Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Doom_Shark]

I know this has been posited before, but that was a while ago (before Renaissance, and I think before Nocturne). I'm also too lazy to hunt down the thread. Dominion setup rules have gotten pretty complex over the years, so I'm going to do this in parts. As I keep the running totals, I'll be correcting rounding errors by just adding the formulas together in Wolfram Alpha, rather than adding the rounded numbers. With all that out of the way, let's do some math!

First, let's do just the kingdom cards. There's a total of 376 kingdom cards (including the removed cards, since they are still techincally possible dominion games). So, without any landscapes, and excluding different possibilities for Young Witch's bane card or shenanigans with the Black Market deck, that gives us 376!/(10! * 366!), or 13.8 quintillion kingdoms, rounded to the nearest 100 quadrillion.

Now, for each of them that has prosperity, there's actually two kingdoms, one with platinum/colony and one without. So how many have prosperity? If we subtract the 25 prosperity kingdom cards from the 376, that leaves us with 351 cards. Using that, we calculate the number of games that don't have prosperity (351!/(10! * 341!)) and subtract that from our 13.8 quintillion to get 6.9 quintillion kingdoms that include platinum/colony.
Running total: 20.7 quintillion kingdoms

Now we add shelters. We do the same thing we did for prosperity, but with dark ages. 376 total cards - 35 Dark Ages cards = 341 non-Dark-Ages cards. 13.8 quintillion kingdoms - (341!/10! * 331!) kingdoms without Dark Ages gives us 8.7 quintillion kingdoms with shelters.
Running total: 29.4 quintillion kingdoms

Now the math starts to get weird, because we need to add the kingdoms that have both platinum/colony and shelters. If we take the number of games without Prosperity and subtract the number of games without Dark Ages or Prosperity, that leaves us with the number of games with Dark Ages and not Prosperity. Subtract that from the 8.7 quintillion total kingdoms with Dark Ages, and we get 4.2 quintillion kingdoms with both.
Running Total: 33.5 quintillion kingdoms

Now we start with landscapes. Each of those kingdoms could be paired with any one of the 115 landscapes. So we multiply that 33.5 quintillion by 116 (adding one for the "no landscape option).
Running Total: 3.9 sextillion kingdoms (rounded to the nearest 100 quintillion)

Now because ways are weird, there's gonna be two parts to having two landscapes. First, we go back to the 33.5 quintillion and multiply that by all the possible combinations of two non-Way landscapes (95!/2 * 93!), giving us 149.8 sextillion kingdoms. That's quite a jump!
Running Total: 153.7 sextillion kingdoms
Note: At this point, Wolfram Alpha broke when I tried to have it run and add all the formulas. So much for correcting rounding error.

Finally, two landscapes, one is a Way. We take our 33.5 quintillion again, multiply by 20 Ways, and then multiply by 95 other landscapes, giving 63.7 sextillion kingdoms.

Grand Total: 217.4 sextillion possible kingdoms

[Gherald]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

[silverspawn]

In mathematical notation: 2.174 * 10^{23}. I.e., a number with 24 digits.

[faust]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

[Dominionaer]

There's a total of 376 kingdom cards (including the removed cards,

I count 366 (truth: table sheet does)

[Doom_Shark]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

The issue with that is you'd have to figure out how to control for what 2's and 3's are already in the kingdom, which makes it hard. Like I said at the beginning: I'm lazy.

[Doom_Shark]

There's a total of 376 kingdom cards (including the removed cards,

I count 366 (truth: table sheet does)

Ah, you're right, I somehow added 10 cards. That's what I get for doing complex math at 2 am.

[GendoIkari]

As of Prosperity edition rulebook, it is officially legal to use Platinum and Colony in games without any Prosperity cards; the rule is "You can include them whenever you want to; they are always used together." So that makes it simpler; the second step can simply be to multiple the total number by 2.

As far as I know, the same update has not happened for Dark Ages and Shelters, though.

One issue with this question is deciding what counts as a different Kingdom. Does the order of Knights count? Various things like that are unclear.

Also, landscapes are not limited to 2 per game.. that's a suggestion, but not the rule. So you have to multiply the non-landscape number by 2^115, or 41,533,837,500,000,000,000,000,000,000,000,000.

That's not counting Obelisk or Way of the Mouse choices.

[Oyvind]

I know this has been posited before, but that was a while ago (before Renaissance, and I think before Nocturne). I'm also too lazy to hunt down the thread. Dominion setup rules have gotten pretty complex over the years, so I'm going to do this in parts. As I keep the running totals, I'll be correcting rounding errors by just adding the formulas together in Wolfram Alpha, rather than adding the rounded numbers. With all that out of the way, let's do some math!

First, let's do just the kingdom cards. There's a total of 376 kingdom cards (including the removed cards, since they are still techincally possible dominion games). So, without any landscapes, and excluding different possibilities for Young Witch's bane card or shenanigans with the Black Market deck, that gives us 376!/(10! * 366!), or 13.8 quintillion kingdoms, rounded to the nearest 100 quadrillion.

Now, for each of them that has prosperity, there's actually two kingdoms, one with platinum/colony and one without. So how many have prosperity? If we subtract the 25 prosperity kingdom cards from the 376, that leaves us with 351 cards. Using that, we calculate the number of games that don't have prosperity (351!/(10! * 341!)) and subtract that from our 13.8 quintillion to get 6.9 quintillion kingdoms that include platinum/colony.
Running total: 20.7 quintillion kingdoms

Now we add shelters. We do the same thing we did for prosperity, but with dark ages. 376 total cards - 35 Dark Ages cards = 341 non-Dark-Ages cards. 13.8 quintillion kingdoms - (341!/10! * 331!) kingdoms without Dark Ages gives us 8.7 quintillion kingdoms with shelters.
Running total: 29.4 quintillion kingdoms

Now the math starts to get weird, because we need to add the kingdoms that have both platinum/colony and shelters. If we take the number of games without Prosperity and subtract the number of games without Dark Ages or Prosperity, that leaves us with the number of games with Dark Ages and not Prosperity. Subtract that from the 8.7 quintillion total kingdoms with Dark Ages, and we get 4.2 quintillion kingdoms with both.
Running Total: 33.5 quintillion kingdoms

Now we start with landscapes. Each of those kingdoms could be paired with any one of the 115 landscapes. So we multiply that 33.5 quintillion by 116 (adding one for the "no landscape option).
Running Total: 3.9 sextillion kingdoms (rounded to the nearest 100 quintillion)

Now because ways are weird, there's gonna be two parts to having two landscapes. First, we go back to the 33.5 quintillion and multiply that by all the possible combinations of two non-Way landscapes (95!/2 * 93!), giving us 149.8 sextillion kingdoms. That's quite a jump!
Running Total: 153.7 sextillion kingdoms
Note: At this point, Wolfram Alpha broke when I tried to have it run and add all the formulas. So much for correcting rounding error.

Finally, two landscapes, one is a Way. We take our 33.5 quintillion again, multiply by 20 Ways, and then multiply by 95 other landscapes, giving 63.7 sextillion kingdoms.

Grand Total: 217.4 sextillion possible kingdoms

There are 366 different Kingdom piles (including the removed cards), not 376:
Dominion 26(+6)
Intrigue 26 (+6)
Seaside 26
Alchemy 12
Prosperity 25
Cornucopia 13
Hinterlands 26
Dark Ages 35
Guilds 13
Adventures 30
Empires 24
Nocturne 33
Renaissance 25
Menagerie 30
Promos 10
Totals 354(+12)

Besides, there are no hard and fast rules regarding the number of landscapes, so you can potentially have anything from zero to 115 landscapes in any given game. DXV recommends using two in every game, and at most one Way. Using his recommended way of doing things will reduce the number of options considerably. As you're including the 'retired' cards, you're not following his recommendations, so you might as well include every possible combination of landscapes as described.

[Oyvind]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, IMHO (the kingdoms contain the exact same cards). Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six, so be sure to take that into account as well.

[hhelibebcnofnena]

What about Druid? Do different Druid Boons count as a different kingdom?

[Oyvind]

What about Druid? Do different Druid Boons count as a different kingdom?

Yeah, probably. There are 1,320 different combinations of those, by the way.

[faust]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

[hhelibebcnofnena]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

Is it just the top knight which changes the kingdom, then?

[Oyvind]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

I agree that it plays vastly different from other Knights games, but you're talking about different experiences, not different kingdoms. Every shuffle makes each game unique as well, so those should also count, if that's the case. For every shuffle during the game, by the way. Would you say that identical kingdoms are different for two players and six players? Maybe we should multiply everything by five to take that into consideration as well? I don't know. Maybe OP can give us some more guidelines.

[faust]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

I agree that it plays vastly different from other Knights games, but you're talking about different experiences, not different kingdoms. Every shuffle makes each game unique as well, so those should also count, if that's the case. For every shuffle during the game, by the way. Would you say that identical kingdoms are different for two players and six players? Maybe we should multiply everything by five to take that into consideration as well? I don't know. Maybe OP can give us some more guidelines.

I mean, I can argue on a more technical level as well that the kingdoms are different because not all cards are in the same place in both kingdoms.

There's an argument to be made to multiply everything by five, yes, since the Curse pile contains a different number of cards in each case. However I usually only think of Dominion through a 2-player lens, so I'm happy enough to ignore that.

[GendoIkari]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

I agree that it plays vastly different from other Knights games, but you're talking about different experiences, not different kingdoms. Every shuffle makes each game unique as well, so those should also count, if that's the case. For every shuffle during the game, by the way. Would you say that identical kingdoms are different for two players and six players? Maybe we should multiply everything by five to take that into consideration as well? I don't know. Maybe OP can give us some more guidelines.

I think one valid definition is "what set of cards and card-shaped things are available this game"? Available meaning "could possibly be available depending on how the game plays out"; so the bottom Knight is considered "available". Under this definition; order of Knights and Black Market cards don't count. Obelisk pick doesn't count. Way of the Mouse and Druid choices do count. Which Ruins are in the game count, but not which order they are in.

[Doom_Shark]

And here I was hoping for some precise combinatorials involving Young Witch's bane and Way of the Mouse, as well as an adjustment factor for kingdoms that include both.

I am disappoint.

Don't forget Obelisk, the ordering of the Knights pile, and the ordering and composition of the Ruins pile.

The ordering of the Knights pile or the Ruins pile doesn't make different kingdoms, though. Yet, the composition of the Ruins pile do. This will also vary depending on the number of players from two to six.

I'd argue that a Kingdom where Dame Anna is the top Knight plays vastly differently from one where Dame Josephine is, and thus it makes sense to call these different kingdoms.

For the Ruins pile of course it matters less, though occasionally it is relevant whether the top Ruins is RUined Market.

I agree that it plays vastly different from other Knights games, but you're talking about different experiences, not different kingdoms. Every shuffle makes each game unique as well, so those should also count, if that's the case. For every shuffle during the game, by the way. Would you say that identical kingdoms are different for two players and six players? Maybe we should multiply everything by five to take that into consideration as well? I don't know. Maybe OP can give us some more guidelines.

I think one valid definition is "what set of cards and card-shaped things are available this game"? Available meaning "could possibly be available depending on how the game plays out"; so the bottom Knight is considered "available". Under this definition; order of Knights and Black Market cards don't count. Obelisk pick doesn't count. Way of the Mouse and Druid choices do count. Which Ruins are in the game count, but not which order they are in.

I'm inclined to agree with this definition. I'm currently working out the formulas with pen and paper. The question now is: what to do about the Black Market deck? On the one hand, there's what should technically be done, which is just one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). On the other, there's how most people play it: with a deck of only 25-30 cards. Thoughts?

[Wizard_Amul]

I'm inclined to agree with this definition. I'm currently working out the formulas with pen and paper. The question now is: what to do about the Black Market deck? On the one hand, there's what should technically be done, which is just one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). On the other, there's how most people play it: with a deck of only 25-30 cards. Thoughts?

Actually, that's not what should technically be done. Here is text from the official FAQ on the card:

"To use the Black Market Kingdom card, you must create a Black Market deck before starting the game. The Black Market deck is made up of Kingdom cards that are not in the Supply of the current game. The players should agree before the game which cards will be used to create the Black Market deck (for example, you could agree to use one of every Kingdom card you own that is not a part of the Supply). It is recommended that the Black Market deck contain at least 15 Kingdom cards, with no duplicates."

The 15 Kingdom cards is a recommended minimum only. I would say the Black Market deck technically has to have between 3 cards minimum (you need to be able to reveal 3 cards for a choice) and one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). Thus, I would think you need to account for all the possibilities of all Black Market deck sizes and Black Market deck compositions.

[GendoIkari]

I'm inclined to agree with this definition. I'm currently working out the formulas with pen and paper. The question now is: what to do about the Black Market deck? On the one hand, there's what should technically be done, which is just one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). On the other, there's how most people play it: with a deck of only 25-30 cards. Thoughts?

Actually, that's not what should technically be done. Here is text from the official FAQ on the card:

"To use the Black Market Kingdom card, you must create a Black Market deck before starting the game. The Black Market deck is made up of Kingdom cards that are not in the Supply of the current game. The players should agree before the game which cards will be used to create the Black Market deck (for example, you could agree to use one of every Kingdom card you own that is not a part of the Supply). It is recommended that the Black Market deck contain at least 15 Kingdom cards, with no duplicates."

The 15 Kingdom cards is a recommended minimum only. I would say the Black Market deck technically has to have between 3 cards minimum (you need to be able to reveal 3 cards for a choice) and one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). Thus, I would think you need to account for all the possibilities of all Black Market deck sizes and Black Market deck compositions.

I think a Black Market deck with only 1 or 2 cards is definitely within the rules. There's no issue with needing to reveal 3 when there are less than 3, that can happen normally after lots of cards have been bought from the Black Market. You do as much as you can. Whether or not Black Market with 0 cards is legal is a different question. I think the answer is yes, but an argument can be made that if it contains 0 cards, it doesn't meet the criteria of being "made up of Kingdom cards that are not in the Supply of the current game".

[mxdata]

I'm inclined to agree with this definition. I'm currently working out the formulas with pen and paper. The question now is: what to do about the Black Market deck? On the one hand, there's what should technically be done, which is just one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). On the other, there's how most people play it: with a deck of only 25-30 cards. Thoughts?

Actually, that's not what should technically be done. Here is text from the official FAQ on the card:

"To use the Black Market Kingdom card, you must create a Black Market deck before starting the game. The Black Market deck is made up of Kingdom cards that are not in the Supply of the current game. The players should agree before the game which cards will be used to create the Black Market deck (for example, you could agree to use one of every Kingdom card you own that is not a part of the Supply). It is recommended that the Black Market deck contain at least 15 Kingdom cards, with no duplicates."

The 15 Kingdom cards is a recommended minimum only. I would say the Black Market deck technically has to have between 3 cards minimum (you need to be able to reveal 3 cards for a choice) and one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). Thus, I would think you need to account for all the possibilities of all Black Market deck sizes and Black Market deck compositions.

I think a Black Market deck with only 1 or 2 cards is definitely within the rules. There's no issue with needing to reveal 3 when there are less than 3, that can happen normally after lots of cards have been bought from the Black Market. You do as much as you can. Whether or not Black Market with 0 cards is legal is a different question. I think the answer is yes, but an argument can be made that if it contains 0 cards, it doesn't meet the criteria of being "made up of Kingdom cards that are not in the Supply of the current game".

If you really want to push that argument, the plural "cards" would also rule out a 1-card deck

[Wizard_Amul]

I'm inclined to agree with this definition. I'm currently working out the formulas with pen and paper. The question now is: what to do about the Black Market deck? On the one hand, there's what should technically be done, which is just one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). On the other, there's how most people play it: with a deck of only 25-30 cards. Thoughts?

Actually, that's not what should technically be done. Here is text from the official FAQ on the card:

"To use the Black Market Kingdom card, you must create a Black Market deck before starting the game. The Black Market deck is made up of Kingdom cards that are not in the Supply of the current game. The players should agree before the game which cards will be used to create the Black Market deck (for example, you could agree to use one of every Kingdom card you own that is not a part of the Supply). It is recommended that the Black Market deck contain at least 15 Kingdom cards, with no duplicates."

The 15 Kingdom cards is a recommended minimum only. I would say the Black Market deck technically has to have between 3 cards minimum (you need to be able to reveal 3 cards for a choice) and one of everything (with the exceptions of only one card from each split pile, one knight, and one castle). Thus, I would think you need to account for all the possibilities of all Black Market deck sizes and Black Market deck compositions.

I think a Black Market deck with only 1 or 2 cards is definitely within the rules. There's no issue with needing to reveal 3 when there are less than 3, that can happen normally after lots of cards have been bought from the Black Market. You do as much as you can. Whether or not Black Market with 0 cards is legal is a different question. I think the answer is yes, but an argument can be made that if it contains 0 cards, it doesn't meet the criteria of being "made up of Kingdom cards that are not in the Supply of the current game".

Sure, that is true, actually. I revise my statement to say that a 1 card black market deck is within the rules. A 0 card deck could be seen as a number theory question; I would say a 0 card black market deck is like an empty set, and since the empty set is technically a subset of every set, you can't do a 0 card black market deck (i.e., the empty set is in the Supply of the current game, because the empty set is in the Supply of every game).

[sudgy]

A 0 card deck could be seen as a number theory question; I would say a 0 card black market deck is like an empty set, and since the empty set is technically a subset of every set, you can't do a 0 card black market deck (i.e., the empty set is in the Supply of the current game, because the empty set is in the Supply of every game).

Funny, I thought of it mathematically as well and came up with the opposite answer.  The rules don't say that the Black Market deck is a subset of the non-Supply cards, but that all cards in the Black Market deck are not in the supply.  Because the statement "∀ x ∈ ∅, P x" is true for all propositions P, I think a 0-card black market deck is fine.

[Wizard_Amul]

A 0 card deck could be seen as a number theory question; I would say a 0 card black market deck is like an empty set, and since the empty set is technically a subset of every set, you can't do a 0 card black market deck (i.e., the empty set is in the Supply of the current game, because the empty set is in the Supply of every game).

Funny, I thought of it mathematically as well and came up with the opposite answer.  The rules don't say that the Black Market deck is a subset of the non-Supply cards, but that all cards in the Black Market deck are not in the supply.  Because the statement "∀ x ∈ ∅, P x" is true for all propositions P, I think a 0-card black market deck is fine.

I hadn't heard of this last statement before, so thanks. You're also right that the way the card is worded makes it the opposite answer. After thinking about it conceptually also, it seems like a 0 card deck should be valid; if you have a pile of cards in the supply, and then you have a pile of cards not in the supply that you are picking from, picking nothing from the "cards not in the supply" pile is a valid choice (i.e., the empty set is a subset of the "cards not in the supply" pile because the empty set is a subset of every set).

[Doom_Shark]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

[GendoIkari]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

I would agree.

[hhelibebcnofnena]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

I would agree.

The best way to calculate that would probably be to calculate the # of kingdoms, pretending Young Witch wasn't even a card, and then double-count all the kingdoms with cards costing or .

[GendoIkari]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

I would agree.

The best way to calculate that would probably be to calculate the # of kingdoms, pretending Young Witch wasn't even a card, and then double-count all the kingdoms with cards costing or .

Not following... the Young Witch Kingdoms would have 11 piles instead of 10; which would do a lot more than just double count wouldn't it?

[Doom_Shark]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

I would agree.

The best way to calculate that would probably be to calculate the # of kingdoms, pretending Young Witch wasn't even a card, and then double-count all the kingdoms with cards costing or .

Way ahead of you there. Black Market is hard though. Busting out integrals for that so I don't have to write a different equation for each possible size. Still working out the various formulas, by the way.

Not following... the Young Witch Kingdoms would have 11 piles instead of 10; which would do a lot more than just double count wouldn't it?

The eleventh pile is Young Witch herself. So if we get a bunch of 10-card kingdoms, all of which have at least one or card, then add Young Witch to that kingdom, it becomes a valid kingdom.

[Doom_Shark]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

[sudgy]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

That is definitely not true in general:

∫₁⁵ x dx = 12
1 + 2 + 3 + 4 + 5 = 15

[crj]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

Ignoring the set-theoretical arguments for a moment, this doesn't make sense in real-world terms. A kingdom containing Young Witch, Ambassador and Fortune Teller will play out very differently depending on which of the latter two is the Bane.

[GendoIkari]

So, I'm going to assume that since we don't care about ruins order, knights order, or which card obelisk picks, we also don't care which card specifically is the bane, only that there are 11 kingdom cards including young witch, and at least one of them costs $2 or $3.

Ignoring the set-theoretical arguments for a moment, this doesn't make sense in real-world terms. A kingdom containing Young Witch, Ambassador and Fortune Teller will play out very differently depending on which of the latter two is the Bane.

Yes but how the game will play out isn’t the question being asked. There’s various definitions of “different kingdom”, but people in general here seem to be going with a question of which cards are available. If you include which card is the bane, then you should also include the order of Knights, order of Black Market deck, Obelisk choice, etc.

[Doom_Shark]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

That is definitely not true in general:

∫₁⁵ x dx = 12
1 + 2 + 3 + 4 + 5 = 15

Ok, is there a way to do that where I don't have to individually calculate (or rather, type into wolfram alpha) f(x) for each integer value in the range?

[sudgy]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

That is definitely not true in general:

∫₁⁵ x dx = 12
1 + 2 + 3 + 4 + 5 = 15

Ok, is there a way to do that where I don't have to individually calculate (or rather, type into wolfram alpha) f(x) for each integer value in the range?

The best thing to do would be to look for a closed-form solution of the sum, which is nontrivial.  Wolfram Alpha can sometimes figure it out for you.  Actually, you could just type in your formula and tell it to do a sum.  That should work as well.

[Doom_Shark]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

That is definitely not true in general:

∫₁⁵ x dx = 12
1 + 2 + 3 + 4 + 5 = 15

Ok, is there a way to do that where I don't have to individually calculate (or rather, type into wolfram alpha) f(x) for each integer value in the range?

The best thing to do would be to look for a closed-form solution of the sum, which is nontrivial.  Wolfram Alpha can sometimes figure it out for you.  Actually, you could just type in your formula and tell it to do a sum.  That should work as well.

So it did! Cool. Back to work.

[Dominionaer]

What about Druid? Do different Druid Boons count as a different kingdom?

Yeah, probably. There are 1,320 different combinations of those, by the way.

?
(12 choose 3) = 220

[Doom_Shark]

Ok, here's the formulas I have so far:

Our starting point is the 366 kingdom cards, minus Druid, the three Looters, Young Witch, and Black Market, so 360 cards. Let's let A_n=360!/(n!(360-n)!), so that I can use variables instead of writing out the formula each time I need to grab a different number of cards from this set. So, without those 6 cards (which will be added back in later), we start with A₁₀ kingdoms.

To add Platinum and Colony, as well as Shelters, we'll do the same as in the op, but corrected with the new numbers. Removing Prosperity leaves us with 335 cards, so let B_n=335!/(n!(335-n)!). We've already taken out 3 Dark Ages cards, so we just need to cut the other 32, leaving us with 328 cards, so let C_n=328!/(n!(328-n)!), and we'll let D_n=303!/(n!(303-n)!), or the kingdoms without either. This gives us the follwoing:
Platinum/Colony = A₁₀-B₁₀
Shelters = A₁₀-C₁₀
Both = A₁₀-B₁₀-(C₁₀-D₁₀), or A₁₀-B₁₀-C₁₀+D₁₀
Add all of these to our initial A₁₀ and combine like terms: 4A₁₀-2B₁₀-2C₁₀+D₁₀, or 2(2A₁₀-B₁₀-C₁₀)+D₁₀

Now, to add Druid, we'll calculate the number of 9-card kingdoms (the 10th card is Druid) and multiply by the number of possible combinations of set aside boons. 12!/(3!*9!)=220, so for each kingdom with Druid, there are actually 220 kingdoms. To get the 9-card kingdoms, we'll use the above total formula, using subscript 9 instead of 10. Since Druid costs , any kingdom that has it can also add Young Witch to make a new, valid kingdom, so we'll multiply all of this by 2 again. 2*220=440, and multiply to get:
Kingdoms with Druid = 440(2(2A₉-B₉-C₉)+D₉), or 880(2A₉-B₉-C₉)+440D₉

Adding Looters gets a little tricky. We're going to limit our scope to 2-player games for this, partly because that's all people care about here, and partly because it makes the math MUCH easier. In a 2-player game, there is a pile of 10 ruins. Since we don't care about probability of any given ruins pile, and there just happen to be 10 copies of each of the 5 ruins, that gives us 5¹⁰ possible ruins piles. However, that also accounts for the order of the pile. We could do 50!/(10!40!), but that assumes 50 unique ruins, and is several orders of magnitude larger than 5¹⁰, so until I figure out how to ignore the order of the pile, 5¹⁰ it is. Since all games with Looters automatically have a chance for shelters, we can ignore the shelters combination formulas and just use 2A_n-B_n, where n is the number non-Looter cards in the kingdom, and then multiply everything by 2. So:
One Looter = 6*5¹⁰(2A₉-B₉)
Two Looters = 6*5¹⁰(2A₈-B₈)
Three Looters = 2*5¹⁰(2A₇-B₇)

To add Druid to the Looters, we reduce all the subscripts by 1, and multiply by 440 (because why not double count for Young Witch while we're at it?)
One Looter = 2,640*5¹⁰(2A₈-B₈)
Two Looters = 2,640*5¹⁰(2A₇-B₇)
Three Looters = 880*5¹⁰(2A₆-B₆)

Now for the rest of the kingdoms with Young Witch, we're going to end up making some more new variables. There are 108 kingdom cards that cost or . However, we've already removed Druid and Black Market, which cost and respectively, leaving 106. 360-106=254, so let E_n=254!/(n!(254-n)!).

Next, we have to double count all those that include Prosperity, which means creating yet another variable for the kingdoms with neither Prosperity nor or cards. 3 Prosperity cards cost (Loan, Trade Route, and Watchtower) and have already been removed in our 254. So subtracting the other 22 leaves F_n=232!/(n!(232-n)!).

Now we need the ones with Dark Ages. The set has 3 cards costing and 6 cards costing . Take the 32 from earlier, subtract those 9, we get 23 Dark Ages cards still included. 254-23=231, so let G_n=231!/(n!(231-n)!).

Finally, we create one more variable that includes none of these. So we take our 232, subtract our 23, and we get 209, so let H_n=209!/(n!(209-n)!). So all of the formulas for kingdoms with Young Witch, excluding those with Druid, Looters, or Black Market, are as follows:
Just Young Witch = A₁₀-E₁₀
Young Witch and Platinum/Colony = A₁₀-B₁₀-(E₁₀-F₁₀), or A₁₀-B₁₀-E₁₀+F₁₀
Young Witch and Shelters = A₁₀-C₁₀-(E₁₀-G₁₀), or A₁₀-C₁₀-E₁₀+G₁₀
Young Witch and Both = A₁₀-B₁₀-(C₁₀-D₁₀)-(E₁₀-H₁₀), or A₁₀-B₁₀-C₁₀+D₁₀-E₁₀+H₁₀
Add them up and combine like terms: 4A₁₀-2B₁₀-2C₁₀+D₁₀-4E₁₀+F₁₀+G₁₀+H₁₀, or 4(A₁₀-E₁₀)-2(B₁₀+C₁₀)+D₁₀+F₁₀+G₁₀+H₁₀

Now we add Young Witch to the Looters, or rather, looters to Young Witch. Once again, we can ignore the Shelters component and just multiply by two.
One Looter = 6*5¹⁰(A₉-B₉-E₉+F₉)
Two Looters = 6*5¹⁰(A₈-B₈-E₈+F₈)
Three Looters = 2*5¹⁰(A₇-B₇-E₇+F₇)

We finally arrive at the dreaded Black Market, and here is where things get even more dicey than the Ruins situation above. Allowing for a 0-card Black Market deck based on the brief number theory discussion earlier is simple enough; I could have even included it in all the other formulas above, but that would have confused me when I got to this step, so instead we'll go through several of the above formulas and do two things to them:
1) Reduce the subscript by 1
2)Double count for Young Witch, since Black Market costs (skip this step for formulas that include Druid, since we've already done that there)
Starting point, Platinum/Colony, and Shelters = 4(2A₉-B₉-C₉)+2D₉
Druid = 880(2A₈-B₈-C₈)+440D₈
One Looter = 12*5¹⁰(2A₈-B₈)
Two Looters = 12*5¹⁰(2A₇-B₇)
Three Looters = 4*5¹⁰(2A₆-B₆)
Druid and one Looter = 2,640*5¹⁰(2A₇-B₇)
Druid and two Looters = 2,640*5¹⁰(2A₆-B₆)
Druid and three Looters = 880*5¹⁰(2A₅-B₅)

Next, we get kingdoms that actually have a Black Market deck. For a Black Market deck of n cards, where 0<n<351, and neither the deck nor the kingdom includes Druid, Looters, or Young Witch, there are 360!/(n!(360-n)!)*(360-n)!/(9!(351-n)!) possible kingdoms. That simplifies to 360!/(9!n!(351-n!)). Getting the exact sum ∑_n=1³⁵⁰ 360!/(9!n!(351-n!)) in wolfram alpha expends more computation time than the free version allows, but that doesn't matter because it output it does give me is this:

5.4142756889123355558613982831116653273300579874856051... × 10^757

which is a number so insanely huge that all the other previous math is essentially pointless now. Good job Black Market.

This is as far as I've gotten. That number for Black Market does not account for: which of the two options is chosen from each split pile, which knight is chosen, druid at all, looters at all, shelters, or platinum and colony. Which means that number can only get even crazier. I haven't put the energy into figuring out those formulas; that is also the only one I've put through Wolfram Alpha, but I think we can agree it's pointless running any of the others if we're going to go for all possible Black Market decks.

[ghostofmars]

Wait, the integral thing: will that actually get me what I want? If I integrate f(x) from say, 1 to 300, will I get the same result as calculating f(1) + f(2) + f(3) + ... + f(300)? I barely remember AP calculus AB, so I'd love to hear from someone better at calculus than I am.

Euler–Maclaurin formula

[Gherald]

In other news, there are 10^80 atoms in the Universe

[GendoIkari]

In other news, there are 10^80 atoms in the Universe

I love how huge the universe is when compared to earth; yet how tiny it is compared to math.

[Doom_Shark]

For a Black Market deck of n cards, where 0<n<351, and neither the deck nor the kingdom includes Druid, Looters, or Young Witch, there are 360!/(n!(360-n)!)*(360-n)!/(9!(351-n)!) possible kingdoms. That simplifies to 360!/(9!n!(351-n!)). Getting the exact sum ∑_n=1³⁵⁰ 360!/(9!n!(351-n!)) in wolfram alpha expends more computation time than the free version allows, but that doesn't matter because it output it does give me is this:

5.4142756889123355558613982831116653273300579874856051... × 10^757

which is a number so insanely huge that all the other previous math is essentially pointless now. Good job Black Market.

So it turns out, when I typed out the simplified version of this equation (and subsequently copied it into Wolfram Alpha), I ended up putting the "!" on the wrong side of a set of parentheses, giving me "(351-n!)" rather than "(351-n)!" which means that the number is wrong. Summing up the corrected formula does not exceed the computation time limit, and so using scientific notation and rounding to two decimal places, we actually get 1.16 × 10¹²³. Still insanely huge, and still several times the estimated number of atoms in the Universe, just significantly less so.

[hhelibebcnofnena]

and still several times the estimated number of atoms in the Universe

I would say that 10^43 is quite a bit more than several. I might say it was several orders of magnitude (I assume that was what you meant).

[Doom_Shark]

and still several times the estimated number of atoms in the Universe

I would say that 10^43 is quite a bit more than several. I might say it was several orders of magnitude (I assume that was what you meant).

Yes, it is what I meant, thank you.