Compare maxima of two Gaussian samples Suppose $X$ and $Y$ are both normally distributed, with $X \sim \mathcal{N}(0,1)$ and $Y \sim \mathcal{N}(c,1),$ where $c > 0$. Consider $n$ independent draws of both $X$ and $Y$. As $n \rightarrow \infty,$ what is the probability that the sample maximum of the draws of $Y$ is greater than the maximum for $X$?
 A: I found the answer, thanks mainly to whuber's comment. Let $M_y$ and $M_x$ denote the sample maxima of $Y$ and $X$, respectively. Scaled versions of $M_y$ and $M_x$ are both Gumbel distributed [source 1], and the difference between two Gumbels is logistic [source 2]. Specifically, we have
$$\sqrt{2 \ln n}~(M_y-M_x-c) \stackrel{d}{\rightarrow} \mathcal{L}(0,1),$$
where $\mathcal{L}(0,1)$ denotes the logistic distribution with location $0$ and scale (standard deviation) $1$. From the usual approximation, we hence have 
$$(M_y-M_x) \approx \mathcal{L}\left(c,\frac{1}{\sqrt{2 \ln n}}\right).$$ Using the cdf of the logistic distribution [source 2], we get
$$P(M_y > M_x) \approx \frac{\exp\left(\sqrt{2\ln n}\times c\right)}{1+\exp\left(\sqrt{2\ln n}\times c\right)}.$$
This approximation formula implies that $P(M_y > M_x)$ does converge to $1$, but at a very (!) slow rate. For example, if $n$ equals one million and $c = 0.1$, the probability is only $62.84$ percent. 
References:
Source 1 - 
http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/sfehtmlnode90.html 
Source 2 - 
http://en.wikipedia.org/wiki/Logistic_distribution
