Prove ranks are uniformly distributed 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.
 A: Suppose $Y_1 = X_2$, $Y_2=X_3$, and $Y_3 = X_1$.  If you can show that the joint distribution of the vector $(Y_1,Y_2,Y_3)$ is the same as the joint distribution of the vector $(X_1,X_2,X_3)$, then it follows that the probability that $X_1$ has a certain rank is the same as the probability that $Y_1$ has that rank.  But the latter is the probability that $X_2$ has that rank.  Thus the probability that $X_1$ has a certain rank is the same as the probability that $X_2$ has that rank.  As with $1$ and $2$, so also with the others.  Thus
$$
\Pr(R_1 = j) = \Pr(R_2=j)=\cdots=\Pr(R_n=j).
$$
Since these are mutually exclusive and exhaustive events, each must have probability $1/n$.  Notice that we didn't need to know the value of $j$ for this.  Therefore
$$
\Pr(R_1 = 1) = \Pr (R_1=2) = \Pr(R_1=3) = \cdots = \Pr(R_1=n) = \frac1n,
$$
and similarly with any of the other indices besides $1$.
So your first guess was right, but one can say a few things to demonstrate that it's right, and call that a proof.
A: Michael Hardy's proof doesn't use the IID assumption but rather assumes exchangeability, and hence is more general.
If we assume IID, here's how you can show the rank of a given random variable is discrete uniform. Showing the rank of $X_n$, denoted by $R_n$, is discrete uniform will suffice.
Define $R_n$ as follows:
$$
R_n = \sum_{i=1}^n I(X_i \leq X_n) = 1 + \sum_{i=1}^{n-1}I(X_i \leq X_n).
$$
Note that $I(\cdot)$ is an indicator function, and $I(X_i \leq X_n) = I(F(X_i) \leq F(X_n))$, where $F$ is the distribution function of $X_n$, which is the distribution function of any arbitrary $X_i$, using IID. By the probability integral transform, $F(X_i) \overset{d}{\equiv} U_i \sim \mathrm{Uniform}(0,1)$. Thus,
$$
R_n = 1 + \sum_{i=1}^{n-1}I(U_i \leq U_n).
$$
Now, to show that $R_n$ follows a discrete uniform on $\{1,\ldots,n\}$, we must show $\mathrm{P}(R_n=r)=1/n$—i.e., derive the probability mass function of $R_n$. Note that $I(U_i \leq U_n)$ and $I(U_j \leq U_n)$ for $i\neq j$ are not independent because they both have $U_n$. Therefore, even though we know $I(U_i \leq U_n)$ follows a Bernoulli distribution with probability $1/2$, we can't use that directly.
Instead, we condition on $U_n$ and use the law of total expectation as follows:
\begin{align}
\mathrm{P}(R_n=r) &= \mathrm{E}\left\{\mathrm{P}\left(\sum_{i=1}^{n-1}I(U_i \leq U_n) =r-1\mid U_n\right)\right\}.
\end{align}
Given $U_n$, $I(U_i \leq U_n) \sim \mathrm{Bernoulli}(U_n)$ and $\sum_{i=1}^{n-1}I(U_i \leq U_n) \sim \mathrm{Binomial}(n-1, U_n)$. Therefore,
\begin{align}
\mathrm{P}(R_n = r) &= \mathrm{E}\left(\binom{n-1}{r-1}U_n^{r-1}(1-U_n)^{n-r} \right)\\
&= \binom{n-1}{r-1}\int_0^1 u^{r-1}(1-u)^{n-r+1-1} \,\mathrm{d}u\\
&= \binom{n-1}{r-1} \mathrm{B}(r,n-r+1) \\
&= \binom{n-1}{r-1} \frac{(r-1)!(n-r)!}{n!} = \frac{(n-1)!}{(r-1)!(n-r)!}\frac{(r-1)!(n-r)!}{n!}\\
&= \frac{1}{n}.
\end{align}
QED.
