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You are just asking to compute $p=P(X>Y)$, where $(X,Y)$ has the bivariate normal distribution with given $EX=\mu_1$, $EY=\mu_2$, $Var\,X=\sigma_1^2:=\Sigma_{1,1}$, $Var\,Y=\sigma_2^2:=\Sigma_{2,2}$, and $\rho:=corr(X,Y)=\Sigma_{1,2}/(\sigma_1 \sigma_2)$. It is easy to see that this probability is $$p=\frac{1}{2} \text{erfc}\Big(\frac{\mu _2-\mu _1}{\sqrt{2} \sqrt{\sigma _1^2+\sigma _2^2-2 \rho \sigma _1 \sigma_2}}\Big) =\Phi\Big(\frac{\mu _1-\mu _2}{ \sqrt{\sigma _1^2+\sigma _2^2-2 \rho \sigma _1 \sigma_2}}\Big),$$ where $\Phi$ is the standard normal cdf. We have $p\in(\frac12,1)$ iff $\mu_1>\mu_2$.