We examine as set of independent normal vectors: $$ \forall i \in [N]\triangleq \{1,\dots,N\}:\,\mathbf{x}_{i}\sim\mathcal{N}\left(0,\mathbf{I}_{d}\right)$$

For any $\epsilon>0$ and $K\leq N$, we define the event $ \mathcal{A}_{K,\epsilon}$ that the mutual coherence is lower bounded by $\epsilon$ $$ \gamma (S) \triangleq \max_{i,j\in S:\,i\neq j}\frac{\mathbf{x}_{i}^{\top}\mathbf{x}_{j}}{\left\Vert \mathbf{x}_{i}\right\Vert \left\Vert \mathbf{x}_{j}\right\Vert }>\epsilon \,.$$ on any subset $S\subset [N]$ of size $|S|=K$.

**Question**: I'm looking for as tight as possible upper bound on $P(\mathcal{A}_{K,\epsilon})$, the probability for this event.

So far, the best I could think of was dividing $[N]$ into $\lfloor N/K \rfloor$ different subsets $S_i$ of size $|S_i|=K$, so we can bound $P(\mathcal{A}_{K,\epsilon})\leq \prod_i P \left( \gamma \right(S_i \left) > \epsilon \right) \, .$ Using standard mutual coherence bounds on each subset $S_i$, I get $P(\mathcal{A}_{K,\epsilon}) \leq 2K^{2N/K}\exp\left(-\frac{Nd\epsilon^{2}}{24K}\right) $. However, I was wondering if there is a way to improve the exponential part in this upper bound.

Thanks in advance!