Warhammer Fantasy Battle Reporter Forum

Had a discussion with our regular gaming group over an issue that's never yet come up, and likely never will. But enjoying a friendly argument as much as we do, tried to come up with an answer.

Scenario:
One tries to cast a spell that has it's cost raised to 15+ (through drain magic, for example)

The dice roll comes up 6,6,2
That totals 14, so the actual cost of the spell hasn't been met, yet there are two 6s rolled. Is it successful or not?

I tend to think not since the entry for Irresistible Force (on p.107 of the rulebook) says "When rolling to cast a spell, any result of two or more unmodified 6s means that the spell has been cast with irresistible force." To me that means that you have to check and see if the spell has been cast first (was casting value met) then if it was met you can check to see if it's possible to be dispelled. However, because of the wording of that sentence, I can understand why people may think that as long as there are two 6s it is successful.

Your thoughts?
-kpasaman (Starving Crazed Weasels)

A roll of double 6s will ALWAYS cast, with no chance to dispell (dispell scrolls aloud?). No matter if you needed 15 and rolled just 2 sixes (making a total of 12) it would stil auto-cast.

the phrase from the rulebook is

"when rolling to cast a spell, any result of two or more unmodified 6s means that the spell has been cast with irresistible force"

the key word in that is ANY. generally GW are pretty good with stating things that may effect the result.

so in my mind regardless if it beats the casting level or not the fact there are two 6s means it's cast with irresistable force

I believe the question has been answered thorough enough? Closing topic