Probability of gaussian being smaller than multiple other gaussians Assuming I have 3 normally distributed random variables with different variance and mean: 
$ X\sim N(\mu_1, \sigma_1^2) $,  $ Y\sim N(\mu_2, \sigma_2^2) $ , $ Z\sim N(\mu_3, \sigma_3^2) $
According to this post, I know how to calculate the probability $P(X<Y)$.  
Now I'm interested in the probability that X is smaller than the two other distributions and I calculate this as
$$ p_x = P((X<Y) \cap(X<Z)) = P(X<Y) \times P(X<Z) $$ by multiplying two $ erfc $ functions mentioned in the link. I can do the same by comparing if $Y$ is smaller than $Z$ and $X$. Eventually I get $p_y$ and $p_z$. 
My questions:
1. Is that process correct?
2. If yes, I thought that $ p_x + p_y + p_z = 1$ but I get a lower probability when I try examples.
 A: In general, your answer is not correct because $P(X<Y)$ and $P(X<Z)$ are not independent and therefore you can't multiply them to get the probability of the intersection.
I'll show a counterexample.
Let $X$, $Y$ and $Z$ be independent random variables with the same distribution. Then all possible permutations are equally probable:
$$P(X<Y<Z)=P(X<Z<Y)=P(Y<X<Z)=P(Y<Z<X)=$$
$$=P(Z<X<Y)=P(Z<Y<X)=\frac{1}{6}$$
You can see that:
$$P(X<Y)=P(X<Z)=\frac{1}{2}$$
But:
$$P(X<Y\text{ and }X<Z)=P(X<Y<Z)+P(X<Z<Y) =\frac{1}{3} \ne$$
$$ \ne P(X<Y) · P(X<Z)=\frac{1}{4}$$
A: Let $C = \{X_1 < \min\{X_2,X_3\}\} = \{(X_1 < X_2)\cap (X_1 < X_3)\}$. Even if the $X_i$ are independent random variables (a condition that the OP has not included), the events $(X_1 < X_2)$ and $(X_1 < X_3)$ are not independent events, and so $P(C) \neq P((X_1 < X_2)P(X_1 < X_3)$. However, continuing to assume independence, 
\begin{align}P(C\mid X_1 = a) &= P\{X_1 < \min\{X_2,X_3\}\mid X_1=a\}\\
&= P\{(X_1 < X_2)\cap (X_1 < X_3)\mid X_1=a\}\\
&= P\{(a < X_2)\cap (a < X_3)\}\\
&= P\{a < X_2\}P\{a < X_3\}\\
&= \left[1-\Phi\left(\frac{a-\mu_2}{\sigma_2}\right)\right]\left[1-\Phi\left(\frac{a-\mu_3}{\sigma_3}\right)\right]\end{align}
and so we can write
\begin{align}P(C) &= \int_{-\infty}^\infty P(C\mid X_1 = a)f_{X_1}(a)
\,\mathrm da\\
&= \int_{-\infty}^\infty\left[1-\Phi\left(\frac{a-\mu_2}{\sigma_2}\right)\right]\left[1-\Phi\left(\frac{a-\mu_3}{\sigma_3}\right)\right]\frac{e^{-(a-\mu_1)^2/2\sigma_1^2}}{\sigma_1\sqrt{2\pi}}\,\mathrm da.\end{align}
To the best of my knowledge, there is no closed-form expression for the value of this integral, and so numerical integration, or simulation, must be used to determine the numerical value of $P(C)$.
