Numerical superiority necessary to beat in $L^\infty$ a population one standard deviation ahead Suppose $m$ independent random variables $X_i$ have the distribution $\mathcal{N}(0, 1)$, and $n$ independent random variables $Y_j$ (also independent of the $X_i$) have the distribution $\mathcal{N}(1, 1)$.
For which $m$ and $n$ is the event $\max(X_i)\ge\max(Y_j)$ more likely than $\max(X_i)<\max(Y_j)$?

We get an analytical expression for the probability from
\begin{align}
P[x \ge \max (X_i)] & = \Phi (x)^m \\
P[x = \max (X_i)] & = m\Phi (x)^{m - 1}\phi (x) \\
P[x = \max (X_i) \ge\max (Y_j)] & = m\Phi (x)^{m - 1}\phi (x)\Phi (x - 1)^n\\
P[\max (X_i) \ge\max (Y_j)] & = \int_{-\infty}^{\infty} m\Phi (x)^{m - 1}\phi (x)\Phi (x - 1)^n dx
\end{align}
I don't see how to simplify that integral, or figure out its asymptotic behavior.
In any case, this gives the following table of $n$'s with the minimal $m$ for which $\max (X_i) \ge\max (Y_j)$ is more likely than $\max (X_i) < \max (Y_j)$:
\begin{matrix}
m & n \\
4 & 1\\
11 & 2\\
29 & 4\\
78 & 8\\
205 & 16\\
535 & 32\\
\end{matrix}
Update: The data for $n \le 51$ is now in the Online Encyclopedia of Integer Sequences as A348913.
