The formula for such a problem

Considering some event with success probability p, the formula for the probability P of at least one success over n trials is:

P = 1 - (1 - p)^n

This formula stems from the intuition that to get at least one success is to NOT get all failures, and the probability of getting all failures over n trials is given by (1-p)^n. 

Addressing the specific problem, this provides another understanding of why the "doubles" approximation is more accurate for small probabilities. For n=2, this formula reduces to

P = 1 - (1 - 2p + p^2) = p*(2 - p), where p is the probability for one success. If p is small, 2-p is approximately equal to 2, and the success over two trials is approximately 2p.

This formula also lets us examine whether the same approximation works for n=3.

P = 1 - (1 - 3p + 3p^2 - p^3) = p*(3 - 3p + p^2), which is approximately 3p for small p.

We can safely generalize -- the doubles approximation holds, at least to some error, for any n.