A problem on continuous i.i.d random variable Let $X_1, X_2, X_3, X_4$ be i.i.d continuous random variables with a common distribution function $F$. How to prove that "all the 4! possible orderings of $X_1, \dots, X_4$ are equally likely" without calculating probability of any ordering by integrating the joint density function?
I guess the "exchangablity" property of i.i.d random variable needs to be used. But this property is proved by integrating the joint density function. 
Another thing I also want to know the geometric intuitive solution of the problem.
 A: The answer is that all orderings are correct "by symmetry". In other words if you were to relabel all of your variables $X_1 => X_2, X_2 => X_3$, etc, you would still have the same probability for the ordering, hence all orderings are equally probable. 
Edit: Look at it like this: 
Assume that the ordering $X_1, X_2, X_3, X_4$ has probability $p$. 
Take another ordering, say $X_4, X_2, X_3, X_1$ and let its probability $= q$.
Now since the random variables are from the same distribution 
we could, WLOG, have labelled $X_1$ as $X_4$ and $X_4$ as $X_1$. 
Hence $P(X_4, X_2, X_3, X_1) = p$ and $p=q$
A: Since our original poster says he can't understand "phcoding"'s answer, I'll try my own rephrasing of it.
The random variables $X_1, X_2, X_3, X_4$ are i.i.d.  How do we know that it is just as probable that
$$
X_4<X_2<X_3<X_1
$$
as it is that
$$
X_1<X_2<X_3<X_4\text{ ?}
$$
So let
$$
W_1=X_4,\qquad W_2=X_2,\qquad W_3=X_3,\qquad W_4=X_1.
$$
The question becomes: How do we know it is just as probable that
$$
W_1<W_2<W_3<W_4
$$
as that
$$
X_1<X_2<X_3<X_4\text{ ?}
$$
The answer is that the quadruple $(X_1,X_2,X_3,X_4)$ has the same probability distribution as the quadruple $(W_1,W_2,W_3,W_4)$.
