Talk:Thunderbolt/@comment-95.245.99.202-20130610094252/@comment-4719392-20130615165837

Hang on. Assuming that each arrow has an independent 30% chance of exploding, let's call A the event "the first arrow explodes" and B the event "the second arrow explodes". We know that Multishot chance is 90% at max and the chance of explosion is 30% at max. So…

P(A) = P("explosion") = 0.3 = 30%

P(B) = P("multishot") × P("explosion") = 0.9 * 0.3 = 0.27 = 27%

That way we also have that the chance of the first arrow not exploding is 70% (1 - 0.3) and the chance of the second arrow not exploding or no second arrow at all (i.e. no explosion related to the second arrow) is 73% (1 - 0.27). Let's call these events A' and B', respectively (they're the opposite or complement of their respective events).

A and B are indepedent events (by the initial assumption), since it doesn't matter for the second arrow if the first arrow exploded or not (and vice-versa). The same applies for A' and B'. Because of that:

P(A ∩ B) = P(A) × P(B), where P(A ∩ B) means the probability of A and B occuring at the same time. That way:

P(A ∩ B) = P(A) × P(B) = 0.3 × 0.27 = 0.081 = 8.1% (chance of both arrows exploding)

P(A' ∩ B') = P(A') × P(B') = 0.7 × 0.73 = 0.511 = 51.1% (chance of no explosion)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.3 + 0.27 - 0.081 = 0.489 = 48.9% (chance of any explosion)

And finally the chance of 1 explosion only is 0.489 - 0.081 = 0.408 = 40.8%.

Generally speaking, if we call P(T) the Thunderbolt chance and P(M) the multishot chance, the probability of any explosion would be:

P(T) + P(T)×P(M) - P(M)×P(T)²