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1676

Puzzles and Challenges / Re: One card - big money

« on: January 04, 2012, 03:18:43 am »

6XTR->Village gives at best 7 actions net, with maximum TR chaining. Also, I wouldn't TR villages straight off, I'd TR TRs into TRs into TRs villages and whatnot.

Perhaps my notation was not clear. I wanted all 6 TR to be chained, so that you can double 6 villages, which nets you 11 actions (6x +2actions - 1 to play the first TR). In the solution I suggested, I played chain 10xTR into 10xKC which yields 9xTR actions and 20x KC actions.
Nevertheless I stand by my point that a golem into TR or villages is not more efficient than TR the villages directly. A golem gives you at most 1 extra action (or 3/5 actions if you TR/KC it), which is exactly the same as a village.

1677

Puzzles and Challenges / Re: One card - big money

« on: January 03, 2012, 08:08:21 am »

Please read post 13. You should be able to do much better than that.

I think KC golem into TR -> village is not a good idea:
KC -> golem -> 6x TR-> village nets 11 actions
6xTR -> village also nets 11 actions and you didn't use a KC and a golem

What I would do
TR-KC chain -> 9 (2x) and 20 (3x) plays
10 x KC-tribute = 120
10 x KC-university = 60 (and gain rest of villages)
9 x TR-city = 36
golem -> city, bazaar = 3
10x shanty town = 10
10x mining village = 10
10x nobles = 10
10x farming village = 10
10x fishing village = 10
10x worker's village = 10
village, festival, shanty town, walled village, trusty steed, inn, border village, hamlet, crossroads, tactician (last turn) = 11
the one action I have every turn is spent to start of the TR-KC chain
=> 290 actions

1678

Puzzles and Challenges / Re: 13 cards in hand

« on: January 03, 2012, 03:40:26 am »

In the last turn I played a Tactician.
My opponent played a Margrave in his turn, to which I replied by revealing 3 Horse Traders.
Then in my turn I have
3 cards in hand
+5 from Tactician
+2 from HT
+3 HT
= 13 (7 copper, 3 estate, 3HT)

edit: As Anon79 pointed out, my numbers didn't add up 

1679

Puzzles and Challenges / Re: Buying the last province

« on: December 11, 2011, 04:52:32 pm »

Just to be clear:
This is his hand at the start of his turn?
He has no duration cards in play?

Solution without possession:
Province + 10x embargo = -4
Fool's Gold > Gold
8x Fairgrounds -> -16
---
-20

1680

Puzzles and Challenges / Re: Buying the last province

« on: December 11, 2011, 04:41:01 pm »

Solution if he is possessed:
He buys the province +6
+7 Gardens
+8 Silk Road
+16 Fairgrounds
----
37

1681

Puzzles and Challenges / Re: The World's most precious Copper

« on: December 07, 2011, 06:02:41 am »

I think one can manage it this way:
11 Kingdom piles:
King's Court
Throne Room
Tribute
Coppersmith
Vault
Secret Chamber (as Bane to Young Witch)
Iron Works
Golem
Horse Trader
Scrying Pool
Black Market

This should raise my previous solution by 30 playing
Golem + Copper -> HT + SP
if we allow to give the copper to my opponent with an Ambassador(?). It might even work without, because we can get a lot of different card drawers from the BM.

As Davio mentioned Harvest could add one additional coin.

1682

Puzzles and Challenges / Re: The World's most precious Copper

« on: December 07, 2011, 05:38:27 am »

In that case, does HT count? It's similar to the above apart from you can't discard only the Copper.

If you have only copper in your hand, you could discard only it to HT.

Golem + copper in hand into HT and Lab

1683

Puzzles and Challenges / Re: The World's most precious Copper

« on: December 07, 2011, 05:33:22 am »

I think the highest number is achieved like that
gain the copper with IW: +1
pass it to opponent with masquerade
discard from his hand
TR + 10x KC gives 20 KC'ed actions
10xKC tribute +60
10xKC coppersmith +30
get it back with another masquerade
9x TR vault + 1x normal vault: +19
10x Secret Chamber +10
and finally play it +1
= 121

edit: replace venture (d'oh) with vault

1684

Puzzles and Challenges / Re: ABC

« on: November 29, 2011, 09:48:18 am »

I have 9 turns in the hard mode

open chancellor/talisman
discard deck after every turn
turn3: buy quarry, gain another
turn4: buy contraband
turn5: buy 2xhighway, gain 2 more
turn6: buy 2xhighway, gain 2 more
turn7: buy 2xgreat market, gain 2 more
turn8: buy/gain remaining GM, HW
turn9: play GM and HW, buy all colonies

1685

Simulation / Re: Dominion Deck Calculator

« on: November 28, 2011, 12:23:06 pm »

I think what Qvist is going for is the average value of a hand, probably also the average amount of cards drawn (which determines the cycling).

Assuming this I start with the no memory approximation, i.e., all cards start as completely random draw pile before drawing a new hand. The chances to draw a particular hand are

#possible permutations x product( #individual cards ) x 5! / ( total number of cards in the deck )!

Example for the 2 Smithy, 1 Village, 7 Victory/Treasure deck:

draw 4(Vi/Tr) cards and 1 Smithy:
5 possible permutations
* 2 (Smithy) * 7 * 6 * 5 * 4 (Vi/Tr)
* 5!/10! = 27.78%

draw 3(Vi/Tr) cards a Smithy and a Village:
20 possible permutations
* 2 (Smithy) * 1 (Village) * 7 * 6 * 5 (Vi/Tr)
* 5!/10! = 27.78%

here's the complete list (X = Vi/Tr, S = Smithy, V = Village)
A) X X X X X =  8.33%
B) S X X X X = 27.78%
C) V X X X X = 13.89%
D) S V X X X = 27.78%
E) S S X X X = 13.89%
F) S S V X X =  8.33%

Now I determine the average value
A trivial = 5*X
B simple: 4*X + 3*(3*X+2*0)/5 (village and smithy are dead cards)
D I will always draw all X -> 7*X
E simple: 3*X + 0 + 3*(4*X+1*0)/5
F simple: 7*X (it would be larger, but there are no more cards)

this leaves C - two cases:
1) 40% draw second smithy -> 7*X
2) 60% draw X -> 5*X

Now you just have to multiply the chances for a particular starting hand with the associated gain to predict the average hand of this deck:
[ 8.33% * 5 + 27.78% * 5.8 + 13.89% * (0.6*7 + 0.4*5) + 27.78% * 7 + 13.89% * 5.4 + 8.33% * 7 ] * X = 6.17 * X
So if all X were copper, I'd have in average 6.17 coins.

Note the importance of the 'no memory' approximation: If you want to go beyond, you'd have to keep track of which cards have been played already. E.g., the case A and C2 which both give 5*X this turn have a significant difference in the outcome for the next turn! An advanced scheme could expand the probability tree until the next reshuffle...

1686

Puzzles and Challenges / Re: ABC

« on: November 28, 2011, 05:24:45 am »

I think I managed 8 turns (nice puzzle btw)

Abbrevations (keeping the alphabetic order):
BO: Border Village
BR: Bridge
CA: Caravan
CH: Chancellor
CR: Council Room
HW: Highway
in addition to these actions I'll use contraband (CO)

The CH is used to discard the draw pile, if no actions are played afterwards

notation:
Turn X:
actions & treasured played
available coins (#buys) cost difference - buys and gains

open CA + CH

Turn 3:
CA, CH + 4x copper
6 (1) 0 - buy BO, gain BR

Turn 4: (draw 1 card from CA)
BO, BR, CH + 4x copper
7 (2) -1 - buy BO, gain BR, buy silver

Turn 5:
2xBO, 2xBR, CA, CH + copper + silver
7 (3) -2 - buy BO, gain BR, buy CO

Turn 6: (draw 1 card from CA)
3xBO, 3xBR, CH + CO + silver
10 (5) -3 - buy 3xBO, gain 3xCR, buy CA

Turn 7:
6xBO, 3xBR, 2xCA, CH, 3xCR + CO + silver + 7x copper
17 (8) -3 - buy BO, gain HW, buy 7xHW
As CH is not played this turn, the top cards of the draw pile are two estates

Turn 8 (draw 2 cards from CA):
7xBO, 3xBR, CH, 3xCR, 8xHW + CO + silver + 7x copper
17 (8) -11 - buy 8x colony


edit: change of abbrevation

1687

Simulation / Re: Dominion Deck Calculator

« on: November 16, 2011, 07:59:21 am »

I wrote a small AI once, which was based on a similar deck evaluation. I went in a slightly different direction, as I wanted to make buy choices before implementing cards. My goal was to write a big money bot which can compete with the BMU strategy without manually setting the cutoffs.
Perhaps a short description of the program may help you (or someone else) to develop a better AI. The concept was the following:
The deck quality is the number of victory points, which will be achieved until the end of the game, if I switch to full greening at a certain point.

To illustrate this idea a little bit:
Suppose it is early in the game and I draw 8 coins. Is it better to buy gold or a province? If I buy gold, my buying power will increase and I will get more provinces later on, however I have less victory points now, which I have to compensate for in later rounds. So if my current score + the points I will get on average when buying gold is larger than my current score + 6 + the points I will get on average with the province buy, I should prefer gold to province.

Details of the implementation:
As I limited my study to BM strategy, the average coin value of the deck is trivial to determine. To calculate the chance with which I can buy a province, I made the following assumptions:
1) consider always the full deck
2) the propability distribution is gaussian
3) the deck quality is constant after this buy
4) consider only province endings
All assumptions drastically simplify the calculations, though one could image more advanced schemes, which get rid of them. The variation of the distribution is simply determined under the 2nd assumption.
To decide, which card to buy, we compare all possible resulting decks (which are few for BM) and check their deck quality, which is given as the average points they will score till the end of the game.
The propability to draw x coins is
p(x) = 0.5 * erfc( (x - avg)/sqrt(2*sigma^2) )
so that the chances for province p_P = p(8), duchy p_D = p(5)-p(8), and estate p_E = p(2)-p(5) are easiliy evaluated. The game will end when all provinces are gone, so in average after
t = province left / sum over p(8) of all players
turns. My final score is my current score + t * ( 6*p_P + 3*p_D + p_E ). I chose the buy with the largest final score.

I've tested this bot against BMU and it loses 42:54, which is propably reasonable result considering all the approximations made.

1688

Simulation / Re: Help decide on a TR/KC priority list

« on: October 26, 2011, 08:10:43 am »

I have a question as to how the "wait" works in practice.
Let's say I have the following hand
KC,Lab,Witch
this could be played
KC -> Lab -> Witch
or
Lab -> KC -> Witch
which would the simulator choose?
I would usually choose the latter one, except for very action heavy decks.

Do I understand it correctly, that by using the "normal" action selection for the KC, the +2 actions would be chosen before the +1 action/+x cards. If this is the case, this is most likely not what you want. KC on Lab is much better than KC on Village.

I think you can also improve the Horn of Plenty treasurePriority, as the current version cannot gain any 2 cost card. Furthermore the order of Bank, HoP should probably depend on the buyPriority list. Only if playing the Bank earlier allows me to buy a better card, I should play it before the HoP. Otherwise, the other way is better, as it increases the value of the Bank.

1689

Puzzles and Challenges / Re: Big hands [lol]

« on: August 30, 2011, 12:11:50 pm »

I think, I've optimized the solution further
I play cards to give me 10 Possessions, 3 TR, 4 KC, 1 Outpost in hand, furthermore I have two remaining actions
TR+TR+TR+KC+KC+KC+KC => 2x 2xPossession + 8x 3xPossession = 28x Possession
+Outpost
In my Outpost turn I manage to draw again 10 Possessions, 3 TR, 4 KC and get another 28x Possession turns.

Now the second player has 56x possessed turn + 1x original turn. In each of these he plays
Village+TR+KC+KC+KC+KC+KC => 10x 3xCouncil Room
+Outpost
where he again plays 30x the Council Room.

This amounts to a total of 57x2x30=3420 cards


PS: I found a infinite card solution
(1)I play KC+Possession
(2)He plays Possession
back to (1) until infinite future Possession turns are accumulated. Then he plays the KC+CR+Outpost combination (see above) and I gain 60xinfinite cards

1690

Puzzles and Challenges / Re: Big hands [lol]

« on: August 30, 2011, 07:51:03 am »

Cards in the Kingdom
King's Court
Wharf
Laboratory
Torturer
Tactician
Golem
Horse Trader
Caravan
Throne Room
Black Market
Haven (bane card for Young Witch in Black Market)

I didn't manage to put the Council Room in

Your deck consists of
10x Tac, 10x Wha, 9x KC, 9x HT, 9x TR, 9x Car, 9x Hav, 9x Gol
Opponents deck consists of
cards which give +2 actions (from BM), 9xTor, 1x Council Room(from BM)

I play
TR - TR - KC - KC - Wha
and then more KC as long as I've cards and then a single Wha to draw more cards. When all KC are played I play all the remaining Wha. I still have 8x "play three times" left afterwards, which I use for Car.
Then I play the remaining TR on the branch of the 2nd TR which gives 8x double actions which I use for 1 Car and 7 Hav. The remaining 2 Hav are played as single actions.
Now I start playing Gol which draws into Lab, Tac. Playing the Tac first and then the Lab I play all Tac.

My opponent then plays 1 Council Room and 9 Torturer for 6 of which I set aside a Horse Trader and gain a Curse for all of them.

In the next round I get:
50 = 10x5 cards from the Tacticians
60 = 10x3x2 cards from KC+Wharf
24 = 8x3 cards from KC+Caravan
2 cards from TR+Caravan
14 = 7x2 cards from TR+Haven
2 cards from the Havens
12 = 6+6 cards from the Horse Traders
9 Curses
---
173 cards

1691

Puzzles and Challenges / Re: Cleanup on (a)Isle 4

« on: August 05, 2011, 03:15:44 am »

By my count, you've only cleared out 5 of the 7 copper there.  Nice, though.

You are right , but I think I can modify the 7th and 8th turn to compensate

Opening: Iron Works (IW), Chancellor
Turn 3: play IW -> gain Throne Room (TR), play Chancellor, flip deck, +2 copper  -> buy TR
Turn 4: TR TR IW(x2) -> gain Island(x2), Chancellor(x2), flip deck, +3 copper -> buy Kings Court (KC)
Turn 5: TR KC -> IW(x3) -> gain Island(x2) and Smithy,Chancellor(x3),flip deck,+1copper -> buy KC
Turn 6: TR TR KC KC -> Smithy(x3), Island(x3)[3xEstate], Island(3x)[3xcopper], IW(x3) -> gain Island(x3), Chancellor(x2), flip deck
Turn 7: TR, KC, KC, Smithy(x3), Island(x3)[3xcopper], Island(x3)[copper,Chancellor,TR], Island(x3)[IW]
Turn 8: KC, Island(x3)[Smithy,KC,TR]
Turn 9: Island[KC]

1692

Puzzles and Challenges / Re: Cleanup on (a)Isle 4

« on: August 04, 2011, 08:59:19 am »

It seems likely that the most efficient way to do this involves ironworks.

I found a solution in 9 turns based on this idea:
Opening: Iron Works (IW), Chancellor
Turn 3: play IW -> gain Throne Room (TR), play Chancellor, flip deck, +2 copper  -> buy TR
Turn 4: TR TR IW(x2) -> gain Island(x2), Chancellor(x2), flip deck, +3 copper -> buy Kings Court (KC)
Turn 5: TR KC -> IW(x3) -> gain Island(x2) and Smithy,Chancellor(x3),flip deck,+1copper -> buy KC
Turn 6: TR TR KC KC -> Smithy(x3), Island(x3)[3xEstate], Island(3x)[3xcopper], IW(x3) -> gain Island(x3), Chancellor(x2), flip deck
Turn 7: KC, KC, Smithy(x3), Island(x3)[2xcopper,IW], Island(x3)[Chancellor,2xTR]
Turn 8: KC, Island(x3)[Smithy,KC,Island]
Turn 9: Island[KC]

1693

Puzzles and Challenges / Re: Impossible score, version two

« on: August 01, 2011, 11:43:20 am »

I get 1515 in the hard mode as highest possible score:

+60 copper
 40 silver
 30 gold
 12 platinum
 16 potions
 5 curses
 14 estate
  8 duchy
  8 province
  8 colony
  8 garden
  8 GH
  8 island
  8 duke
  1 harem
  8 nobles
  8 vineyard
  8 fairgrounds
 10 Ambassador
 10 YW
 10 BM
 10 Bridges
122 further cards

---
420 cards, 172 action cards, 144 differently named cards
42*8(garden) + 57*8(vineyard) + 57*8(fairgrounds) + 8*11(duchy/duke) + 8*10(colony) + 8*6(province) + 8*(2+2+1)(island,GH,nobles) + 2(harem) + 14(estate) - 5(curses) = 1515
The opponent gets an Ambassador to put all his cards back in the supply. Then I play Masquerade to exchange his Ambassador with a Curse.

1694

Puzzles and Challenges / Re: Blooki's Puzzle #1 - Sad Saboteur

« on: July 29, 2011, 07:24:59 pm »

I am curious if you can get more with Black Market.  You banned it in the puzzle rules, but it doesn't change the puzzle that much.

You wouldn't use Black Market for the hero, but for the villain. If he played it before the saboteur he could add several Quarries and hence not trash any of your action cards. Then you can go for Goons and King's Court

I think it is possible to further optimize my solution.
At the moment I'm not really limited by coins, but by the number of buys, so I don't need the Apothecary. In addition, it might be better to replace one of the Throne Rooms of the villain with one of the hero's Worker Villages.
9xTR + 9x WV => +18 buys, +18 cards, +35 actions
8xScrying Pools => +8 cards
9xCoppersmith => -9 actions
8xTournament => +8 cards, +8$
7xGreat Hall => +7 cards
Followers => +2 cards, -1 action, +1 estate
Trusty Steed => +2 cards, +4x silver, -1 action
Bag of Gold => +1 gold
Diadem => 2$ + 25 unused actions = 27$
If I have not miscounted, I could still have 45 copper and 1 potion in the hand, adding up to sufficient money to buy everything I have buys for. The spot of the Apothecary would go to the Island.

19 buys = 8 colony, 6 province, 1 island, 1 great hall, 1 garden, 1 vineyard, 1 action card

Total cards: 230
Total actions: 81
Garden      8 x 23 = 184
Vineyard    8 x 27 = 216
Colony      8 x 10 =  80
Province    6 x  6 =  36
Island      8 x  2 =  16
Estate     11 x  1 =  11
Great Hall  8 x  1 =   8
------------------------
                     551

PS: I just realized, I could do even better. If I replace the Great Hall with Pawn (+1 card/+1 buy) to get more buys. I think I could buy all Duchies then. I haven't calculated it precisely, but I think the points would increase by 18, 21, or 24, depending on how much action cards have to be bought to satisfy the vineyards.

1695

Puzzles and Challenges / Re: Blooki's Puzzle #1 - Sad Saboteur

« on: July 29, 2011, 05:56:46 am »

I've achieved 526vp. I hope I've not missed anything.
The deck of the villain contains at least:
1 Province + 1 Saboteur + 1 Tournament + 1 Gold + 7 Copper + 3 Estate + 2 Throne Rooms + 1 Princess

In the kingdom the following cards are present (in brackets the number the available/ the number the hero has before the last turn)
Colony (8/0)
Province (6/0)
Duchy (8/0)
Estate (1/10)
Copper (1/52)
Silver (1/39)
Gold (29/0)
Potion (1/15)
Tournament (1/8)
Scrying Pool (1/9)
Throne Room (0/8)
Worker Village (0/10)
Coppersmith (1/9)
Apothecary (1/9)
Saboteur (1/8)
Garden (1/7)
Vineyard (1/7)
Young Witch (1/9)
+ bane card: Great Hall (1/7)
+ prizes (0/4)

After the villain successfully trashed the province, the hero has the following cards in his hand:
4x Estate + 1x Scrying Pool

He plays Scrying Pool and draws all of his Action cards in his hand.
Playing 8x Throne Room, he doubles the effect of 8 of his Worker Villages, the other two worker villages are played without a Throne Room
=> total 19 buys and a lot of actions (34) and several cards (18)
He proceeds with the 9x coppersmiths making all copper worth 10. Playing the Apothecaries he draws enough coppers and potions to buy everything he want:
8x colony, 6x province, 1x garden, 1x vineyard, 3x duchy
In total he has 230 cards making each garden worth 23, 77 action cards so that each vineyard is worth 25 points.

Garden      8 x 23 = 184
Vineyard    8 x 25 = 200
Colony      8 x 10 =  80
Province    6 x  6 =  36
Duchy       3 x  3 =   9
Estate     10 x  1 =  10
Great Hall  7 x  1 =   7
------------------------
                     526

1696

Puzzles and Challenges / Re: Who has which cards?

« on: July 28, 2011, 03:35:54 am »

If the assumption I made in the previous post is incorrect, I submit the following, based on ghostofmars' partial solution:


5. Berta plays Tactician on her previous turn, drawing Throne Room, Ambassador, 4 Copper, and 4 Gold.
 - She plays Throne Room on her Ambassador, returning 4 copper to the supply; Adam and Charlie are protected by Moat/Lighthouse.
 - She buys 2 gold, gaining 2 curses from a previously played Embargo.

Prizes are kingdom cards.

Your solution has does not satisfy

5.   Another remarkable turn is executed by Berta. At the start of the clean-up phase of that turn, there are only 3 kingdoms cards, all of which are different, and no basic cards in her play area. [...]

I think, I found a loophole in the puzzle:
If B played a Quarry in turn 3, the prize of Black Market and Masquerade drops to 1, so that we can produce any possible solution, as long as there is only 1 action card which costs 2,5,6,8 in the beginning. The rest of the necessary cards are gained by Black Market and then using Masquerade to pass the card to the appropriate player.

1697

Puzzles and Challenges / Re: Who has which cards?

« on: July 25, 2011, 09:34:27 am »

@mischiefmaker
I think your solution doesn't fulfill

5.   [...]However, at the end of his turn, the total number of cards (not piles) in the supply is still the same as in the beginning of her turn.

My partial solution for hint (5)
B has Tactician in play. The 10 Cards in her hand include Throne Room, Ambassador, 4x copper and cards worth 10 coins. She plays the Throne Room and the Ambassador to return the 4 copper to the supply. The other players are protected by a Moat(A) and a Lighthouse(C). She uses the 10 coins to buy 2x Mint and trashes her money cards. As C played an Embargo on the Mint, B gains two curses.

Already 7 cards are fixed, now I use mischiefmaker's Bureaucrat-Young Witch trick to resolve hint (4)
C played Native Village and 2x Bureaucrat. A puts two victory cards back on his empty deck. Playing the Young Witch he draws these two cards and discards them. Native Village handles the third hint, too.
The rest of the hints reveal that Peddler is one of the remaining cards which was bought by C, which fixes all cards.

However, I do not know how to interpret hint 9+10. If A and C have only one victory card left in their deck, then, they would need some trashing e.g. by Bishop. If they bought exactly one additional victory card (to the 3 estates they have), the gap between Estate, Duchy and Province is not one which could be overcome by addition of a curse. In either case an additional card would be necessary which doesn't fit in the current solution.

1698

Puzzles and Challenges / Re: [Math] How many games and years?

« on: July 18, 2011, 11:36:04 am »

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.