Dominion Strategy Forum
Dominion => Dominion General Discussion => Topic started by: Skumpy on April 17, 2018, 03:48:09 am
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1. Temple scores a point when you play it
2. If you are down by 6 with 1 province left, you can play a temple, then play 8 coins to buy the last province and win.
That is all.
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This is provided you can spare a card that you feed to the temple and still have $8 for the Province.
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This is provided you can spare a card that you feed to the temple and still have $8 for the Province.
Just throw it an Estate.
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Quick Dominion tip: If you are down by fewer points than you can acquire while ending the game, acquire them and end the game.
This way, you will win the game, which makes it impossible for you to lose.
#lifehacks
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This way, you will win the game, which makes it impossible for you to lose.
Dominion has a move that makes it impossible to lose? Man, what a broken game.
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This is provided you can spare a card that you feed to the temple and still have $8 for the Province.
Just throw it an Estate.
This made my day.
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This way, you will win the game, which makes it impossible for you to lose.
Dominion has a move that makes it impossible to lose? Man, what a broken game.
alls you gotta do is buy silver, or something better than silver
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This way, you will win the game, which makes it impossible for you to lose.
Dominion has a move that makes it impossible to lose? Man, what a broken game.
alls you gotta do is buy silver, or something better than silver
Pro-tip: On a coins-provided-per-coin-cost basis, Copper is infinitely better than Silver.
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This way, you will win the game, which makes it impossible for you to lose.
Dominion has a move that makes it impossible to lose? Man, what a broken game.
alls you gotta do is buy silver, or something better than silver
Pro-tip: On a coins-provided-per-coin-cost basis, Copper is infinitely better than Silver.
So is Transmute though.
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You can't divide by zero.
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You can't divide by zero.
That's a fact.
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You can't divide by zero.
But you can divide by arbitrarily small numbers tending down to zero. So, if you look at card-cost-per-coin-produced on a scale running from silver to copper, it gets better and better as you approach copper. Taking the limit, you can then say that copper is better than silver on this basis, even though it can't be evaluated by itself to have a value on this scale.
Of course, this normally not the right way to assess the strength of treasures, and copper should not in general be considered to be better than silver. I'm aware that was the joke, but I'm just pointing it out in case any fairly new players take it seriously.
That said, there are situations where this is a reasonable way to assess treasure strength, and where the cheaper standard treasures are therefore the better ones. If you will only take one more turn, and you have more 100% (or very close to) reliability overdraw than you have buys that you will use to buy treasures, then, in fact, you do want to evaluate treasure in this way. Adding four silvers to a hugely overdrawing deck is generally better than adding two golds, if the game will end soon. You're unlikely to have got into that situation without having missed an option to build more intelligently earlier, but it might happen. Just the other day I was playing IRL with my brother on a board with Ranger and Wharf for draw, but Royal Carriage was the only splitter (Cartographer and Menagerie the only other +action for a pseudo-village when replayed) and there was absolutely no trashing, so this was a very dicey kind of engine situation that would fall apart quickly when greening. He ended up saying "I would have won right now if I'd bought just one copper with one of my spare buys". Sometimes, particularly for megaturn similar very abruptly greening engines with overdraw, the total economy in your deck is much more important than the density of that economy. Interestingly, that was a game where coppersmith could have been absolutely epic. RIP coppersmith.
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Adding four silvers to a hugely overdrawing deck is generally better than adding two golds, if the game will end soon.
But is it better than adding two Golds and two Coppers?
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But you can divide by arbitrarily small numbers tending down to zero."
Or up to zero, where your result will tend towards minus infinity (meaning it would be infinitely worse than all other Treasure Cards not costing zero). That's the reason you cannot divide by a number tending towards zero without spesifying the direction, and also the reason why dividing by zero would make no sense. Luckily it's prohibited. Remember that the coin cost of Copper is as close to -1 as to 1.
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Adding four silvers to a hugely overdrawing deck is generally better than adding two golds, if the game will end soon.
But is it better than adding two Golds and two Coppers?
I'd say no. I can't think of a way in which it makes a difference, assuming one is just using the treasure to produce $. If TfB is involved for payload then it will vary.
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But you can divide by arbitrarily small numbers tending down to zero."
Or up to zero, where your result will tend towards minus infinity (meaning it would be infinitely worse than all other Treasure Cards not costing zero). That's the reason you cannot divide by a number tending towards zero without spesifying the direction, and also the reason why dividing by zero would make no sense. Luckily it's prohibited. Remember that the coin cost of Copper is as close to -1 as to 1.
Cards can't cost less than zero, though, so at least in the context of Dominion, 1/0 only gives one answer.
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Cards can't cost less than zero, though
There are no cards that cost less than zero, but that doesn't mean it isn't possible.
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Coins_produced(Cost_treasure) = Fibonacci(Cost_treasure/3+1)
Return_Copper = d(Coins_produced)/d(Cost_treasure) (Cost_treasure = 0) = Fibonacci’(1)/3
Now you only need to find a continuous expression for the Fibonacci series to figure out the return on investment of copper. Should be easy enough.
EDIT: messed out my math
EDIT: forgot the f(0)/x term, so my math makes no sense. Meh.
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Coins_produced(Cost_treasure) = Fibonacci(Cost_treasure/3+1)
Return_Copper = d(Coins_produced)/d(Cost_treasure) (Cost_treasure = 0) = Fibonacci’(1)/3
Now you only need to find a continuous expression for the Fibonacci series to figure out the return on investment of copper. Should be easy enough.
EDIT: messed out my math
EDIT: forgot the f(0)/x term, so my math makes no sense. Meh.
In case anyone needs a visual representation to see how the math breaks down on this one:
(https://i.imgur.com/f0Ztzjy.jpg)
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Well, I believe Fibonacci'(1) = 0 anyway.
(At least, if we use the continuous function I found on Wikipedia,
Fibonacci(x) = (φx–cos(xπ)φ–x)/√5
anyway)
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Waaaait, maybe I'm wrong, and F'(1) = log(φ)/√5.
Can someone check this?
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I get ln(phi)/sqrt(5)*(phi+1/phi) with the expression you gave. The expression I knew is not differentiable...
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I get ln(phi)/sqrt(5)*(phi+1/phi) with the expression you gave. The expression I knew is not differentiable...
You sure it's (phi+1/phi) and not (phi–1/phi)? The minus sign comes from the cos(π) = –1.
(And then (phi–1/phi) = 1, so it disappears.)
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On the second term, there's the original -1, another -1 because of the cos, and another -1 because the derivative of phi^(-n) is -ln(phi) * phi^(-n). That's three minuses total. So you are right! I don't know how I miscounted.
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Okay so the return on investment of Copper is about 0.07.
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I am glad we could settle that.
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I'm just glad the thread reverted back to English after that weird foreign language with funny symbols.
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But you can divide by arbitrarily small numbers tending down to zero."
Or up to zero, where your result will tend towards minus infinity (meaning it would be infinitely worse than all other Treasure Cards not costing zero). That's the reason you cannot divide by a number tending towards zero without spesifying the direction, and also the reason why dividing by zero would make no sense. Luckily it's prohibited. Remember that the coin cost of Copper is as close to -1 as to 1.
Cards can't cost less than zero, though, so at least in the context of Dominion, 1/0 only gives one answer.
No, it doesn't give an answer, because it's impossible to divide by 0 (and to use your own argument, sort of, we're not having cards with fractions in their cost in Dominion, so the closest we can come is to divide by one, which, of course, doesn't help us immensely).
That being said, an often overlooked fact is that three plays of Copper gives the same net effect as three plays of Silver and three plays of Gold.
Let's say you buy a:
Copper - Cost 0. After one play, you've gained 1 in value. After the second play, you've gained 2, After the third play, you've gained 3.
Silver - Cost 3. After one play, you've lost 1. After the second play, you've gained 1. After the third play, you've gained 3.
Gold - Cost 6. After one play, you've lost 3. After the second play, you haven't lost nor gained anything. After the third play, you've gained 3.
Platinum - Cost 9. After one play, you've lost 4. After the second play, you've gained 1. After the third play, you've gained 6.
In a vacuum, this means that it's not worth it to buy a Silver or a Gold instead of Copper, unless you get to play it at least four times. Platinum is profitable after three Plays, though. With fewer plays, Copper is the best of them. There are always edge-cases, and yes there are a lot of them, especially here. Still, this is something that you should probably know about.
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The real problem with all of this "dividing by zero" discussion is that the buy required is a part of the cost. You can't just buy the whole Copper pile, which "dividing by zero" would suggest. With this proper accounting, Copper costs $0 and 1 buy, Silver costs $3 and 1 buy, Gold costs $6 and 1 buy, and because the exchange rate between buys and coins varies from kingdom to kingdom, you can't compare those in a vacuum. By contrast, Delve costs $2 and 0 buys, which means that Delve is actually less than 1/3 the price of Gold. And if there were a way to reduce its cost to $0, you could indeed gain the entire Silver pile, the finite-supply equivalent of dividing by zero. I guess it's good there isn't!
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Does any mathsy person know what would happen if I tried to divide by Potion?
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Does any mathsy person know what would happen if I tried to divide by Potion?
Simple: ($6 + 2P)/P = 2 with remainder $6. In other words, if you have $6 and 2 Potions, you can buy 2 Transmutes and still have $6 leftover.
Except my point is that you also need the buys: ($6 + 2P + 1 buy)/(P + buy) = 1 with remainder $6 + P. In other words, if you have $6 and 2 Potions, but only 1 buy, you can only buy one Transmute.
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Does any mathsy person know what would happen if I tried to divide by Potion?
Simple: ($6 + 2P)/P = 2 with remainder $6. In other words, if you have $6 and 2 Potions, you can buy 2 Transmutes and still have $6 leftover.
Except my point is that you also need the buys: ($6 + 2P + 1 buy)/(P + buy) = 1 with remainder $6 + P. In other words, if you have $6 and 2 Potions, but only 1 buy, you can only buy one Transmute.
But on a coins-provided-per-cost basis how does Philosophers Stone compare to Silver??
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Does any mathsy person know what would happen if I tried to divide by Potion?
Simple: ($6 + 2P)/P = 2 with remainder $6. In other words, if you have $6 and 2 Potions, you can buy 2 Transmutes and still have $6 leftover.
Except my point is that you also need the buys: ($6 + 2P + 1 buy)/(P + buy) = 1 with remainder $6 + P. In other words, if you have $6 and 2 Potions, but only 1 buy, you can only buy one Transmute.
But on a coins-provided-per-cost basis how does Philosophers Stone compare to Silver??
We might need to use imaginary numbers for this. So 1/P = -P?
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
The real question is why you had Coppers in your hand when you were buying a card.
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
The real question is why you had Coppers in your hand when you were buying a card.
Why not? Autobuy is easier than autoplaying Treasures, and you don't want to give extra information to your opponent.
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
The real question is why you had Coppers in your hand when you were buying a card.
Why not? Autobuy is easier than autoplaying Treasures, and you don't want to give extra information to your opponent.
Pretty sure there's not much your opponent can do with that extra info if you ended the game on that turn!
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
The real question is why you had Coppers in your hand when you were buying a card.
Why not? Autobuy is easier than autoplaying Treasures, and you don't want to give extra information to your opponent.
Pretty sure there's not much your opponent can do with that extra info if you ended the game on that turn!
Autobuy is still easier than autoplaying Treasures on your last turn, however.
Mic didn't autobuy in the game jonts is referring though, he clicked his Treasures one by one.
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If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
The real question is why you had Coppers in your hand when you were buying a card.
Why not? Autobuy is easier than autoplaying Treasures, and you don't want to give extra information to your opponent.
Pretty sure there's not much your opponent can do with that extra info if you ended the game on that turn!
Autobuy is still easier than autoplaying Treasures on your last turn, however.
Mic didn't autobuy in the game jonts is referring though, he clicked his Treasures one by one.
I hadn't gotten that jonts was talking about an actual game that people knew about.
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Does any mathsy person know what would happen if I tried to divide by Potion?
Simple: ($6 + 2P)/P = 2 with remainder $6. In other words, if you have $6 and 2 Potions, you can buy 2 Transmutes and still have $6 leftover.
Except my point is that you also need the buys: ($6 + 2P + 1 buy)/(P + buy) = 1 with remainder $6 + P. In other words, if you have $6 and 2 Potions, but only 1 buy, you can only buy one Transmute.
But on a coins-provided-per-cost basis how does Philosophers Stone compare to Silver??
Define Usefulness as a function that looks like (return)/(investment). So Silver is ($2 * plays) / ($3 + 1buy), and Stone is ((Deck / 5) * plays)/($3 + 1P + 1buy).
So Usefulness(Stone)/Usefulness(Silver), is [((Deck / 5) * StonePlays) * ($3 + 1buy]) / [($3 + 1P + 1buy) * ($2 * SilverPlays)]
The problem is the value of Stone varies, especially when you play it, which makes this mess difficult to to anything with. I wish I knew LaTeX - this would be much easier.