Create a clan question
10 May 2012 15:25 #29923
by Oortje
Have a cold long night,
Jeroen van Oort
Dutch National Coördinator
Create a clan question was created by Oortje
If you spoof a card, the card cost 1 extra pool or in case of blood cost an extra blood.
If you spoof a sybils tongue would it cost 1 pool + xblood or x+1 blood. I was wondering since the first card does not cost any blood.
If you spoof a sybils tongue would it cost 1 pool + xblood or x+1 blood. I was wondering since the first card does not cost any blood.
Have a cold long night,
Jeroen van Oort
Dutch National Coördinator
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10 May 2012 15:53 #29925
by jamesatzephyr
Since it's an unofficial, unsanctioned format, check with the organiser.
However, following the usual CaC rules, Sibyl's Tongue has a blood cost of X, so it has a blood cost, so increase that by one.
Replied by jamesatzephyr on topic Re: Create a clan question
If you spoof a sybils tongue would it cost 1 pool + xblood or x+1 blood. I was wondering since the first card does not cost any blood.
Since it's an unofficial, unsanctioned format, check with the organiser.
However, following the usual CaC rules, Sibyl's Tongue has a blood cost of X, so it has a blood cost, so increase that by one.
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10 May 2012 17:58 #29940
by BenPeal
Replied by BenPeal on topic Re: Create a clan question
Here's Jussi Hattara's version of the CAC rules, the most up-to-date version I'm aware of:
www.tenerdo.org/cac.html
In those rules, spoofing a card with a blood cost increases the blood cost by 1 and does not add a pool cost.
www.tenerdo.org/cac.html
In those rules, spoofing a card with a blood cost increases the blood cost by 1 and does not add a pool cost.
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10 May 2012 18:19 #29942
by jhattara
Jussi Hattara 
Webmaster Extraordinaire 
Finnish
Politics!
Replied by jhattara on topic Re: Create a clan question
Cards that have blood cost increase blood cost, other cards increase pool cost.




Finnish

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11 May 2012 14:58 #30034
by Pascal Bertrand
Replied by Pascal Bertrand on topic Re: Create a clan question
LSJ's ruling on Lord Aaron Wesley Wilshire is that the Methuselah playing the card chooses.
I'll link the reference later. Best to do is: check with organiser.
I'll link the reference later. Best to do is: check with organiser.
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11 May 2012 15:29 - 11 May 2012 15:35 #30038
by jamesatzephyr
This is set at design time, though, surely?
For example, if you decided to spoof Leathery Hide, it has no blood cost, so it'd increase by 1 pool (using the default CaC rules). (And this would almost certainly mean you didn't choose to spoof it!) However, if you faced a deck playing Terror Frenzy, the card would now have a blood cost. But I don't think anyone would argue that would remove the pool cost you'd added to it.
The fact that Sibyl costs X blood (even though X can be zero) suggests that at design time, you should be bumping it to cost X+1.
Similarly, if I played Masochism - which actually reduces blood costs - I don't think a spoofed card with a now non-existent blood cost would pick up an extra cost of a pool.
Replied by jamesatzephyr on topic Re: Create a clan question
LSJ's ruling on Lord Aaron Wesley Wilshire is that the Methuselah playing the card chooses.
This is set at design time, though, surely?
For example, if you decided to spoof Leathery Hide, it has no blood cost, so it'd increase by 1 pool (using the default CaC rules). (And this would almost certainly mean you didn't choose to spoof it!) However, if you faced a deck playing Terror Frenzy, the card would now have a blood cost. But I don't think anyone would argue that would remove the pool cost you'd added to it.
The fact that Sibyl costs X blood (even though X can be zero) suggests that at design time, you should be bumping it to cost X+1.
Similarly, if I played Masochism - which actually reduces blood costs - I don't think a spoofed card with a now non-existent blood cost would pick up an extra cost of a pool.
Whenever this vampire plays a card, you may remove X rush counters from this card to reduce that card's blood cost by X.
Last edit: 11 May 2012 15:35 by jamesatzephyr.
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