Probability of $x_1<\dots < x_n$ with symmetric multivariable density Let $f(x_1, \dots, x_n)$ be the p.d.f. of variables $X_1, \dots, X_n$, where $f(x_1, \dots, x_n)>0$ if $X_i \in D, i=1, \dots, n$ for some set $D$, $0$ elsewhere. Suppose $f$ is symmetric, that is, its value is the same for any permutation of the variables.
Is $Pr(X_1 < \dots, X_n)=\frac{1}{n!}$?
My reasoning is that $f$ being symmetric, the above probability is equal for any permutation of the inequalities, and the union of all such permutation is dense in the variable space. Is this correct? Are these conditions alone sufficient for the conclusion?
 A: Yes I think your reasoning is good.
They are called exchangeable random variables
https://en.wikipedia.org/wiki/Exchangeable_random_variables
A: This is the correct idea but we have to be careful about the argument of density in the variable space and the meaning of a union of permutation. Here is a way to formalize this: let 
$$
D_{\neq}:=\left\{(x_1,\dots,x_n)\in D^n\mid x_i\neq x_j\mbox{ if }i\neq j\right\}.
$$
Then 
$$
D_{\neq}=\bigcup_{\sigma\in\mathcal S_n}D_{\sigma},D_{\sigma}:=\left\{(x_1,\dots,x_n)\in D^n\mid x_{\sigma(1)}<\dots<x_{\sigma(n)}\right\}
$$
and the union is disjoint. It follows that 
$$
\mathbb P((X_1,\dots,X_n)\in D_\neq)=\sum_{\sigma\in\mathcal S_n}
\mathbb P((X_1,\dots,X_n)\in D_{\sigma})
$$
and by symmetry of $f$, the events $\{(X_1,\dots,X_n)\in D_{\sigma}\}$ have all the same probability, namely, $P(X_1<\dots<X_n)$. 
Then it remains to show that $\mathbb P\left((X_1,\dots,X_n)\in D\setminus D_\neq \right)=0$. To do so, use the inclusion
$$
\{(X_1,\dots,X_n)\in D\setminus D_\neq\}\subseteq \bigcup_{1\leq i<j\leq n}
\{X_i=X_j\}
$$
and the fact that $(X_i,X_j)$ has a density guarantees that $\{X_i=X_j\}$ has probability zero.
