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##### Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« on: May 23, 2017, 09:12:39 am »
+6

Motivation

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.
« Last Edit: May 23, 2017, 11:14:00 am by jonaskoelker »
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#### Limetime

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #1 on: May 23, 2017, 10:06:29 am »
0

It is less about money dencity and more about being able to spike certain price points.
Also resiliency when greening is important.
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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #2 on: May 23, 2017, 11:09:58 am »
+1

It is less about money density and more about being able to spike certain price points. Also resiliency when greening is important.
Is that an argument against any specific point, against all my points or against the use of money density in general? I'll happily concede that money density is at best a low-resolution pointer in maybe somewhere not far away from the right direction, rather than a tool you can use to 100% determine both your strategy and tactics. Is there any of my money density calculations you think lead more away from the right answer than towards it?

Of course, hitting \$8 is more than ~14.3% better than hitting \$7, that I agree with. However, getting to \$8 correlates with increasing your money density, so I think the concept is not completely useless. Complication: given that you'll have treasures of varying money amounts in BM, selection is also useful (such as from Embassy); that is, on top of just the average you also have to consider the variance.

I think resilience while greening means sustaining your ability to green, which again correlates with having a high money density (and keeping it high, e.g. such as with Hoard).

Agree/disagree?
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#### DG

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #3 on: May 23, 2017, 11:44:00 am »
+2

Of course, hitting \$8 is more than ~14.3% better than hitting \$7, that I agree with. However, getting to \$8 correlates with increasing your money density, so I think the concept is not completely useless.

Since you are comparing two situations where you believe the density difference to be narrow, the difference in spiking \$8 hands may be more important. Anyone who has played with masterpiece a lot will tell you that it typically produces a lot of 7 coin hands. I'll offer you a simpler rule of thumb that doesn't need as much maths.

In any simple situation where it is bad to add a copper to a deck, adding one gold on its own to a deck is better than adding a copper and a gold, which is better than adding two silvers.
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#### Limetime

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #4 on: May 23, 2017, 09:44:39 pm »
0

It is less about money density and more about being able to spike certain price points. Also resiliency when greening is important.
Is that an argument against any specific point, against all my points or against the use of money density in general? I'll happily concede that money density is at best a low-resolution pointer in maybe somewhere not far away from the right direction, rather than a tool you can use to 100% determine both your strategy and tactics. Is there any of my money density calculations you think lead more away from the right answer than towards it?

Of course, hitting \$8 is more than ~14.3% better than hitting \$7, that I agree with. However, getting to \$8 correlates with increasing your money density, so I think the concept is not completely useless. Complication: given that you'll have treasures of varying money amounts in BM, selection is also useful (such as from Embassy); that is, on top of just the average you also have to consider the variance.

I think resilience while greening means sustaining your ability to green, which again correlates with having a high money density (and keeping it high, e.g. such as with Hoard).

Agree/disagree?
This really isn't an argument, I was just saying that money density is definitely not the best metric to gauge your deck's performance.
Hitting certain price points is best done by high value treasure.
Resiliency is best done by having a million treasures.
Your point about trashing estates is on point because every estate slows down growth (that estate will hurt you every single time you draw it) and is bad at making sufficient treasures collide to hit certain price points.
You can throw out everything I just said when you are playing an engine.
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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #5 on: May 24, 2017, 08:14:30 am »
0

Since you are comparing two situations where you believe the density difference to be narrow, the difference in spiking \$8 hands may be more important.

(Emphasis added.) The general rule I'm trying to extract from between the lines seems like it should have an implication roughly like this: if treasure-bundle A lets you hit \$8 two turns out of every three, that's better than treasure-bundle B that lets you hit \$7 consistently, even if the money density with bundle B is higher. Is that roughly what you were getting at? (I agree with the assessment, FWIW.)

In any simple situation where it is bad to add a copper to a deck, adding one gold on its own to a deck is better than adding a copper and a gold, which is better than adding two silvers.

Let's math it! I already covered Gold vs. 2xSilver, which hinges on money density being above or below one, so the math already agrees with that part of your statement. Let's do Gold (m1=3,c1=1) vs. Gold+Copper (m2=4,c2=2 and dc=2-1=1):

Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+3)*1 > (c+1)(4-3)<=> m + 3 > c + 1<=> m > c - 2`
So Gold+Copper is only good when money density is less than 1-(2/c) (~= 1). Obviously Gold+Copper impacts money density the same as 2xSilver; the difference is in the variance. What makes Gold+Copper better than 2xSilver? FWIW, I interpret Limetime as being in agreement:

Hitting certain price points is best done by high value treasure.

The first and obvious counterclaim is that the infinite silver deck hits province every turn, whereas the half-gold/half-copper deck sometimes doesn't hit province. In defense of Gold+Copper, it sometimes hits Province even after a Militia, where infinite silver doesn't; more variance makes selection/sifting better and selective card-loss less bad. So, uhm, I'm not sure that there's a clear-cut unequivocal truth here?

Future work: do combinatoric math or numerical simulations to assess the provincing rate of (GSSCE)xN vs. (SSSSE)xN and the like.

I conjecture that to reach a price point of x at least once out of k hands, if x is higher than your money density you want more variance and if x is less than your money density you want less variance. [Extreme case is no variance, in which case my conjecture is obviously true; and *mumble-mumble* by induction *hand-waving* it WLOG *appeal to authority* also applies in the general case.]

Quote
I was just saying that money density is definitely not the best metric to gauge your deck's performance.

Agree.

Quote
Resiliency is best done by having a million treasures.

Let's do more math. The change in money density when adding dm money and dc cards:

Code: [Select]
`   (m+dm)/(c+dc) - m/c= c(m+dm)/c(c+dc) - m(c+dc)/c(c+dc)= (c*m + c*dm - c*m - m*dc)/c(c+dc)= (c*dm - m*dc)/c(c+dc)= dm/(c+dc) - m*dc/(c^2+c*dc)= dm/(c+dc) - m/(c + c^2/dc)`
When adding a green card, dm=0 and dc=1, so we get
Code: [Select]
`= 0/(c+1) - m/(c+c^2/1)= -m/c(c+1)`
The c and c^2 terms in the denominators above suggest that the more cards you already have, the smaller the impact of adding more cards (duh). This suggests, at least at first blush, that resilience comes from a big deck, no matter its composition. A big, good deck becomes almost-as-good when greening, a small good deck deteriorates quickly, and a big bad deck only gets as much worse as an equally big good deck. (Once again, in terms of money density. The discreteness of strategically important price thresholds are not considered here.)

Which means that what I said ("resilience correlates with money density") is misleading; it's probably true in practice, since both money density and total number of cards go up over time (and in tandem), but it's the card part that's the stabilizing factor. As to what you said, a million anything, not just treasures, will make a deck very stable to greening, but of course a million bad cards will make your deck stably bad ;-)

Also, the math backs up the commonly held view that thinning, as distinct from un-junking, is not important in Big Money. BM prefers 20xSilver and 3xCurse+2xEstate to 4xSilver and 1xCurse (where an engine has the opposite preference), because the bigger deck suffers less while greening.

Quote
Your point about trashing estates is on point because every estate slows down growth (that estate will hurt you every single time you draw it) and is bad at making sufficient treasures collide to hit certain price points.
You can throw out everything I just said when you are playing an engine.

Cool.
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#### aku_chi

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #6 on: May 24, 2017, 09:43:29 am »
+3

Cool formula.  In situations where it's valuable to assess money density, target money density (i.e. what money density you expect to have while greening) can be a more important metric than current money density.  Let's see what the results are for some common situations with Delve, Trade, and Bonfire.

Delve

Let's compare Gold vs. 3xDelve, i.e. m1=3,c1=1 vs. m2=6,c2=3 and dc=c2-c1=2
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+3)*2 > (c+1)*(6-3)  [insert values of variables]<=> m > (3/2)(c-1)`So Gold is better (in terms of money density) only when money density m/c is greater than ~1.45.  This implies that Delve x3 is better in a single Province situation where you expect to have a handsize of 6 or more, Gold is better in a single Province situation where you expect to have a handsize of 4 or less.  With a handsize of 5, Gold will make it slightly easier to reach Province, but Delve x3 will help more when your density is lower and provide more resiliency to greening.

Let's evaluate trashing two Estates (and/or Curses) with Trade vs. buying Gold, m1=4,c1=0, and m2=3,c2=1 giving dc=1-0. Trade is best when:
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+4)*1 > (c+0)(3-4)<=> m > -c - 4`Unsurprisingly, it's a blowout.  Always Trade your Estates, kids.

What if you only have one Estate in hand?  Consider trashing one Estate and one Copper with Trade vs. buying Gold, m1=3,c1=0, and m2=3,c2=1 giving dc=1-0.  Trade is best when:
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+3)*1 > (c+0)(3-3)<=> m > -3`Another blowout - this time a little surprising (to me).  Always Trade your Estates, kids.

And finally, consider trashing two Copper with Trade vs. buying Gold, m1=2,c1=0, and m2=3,c2=1 giving dc=1-0.  Trade is best when:
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+2)*1 > (c+0)(3-2)<=> m > c - 2`You should favor Trading two Coppers over Gold if your (target) money density is >= 1.  This should hold true even if you're playing Smithy, Envoy, or Embassy BM.  It turns out Trade is a great event!

Bonfire

Finally, Bonfire.  When does it make sense to trash two Coppers in a money strategy?  Bonfire Coppers vs. buying Silver, m1=-2,c1=-2, and m2=2,c2=1 giving dc=3. Bonfire is best when:
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m-2)*3 > (c-2)(2+2)<=> m > (4/3)(c - 1/2)`Bonfiring Coppers will increase your money density if your (target) money density (m/c) is greater than ~1.3.  Here's a situation where using only the target money density will get you to make some bad plays.  Opening Bonfire over Silver in most money strategies is a bad idea, because it makes it harder for you to get enough money for Gold.  But, once you've bought your first Gold (or so), buying Bonfire to trash two Coppers when you would otherwise buy a Silver can be a smart move, especially if you expect your handsize to be 5 or fewer.

Edit: I had an error in my Bonfire math.
« Last Edit: May 24, 2017, 10:00:19 am by aku_chi »
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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #7 on: May 25, 2017, 08:28:43 pm »
0

Cool formula.  In situations where it's valuable to assess money density, target money density (i.e. what money density you expect to have while greening) can be a more important metric than current money density.  Let's see what the results are for some common situations with Delve, Trade, and Bonfire. [...]

Thanks, I'm glad you like it, and your applications are cool. And I agree: m and c can refer to the money density of any deck, either the one you have now or the one you're intending to build or... anything else, although those two seem the most relevant.

[Early Bonfire: h]ere's a situation where using only the target money density will get you to make some bad plays.

Yup, similar to my example with trashing a Copper vs. gaining a Silver: the move only takes you towards your desired money density when you're already sufficiently close. I think experience and understanding will be better than math at telling you when to Bonfire (etc.) a little early, when it's too early, when it's too late to matter, etc.

[Trade on Estate/Copper: a]nother blowout - this time a little surprising (to me).

Cool. It sounds like the math had an impact, which I consider to be successful math :-)

Let's extract another general rule: when total money is the same but on fewer cards, money density is higher. In math, assume m1=m2 and c1 < c2 (implying dc > 0); then
Code: [Select]
`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m+m1)*dc > (c+c1)*0<=> (m+m1)*dc > 0<=> (m+m1) > 0`So when your total amount of (additional) money is positive, you want it on as few cards as possible. By that logic, Silver+Silver > Estate+Copper+Gold.

Even more generally, assume m1 >= m2 and c1 <= c2, implying dc >= 0; then
Code: [Select]
`    (m+m1)*dc >= (c+c1)(m2-m1)<=> (m+m1)*dc + (c+c1)(m1-m2) >= 0which is true if (m+m1) >= 0 and (c+c1) >= 0, which they always are.`That is, if you have more money (or the same amount) in fewer cards (or the same amount), you're better off.

(Until you start greening, that is: then you want a big deck for reasons already explored. But for each incremental change the above logic is probably strategically valid most of the time.)
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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #8 on: May 25, 2017, 09:00:56 pm »
+2

Specific money densities for <treasure>-flood, now with terminal draw

Just for fun, let's do some more money density calculations on particular decks. This time, Smithy/silverflood. You're playing 7-card hands and want to hit \$8 with a deck of estate, copper and silver. In fact, let's do it generically first; you want a money density of t:
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`    (c + 2s)/(c+e+s) >= t<=> c + 2s >= t(c + e + s)<=> (2-t)s >= (t-1)c + t*e<=> s >= c*(t-1)/(2-t) + e*t/(2-t)  [assuming 2-t > 0, or equivalently t < 2]`
For t=1.6, I have already calculated this; for t = 8/7 ~= 1.14, we get
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`<=> s >= c*(1/7)/(6/7) + e*(8/7)/(6/7) = (1/6)c + (8/6)e = (1/6)c + (4/3)e`So you need to add ~5.16 silvers before Smithy hands can hit Province. And indeed, SCCCCCC and SSSSEEE are Province hands.

With a gold flood, we get
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`    (c+3g)/(c+e+g) >= t<=> (3-t)g >= (t-1)c + t*e<=> g >= c*(t-1)/(3-t) + e*t/(3-t) = c*(1/7)*(13/7) + e*(8/7)/(13/7) = c*1/13 + e*8/13 [assuming t < 3]`Yay for coprime numbers; 3*8/13 = 24/13 ~= 26/13 = 2, so add two+ golds per three blanks and ~half a gold to cover for the coppers, when you're playing 7-card hands. (Let's round up and say three golds compensates for the initial deck.)

Province hand sanity check: with 7 Copper and 4 blanks we need 7/13 + 32/13 = 39/13 = 3 Golds to meet our targets; and we can partition such a deck into GGCCEEE and GCCCCCE which are two province hands.

For six-card hands (Masquerade, Moat, etc.), the target is 8/6=4/3 ~= 1.33 and the silver and gold flood coefficients are:
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`s >= c*(t-1)/(2-t) + e*t/(2-t) = c*(1/3)/(2/3) + e*(4/3)/(2/3) =   c*1/2 + e*2   (= 3.5 + 6 = 9.5)g >= c*(t-1)/(3-t) + e*t/(3-t) = c*(1/3)/(5/3) + e*(4/3)/(5/3) =   c*1/5 + e*4/5   (= 7/5 + 12/5 = 19/5 ~= 4)`
For eight-card hands (Council Room, Hunting Grounds), the target is 1 and you need one silver or half a gold per estate, and nothing for the coppers.

Corollary: for five card hands, greening with a single Hoard in play moves you away from your target (towards exactly \$1.5 is away from "at least \$1.6"), but for any larger hand size, i.e. any kind of terminal draw BM, Hoard-ing moves you towards your target money density. Said another way: buying a Duchy with a Hoard in play is not greening-with-slower-degradation, it's building! (Caveat: you don't have you Smithy in every starting hand, of course...)

Let's also do the math for hand size attacks; for 4-card hands, t=2
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`s >= c*(t-1)/(2-t) + e*t/(2-t) = c/0 + e*2/0 = :( [we violated the assumption that t < 2]g >= c*(t-1)/(3-t) + e*t/(3-t) = c*1/1 + e*2/1 = 7+3*2 = 13`So for Urchin, no amount of silver can bring you to your target money density, even though SSSS is a Province hand; I guess this highlights what average means and doesn't mean. A very large amount of Gold can bring your money density to 2.

For 3-card hands (Militia), t=8/3 ~= 2.33
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`g >= c*(5/3)/(1/3) + e*(8/3)/(1/3) = c*5 + e*8 (= 35+24 = 59)`So if you add 59 Golds (there aren't that many) to your initial deck, it has a money density (average treasure value) sufficient to buy you a Province after getting Militia'ed. I recommend against using this strategy ;-)

Unlike all other money density math I've done, for reduced hand sizes there's no way of dividing your deck into n-card Province hands, even though the deck has the right proportions of high-valued treasures to coppers and blanks. But the reduced hand-size math is misleading: you don't discard two random or average cards, you discard your two worst cards. Which means that a high-variance deck does better than a low-variance deck, and averages alone are much more insufficient than usual.

But, a simple analysis maybe points somewhere in the right direction: you can overcome Urchin attacks if you have four silvers per initial junk card (or 40 silvers in total), and you can overcome Militia if you have three golds per two initial junk cards, or 15 golds in total. [This assuming really good shuffle luck.] If you're never hitting \$12 and 2 buys, that's at least 15 turns worth of building before you can start greening.

So, the lesson: don't play Big Money with small hands.
« Last Edit: May 26, 2017, 09:17:19 am by jonaskoelker »
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#### Rabid

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #9 on: May 26, 2017, 08:33:13 am »
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"you can overcome Minion attacks if you have four silvers per initial junk card"

Minion gives a 4 card random hand.
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Twitch
1 Day Cup #1:Ednever

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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #10 on: May 26, 2017, 09:23:20 am »
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"you can overcome Minion attacks if you have four silvers per initial junk card"

Minion gives a 4 card random hand.
You are correct. I edited my post to say Urchin instead of Minion. It was late when I wrote that post, I first put in Urchin, then updated to Minion to refer to an earlier set (likely known by a larger fraction of players), and then forgot to cross-check against assumptions elsewhere in the text. I kno how teh card work, I swears!

[Against Minion the first chunk of math holds: there you want a money density of 2, and no amount of silvers can get you there; at most you can get arbitrarily close, pretending the silver supply is infinite. But due to not only drawing average hands, of course sometimes you will still hit Province in a Minion-attacked silver flood. This does not imply that Silver counters Minion ]
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##### Re: Money density math: Gold/2xSilver/Cache, Silver/Peddlers, trashing estates
« Reply #11 on: May 26, 2017, 12:23:52 pm »
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In the thread on simulating two-card combos, there was some discussion about Bonfire/Jack vs. Counterfeit/Jack, with simulator results reported.

Let's do some math: Bonfire, m1=-2,c1=-2 vs. Counterfeit, m2=-1,c2=-1, dc = (-1)-(-2) = 1
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`    (m+m1)*dc > (c+c1)(m2-m1)<=> (m-2) > (c-1)((-1)-(-2))<=> m - 2 > c - 1<=> m > c + 1`So each purchase of Bonfire does more for you deck than each play of Counterfeit, when they both trash Copper, so long as your (current or target) money density is greater than 1.

Note that purchasing Bonfire costs \$3 whereas playing Counterfeit gives you \$2 (above the \$1 you would've gotten from playing the trashed copper normally); Bonfire is available on every turn whereas Counterfeit is only available on some of your turns; and Counterfeit can sometimes be used on Silver and Gold with benefit.

So, the math alone agrees with the simulator results, the paragraph of commentary suggests that the math isn't the whole story, and the simulator results suggest that the costs and non-copper-trashing effects don't alter the conclusion produced by the math. That is, the money density model produced the right answer even though it left some potentially important aspects of the game un-modeled.
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