# Dominion Strategy Forum

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### AuthorTopic: A quick Temple tip  (Read 1917 times)

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#### pacovf

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##### Re: A quick Temple tip
« Reply #25 on: April 18, 2018, 10:47:11 pm »
0

I am glad we could settle that.
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##### Re: A quick Temple tip
« Reply #26 on: April 19, 2018, 12:17:17 am »
+3

I'm just glad the thread reverted back to English after that weird foreign language with funny symbols.

#### Oyvind

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##### Re: A quick Temple tip
« Reply #27 on: April 19, 2018, 04:12:56 am »
+3

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.

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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#### samath

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##### Re: A quick Temple tip
« Reply #28 on: April 19, 2018, 11:28:07 am »
0

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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#### Gazbag

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##### Re: A quick Temple tip
« Reply #29 on: April 19, 2018, 12:22:24 pm »
+2

Does any mathsy person know what would happen if I tried to divide by Potion?
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#### samath

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##### Re: A quick Temple tip
« Reply #30 on: April 19, 2018, 12:33:12 pm »
+2

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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#### Gazbag

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##### Re: A quick Temple tip
« Reply #31 on: April 19, 2018, 01:12:28 pm »
0

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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#### Commodore Chuckles

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##### Re: A quick Temple tip
« Reply #32 on: April 19, 2018, 04:58:16 pm »
+1

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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#### jonts26

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##### Re: A quick Temple tip
« Reply #33 on: April 22, 2018, 03:59:58 pm »
+7

If you end the game by buying the last cemetery, don't trash your coppers and lose your fountain points.
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#### GendoIkari

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##### Re: A quick Temple tip
« Reply #34 on: April 22, 2018, 05:25:42 pm »
0

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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#### Awaclus

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##### Re: A quick Temple tip
« Reply #35 on: April 22, 2018, 05:40:17 pm »
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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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#### GendoIkari

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##### Re: A quick Temple tip
« Reply #36 on: April 22, 2018, 05:43:09 pm »
0

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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#### Awaclus

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##### Re: A quick Temple tip
« Reply #37 on: April 22, 2018, 06:10:51 pm »
0

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.
« Last Edit: April 22, 2018, 06:21:05 pm by Awaclus »
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#### GendoIkari

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##### Re: A quick Temple tip
« Reply #38 on: April 22, 2018, 07:55:17 pm »
0

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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#### KingPeter

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##### Re: A quick Temple tip
« Reply #39 on: April 22, 2018, 08:23:42 pm »
0

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.
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