Poison Pill -clarification

Oh yeah misread that part. Well that would then make it fairly weird play on C's behalf, giving someone else a free victory point.

Klaital wrote:

Oh yeah misread that part. Well that would then make it fairly weird play on C's behalf, giving someone else a free victory point.

Then again, we don't have the full picture. I suppose C were trying to make the vote fail, but the other methuselah finally decided to oust A

Same with Ancient Influence. PP hits the acting Meth for five pool.

yappo wrote:

Same with Ancient Influence. PP hits the acting Meth for five pool.

Possibly less, if the relevant Methuselah has less than five pool.


For example, Parity Shift for 5 against a Methuselah who has 2 pool is legal. (As in, there are five players in the game, so Parity Shift attempts to take 5 pool.) That's confirmed in [LSJ 20070525]

In that situation, Poison Pill will hit for 2 pool, not the 5 pool that was tried for. [LSJ 20090909]


So it appears that the rule is that Poison Pill notes how many counters you actually lose, rather than how many the effect wants to hit you for (which matters when ousted).


I can't find anyone asking the question directly, which is odd as I would have expected it to come up.

I think that ruling might only apply to Parity Shift, because it is not causing a pool burn but specifically targeting "X of his or her pool" so it can not affect negative pool. for the same reason, parity shifting someone who has 2 pool means you can only allocate 2 pool to yourself.

but you can use Reckless Agitation to burn 5 pool from someone who has 2, and if someone wants to play Life Boon they must still cover the -3 deficit.

Correct answers. If the card reads "gains X and burn Y", you effectively burn Y (notice most of the cards that actually have players gain and lose pool are worded so that they first gain and then lose).