Suppose I sample $k$ i.i.d. elements $x_{1i},...,x_{1k}$ from a normal $\mathcal{N}(0,\sigma^2_1)$ distribution, and I sample a separate $k$ i.i.d. elements $x_{2i},...,x_{2k}$ from a normal $\mathcal{N}(0,\sigma^2_2)$ distribution, with $\sigma_1 \ne \sigma_2$.
What is the probability that $\min_i(x_{1i}) > \max_i(x_{2i})$?
That is, what is the probability that all of the elements in the first sample are greater than all of those in the second sample?
I know when $\sigma_1 = \sigma_2$ this reduces to a combinatorics problem.
But what about when the variances of the two groups are unequal? I've tried formulating this using the extreme-value distribution, but have had minimum luck. For example, if we let $m_1=\min(x_{1i})$ and $M_2=max(x_{2i})$ we can write:
$$P(m_1>M_2) = \int_a P(m_1>a \mid M_2=a)\, f_{M_2}(a)$$
And then $f_{M_2}(a)$ can be approximated by pdf of the extreme value distribution, and $P(m_1>a \mid M_2=a)$ can be approximated by the cdf of the extreme value distribution. But this gets really messy and I can't find a closed-form solution. I'm hoping there's a simpler and more intuitive approach to take.
Any suggestions or thoughts? Thanks!