I encountered a problem recently,
Let $f(x_1,x_2,x_3)=e^{-(x_1+x_2+x_3)},0<x_1,x_2,x_3<\infty$ be the joint pdf of random variables $X_1,X_2,X_3$.
Find $P(x_1<x_2<x_3)$.
I want to state that the function is some sort of "symmetrical" regarding $x_1,x_2,x_3$. As a result, the function would not biased toward $x_1,x_2$ or $x_3$, and the number of combinations of order is 6(=$3!$). Therefore the answer is $1/6$.
However, I can't come up with a formal way to write down these notions. The only way I can come up with is using the concept of contraposition, but I think that was a little bit too verbose.
Is there a general way to approach this kind of symmetrically-structured problems?
You just need to demonstrate that $\mathbb{P}(x_i < x_j < x_k)$ are equivalent for any permutation $i, j, k$ in $\{1, 2, 3\}$. This is simple because the probability density is symmetric in $x_1, x_2, x_3$: swapping the values of $x_m$ and $x_n$ yields the same density (because addition is commutative) but, supposing no two $x$ are equal, this swapping yields a different ordering (of which there are $6$ total by simple counting).
There is one additional step you need to take before you can conclude the probability of any specified strict ordering is $\frac{1}{6}$, which is showing that $\mathbb{P}(x_i = x_j) = 0$ for any $i \ne j$. This should be straightforward.