Talk:Thunderbolt/@comment-24.23.135.165-20141103015754/@comment-64.139.110.145-20141103161543

Good question. I think it should be 49.2% btw.

The calculation is 1- (1-0.3)^1.9 = 0.4922

(probability of failure)^(number of events) = probability of only failures (no explosions)

1 -  (probability of failure)^(number of events) = the opposite, or probability of at least one nonfailure (>=1 explosion).

My initial thought was that the 58.5% represented the expected number of explosions per shot (including the possibility of a double explosion), which is simply represented by:

0.3 * 1.9 = 0.57

However, that number is also wrong.