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Subject: Winning conditions and scoring rss

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Hank Panethiere
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Just making sure I'm correct.

The game round ends if there are four order cards in the pool or if someone reveals or declares order. These are the only end of round winning conditions, correct?

Is scoring cumulative? In the rules it says you get 3 points for having three order cards in your hand. It also says you get five points for declaring order and four for revealing. Doesn't this fulfill two scoring conditions (3 order cards and declaring or revealing order?) thus scoring 8 points (3 + 5) or 7 points (3 + 4)?

When would you have 3 order cards in your hand and have the round end and score only 3 points?

Thanks for any clarifications.
 
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Eric Engstrom
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The round ends if:

A) All 13 spectrum sets are on the table
B) All 4 order cards are in the pool (the surrender of order)
C) Player declares order on their turn (3 order cards and no others in hand)
D) Player reveals order off their turn (3 order cards and no others in hand)
E) One player has no cards in their hand

I'm pretty sure scoring is not cumulative. If you Declare order on your turn (3 order cards and no others) you get 5 points. This is the hardest to do. If you reveal order off turn, you get 4 points. If the round ends and you have 3 order cards in your hand (plus other cards most likely), you just get 3. If you have all 4, you get zero.

As mentioned above, you can only declare/reveal order if you have no other cards in hand. So, if round ends and you have 3 order cards plus others, you get the 3 points.

Interesting enough, this seems to imply that if you have only 4 order cards in hand, you can neither declare nor reveal order.

Funny story, I tried to end a round once by taking a chaos card from the pool, discarding my 3 order cards, filling the pool with all 4 (surrender of order). I had forgotten that a Chaos card reduces your round score to zero. First game and I learned my lesson.
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Francois Valentyne
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Good answer, Eric, you've done all the work for me!

To be clear and simple about having 4 Order cards in your hand: you must get rid of the 4th card if you want to score the Order set in any way.

F>
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Hank Panethiere
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Thanks to both of you!

When the fountain stack is empty you just keep playing correct?
 
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Francois Valentyne
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That's right. Getting through the Fountain draw stack does not end the round.
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Joshua Ostrander
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bungeeboy wrote:
The round ends if:
...
I'm pretty sure scoring is not cumulative. If you Declare order on your turn (3 order cards and no others) you get 5 points. This is the hardest to do. If you reveal order off turn, you get 4 points. If the round ends and you have 3 order cards in your hand (plus other cards most likely), you just get 3. If you have all 4, you get zero.
...


So just to be clear if you meet any of the scoring conditions with Order then you score nothing for any spectrum sets you have collected. Therefore you either score for order or spectrum, but never both.

Also if you happen to end with all 4 order cards in your hand you score zero for the round just like if you had a chaos card.

Are both points above correct? (we have been playing you score for order and spectrum)
 
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Francois Valentyne
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Sorry I missed the implication that scoring Order negates Spectrum points in Eric's contribution. This is incorrect. You score both. Consider the Order points (3,4 or 5) as bonus points. And having all 4 Order cards just means you score 0 bonus points.
Cheers!
F>
 
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Joshua Ostrander
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oize wrote:
Sorry I missed the implication that scoring Order negates Spectrum points in Eric's contribution. This is incorrect. You score both. Consider the Order points (3,4 or 5) as bonus points. And having all 4 Order cards just means you score 0 bonus points.
Cheers!
F>


Ah ok, that's what I thought. Thanks for the clarification!
 
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Eric Engstrom
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oopsie yes, you always score your spectrum sets.

Of course, holding a chaos card nets you a total of 0 for the whole round of play. yuck!
 
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