Let $k_1,\dots,k_m\in\mathbb{R}^d$, denote by $\mathcal{D}:=\mathcal{N}\left(0,\frac{1}{d}I_d\right)$ and $[n]= \{ 1,\dots,n \} $. I am interested in the following probability:
$P_{ x_1,\dots,x_n\sim \mathcal{D}}\left(\forall i\in[m] , ~j\in [n] , \langle x_1,k_i\rangle > \langle x_j,k_i\rangle \right) $
In other words, given $m$ vectors, what is the probability that drawing $n$ vectors from a Gaussian distribution, the first one will have the largest correlation with all these vectors?
For $m=1$ this probability is $\frac{1}{n}$. This is because the draw of the $x_j$'s is symmetric, so we can also choose $k_1=e_1$. In this case from the symmetry of the problem, the probability that the first coordinate of the first vector is larger than the first coordinate of the other vectors is equal for all of them.
If the $k_i$'s are orthogonal then I think I can show by a similar symmetric argument that this probability is $\left(\frac{1}{n}\right)^m$.
I'm seeking a more general result for any given $k_i$'s. Also, if it can be shown for $\mathcal{D} = \mathcal{U}(\mathbb{S}^{d-1})$ (uniform distribution over the sphere) it is also interesting.