Theorem. If $X_1$, ..., $X_N$ are i.i.d. random points in $R^d$ whose distribution is symmetric with respect to $0$ and assigns measure zero to every hyperplane through $0$, then $$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\frac{1}{2^{N-1}}\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$
The proof is straightforward. Let $\mu$ be the distribution of $X_k$, and set $$f(x_1,\dots,x_N) = \begin{cases} 1, & \mbox{if } x_1,\dots,x_N\ \mbox{ lie in an open halfspace of \mathbb R^d with 0 in the boundary}, \newline 0, & \mbox{else} mbox{else.} \end{cases}.$$ end{cases}$$Then due to the invariance of \mu under reflection in the origin, we have that$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\int_{\mathbb R^d}\dots \int_{\mathbb R^d} \frac{1}{2^n}\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)\ frac{1}{2^N}\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)\ \mu(dx_1)\dots\mu(dx_N).$$Now, the sum$$C(N,d)=\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)$$can be interpreted as the number of connected components of the set \mathbb R^d\backslash (H_1\cup\dots\cup H_N) induced by the hyperplanes H_1, ..., H_N through 0 which are in general position. But there is a classical calculation going back to to Steiner and Schläfli, which shows that$$C(N,d)= 2\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$2 added 1006 characters in body This is a classical and essentially geometric problem. In fact, the answer does not depend on the distribution of the points (as long as the distribution is centrally symmetric). The following result is due to Wendel (link). Theorem. If X_1, ..., X_N are i.i.d. random points in R^d whose distribution is symmetric with respect to 0 and assigns measure zero to every hyperplane through 0, then$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\frac{1}{2^{N-1}}\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$The proof is straightforward. Let \mu be the distribution of X_k, and set$$ f(x_1,\dots,x_N) = \begin{cases} 1, & \mbox{if } x_1,\dots,x_N\ \mbox{ lie in an open halfspace with $0$ in the boundary}, \newline 0, & \mbox{else} \end{cases}.$$Then due to the invariance of \mu under reflection in the origin, we have that$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\int_{\mathbb R^d}\dots \int_{\mathbb R^d} \frac{1}{2^n}\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)\ \mu(dx_1)\dots\mu(dx_N).$$Now, the sum$$C(N,d)=\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)$$can be interpreted as the number of connected components of the set \mathbb R^d\backslash (H_1\cup\dots\cup H_N) induced by the hyperplanes H_1, ..., H_N through 0 which are in general position. But there is a classical calculation going back to to Steiner and Schläfli, which shows that$$C(N,d)= 2\sum\limits_{k=0}^{d-1}{N-1 \choose k}.$$1 This is a classical and essentially geometric problem. In fact, the answer does not depend on the distribution of the points (as long as the distribution is centrally symmetric). The following result is due to Wendel (link). Theorem. If X_1, ..., X_N are i.i.d. random points in R^d whose distribution is symmetric with respect to 0 and assigns measure zero to every hyperplane through 0, then$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\frac{1}{2^{N-1}}\sum\limits_{k=0}^{d-1}{N-1 \choose k}.