense of increasing your money density the most (or equivalently, in the sense of having the highest post-change money density) when d_1 > d_2, that is, when (m+m1)/(c+c1) > (m+m2)/(c+c2). Rewriting,
Code: [Select]
    (m+m1)/(c+c1) > (m+m2)/(c+c2)
<=> (m+m1)(c+c2) > (m+m2)(c+c1)
<=> m*c + m1*c + m*c2 + m1*c2 > m*c + m2*c + m*c1 + m2*c1  [distributive law]
<=> m*(c2-c1) + m1*c2 > c*(m2-m1) + m2*c1                  [subtract (m*c + m*c1 + c*m1) on both sides]

It seems obvious that if you add the same number of cards, the bundle with the most money is better. Let's sanity-check the math, letting c1=c2:
Code: [Select]
    m*(c2-c1) + m1*c2 > c*(m2-m1) + m2*c1
<=> m*0 + m1*c1 > c*(m2-m1) + m2*c1  [apply c1=c2]
<=> c*(m1-m2) + c1*(m1-m2) > 0       [move stuff around]
<=> (c+c1)(m1-m2) > 0
<=> (m1-m2) > 0                      [assuming c+c1 > 0]
<=> m1 > m2
In words, bundle 1 is better when it has the most money. Obvious truth is obvious, and the math passed a simple sanity check.

Next, let's assume (WLOG) that the second bundle has more cards; c2=c1+dc (dc > 0);
Code: [Select]
    m*(c2-c1) + m1*c2 > c*(m2-m1) + m2*c1
<=> m*(c1+dc-c1) + m1*(c1+dc) > c*(m2-m1) + m2*c1
<=> m*dc + m1*dc > c*(m2-m1) + m2*c1 - m1*c1
<=> (m+m1)*dc > (c+c1)(m2-m1)
(Actually I never use the assumption that dc > 0, so this formula also works when c2 < c1.)

Comparing some specific treasure bundles

Let's use this to compare Gold vs. 2xSilver, i.e. m1=3,c1=1 vs. m2=4,c2=2 and dc=c2-c1=1
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m+3)*1 > (c+1)*(4-3)  [insert values of variables]
<=> m > c - 2
So Gold is better (in terms of money density) when total money exceeds the number of cards minus two; in that case, your money density m/c is greater than (c-2)/c = 1-(2/c). This threshold is always less than 1, it approaches 1 as your deck grows larger, and it's 0.8 on turn one; your actual money density is 0.7, so if you can muster $6 and two buys on turn one, go for 2xSilver over Gold (if all you care about is money density right then and there; there are of course other considerations).

With more practical relevance: favor 2xSilver only when money density applies and you have low money density (i.e. a slog), and then probably only some of the time, depending on the particular kingdom.

(In engines, my choices are determined more by engine capacity math than by money density considerations; money density stops being a useful concept when you draw a constant fraction of your deck, e.g. 1, rather than a constant number of cards.)

Let's compare Gold to Cache: m1=3,c1=1 vs. m2=5,c2=3 and dc=3-1=2:
Code: [Select]
    (m+3)*2 > (c+1)*(5-3)
<=> 2m + 6 > 2c + 2
<=> m > c - 2
Oh look, it's the same condition. Geronimoo's simulation of a Duke/Duchy slog comes to the conclusion that you want Cache over Gold. It seems plausible that money density would be around 1 (which is around 1-(2/c)) if you green early and often (and you don't need much more in Duchy/Duke), so I consider this a confirmation of what the math predicts. (In slogs you draw a constant number of cards rather than a constant fraction of your deck, so the concept of money density is sensible in the first place.)

Just for completeness, let's compare 2xSilver to Cache, i.e. m1=4,c1=2 vs. m2=5,c2=3 and dc=3-2=1:
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m+4)*1 > (c+2)(5-4)
<=> m + 4 > c + 2
<=> m > c - 2
Hey, what do you know; it's the same condition. So whenever m > c - 2 (mediocre-or-better money density) we have Gold > 2xSilver > Cache, and whenever m < c - 2 (very low money density) we have Cache > 2xSilver > Gold.

Comparing treasures and cantrip money

Next, let's use the generic formula to compare silver and some flavor of cantrip copper. Intuitively, a silver in hand gives you $2 where a cantrip copper gives you $1 plus your money density (in expectation), so we should expect cantrip coppers to be better when money density exceeds 1. Let's do the math, and remember that we don't count cantrips as cards because they draw replacements when you play them; for cantrip copper, m1=1,c1=0 and for silver, m2=2,c2=1 giving dc=1-0. Cantrip copper is best when:
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m+1)*1 > (c+0)(2-1)
<=> m > c - 1
When m is an integer, m > c - 1 is the same as m >= c, in which case money density is at least one. So the money density analysis matches the payload-when-in-hand analysis perfectly.

Of course, there are considerations other than money density: if you're playing BM with terminal draw, drawing cantrips dead is worse than drawing live treasures. And no two cantrip coppers are the same; what they do for you depends on what else your deck is doing.

Let's also consider Grand Market (m1=2,c1=0), compared to Gold (m2=3,c2=1). Intuitively it should be the same, GM is better when 2+density > 3, but let's check when GM is best:
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m+2)*1 > (c+0)(3-2)
<=> m > c - 2
That's not the same result, though. Let's do a thought experiment: you deck is 20 copper, plus either a Gold or a Grand Market. On the turn where you draw your non-copper, your payload will be 1x2+5x1 or 1x3+4x1, in both cases $7; but because GM draws you a card, you will see the higher payload ever so slightly more often. So it makes sense that the cutoff shouldn't be m/c=1.

But then, why is the threshold different for Peddler/Silver vs. GM/Gold? Because one Gold can outweigh two estates (in terms of hitting m/c = 1) where a silver can only outweigh one. *Vigorous handwaving*

Just for fun, let's compare cantrip copper (m1=1,c1=0) to Gold (m2=3,c2=1 and dc=1-0=1):
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m+1)*1 > (c+0)(3-1)
<=> m > 2c - 1
So cantrip copper is best when money density is at least $2/card, way above what you need for Province, close to enough to what you need for a Colony. (Reasonable, since intuitively $1 plus a card is better than $3 when a card is better than $2.)

Specific money densities and the importance of trashing estates

So, that was some math to help you evaluate particular money bundles vs. each other, not assuming anything about the rest of the deck, except the applicability of money density in the first place. Now I want to consider some specific (somewhat contrived) decks.

The first will be the silver flood: it has c coppers, e estates and s silvers. Its total money m is c+2s, and the total number of cards # = c+e+s, so money density is (c+2s)/(c+e+s). If you want to hit a money density of $1.6 so you can buy a province, how many silvers do you need?
Code: [Select]
    (c+2s) / (c+e+s) >= 1.6
<=> c + 2s >= 1.6c + 1.6e + 1.6s
<=> 0.4s >= 0.6c + 1.6e
<=> s >= (10/4*6/10)c + (10/4*16/10)e
<=> s >= 1.5c + 4e
So for every two coppers you need three silvers (that's a province hand right there: SSSCC); for every blank you need four silvers (that's also a province hand: SSSSE). Said another way: for every estate you trash, that's four fewer silvers you need to buy. Also, trashing a copper moves you closer to $1.6/card more than adding a silver does.

Let's re-check that statement with the math from earlier: when is trashing a copper (m1=-1, c1=-1) better than adding a silver (m2=2, c2=1 with dc=1-(-1)=2):
Code: [Select]
    (m+m1)*dc > (c+c1)(m2-m1)
<=> (m-1)*2 > (c-1)(2-(-1))
<=> 2m - 2 > 3c - 3
<=> 2m > 3c - 1
<=> m > 1.5c - 0.5
So the above statement needs a caveat: trashing a copper is only better than adding a silver when you're already quite close to $1.6.

Let's compare a silver flood to a gold flood; same $1.6 target:
Code: [Select]
    (c+3g) / (c+e+g) >= 1.6
<=> c+3g >= 1.6c + 1.6e + 1.6g
<=> 1.4g >= 0.6c + 1.6e
<=> g >= (10/14*6/10)c + (10/14*16/10)e
<=> g >= 3/7c + 8/7e
So you need one gold (and a bit) per estate, and one gold per two-and-a-bit coppers. A province hand would be GGCCE, and we have 2 = 2*3/7 + 8/7, so the math survives a province-hand sanity check. Every estate you trash is one less gold you need to buy.

This suggests that if you're playing Big Money, trashing your estates is good; it means you can start greening that much sooner.  Of course, -3VP also means you'll have to green more, i.e. longer, but ramping up your money density sooner also means you'll hit $1.2 sooner, which is the Gold breakpoint; the upshot: trashing estates should help you more than the linear formula suggests. To know exactly how much, and whether it outweighs the -3VP, running simulations is probably useful.

The wiki entry on DoubleJack suggests that Jack is fast because it gains you silvers. But Ironworks can gain you a silver too; where Jack draws you a card, most often a copper the first few times, Ironworks "draws" you a virtual copper when you gain a silver. So why isn't DoubleIronworks a BM strategy to be reckoned with, only slightly less good than DoubleJack (ignoring attacks of course)? My money, and my math, is on the fact that trashing one estate brings you four silvers (or one gold) closer to the $1.6 threshold, and Ironworks doesn't trash estates.

This also suggests that on some wonky boards, Hermit can be a poor man's Jack in BM: it gains you Silver and trashes your estates; it fails to draw you cards, but maybe you can transition via Madman into a vanilla terminal draw BM build once Hermit has cleared out your estates, if there's some decent enough terminal draw on the board. Is the +$1 of Ironworks outweighed by trashing estates? Based on my math I would think so. Todo/future work: running simulations to compare various silver gainers and estate trashers (e.g. Hermit vs. Ironworks), to better understand the importance of each aspect of Jack.

TL;DR Buy many small treasures when treasure density is low, buy cantrip +$x over a $y treasure when money density exceeds y-x, and trashing estates is really frigging good. While this might not be news, I like having the math to back up more word-heavy arguments.

Rules Questions / Effect of Outpost in a Possession'ed turn?
« on: May 11, 2017, 06:39:34 am »
Setup: two-player game between Alice and Bob.

Alice plays Possession on her turn. Next turn is Bob's, possessed by Alice. On that turn, Bob/Alice plays Outpost. What happens next?

My guess:

If Outpost didn't say "This can’t cause you to take more than two consecutive turns", then Bob would take a 3-card turn followed by Bob taking a normal turn.  (Then, barring any turn-order manipulation it would be Alice's turn, then Bob again, etc.)

However, this would mean that Bob takes three consecutive turns: one possessed by Alice, one introduced by his own Outpost, and one normal turn.

Since Outpost rules out this particular outcome, one (or more?!) of the turns must be omitted.

I think logic would disappear in a puff of divinity if one were to skip the possessed turn, since the need to skip a turn only arises during that turn---general rule: turns that have already begun can not be skipped?

I also think it would be inconsistent with "do as much as you can" to skip the extra turn introduced by Outpost.

Hence, by process of elimination I conclude that Bob's normal turn would be skipped (and only that turn).

Do people agree/disagree with my conclusion? With my way of reasoning?

I recently played several games on the following kingdom:

(Cellar, Merchant, Harbinger, Village, Militia, Throne Room, Moneylender, Council Room, Festival, Artisan)

Here's my analysis: there's Village and Festival for +action, Council Room for +cards and +buy and Moneylender for thinning Coppers, so an engine is definitely possible. Militia (semi-)nullifies the benefit to my opponents of Council Room, and seriously hobbles Big Money.  The only support for Big Money is Council Room, with no good way of following it up with a Militia, so... engine is probably likely to beat BM.

I recall one game going all-out for the engine, and another game going almost-all-out BM (with a few Festivals as a $5 Silver, and perhaps one or two terminals; probably an opening Militia.)  I lost both games.

In the only game I won, my final deck consisted of Throne Room, Harbinger, Artisan, Militia, Moneylender, 2xCouncil Room and 6xFestival, 2xGold, 1xSilver, 1xCopper plus a big bunch of green. I remember playing multiple terminals in a few turns (3 or 4 maybe), drawing many cards in slightly fewer turns (2 or 3), perhaps with a single double-Province turn, but I never got close to drawing deck.

If I understand the "Good Stuff" category correctly, it sounds like my winning deck belongs there.

My questions to the community: which strategy do you think is best in this kingdom, and why?  Is it the Good Stuff deck?  What are the opening 3-6 buys for the best strategy, what's it's final composition, and what are the priorities in building and playing it? Any noteworthy subtleties?

If Good Stuff is best, what's the reason? I guess I know the technically correct answer, "the combined effect of all the variables makes it so", but I would guess there's a lesson which transfers to other kingdoms.  Is it the high cost of the drawer and gainer which makes Good Stuff outperform the engine, and the presence of Militia, and the ability to play it almost reliably, which makes it outperform Big Money?  How important is the +$2 from Festival?

My sense is that packing a lot of Festivals made me able to have a bunch of "play $8, buy Province" turns plus a few "play a lot of actions, Militia you, gain a Festival with Artisan, buy a Province and a Dunchy" turns; playing Big Money only allows for the Province turns, and putting in more villages and terminal draw didn't buy me very much in my experience.  Is that an indication that I misplayed stuff, or is this roughly how you would expect things to play out?

Help! / How to engine with Vil/Smith/Lib/Lab/Festi/CouncilR/MoLender?
« on: April 14, 2017, 03:17:26 pm »
I played the following kingdom (2nd ed. base set): Merchant, Village, Workshop, Moneylender, Smithy, Bandit, Council Room, Festival, Laboratory, Library.

I opened Moneylender/Workshop (seemed sane at the time -_-), trashed one or two coppers, gained some Villages and Smithies because I couldn't afford the good parts.  By the time my deck could draw itself it was about 30 cards thick, and my turns were super long, contrary to my expectations.  My opponent suggested I should sleep on the couch for taking all that time ;-)

I won, but I think I would've lost to a half-way competent Smithy/BM bot, so there's something I'm totally failing to understand about Dominion. Hence, "help!"

My own thoughts about the board: the only thing that isn't +action/cards/money/buy is the gain from Workshop (which is ~= -1 action -1 card +$4 +1 buy), and the trashing from Bandit. The engine parts are great, so the obvious play seems to be an engine that payloads a bunch of money and buys all the green cards.  Moneylender thins out the copper, sadly the Estates stay.  Festival is clearly the best village, Library is awesome drawing along with it, perhaps supplemented by some Smithies, yes?  Throw in a Bandit for the Gold, or does it make kicking off too hard?  Do it for the attack, but only if the opponent is playing something treasure-heavy, e.g. Smithy/BM?

Once up and running, I think the engine plays itself; what I would like to get better at is putting it together.  I think what went wrong in my game is that I trashed too much money to Moneylender, which made me never get to five until waaaaaay too late.  So don't do that.  Instead get more money by drawing it with Smithy and playing Festival for a while, yes?  Open Smithy/Silver, then 60/40 of Festival/Library ASAP?

I ramble.  To bullet-list my main questions:
  • What should the end-product look like?  Which mix of drawing cards?  Do you include a Bandit?
     Only if opponent is Smithy/BM-ish?
  • How do you get there?  In particular, what should your first, say, six to ten buys be?  When do you get Moneylender? On the first $4 turn after the opening?

(IRL game, so no log)

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