Probability that a normal random variable is greater than the maximum of $n$ i.i.d. random variables

Question

Let $M_n = \max\{X_1,\ldots,X_n\}$, where $X_1,\ldots,X_n$ are i.i.d. random variables. We know about $M_{n}$ from the extreme value theorem and know it's mean and variance. Let $Y \sim \mathcal{N}(\mu,\sigma^2)$ independent of $M_{n}$. What can we say about $P(Y \geq M_{n})$?

Answer 1

This depends on the distribution of the $X_i$. If they have the same law as $Y$, then the answer is $1/(n+1)$ by symmetry.

Answer 2

Learner (https://math.stackexchange.com/users/48763/learner), Finding probability $P(X<Y)$, URL (version: 2012-12-18): https://math.stackexchange.com/q/261078

Virtually identical to this.