The original question is
Let $X_1,\cdots,X_n$ be iid continuous random variables. Show that $\mathbb{P}\left\{X_n > X_1 > X_2, X_3,\cdots,X_{n-1} \right\}=\frac{1}{n(n-1)}$.
I get that the argument goes like this: There are totally $n!$ order for $X_1,\cdots,X_n$. Among those orders, to get an order such that $X_n > X_1 > X_2 \text{ and } X_3 \cdots \text{ and }X_{n-1}$, the order of $X_3,\cdots,X_{n-1}$ doesn't matter (so we have $\left(n-2\right)!$ orders). So we have $(n-2)! / n! = 1/n(n-1)$.
The part I don't clearly understand is why can we treat different orders as having equal probability? It seems intuitively convincing (since they are iid?) but how would one rigorously explain it?
This specific problem can be solved explicitly, by doing some conditioning on $X_1$ and integrating and etc. But what if we have a different one? Say $\mathbb{P}\left\{ X_1 < X_2 <\cdots < X_{n-1} > X_n\right\}$. Using combinatoric argument, I can easily show that this is $(n-1)/n!$. I certainly don't want to do a bunch of integration each time…
Best Answer
For any permutation $\sigma$ on $\{1,\cdots,n\}$, let $E_{\sigma} = \{ X_{\sigma(1)} > \cdots > X_{\sigma(n)}\}$. Then by symmetry and continuity of $X_i$'s, we have
$$\mathbb{P}(E_\sigma) = \frac{1}{n!}. $$
Indeed, symmetry tells that the value of $\mathbb{P}(E_\sigma)$ is independent of the choice of $\sigma$. Continuity tells that $\sum_{\sigma} \mathbb{P}(E_\sigma) = 1$.
This is the very reason we can play with the order of indices, since you can convert the problem of computing the probability of any events only in terms of orderings of $X_i$'s into the corresponding counting problem. For instance, in OP's problem,
$$ \mathbb{P}(X_n > X_1 > \max\{X_2,\cdots,X_{n-1}\}) = \sum_{\substack{\sigma(1) = n \\ \sigma(2) = 1}} \mathbb{P}(E_{\sigma}), $$
where the sum on the RHS runs over all the permutations $\sigma$ on $\{1,\cdots,n\}$ such that $\sigma(1) = n$ and $\sigma(2) = 1$. Since there are exactly $(n-2)!$ such permutations, we obtain
$$ \mathbb{P}(X_n > X_1 > \max\{X_2,\cdots,X_{n-1}\}) = \frac{(n-2)!}{n!} = \frac{1}{n(n-1)}. $$
Remark. Notice that the entire argument depends only on the following two obserations
The joint distribution of $(X_1, \cdots, X_n)$ is continuous.
$(X_1, \cdots, X_n)$ is exchangeable, i.e. for any permutation $\sigma$, $(X_{\sigma(1)}, \cdots, X_{\sigma(n)})$ has the same distribution as $(X_1, \cdots, X_n)$.
So the above argument extends without modification whenever these two conditions hold.