Say there are two standard Gaussian random variables $X$ and $Y$. I am trying to evaluate the probability that the larger of the two is selected, given that it is known whether $X$ is positive or negative (the strategy is selecting $X$ if $X$ is positive and selecting $Y$ if $X$ is negative). In equation form this is $$Pr(X-Y>0|X>0) + Pr(X-Y<0|X<0)$$
How can this expression be evaluated? Numerically it appears to be $\frac{3}{4}$, and intuitively this makes sense.
I am also interested in this probability in the more general case, where the strategy involves selecting $X$ if $X-S>0$ and $Y$ otherwise, where $S$ is another independent Gaussian random variable.
The following assumes $X$ and $Y$ are independent.
As Angela Richardson pointed out, you probably actually want to compute $$P(X-Y > 0 \mid X > 0) P(X>0) + P(X-Y <0 \mid X< 0) P(X<0)$$ which is $3/4$. (The quantity in your post is not $3/4$.)
For the first term, it suffices to compute $P(X-Y > 0, X>0)$. (Why?)
Consider the region of the plane that contains $(x,y)$ pairs satisfying $x-y>0$ and $x>0$. Then use rotational symmetry of the vector $(X,Y)$ to compute the probability.
The other term can be handled similarly.