Prove symmetry of probabilities given random variables are iid and have continuous cdf Let $Y_1, Y_2, ...$ be independent and identically distributed random variables in $(\Omega, \mathscr{F}, \mathbb{P})$ s.t. their distributions are continuous and $$F_{Y}(y) := F_{Y_1}(y) = F_{Y_2}(y) = ...$$
For $m = 1, 2, ...$ and $i \le m$, define $$A_{i,m} := (\max\{Y_1, Y_2, ..., Y_m\} = Y_i), B_m := A_{m,m}$$

Is it really that $$P(B_m) = P(A_{m-1,m}) = ... = P(A_{2,m}) = P(A_{1,m})$$
and hence $$P(B_m) = 1/m?$$
I guess $(A_{(i,m)})_{i \le m}$'s pairwise disjoint, for a fixed m and that 
$$\sum_{i=1}^{m} P(A_{(i,m)}) = P(\bigcup_{i=1}^{m} A_{(i,m)}) = 1$$
I don't really understand why all the $P(A_{(i,m)})$'s are equal
For example why is it that $$P(A_{1,2}) := P(Y_1 \ge Y_2) = P(Y_2 \ge Y_1) := P(A_{2,2})$$ Why 1/2 each? Why not 1/4, 3/4? Or even 1, 0? I'm guessing this has something to do w/ independence or continuity.

Assuming the random variables are absolutely continuous (What if the pdfs don't exist? :O),
$$P(B_m) = \int_{\mathbb R} \int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1, ..., Y_n} (y_1, ..., y_n) dy_1dy_2...dy_{n-1}dy_n$$
By independence, we have
$$ = \int_{\mathbb R} \int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1}(y_1) \cdots f_{Y_n}(y_n) dy_1dy_2...dy_{n-1}dy_n$$
$$ = \int_{\mathbb R} [\int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1}(y_1) \cdots f_{Y_{n-1}}(y_{n-1}) dy_1dy_2...dy_{n-1}] f_{Y_n}(y_n) dy_n$$
$$ = \int_{\mathbb R} [F_{Y_1}(y_n) ... F_{Y_{n-1}}(y_n)] f_{Y_n}(y_n) dy_n$$
By identical distribution,
$$ = \int_{\mathbb R} [F_{Y_n}(y_n) ... F_{Y_{n}}(y_n)] f_{Y_n}(y_n) dy_n$$
$$ = \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} f_{Y_n}(y_n) dy_n$$
$$ = [F_{Y_n}(y_n)]^{n-1} F_{Y_n}(y_n)|_{-\infty}^{\infty} - \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} (n-1) f_{Y_n}(y_n) dy_n$$
$$ = [(1)(1) - (0)(0)] - \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} (n-1) f_{Y_n}(y_n) dy_n$$
$$ = 1/n$$
 A: If $Y_1,\dots,Y_n$ are iid and $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$ is a permutation then:
$$F_{Y_{\sigma(1)},\dots Y_{\sigma(n)}}(y_1,\dots,y_n)=F(y_1)\times\cdots\times F(y_n)=F_{Y_1,\dots,Y_n}(y_1,\dots,y_n)$$
where $F$ is the common CDF of the $Y_i$. 
So the random vectors $\langle Y_{\sigma(1)},\dots,Y_{\sigma(n)}\rangle$ and $\langle Y_1,\dots,Y_n\rangle$ have equal distribution.
Consequently the events $\{Y_n=\max(Y_1,\dots,Y_n)\}$ and $\{Y_{\sigma(n)}=\max(Y_1,\dots,Y_n)\}$ have equal probability.
A: An intuitive answer to the question, why all events $B_m$ have probability $\frac{1}{m}$:
Drop your $Y_i$ in the desired interval or on the real line. The values are distinct almost surely. Then some $Y_j$ has the highest value. Given that you have $n$ $Y_i$s, the probability for any $Y_j$ to be the highest value is $\frac{1}{n}$. 
This can also be shown by a combinatorial argument similar to the one in the answer by drhab, but in slightly different notation: 
Take $n$ values $Y_1,...,Y_n$. There are $n!$ permutations, i.e. ways to mix them. Fix $Y_n$ to be the highest value. Then there are $(n-1)!$ permutations for the remaining elements. Hence, $P$("$Y_n$ is max") $=$ $\frac{(n-1)!}{n!} = \frac{1}{n}$. 
