August 24, 2019, 08:38:35 AM

Author Topic: Reducing numbers to zero  (Read 1561 times)

Koz

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Reducing numbers to zero
« on: October 22, 2012, 11:39:02 AM »
I know that an attack cannot have it's dice reduced below one, but what about with Chain Lightning?  You roll one less dice for each successive attack, so does it eventually drop to zero (meaning the attack is over), or does it "bottom out" at one dice and keep bouncing from target to target until there is nothing left in range?  

Also, in another question related to reducing numbers to zero, if you reduce the cost of something (like say with Arcane Ring), you can redude the cost to zero correct?  I didn't see anything in the rules to say you can't, but I want to make sure.

Shad0w

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Re: Reducing numbers to zero
« Reply #1 on: October 22, 2012, 11:46:20 AM »
Quote from: "Koz" post=2439
I know that an attack cannot have it's dice reduced below one, but what about with Chain Lightning?  You roll one less dice for each successive attack, so does it eventually drop to zero (meaning the attack is over), or does it "bottom out" at one dice and keep bouncing from target to target until there is nothing left in range?  

Also, in another question related to reducing numbers to zero, if you reduce the cost of something (like say with Arcane Ring), you can redude the cost to zero correct?  I didn't see anything in the rules to say you can't, but I want to make sure.

This was from another thread but it was basically the same question.
Quote from: "Shad0w" post=1345
Quote from: "Klaxas" post=1317
that isnt what i was asking.  since the minimum dice to roll is 1, can you keep rolling 1 die 1 die 1 die, until you run out of targets or fail to damage one?

Shad0w Wrote:
Correct


So as long as you keep dealing damage and have legal targets chain lightning keeps going. It would bottom out at 1D, until you run out of targets or fail to damage one.

Yes you could in theory make something cost zero just not less than zero. I thought this was in the rule book but I may be mistaken.

I hope that helps.
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