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[dsc]

I play Forge, trashing Alchemist (3P) and Silver (3), aiming to gain Possession (6P). Goko automatically gave me Gold (as it does when there is only one choice at that cost). This seems wrong to me. Yes?

[WanderingWinder]

Forge says 'Cost in coins'. Potions don't count as coins. As it also says exactly, then you must gain a 6-coin-costing card (with no potion cost), and assuming that gold is the only one of those, what happened is correct.

[Gveoniz]

Forge refers to "total cost  in coin" instead of "total cost"


ninja'd by WanderWinder.

[dsc]

Gah. You know what's funny, I read for that wording specifically, and got through "Trash any number of cards from your hand. Gain a card with cost exactly equal--" and stopped, going, "Ah-ha! It doesn't say coins!" And of course, the next clause is "to the total cost in Coins of the trashed cards". Sigh.

Sorry for the waste of time.

[Asper]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

[Awaclus]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

[zporiri]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

neither one of those posts received upvotes...awkward...

[DG]

They don't have up votes because they didn't quote the rulebook.

"potion symbols (on cards from Dominion: Alchemy) are not added, and the card you gain cannot have a potion symbol in its cost."

[mail-mi]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

neither one of those posts received upvotes...awkward...

liar.

[shMerker]

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

First post of all is the perfect answer for the various versions of the word of the other.

[florrat]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

neither one of those posts received upvotes...awkward...

Saying this must be the best way to give a post some upvotes.

[Awaclus]

I see nobody made the clever, upvote-prone prediction that "while the first post allready gave a perfect answer, there will still follow various versions of the same explanation with different words", yet. So i'll just predict that somebody will make that prediction and see what happens.

While the first post already gave a perfect answer, there will still follow various versions of the same explanation with different words.

neither one of those posts received upvotes...awkward...

Saying this must be the best way to give a post some upvotes.

My post received only 2 upvotes... awkward...

Like this, right?

[Compynerd255]

Something I should add to this discussion is that, IMO, Potion is designed to be frustratingly difficult to get. All of these clauses with coins and only coins make it so that the only way to get a card costing Potion is to play the freakin' mostly-dead Potion for it, and that's the only thing that the Potion treasure is useful for (besides trash fodder).

[AJD]

Something I should add to this discussion is that, IMO, Potion is designed to be frustratingly difficult to get. All of these clauses with coins and only coins make it so that the only way to get a card costing Potion is to play the freakin' mostly-dead Potion for it, and that's the only thing that the Potion treasure is useful for (besides trash fodder).

Fun trivia: ways to get a Potion-costing card without having played a Potion, in order of decreasing likelihood.

1. Clone it with Jester.
2. Trash Squire for Familiar or Scrying Pool.
3. Get it passed to you with Masquerade.
4. Steal Philosopher's Stone with Thief.
5. Gain it by an opponent's Ambassador.

...That's it, I think.

[eHalcyon]

Something I should add to this discussion is that, IMO, Potion is designed to be frustratingly difficult to get. All of these clauses with coins and only coins make it so that the only way to get a card costing Potion is to play the freakin' mostly-dead Potion for it, and that's the only thing that the Potion treasure is useful for (besides trash fodder).

Fun trivia: ways to get a Potion-costing card without having played a Potion, in order of decreasing likelihood.

1. Clone it with Jester.
2. Trash Squire for Familiar or Scrying Pool.
3. Get it passed to you with Masquerade.
4. Steal Philosopher's Stone with Thief.
5. Gain it by an opponent's Ambassador.

...That's it, I think.

You could also use Remake, Upgrade, Expand, etc.  And you could say that I would have had to play a Potion beforehand to get the starting card, but then I could apply your #1-5 first.

I think I would put your #2 above Jester cloning, because Jester is much more luck dependent whereas Squire trashing just requires the right board.

[Compynerd255]

Something I should add to this discussion is that, IMO, Potion is designed to be frustratingly difficult to get. All of these clauses with coins and only coins make it so that the only way to get a card costing Potion is to play the freakin' mostly-dead Potion for it, and that's the only thing that the Potion treasure is useful for (besides trash fodder).

Fun trivia: ways to get a Potion-costing card without having played a Potion, in order of decreasing likelihood.

1. Clone it with Jester.
2. Trash Squire for Familiar or Scrying Pool.
3. Get it passed to you with Masquerade.
4. Steal Philosopher's Stone with Thief.
5. Gain it by an opponent's Ambassador.

...That's it, I think.

That's true. But it's still difficult - Jester is finicky and requires opponent's deck inspection, and it just goes down from there. It's much easier to get your hands on fake Coin.

[Asper]

Something I should add to this discussion is that, IMO, Potion is designed to be frustratingly difficult to get. All of these clauses with coins and only coins make it so that the only way to get a card costing Potion is to play the freakin' mostly-dead Potion for it, and that's the only thing that the Potion treasure is useful for (besides trash fodder).

Potions are a little system in itself. If you go for them, don't expect that you can just build the same kind of deck you'd build without them.

If you want Potions to be more attractive, consider a custom setup-rule Donald once mentioned: When drawing cards randomly, if out of your first 8 kingdom cards none has a Potion cost, make sure the remaining 2 aren't with Potion costs either. If one of them has Potion cost, make sure the remaining two have, too. (And as the max number of Potion cards recommended is 5, stop adding potion-costs if you reach 5). This way you make sure that if there's Potions included, there's also more room for them, and you'll never have games where 9 cards are potionless and 1 isn't. Statistically Potions will show up just as often.

[florrat]

Potions are a little system in itself. If you go for them, don't expect that you can just build the same kind of deck you'd build without them.

If you want Potions to be more attractive, consider a custom setup-rule Donald once mentioned: When drawing cards randomly, if out of your first 8 kingdom cards none has a Potion cost, make sure the remaining 2 aren't with Potion costs either. If one of them has Potion cost, make sure the remaining two have, too. (And as the max number of Potion cards recommended is 5, stop adding potion-costs if you reach 5). This way you make sure that if there's Potions included, there's also more room for them, and you'll never have games where 9 cards are potionless and 1 isn't. Statistically Potions will show up just as often.

There is one disadvantage if you do this: you'll get to see each card with a potion cost in it way more often than cards without potion, more than twice than cards without potion cost in it (assuming you have all sets, I'll show the math if anyone is interested). If you want to make sure each card will show up approximately as much, you could do the following:

Draw the first 4 kingdom cards randomly. If there are no potion cost cards in it, make sure there are no potion cost cards in the last 6 as well. If there is at least one card with a potion cost in the first 4, make sure there will be at least two more cards with potion cost in the last 6 cards (so choose the ninth and tenth card randomly from potion cost cards if you do not have enough, yet).

This will make potion cost cards show up about as much as all other cards (only ~10% more often).

[Asper]

Potions are a little system in itself. If you go for them, don't expect that you can just build the same kind of deck you'd build without them.

If you want Potions to be more attractive, consider a custom setup-rule Donald once mentioned: When drawing cards randomly, if out of your first 8 kingdom cards none has a Potion cost, make sure the remaining 2 aren't with Potion costs either. If one of them has Potion cost, make sure the remaining two have, too. (And as the max number of Potion cards recommended is 5, stop adding potion-costs if you reach 5). This way you make sure that if there's Potions included, there's also more room for them, and you'll never have games where 9 cards are potionless and 1 isn't. Statistically Potions will show up just as often.

There is one disadvantage if you do this: you'll get to see each card with a potion cost in it way more often than cards without potion, more than twice than cards without potion cost in it (assuming you have all sets, I'll show the math if anyone is interested). If you want to make sure each card will show up approximately as much, you could do the following:

Draw the first 4 kingdom cards randomly. If there are no potion cost cards in it, make sure there are no potion cost cards in the last 6 as well. If there is at least one card with a potion cost in the first 4, make sure there will be at least two more cards with potion cost in the last 6 cards (so choose the ninth and tenth card randomly from potion cost cards if you do not have enough, yet).

This will make potion cost cards show up about as much as all other cards (only ~10% more often).

I'm interested in the math, if you care. I thought that it wouldn't change the likelihood for potions, but if i'm wrong i'd like to know the right answer.

[florrat]

I'm interested in the math, if you care. I thought that it wouldn't change the likelihood for potions, but if i'm wrong i'd like to know the right answer.

If you have all sets, but no promos, then there are 200 kingdom cards, of which 10 have potion in its cost. If we ignore the bane (which will hardly matter for computations), there are 10 kingdom cards in each game. If we choose the kingdom cards (uniformly) at random, every card shows up on average in 1 in 20 games, i.e. with probability 0.05.

Now let's compute how often potion cost cards appear in your setup rule. If we use standard notation for probabilities (P[A] is probability of A and P[A|B] is probability of A given B)

P[no potion cost card in in the first eight cards] =
P[first card has no potion cost]*P[second card has no potion cost | first card has no potion cost]*P[third card has no potion cost | first two cards have no potion cost]*... =
(190/200)*(189/199)*(188/198)*...*(183/193) = 0.658.

So the probability that there is at least one potion cost card is in the first eight kingdom cards is 0.342. According to the setup rules, in all such games there will be at least 3 potion cost cards, so there will be on average more than 3*0.342=1.02 potion cost cards in each game, which means that each potion cost card appears with probability at least 0.102, more than twice as much as what would happen when you choose cards randomly. The fact that I ignored in this computation the possibility that there are more than 3 cards with potion cost in the game doesn't matter, because that probability is very very small (I think it starts to matter in the 5th digit after the decimal point).

Similarly one can compute that the probability of a potion cost card in the first 4 cards is 0.187, which means that each potion cost card has a probability of 0.0560 of appearing in each game using "my" setup rule, which is just slightly more than the 0.05 when choosing randomly.

[Asper]

I'm interested in the math, if you care. I thought that it wouldn't change the likelihood for potions, but if i'm wrong i'd like to know the right answer.

If you have all sets, but no promos, then there are 200 kingdom cards, of which 10 have potion in its cost. If we ignore the bane (which will hardly matter for computations), there are 10 kingdom cards in each game. If we choose the kingdom cards (uniformly) at random, every card shows up on average in 1 in 20 games, i.e. with probability 0.05.

Now let's compute how often potion cost cards appear in your setup rule. If we use standard notation for probabilities (P[A] is probability of A and P[A|B] is probability of A given B)

P[no potion cost card in in the first eight cards] =
P[first card has no potion cost]*P[second card has no potion cost | first card has no potion cost]*P[third card has no potion cost | first two cards have no potion cost]*... =
(190/200)*(189/199)*(188/198)*...*(183/193) = 0.658.

So the probability that there is at least one potion cost card is in the first eight kingdom cards is 0.342. According to the setup rules, in all such games there will be at least 3 potion cost cards, so there will be on average more than 3*0.342=1.02 potion cost cards in each game, which means that each potion cost card appears with probability at least 0.102, more than twice as much as what would happen when you choose cards randomly. The fact that I ignored in this computation the possibility that there are more than 3 cards with potion cost in the game doesn't matter, because that probability is very very small (I think it starts to matter in the 5th digit after the decimal point).

Similarly one can compute that the probability of a potion cost card in the first 4 cards is 0.187, which means that each potion cost card has a probability of 0.0560 of appearing in each game using "my" setup rule, which is just slightly more than the 0.05 when choosing randomly.

I see you are right. Normally, the average number of potion cards is 0.5 per game. According to your calculation, the chance for a Potion to be under the first 8 is 34% (recalculated that and came to the same conclusion), so in 34% of cases the number of potion cards used is 3 or more. That means on average, it's at least 1. Thank you very much