Probability that random high dimensional vectors are all on the convex hull Say I pick $n$ i.i.d. random standard normal points in $\mathbb{R}^d$.  Roughly, as long as $n$ is much smaller than exponential in $d$, with high probability all points will be on the convex hull.  This is because with high probability they will all be near the radius $\sqrt{d}$ sphere and all almost orthogonal, and thus each point is the furthest in its own direction from the origin.  Let $p(n,d)$ be the failure probability that at least one point is in the interior of the convex hull.
Question: What's the best upper bound on $p(n,d)$ as a function of $n$ and $d$?  I care most about the regime $d \gg 1$, $n \in O(\operatorname{poly}(d))$.
 A: It's not too bad to see that the probability is at most $2n^2 e^{-d/2e}$.  Let $x_1,\ldots,x_n$ be the points.  We will use a union bound, so it is sufficient to examine the probability that $x_1$ is in the convex hull of $x_2,\ldots,x_n$.  This happens if and only if there are $\lambda_j \in [0,1]$ with $\sum \lambda_j = 1$ and $$x_1 = \sum_{j = 2}^n \lambda_j x_j\,.$$
Take an inner product with $x_1$ to see that this implies $$\| x_1 \|_2^2 = \sum_{j = 2}^n \lambda_j \langle x_1, x_j \rangle.$$
Thus $$P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) \leq P( \|x_1\|^2 \leq \max_{j \geq 2} |\langle x_1, x_j \rangle|).$$
If we divide by $\|x_1\|$, the RHS probability bound becomes
$$P\left(\|x_1\| \le \max_{j \ge 2} \left|\left<x_1/\|x_1\|, x_j \right>\right|\right).$$
$\|x_1\|^2 \sim \chi^2_d$ and $\left<x_1/\|x_1\|, x_j\right> \sim N(0,1)$, so from $\chi^2_d$ and $N(0,1)$ tail bounds we have
\begin{align*}
P(\|x_1\| \le t\sqrt{d}) &\le \left(t e^{(1-t^2)/2}\right)^d \\
P\left(\left<x_1/\|x_1\|, x_j\right> \ge t\sqrt{d}\right) &\le \frac{\left(e^{-t^2/2}\right)^d}{t\sqrt{2\pi d}}
\end{align*}
for any $t \in (0,1)$.  Matching the base of the exponents gives
\begin{align*}
t e^{(1-t^2)/2} &= e^{-t^2/2} \\
t &= e^{-1/2} \approx 0.606531
\end{align*}
whence union bounding shows
\begin{align*}P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n))
  &\le (n-1) \left(1 + \frac{1}{\sqrt{2\pi d/e}}\right) e^{-d/2e} \\
  &< 2ne^{-d/2e}
\end{align*}
and so
$$P(\exists~j \text{ s.t. }x_j \in \mathrm{conv}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n)) < 2n^2 e^{-d/2e}.$$
I do not know if this is optimal, but it's worth noting that it's basically the strategy you suggested.  When $n$ is exponentially large in $d$ the probability does not tend to $0$ provided the exponent is big enough, which is where this bound breaks.
