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 A
10 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 = A
10-B
10Shelters = A
10-C
10Both = A
10-B
10-(C
10-D
10), or A
10-B
10-C
10+D
10Add all of these to our initial A
10 and combine like terms: 4A
10-2B
10-2C
10+D
10, or 2(2A
10-B
10-C
10)+D
10Now, 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(2A
9-B
9-C
9)+D
9), or 880(2A
9-B
9-C
9)+440D
9Adding 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 5
10 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 5
10, so until I figure out how to ignore the order of the pile, 5
10 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*5
10(2A
9-B
9)
Two Looters = 6*5
10(2A
8-B
8)
Three Looters = 2*5
10(2A
7-B
7)
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*5
10(2A
8-B
8)
Two Looters = 2,640*5
10(2A
7-B
7)
Three Looters = 880*5
10(2A
6-B
6)
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 = A
10-E
10Young Witch and Platinum/Colony = A
10-B
10-(E
10-F
10), or A
10-B
10-E
10+F
10Young Witch and Shelters = A
10-C
10-(E
10-G
10), or A
10-C
10-E
10+G
10Young Witch and Both = A
10-B
10-(C
10-D
10)-(E
10-H
10), or A
10-B
10-C
10+D
10-E
10+H
10Add them up and combine like terms: 4A
10-2B
10-2C
10+D
10-4E
10+F
10+G
10+H
10, or 4(A
10-E
10)-2(B
10+C
10)+D
10+F
10+G
10+H
10Now 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*5
10(A
9-B
9-E
9+F
9)
Two Looters = 6*5
10(A
8-B
8-E
8+F
8)
Three Looters = 2*5
10(A
7-B
7-E
7+F
7)
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(2A
9-B
9-C
9)+2D
9Druid = 880(2A
8-B
8-C
8)+440D
8One Looter = 12*5
10(2A
8-B
8)
Two Looters = 12*5
10(2A
7-B
7)
Three Looters = 4*5
10(2A
6-B
6)
Druid and one Looter = 2,640*5
10(2A
7-B
7)
Druid and two Looters = 2,640*5
10(2A
6-B
6)
Druid and three Looters = 880*5
10(2A
5-B
5)
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.
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Topic: How many possible kingdoms are there? (Read 9204 times)