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[Davio]

This is more of a math question than a Dominion puzzle.

With regard to every Dominion card in current existence* (not your own custom ones obviously) and complying as much as possible with the rules for choosing cards for the setup (as mentioned in the original rulebooks or addenda), how many different setups for Dominion are there?

And, given these facts:
 - You and your nemesis are both immortal
 - Every game of Dominion lasts exactly 30 minutes
 - You both need no breaks for sleeping or eating

How many years does it take to have played each different setup? (Rounding to the nearest year is fine)

*Cards in existence means cards from: Dominion Base, Intrigue, Alchemy, Seaside, Prosperity, Cornucopia and the Promo Cards.


To get you started: With just the Base set, there are 25 different cards, offering 3,268,760 different setups, or 186 years of uninterrupted Dominion madness!

[WanderingWinder]

I can do it for you relatively quickly if you've got a count of how many kingdom cards there are, and if you assume that for Young Witch, different bane cards don't count as different 'setups' (which is a horrible assumption) and different black market decks don't count as different setups (again a really bad assumption). Those two things would be really difficult to account for, but without them, it shouldn't be much problem.

[Davio]

Well, I wanted to make it challenging, so Black Market and Bane has to be taken into consideration, otherwise it's just a simple X Choose Y.

Also, I am going for completeness.

With regards to Black Market: Assume every possible kingdom card which is not in the regular supply will be in the BM deck.
This has serious consequences, I know, because for every BM-supply (easily calculated), there needs to be an extra Bane Pile, unless Young Witch is in the regular supply, then the Bane Pile wouldn't be "extra".

In retrospect, I think this would be better as a team effort than an actual puzzle.

Consider the Prosperity rule too: With only Prosperity cards, Colony and Platinum are included. With a mixed set, if there is at least 1 Prosperity card, there are 2 setups: 1 with and 1 without Col/Plat.

All in all I counted 134 different kingdom cards from Adventurer to Wharf.
There are 44 Cost $2 or $3 (bane piles)

I can start with:

There are 3.268.760 Prosperity only setups, these are the simplest.

[ghostofmars]

I considered first the total number of possibilities excluding the young witch. Additionally, I considered only one setup for the black market, not the one implemented in isotropic.
I arrive at this equation

                  10-a
   -----         -----
    \    / A \    \     / P \             / N-A-P-1 \
     )  (     )    )   (     )( 2 - d   )(           ) ,
    /    \ a /    /     \ p /        p0   \  10-a-p /
   -----         -----
 a=0,3,4,5       p = 0


where A(a) and P(p) are the total(actually used) number of Alchemy and Prosperity cards, respectively. N is the total number of cards. d_p0 is a Kronecker delta, which is 1 if p = 0.

Explanation for the individual factors:
The last one is simple: After drawing Alchemy and Prosperity cards, there are still N-A-P cards left (-1 for YW) of which (10-a-p) are drawn.
The (2 - d_p0) factor doubles the number of setups if at least one Prosperity card is included. The (^P_p) factor accounts for the number of Prosperity cards draw.
The first term includes the recommendation of the rules to use no or 3 to 5 Alchemy cards.

I think, for the YW (/wo black market interaction) additionally these terms would arise

                      9-a
       -----         -----
        \    / A \    \     / P \             / N-A-P-2 \
T *      )  (     )    )   (     )( 2 - d   )(           ) ,
        /    \ a /    /     \ p /        p0   \   9-a-p /
       -----         -----
     a=0,3,4,5       p = 0


where T is the total number of two or three cost cards.
The formula has only slight differences to the one above:
The prefactor is trivial and accounts for the different bane cards. As I've selected already the YW only 9 further cards remain to be chosen. The upper expression in the last factor changed, as two cards (bane and YW) are removed.

Actually this equation holds only if the bane card can be any 2/3$ card. If one would include the Alchemy restrictions for the bane card the term would be more complicated.

Perhaps someone can calculate the actual numbers and work out the interactions YW with black market.

[rinkworks]

All in all I counted 134 different kingdom cards from Adventurer to Wharf.

So 140 with Wishing Well, Witch, Woodcutter, Worker's Village, Workshop, and Young Witch, then?  ;-)

More seriously, the enormousness of these numbers is what initially intrigued me about Dominion.  Even with just the base set, you could pick a random kingdom and be reasonably confident that no one in the world had ever played it before, nor ever would again.  Actually that might not be strictly true, but certainly it is if you add in even just one of the mini-expansions.

[Davio]

Wow, ghostofmars, you have done some great work.

I had originally posted this as a sort of wondering myself, without even having tried it, but wanting to try it eventually.

I think you can calculate it better than we can, ghostofmars, you just need the numbers.

There are:
- 12 Alchemy cards
- 25 Prosperity cards
- 44 Possible bane cards

[guided]

Well, I'm ready to roll up my sleeves here, but somebody please stop me if I've got this wrong: I see 130 Kingdom cards.

25 base
25 Intrigue
26 Seaside
12 Alchemy
25 Prosperity
13 Cornucopia
4 promo

...and only 34 possible Bane cards. Not sure where the number 44 came from.

The number of Alchemy cards isn't relevant since there is no actual rule regarding the number of Alchemy cards that may be used, only a suggestion in the rulebook.

Assuming all cards outside the Kingdom are always in the Black Market deck does make things a bit simpler, though of course Young Witch complicates things a bit.

[guided]

OK, here's attempt #2 having deleted my previous post, now accounting for choosing which card on the 11-card board is the Bane card. It got to the point where I didn't want to be doing these calculations by hand, so I installed Mathematica

Code: [Select]

In[14]:= nYnBnP = Binomial[103, 10]

Out[14]= 23591276125340

In[15]:= nYnBP = 2*(Binomial[128, 10] - nYnBnP)

Out[15]= 406509756110920

In[7]:= YnP = Sum[n*Binomial[31, n]*Binomial[73, (10 - n)], {n, 1, 10}]

Out[7]= 77801017009100

In[9]:= YPt0 = Sum[n*Binomial[34, n]*Binomial[95, 10 - n], {n, 1, 10}]

Out[9]= 648131869088000

In[10]:= YP = 2*(YPt0 - YnP) - 3*Binomial[104, 9]

Out[10]= 1132419287415240

In[11]:= nYBnP =
 Sum[(n + 1)*Binomial[30, n]*Binomial[73, 10 - n], {n, 0, 10}]

Out[11]= 92303730859340

In[12]:= nYBPt0 =
 Sum[(n + 1)*Binomial[33, n]*Binomial[95, 10 - n], {n, 0, 10}]

Out[12]= 811683895428175

In[13]:= nYBP = 2*(nYBPt0 - nYBnP) - 3*Binomial[103, 9]

Out[13]= 1431231198459370

In[18]:= Kingdoms = nYnBnP + nYnBP + YnP + YP + nYBnP + nYBP

Out[18]= 3163856265979310

In[35]:= Years = AccountingForm[Kingdoms/365.25/48, 13]

Out[35]//AccountingForm= 180461799337.2

It's probably appropriate to use the astronomical Julian year since after 180 billion of them let me tell you that our present calendar years will no longer have any significance in the universe.


Annotation:

nYnBnP: no Young Witch, no Black Market, no Prosperity. 103 possible cards, choose 10 of them, and you're done.

nYnBP: Now we include Prosperity cards. Choose 10 out of 128, exclude Kingdoms with no Prosperity cards (already counted above as nYnBnP), and multiply by 2 since we can choose Colony or no-Colony for each set of 10.

YnP: Young Witch but no Prosperity cards. This gets tricky. There are 31 possible Bane cards and 73 other cards. In addition to Young Witch, we must choose 10 other cards, including "N" possible Bane cards (for N>=1, since we need at least 1 Bane card). For each number N, we choose N out of 31 Bane-eligible cards, (10-N) out of 73 other cards, and then multiply by N to pick which card is the Bane card. Then we sum everything up for values of N from 1 to 10.

YP: Young Witch and Prosperity cards. The toughest case. There are 34 possible Bane cards and 95 other cards, giving us (N * 34cN * 95c(10-N)) possible Kingdoms for each number N from 1 to 10, so we sum these up to get YPt0 (a temporary result). Now we need to subtract out Kingdoms with no Prosperity cards that we already counted (calculated above as YnP), then multiply by 2 to account for the choice of whether to include Colony. But we're not quite done, since we've counted some illegal Kingdoms where the only Prosperity card is the Bane card but we put in Colony anyway: There are 3 Bane-eligible Prosperity cards, and we have Young Witch, and we have 9 other cards chosen from the other 104 non-Prosperity cards, 3 * 104c9.

nYBnP: Black Market, but no Young Witch or Prosperity. There are 30 possible Bane cards (other than Black Market) and 73 other cards. We must pick 10 cards other than Black Market to account for the Bane card (since Young Witch is in the BM deck), and then we can either choose Black Market or one of the other 10 cards as the Bane card. For each number N then there are (N+1) * 30cN * 73c(10-N) Kingdoms, and N can range from 0 to 10. (0 is legal since we can still choose BM as the Bane.)

nYBP: Black Market and Prosperity, but no Young Witch. 33 possible Bane cards (other than Black Market), 95 other cards. So we start with (N+1) * 33cN * 95c(10-N) Kingdoms for each number N from 0 to 10, storing this temporary result as nYBPt0. Then we subtract out any Kingdoms with no Prosperity cards (already calculated as nYBnP) and multiply by 2 to choose whether we have Colonies or not. Finally we subtract out the remaining illegal Colony Kingdoms where the only Prosperity card is the Bane card: Black Market, 3 possible Bane cards, and 9 other cards chosen from the 103 non-Prosperity cards other than Black Market and Young Witch: 3 * 103c9

[Kirian]

It's probably appropriate to use the astronomical Julian year since after 180 billion of them let me tell you that our present calendar years will no longer have any significance in the universe.

New Card:

Heat Death
Action-Attack-Permanent, $8

Trash your hand, your deck, your discard pile, and all cards in play other than this one.  All of your opponents trash their hands, decks, and discard piles.
Take the Curse pile and deal any remaining Curses equally to all players, starting with yourself.
All card costs are increased by $10 for the remainder of the game.

----

In a thread about 180 GY of Dominion games, I think this fits...

[theory]

To make that number more graspable, assuming you could play 1 Dominion game every second, it would take you over 100 million (100,258,696) years to play them all.

[Davio]

This is what keeps me coming back to this game, there are essentially 2 parts to Dominion's success:

- The sheer number of different setups (although somewhat reduced by dominant cards and strategies)
- The short length  of a game, usually around 30mins in real life and 15mins on Isotropic.

Even though we can't try all setups in our lives, games are short enough for us to try a lot of different setups and setups are unique enough to offer something slightly new every game!

Thanks all for joining in this topic and doing the hard work. 

Just a silly note: A year in my book is 365.24 days, not 365.25, but this may change as we go into the millions of years and Earth's speed is changed.

It's a mind boggling number nonetheless.

Fun fact: If you pair up every Chinese with another, it would take them only 225 years to play every setup. Since Chinese people tend to age very well, this is not impossible.

[guided]

Just a silly note: A year in my book is 365.24 days, not 365.25, but this may change as we go into the millions of years and Earth's speed is changed.

Like I said, Julian year: http://en.wikipedia.org/wiki/Julian_year_%28astronomy%29

There will be no Earth to count out solar years within 3 or 4% of our 180 billion year gaming session, I'm afraid!

[DStu]

There will be no Earth to count out solar years within 3 or 4% of our 180 billion year gaming session, I'm afraid!

Probably one should not even extrapolate our current understanding of "time" to that scales. At least backwards it won't work.

[Davio]

Well, the way I see it, "time" becomes more complex, the better we try to understand it, so it's best to steer clear of it.

Even though, if you are playing 50 games of Dominion a day without eating or sleeping (or peeing!), you might start to wonder if you're not wasting your time...

Still, it is best played in real life, I found out yesterday.
I played 3 games with my mom and dad and even though I have played hundreds of games more than they have (was my mom's first time), my dad beat me the 3rd time!

I actually felt happy for him and it was a deserved win, but it goes to show that even after so many games, some setups can still surprise you.

The last setup was Intrigue only: Masquerade, Wishing Well, Saboteur, Ironworks, Scout, Tribute, Courtyard, Bridge, Shanty Town and Conspirator.

I tried some Ironworks shenanigans with Wishing Well and Conspirator, but after my Masquerade got Sabotaged, I couldn't trash more cards for Conspirator. It also showed I'm not that used to 3p anymore, because getting Sabotaged twice as much as with 2p proved it was hard to get anything decent going and it was basically a scramble for everyone.

[guided]

Yeah, on the rare occasions I play 3p & 4p games anymore, I do lose to novices from time to time. All my heuristics are optimized for 2p, so my instincts sometimes fail me in multiplayer.


Though I had an amazing recent game with 2 novices, one other very good player, and myself:

We all opened Pirate Ship, both novices got lucky, and the other two of us got super unlucky. My Ship got discarded 6 shuffles in a row, leaving me with $1 on my mat by the time almost all of the coppers were gone. The novices were at $4 and $5, and the other good player was stuck at $2.

In the meantime I focused on Native Villages, eventually saving 3 of my Coppers on my mat and emptying my draw pile so that I finally saw my Ship in hand for literally the 2nd time in the game. The novices had started thinking about Colonies, buying up treasure again. I played my Ship and was lucky enough to hit a treasure and get up to $2. Next turn, I picked up my Coppers off the mat, played my Ship for $2, and bought a Mountebank. Nobody else had bothered to buy one! hehe...

Native Villages and Pirate Ships were empty, but the other good player was in no position to end the game, and the novices were too focused on Provinces. Eventually I got several Throne Rooms (playing TR/Mountebank most turns) and 8 fully powered Cities, having absolutely destroyed everyone's deck with Mountebank. I ended up with most of the Colony pile for a resounding victory. Game-end condition? I finally succeeded in emptying the entire Curse pile with Mountebank.

It may have been enabled only by very poor play, but it's probably my favorite ever come-from-behind win

[KMueller]

Game-end condition? I finally succeeded in emptying the entire Curse pile with Mountebank.

It may have been enabled only by very poor play, but it's probably my favorite ever come-from-behind win

Another Game-end condition: Two new players thoroughly hated Montebanks. I have introduced a number of friends to this great game, but try to avoid cursing them at first. Maybe I am just too nice...

[guided]

A pair of gamers I love playing other games with refuses to play Dominion with me anymore. I built a Village/Witch/Wharf engine last time.

[Buttons]

YnP: Young Witch but no Prosperity cards. This gets tricky. There are 31 possible Bane cards and 73 other cards. In addition to Young Witch, we must choose 10 other cards, including "N" possible Bane cards (for N>=1, since we need at least 1 Bane card). For each number N, we choose N out of 31 Bane-eligible cards, (10-N) out of 73 other cards, and then multiply by N to pick which card is the Bane card. Then we sum everything up for values of N from 1 to 10.

For YnP, I think you're making this harder than it needs to be. Pick Young Witch as your first card, then pick the Bane card (for which there are 31 possibilities), and finally choose 9 out of the other 103 remaining cards, for a total of 31 * (103 choose 9) options.

Likewise for YP you've got 34 * (128 choose 9) ways to pick a set with Young Witch required and Prosperity allowed. Subtract 31 * (103 choose 9) to remove the ones without Prosperity cards, then double to account for the choice of platinum and colony.

EDIT: And if Platinum/Colony can't occur when the only Prosperity card is the Bane (I'm a little fuzzy on the rules there myself), then you have to be a little more careful, separating the YnP cases into 31*(103 choose 9) and 3*(104 choose 9).

[Deadlock39]

Another (I think) minor error in the calculation. 
If all 10 cards are from Prosperity, there should only be one option for that set (with Colony and Platinum).   

[Buttons]

Another (I think) minor error in the calculation. 
If all 10 cards are from Prosperity, there should only be one option for that set (with Colony and Platinum).   

Good point. Fortunately it doesn't affect the Young Witch, Black Market, or Alchemy calculations, so we just have to subtract (25 choose 10) at the end.

[guided]

Correct, that's an easier way to deal with the Young Witch cases, and you need to subtract out all-Prosperity boards without Colony/Platinum from the final result.