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Author Topic: How many kingdoms?  (Read 743 times)

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NoMoreFun

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How many kingdoms?
« on: January 17, 2017, 12:56:20 am »
0

What's the up to date "how many different games" statistic?

Rules:
* Normal Kingdom rules (10 cards+other set up rules)
* 0-2 Events and Landmarks
* Different Obelisks matter (IMO it's game warping enough what the Obelisk is)
* Platiunum and Colony matter and have the setup rules from Prosperity
* Shelters matter (with rules from Dark Ages)
* Don't include cards removed from Base and Intrigue
* Sizes of victory card piles don't matter (including castles)
* The order of Knights doesn't matter (not sure 100% how to rule in this one)
* The contents of the Ruins pile don't matter (inconsistent with # of players + somewhat irrelevant)
* The contents of the Black Market pile don't matter (would include except rules aren't consistent and it would dramatically increase the number)
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Haddock

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Re: How many kingdoms?
« Reply #1 on: January 17, 2017, 05:35:39 am »
+1

Oh so very very many.
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M Town Wins-Losses (6-2, 75%): 71, 72, 76, 81, 83, 87 - 79, 82.  M Scum Wins-Losses (2-1, 67%): 80, 101 - 70.
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GendoIkari

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Re: How many kingdoms?
« Reply #2 on: January 17, 2017, 02:52:51 pm »
0

It's not that hard of a calculation, with the exception of following the Colony / Shelters rule properly.

C = total number of kingdom card piles
T = total number of kingdom card piles costing or less
E = total number of events + landmarks

Non-Young Witch games:
((C-1) Choose 10) +
((C-1) Choose 10) * (E) +
((C-1) Choose 10) * (E Choose 2)

+

Young Witch games:
(C-1) Choose 9) * T +
(C-1) Choose 9) * T * (E) +
(C-1) Choose 9) * T * (E Choose 2)

An upper bound would be to multiply it by 4 for the Shelters / Colonies. Also, my calculation isn't accounting for the fact that there's fewer bane choices when the Kingdom already has cards that cost or less.

*Edit* I suppose you need to calculate Kingdoms that contain Obelisk separately; those ones are just multiplied by 10.
« Last Edit: January 17, 2017, 02:57:51 pm by GendoIkari »
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Dingan

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Re: How many kingdoms?
« Reply #3 on: January 17, 2017, 03:03:29 pm »
0

http://forum.dominionstrategy.com/index.php?topic=13647

It's not that hard of a calculation
I respectfully disagree.

Lower bound is (number of randomizers) choose 10 = 263 choose 10 = 3.7 * 10^17

which doesn't include Events, Landmarks, Colonies, Shelters, first edition cards, etc.
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GendoIkari

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Re: How many kingdoms?
« Reply #4 on: January 17, 2017, 03:23:02 pm »
0

http://forum.dominionstrategy.com/index.php?topic=13647

It's not that hard of a calculation
I respectfully disagree.

Lower bound is (number of randomizers) choose 10 = 263 choose 10 = 3.7 * 10^17

which doesn't include Events, Landmarks, Colonies, Shelters, first edition cards, etc.

Well adding in the Events/Landmarks is simple enough. It's just determining when to multiply by 2 for Colonies that's not clear. And also my calculation doesn't account for having Young Witch in the Black Market. So yeah, there's various tricky things. Actually, if we're saying that the contents of the Black Market deck aren't considered, then we have to decide if we care about Young Witch in Black Market, or just assume that it's never there.
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faust

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Re: How many kingdoms?
« Reply #5 on: January 18, 2017, 04:38:32 am »
+1

Calculation should still be managable.

Da3- = Number of Dark Ages card randomizers costing 3 or less (10)
Pr3- = Number of Prosperity card randomizers costing 3 or less (3)
Dao = Number of other Dark Ages card randomizers (25)
Pro = Number of other Prosperity card randomizers (22)
O3- = Other card randomizers costing 3 or less (67)
T = total number of card randomizers (264)
E = total number of Events + Landmarks without Obelisk (54)

Derived value:
Oo = card randomizers not in the above categories that are not Young Witch (136)

Let's start with some easy ones:

All Dark Ages cards (always use Shelters), no Obelisk:
((Da3- + Dao) choose 10) * (1 + E + (E choose 2)) = 2.73*10^11

All Dark Ages cards (always use Shelters) with Obelisk:
10 * ((Da3- + Dao) choose 10) * (1 + E) = 1.01*10^11

All Prosperity cards (always use Platinum/Colony), no Obelisk:
((Pr3- + Pro) choose 10) * (1 + E + (E choose 2)) = 4.89*10^10

All Prosperity cards (always use Platinum/Colony) with Obelisk:
10 * ((Pr3- + Pro) choose 10) * (1 + E) = 1.80*10^9

Stopping here for now.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

Mic Qsenoch

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Re: How many kingdoms?
« Reply #6 on: January 18, 2017, 11:23:44 am »
+2

Obelisk multiplies by the number of Action supply piles, not 10.
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faust

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Re: How many kingdoms?
« Reply #7 on: January 18, 2017, 01:17:28 pm »
+2

Obelisk multiplies by the number of Action supply piles, not 10.
Ah, right. Action supply piles are bad. So need to split the above thing in Action/non-Action + Looters.
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Since the number of points is within a constant factor of the number of city quarters, in the long run we can get (4 - ε) ↑↑ n points in n turns for any ε > 0.

GendoIkari

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Re: How many kingdoms?
« Reply #8 on: January 18, 2017, 05:06:17 pm »
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Obelisk multiplies by the number of Action supply piles, not 10.

Oh wow... yeah.. caring about Obelisk makes the calculation WAY more complex. Group the possible Obelisk games based on how many Kingdom piles are action piles?
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