Given four random variables, A,B,C,D, chosen independently from the same normal distribution (with mean $\mu$ and standard deviation $\sigma$), I am trying to solve:
$$P[(2+A+B)>(1+B+C) \cap (2+A+B)>(C+D)] $$
The first thing I am doing is combining each of the terms to form a single variable, using summation of normal distributions:
$X_0 = 2+A+B \sim \mathcal N(2+2\mu, 2\sigma^2) $
$X_1 = 1+B+C \sim \mathcal N(1+2\mu, 2\sigma^2) $
$X_2 = C+D \sim \mathcal N(2\mu, 2\sigma^2) $
Now the probability becomes:
$$P(X_0>X_1 \cap X_0>X_2) $$
similar to this question: What is $P(X_1>X_2 , X_1>X_3,... , X_1>X_n)$?
However, the solution to that problem assumed that each $X$ is chosen independently, which may not apply here. I believe that a dependency would exist between $X_1$ and $X_2$, since they share the same selection C. (the same could be said for $X_0$ and $X_1$, but the variables could cancel out in the comparison so it may not matter. $X_1$ and $X_2$, though, are not compared directly). How do I proceed with this to handle the dependencies?