Assume the $1 \times N$ vector
$\mathbf X = [X_1, X_2, \ldots , X_N]$
contains i.i.d. normal samples such that $\mathbf X$ has a multivariate normal distribution. Now assume another random variable $Y$ also has a normal distribution, independent from the samples in $\mathbf X$, such that $Y$ has a mean and variance not necessarily equal to that of $\mathbf X$.
What is the probability that $Y$ will exceed all the values in $\mathbf X$ concurrently (or jointly)? In other words what is the probability that the value of $Y$ will be the maximum value of $[X_1, X_2, \ldots , X_N, Y]$
In the simplest case $N=1$, the problem reduces to
$p(Y>X_1)$ = $p(Y-X_1>0)$
which can be calculated by convolution of $p(Y)$ and $p(-X_1)$ and integrating the resulting PDF from $0$ to $\infty$. When $Y$ is also i.i.d. w.r.t the samples of $\mathbf X$, the resulting PDF is zero-mean, such that $p(Y>X_1)$ = 0.5. This makes intuitively sense, since the probability that one RV will exceed another i.i.d. RV is 0.5. The probability
$p(Y>X_1, Y>X_2, \ldots, Y>X_N )$
should therefore equal $1/(N+1)$ if $Y$ is i.i.d. w.r.t. $\mathbf X$. When $N>1$, the problem does not seem straightforward (maybe it is?).