Proof that the probability of one RV being larger than $n-1$ others is $\frac{1}{n}$ This is a follow-on from my previous question about samples from a distribution.
Suppose $X_1 \ldots X_{n-1}, X_n$ are random variables all following some fixed distribution $D$. How do I prove that $P(X_n > X_1 \wedge \ldots \wedge X_n > X_{n-1}) = \frac{1}{n}$ (this is true intuitively and experimentally).
 A: Here's a slightly more notational proof, since sometimes people feel squeamish about intuitive proofs like glen_b's comment. (Sometimes for good reason, since it's not necessarily immediately obvious that his proof doesn't apply to discrete distributions.)
Suppose that $X_i$ are distributed iid according to some distribution $D$.
Let $M_i$ be the event that $X_i = \max(X_1, \dots, X_n)$.
Clearly, at least one of the $M_i$ must hold, so
$\Pr(M_1 \cup \dots \cup M_n) = 1$.
But, using the inclusion-exclusion principle
\begin{align*}
\Pr(M_1 \cup \dots \cup M_n)
&= \sum_i Pr(M_i) - \sum_{i < j} \Pr(M_i \cap M_j) + \sum_{i < j < k} \Pr(M_i \cap M_j \cap M_k) - \dots
\end{align*}
If $D$ is continuous, then $\Pr(M_i \cap M_j) = 0$ for all $i \ne j$; all the latter terms drop. Also, since the $X_i$ are identically distributed clearly $\Pr(M_1) = \Pr(M_i)$ for all $i$. Thus
$$ \Pr(M_1 \cup \dots \cup M_n) = \sum_i \Pr(M_i) = n \Pr(M_1) = 1,$$
so $\Pr(M_1) = \frac{1}{n}$.
If $D$ is not continuous, the higher terms don't drop out. Taking @Yair Daon's example where $X_i$ is identically 1, every $M_i$ always holds, and the sum becomes
$$
\Pr(M_1 \cup \dots \cup M_n) = \sum_i 1 - \sum_{i<j} 1 + \sum_{i<j<k} 1 - \dots
= \sum_{k=1}^n (-1)^{k+1} \binom{n}{k} = 1
.$$
A: For absolutely continuous random variables, this has a nice-looking proof.
We have an i.i.d. sample characterized by density $f$ and distribution function $F$. To avoid subscripts, denote $Y \equiv X_{(n-1)}$ the maximum of the subsample of size $n-1$, and $W \equiv X_n$ the $n$-th draw. Being the maximum order statistic, the density function of $Y$ is $f_Y(y) = (n-1)f(y)[F(y)]^{n-2}$. We want to calculate the probability that the $n$-th draw will be maximum (we do not know the values of any draw),
$$P(Y \leq W) = \int_{-\infty}^{\infty} \int_{-\infty}^w f_{WY}(w,y){\rm d}y{\rm d}w$$
$$=\int_{-\infty}^{\infty} \int_{-\infty}^w f(w) f_Y(y){\rm d}y{\rm d}w$$
the decomposition of the joint density due to independence. $f_Y(y)$ is not a simple density, so we change the order of integration
$$P(Y \leq W) =\int_{-\infty}^{\infty} f_Y(y)\int_y^{\infty} f(w) {\rm d}w{\rm d}y$$  $$=\int_{-\infty}^{\infty} f_Y(y)[1-F(y)] {\rm d}y = 1-\int_{-\infty}^{\infty}f_Y(y)F(y) {\rm d}y$$
since we have integrated the density of $Y$ over the whole support. Writing out this density for the remaining integral we have
$$\int_{-\infty}^{\infty}f_Y(y)F(y) {\rm d}y = \int_{-\infty}^{\infty}(n-1)f(y)[F(y)]^{n-2}F(y){\rm d}y $$
$$=\frac {n-1}{n}\int_{-\infty}^{\infty}nf(y)[F(y)]^{n-1}{\rm d}y = \frac {n-1}{n}$$
since the integrand has become the density function of the maximum order statistic from a sample of size $n$, and so integrated over the whole support, equals unity too.
So,
$$P(Y \leq W) = 1- \frac {n-1}{n} = \frac 1n$$
A: Assume that these are continuous random variables then you'll be right on the money. Obviously they must be independent, i.e. i.i.d. in this case.
Think of how many permutations are of the sequence $X_1,\dots,X_n$, where the maximum will end up being at the last position? It's $\frac{n!}{(n-1)!n!}$. There are $n!$ permutations in total. So, your probability is $\frac{1}{n}$.
For discrete probabilities your claim will not be right, as @YairDaon showed you
