We have n IID random variables $X_1, X_2, \ldots, X_n$. Let $R_i$ be $X_i$'s rank in the set $\{X_1, X_2, \ldots, X_3 \}$ when we order from large to small. How to prove $R_i, \forall i \in \{1, 2, \ldots, n\}$, is uniformly distributed on $\{1, 2, \ldots, n\}$?
My first guess is that, for any position $j$ in ordered sequence of $X$s, as $X_1, X_2, \ldots, X_n$ are equally likely to be the $j$th largest,
$\Pr \{ R_i = j \} = \frac 1 n.$
So $R_i$ is uniformly distributed on $\{1, 2, \ldots, n\}$.
Another way to think about it, is to count how many possible cases are there when $R_i = j$. As $X_i$'s position in ordered sequence is fixed at $j$, then we can just permute the rest of $X$s to get all possible ordered sequence. Since there are $n-1$ variables left, there are $(n-1)!$ situations. As for any $R_i = j, 1 \le j \le n$, there are always $(n-1)!$ possible ordered sequences, we can say $R_i$ is uniformly distributed on $\{1, 2, \ldots, n\}$.
Are these 2 proof rigorous?
Edit:
As cardinal pointed out, an additional condition is needed for the proof, and a sufficient such condition is that $X_i$ to be a continuous random variable.