# How to find which of two events -drawn from a normal distribution- is more likely?

I know the probability of event A is given by:

$$\Phi(f(x)+g(y)) - \Phi(f(x)-g(y)),$$

and the probability of event B is

$$\Phi(m(y)+n(x)) - \Phi(m(y)-n(x)).$$ where $$\Phi$$ is the cumulative distribution function of the standard normal.

I want to find the more likely of those two events as a function of $$x$$ and $$y$$. I tried a few different approaches, but I couldn't make it much simpler than this:

$$\Pr(A|x,y) > \Pr(B|x,y) \\ \Leftrightarrow \frac{1}{2 \pi} \int\limits_{ f(x) - g(y) }^{f(x) + g(y) } e^{-t^2/2} dt > \frac{1}{2 \pi} \int\limits_{ m(y) - n(x) }^{ m(y) + n(x) } e^{-t^2/2} dt$$

An obvious sufficient condition for $$\Pr(A|x,y) > \Pr(B|x,y)$$ is $$f(x) - g(y) < m(y) - n(x)$$ and $$f(x) + g(y) > m(y) + n(x)$$. But I can't figure out a tractable necessary condition. I would appreciate any hints!

Edit: I can assume that $$f$$ and $$m$$ are linear, and $$g$$ and $$n$$ quadratic, if that is of any use.