Can probability in the context of a normal distribution be interpreted as a proportion?

Question

I am trying to "hunt down" an error in reasoning that came up during a discussion with one of my colleagues. Let $X \sim N(\mu_x,\sigma^2)$ and $Y \sim N(\mu_y,\sigma^2)$. I am interested in $P(X>Y) - P(X<Y) = 2 P(X>Y) - 1$ and more specifically, in the variance of this quantity. In my view, this variance would be far from straightforward and might not even have a closed form expression. My colleague reasoned that because essentially $P(X>Y)$ is a proportion of observations where one group is better than the other, we can restate the problem in terms of Bernoulli distribution, denote this probability as $\pi$ and express the quantity of interest as $2 \pi - 1$. Then, the variance expression becomes straightforward: $Var (2 \pi - 1) = Var (2 \pi) = 4 Var(\pi) = 4 \frac{\pi(1-\pi)}{N}$ I know that the reasoning is wrong, and several simulations have confirmed this, but I cannot put my finger on where it goes wrong. I wonder if anyone might spot it? Thank you.