Can Meth A call a Parity Shift targeting B, even though X is not equal
to the Meth's in the game, or would Meth B have to be on 5+ pool for
the shift to be legal?
--> J
grail_pbem "at" hotmail.com
Legal. Meth B simply loses 5 pool and is ousted. UNLESS Meth C plays
Life Boon for at least 2 to save Meth B (because Meth B is at -1 pool
when they are ousted). The targeting rules for PS are simply who has
more pool than the calling Methuselah. How much they have is
irrelevant, except as it relates to how much the calling Meth has.
BUT - I am usually wrong on this stuff, so I would wait for James, Jeff
or of course LSJ (whose call tag I put in the subject...)
best -
chris
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Or unless the referendum fails or is canceled.
> The targeting rules for PS are simply who has
> more pool than the calling Methuselah. How much they have is
> irrelevant, except as it relates to how much the calling Meth has.
Well, the question seems to be more on the "can I allocate 5 of B's pool when B
has only 4 pool?" (rather than "can B be targeted?").
The answer is yes, just as you can bleed B for 5 of her pool.
The extra point is lost. The acting Methuselah decides who comes up short.