James Hemsley
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Hi,

I just played the game for the first time last night. It was fun and close.

One aspect of the rules stuck out at me and I was wondering if anyone else had noticed this or if I'm playing incorrectly.

Toward the end of the game it becomes evident that for every corruption card in your hand at the end, you will gain one corruption. Ideally you will try and get rid of them. It seemed to me then, to make sense to take the largest market, pushing your hand above 10 and paying the one corruption, to get rid of multiple corruption cards.

E.g I have 6 cards in my hand. 4 of them are corruption cards. I pick up a market with 7 cards. I have to get rid of 3 (or 2) cards. Hopefully there is less corruption in the pile I picked up, but then I could get rid of 2 corruption cards for the cost of one corruption.

The above makes even more sense if you already have 10 cards. You could pick a pile of 2 or 3, especially if you know that some of them are not corrupt and effectively trade out your corruption cards, so you're not penalized at the end.

Do I understand this correctly?

--James
 
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Adam Smiles
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You are correct. That is a viable way to get rid of unwanted corruption in the end game. Assuming of course that all of the markets are free of tainted cards.

However, I think that a much better strategy would be to not be in a position that the extra corruption matter that much. Then you can use your 6-8 cards in the last hand and build something for more victory points.
 
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Jeff Wood
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Just a note, in your example you got 13 cards in hand (6 + 7 from the market).

You will need to discard 3 for the one corruption to get down to 10 cards, with no option to stop at 11 (much to my own annoyance sometimes. )
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James Hemsley
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Hi Jeff,

I was assuming that you could keep 11 in your hand, because you pay the 1 corruption to get rid of cards over 10, but you also pay 1 corruption for each card over ten. 11 cards in your hand is the sweetspot, since you pay 1 corruption either way.

In my example though, I should have discarded more to get down to 10! That's the whole point!

--James
 
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Barry Figgins
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Iceberg1 wrote:
Hi Jeff,

I was assuming that you could keep 11 in your hand, because you pay the 1 corruption to get rid of cards over 10, but you also pay 1 corruption for each card over ten. 11 cards in your hand is the sweetspot, since you pay 1 corruption either way.

In my example though, I should have discarded more to get down to 10! That's the whole point!

--James


I believe it's all or nothing - you either keep 10 and pay 1, or keep them all and pay 1 corruption each. You can't choose how many to keep.
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Matt Smith
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Barry is correct.

11 cards in hand is considered a sweep spot because, if you can draw a market stall such that you end up with 11 cards, you can keep all 11 and only take 1 corruption. If you were to discard 1 card to get down to 10, you'd still have to take 1 corruption, so it makes sense to keep the 11th card.
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James Hemsley
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So, if I kept more than 11, say 15, I'd take 5 corruption? I understand that. But, I have to discard down to 11, right? Then I only pay 1 corruption; same as if I had 10.

Are you saying you can't discard when you have more than 10? If so, I think I played incorrectly.
 
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Richard Pardoe
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Iceberg1 wrote:
But, I have to discard down to 11, right?

Wrong, if you discard, you must discard down to 10.

So with 15 cards, your options are:

Keep 15 - Pay 1 corruption per extra card (Total of 5)

(or)

Discard down to 10 - Pay 1


Edit - Answering your second question.

You can discard down to 10. The point being made above is that in the special case of 11 cards, your options are:

Keep 11 - Pay 1 corruption per extra card (Total of 1)
(or)
Discard down to 10 - Pay 1

Since the cost is the same, it makes no sense to discard in the special case of holding 11 cards in hand. But for any other number greater than 11, it costs more to hold the cards then to discard. And the player needs to decide which option (keep/discard) makes more sense.

And if you do discard, it is down to 10 and exactly 10.
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    Getting back to your original question, yes, tanking up at the end of the game and paying one corruption to dump a handful of corrupt cards is an excellent strategy. Naturally, it depends on where your hand stands at the end of the game.

             Sag.
 
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Alan Kwan
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11 cards is not really a sweet spot, because you get 1 corruption for keeping 1 extra card, which is the same rate as any higher number of cards. 10 cards is the true sweet spot.

Because of the resource mix required to build stuff, you are often better off keeping the best 10 cards out of 14, than just keeping 11 out of 11. When you need to draw over 10, you usually have some resource in excess which you can afford to discard. So, take a small stall and stay under 10, or take a large stall and discard down, rather than getting to 11, if you can.
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James Hemsley
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I understand now. Thank you for all of your replies!

--James
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