Constant revolution question (LSJ)

[John P.]

abbreviated card text:
he or she must burn X pool and/or cards at random from his or her
hand, where X is the number of counters on this card"

I've seen the ruling that when X=2 or more and discarding is chosen
that all discards must be made prior to replacing.

So when X=8 (unlikely as that is) can a person with a 7 card hand side
choose to burn 8 cards from their hand, losing only 7 cards in the
process?

Or if your hand is less than X you must burn the difference?

I'm assuming that you *must* lose the difference between hand size and
X (where X>hand size) in pool, similar to smiling Jack where I've
always assumed that if you do not have enough blood on your vampires
you burn pool to meet the cost.

Thanks,

John P
Winnipeg

[LSJ]

John P. wrote:
> abbreviated card text:
> he or she must burn X pool and/or cards at random from his or her
> hand, where X is the number of counters on this card"
>
> I've seen the ruling that when X=2 or more and discarding is chosen
> that all discards must be made prior to replacing.

Yes. Also trivially true when X=1.

> So when X=8 (unlikely as that is) can a person with a 7 card hand side
> choose to burn 8 cards from their hand, losing only 7 cards in the
> process?

No. Burning 7 cards is not burning 8 cards. Xe'd have to burn 1 pool and 7 cards
(or 2 pool and 6 cards or some other combination that makes 8).

> Or if your hand is less than X you must burn the difference?

The total cards+pool burned must be X.

> I'm assuming that you *must* lose the difference between hand size and
> X (where X>hand size) in pool, similar to smiling Jack where I've
> always assumed that if you do not have enough blood on your vampires
> you burn pool to meet the cost.

Correct.

[AngryNorway]

>
> So when X=8 (unlikely as that is) can a person with a 7 card hand side
> choose to burn 8 cards from their hand, losing only 7 cards in the
> process?
>

Heh, managed to do it once, killed my prey because of it

Jahn