Statistics is a subject loooong forgotten...

So... how do I set up a formular that returns the chance of 3 dice (misfortune dice), comming up with 2+ of Challenges?

With 2 Dice (for misfortune) it would be, 0,33^2 = 11%, but when you suddenly have a third (or fourth) dice this changes...

Remember, I'm looking for 2 or more, not exactly 2 Challenges.

I have no idea without being very laborious & painstaking but this site seems to work.

http://www.jaj22.org.uk/wfrp/diceprob.html

Yeah, this is quite good. If only I could figure out how it's calculated... I want to use it, so I can see the effect of making Chaos Star count as challenge+(bane/star).

This is gonna keep me sleepless

Silly me, using Fortune Dice gave me the statistics I wanted