Is it true, that the probability for both events is always equal? If yes how to prove it, if not, why not? Imagine I have a real random variable $X$ with some distribution (continuous, discrete or continuous with atoms)
Now Imagine I have i.i.d. copies $X_1,...,X_n$, all independently and equally distributed as $X$
My claim is:
$$\mathbb{P}(X_2>X_1)=\mathbb{P}(X_2<X_1)$$
My secondy claim is the following:If I order them by size, so that $X_{(1)}<X_{(2)}<\ldots<X_{(n)}$ and I define the interval $I_n=[X_{(1)},X_{(n)}]$; Then I claim:
$$\mathbb{P}(X_{n+1}<X_{(1)})=\mathbb{P}(X_{n+1}>X_{(n)})$$
So the probability that the $n+1$-th number exceeds the interval on the left equals the probability it exceeds on the right
I guess the first one is true, but the second one not;
E.g. Assume X can take the value 0 and 1; and assume $n=3$; Then
$$\mathbb{P}(X_3>{X_1,X_2})=\mathbb P (X_3=1)\mathbb P (X_2=0)\mathbb P (X_1=0)=\mathbb P (X=1)\mathbb P (X=0)\mathbb P (X=0)$$
but also 
$$\mathbb{P}(X_3<{X_1,X_2})=\mathbb P (X_3=0)\mathbb P (X_2=1)\mathbb P (X_1=1)=\mathbb P (X=0)\mathbb P (X=1)\mathbb P (X=1)$$
which is gernerally not the same; But What I am wondering is if there are simply conditions that it would become true
 A: Let $Y_n =\min(X_1,X_2 \cdots X_n)$, and let $y_n=\sum_{i=1}^n[X_i=Y_n]$ count the number of elements that attain that minimum.
Analogously, let $Z_n$ and $z_n$ be the maximum and maximum-count.
Then, by symmetry $P( X_{n} = Y_n \wedge y_n=1)=P(X_n=Y_n) P(y_n=1 \mid X_n=Y_n)=\frac{1}{n} P(y_n=1)$
Then, esentially you are asking if $P(y_n=1)=P(z_n=1)$ , that is, if the probability of having a single maximum equals the probability of having a single minimum. This is not true in general.
It's true for a continuous variable (continuous CDF) because in that case the probability of a having a single extrema equals $1$. It's also true for a symmetric (around the median) random variable.  I'm not sure if there's a simple characterization for its CDF to be true in general.
Added:
Let $F(x) = P(X \le x)$ be the CDF, and let $p(x)= F(x) - F(x^-)$. 
Then the probability of having a single minimun in $n+1$ realizations equals
$$A=p(y_{n+1}=1)= \int \left(\frac{1-F(x)}{1-F(x^-)}\right)^n dF(x)=
\int \left(1-\frac{p(x)}{1+p(x)-F(x)}\right)^n dF(x) \tag{1}$$
Similarly, for the maximum:
$$B=p(z_{n+1}=1)= \int \left(\frac{F(x^-)}{F(x)}\right)^n dF(x) =
\int \left(1- \frac{p(x)}{F(x)}\right)^n dF(x) \tag{2}$$
If $F(x)$ have finite discontinuities at $x_i$, $i=1,2\cdots k$ (perhaps the result is also valid for more general settings), we can write $F(x)=F_c(x) + \sum_i p(x_i)u(x-x_i)$ where $F_c(x)$ is continuous and $u(\cdot)$ is the unit-step function. Then
$$\begin{align}
A &=\sum_i p(x_i) \left(1-\frac{p(x_i)}{1+p(x_i)-F(x_i)}\right)^n +F_c(+\infty)\\
&=1- \sum_i p(x_i)\left[1- \left(1-\frac{p(x_i)}{1+p(x_i)-F(x_i)}\right)^n \right]\tag{3}
\end{align}
$$
$$\begin{align}
B&=\sum_i p(x_i) \left(1- \frac{p(x_i)}{F(x_i)}\right)^n +F_c(+\infty)\\
&=1- \sum_i p(x_i)\left[1- \left(1- \frac{p(x_i)}{F(x_i)}\right)^n 
\right] \tag{4}
\end{align}$$
Of course, $A=B=1$ if $F(x)$ is continuous. Also, $A=B$ if the probability (both the continuous and the discrete parts!) is symmetric. There's not much more to say in general, I think...
A: You're asking whether $P(X > \max({X_1, \ldots, X_n})) = P(X < \min({X_1, \ldots, X_n}))$, where $X_1, \ldots, X_n$ and $X$ are all independent and distributed the same.
This is true in some distributions and false in others. For example if the $X_i$s are sampled uniformly at random from $[0,1]$ then the probabilities would be the same (due to symmetry).
In general, for any distribution over $[a,b]$ which is symmetric about $(a+b)/2$, you would expect this to be true.
Also, I believe this is true for any continuous distribution over $(a,b)$ (i.e. such that the probability of any two variables receiving exactly the same value is 0). To see this, note that $P(X > \max({X_1, \ldots, X_n})) = \frac{1}{n+1} = P(X < \min({X_1, \ldots, X_n}))$. This is because we can first generate $X, X_1, \ldots, X_n$, then order $X_1, \ldots, X_n$, without affecting the probabilities, and any of these are equally likely to be the maximum or the minimum.
To see when this could be false, suppose that the $X_i$s are chosen as follows: with probability $1/2$, set $X_i = 10$, otherwise sample $X_i$ uniformly from $[0,1]$. Then $P(X > \max({X_1, \ldots, X_n}))$ will approach $0$ exponentially fast, since it is impossible for $P(X > 10)$ to happen, but $P(X < \min({X_1, \ldots, X_n}))$ is roughly proportional to $1/n$. 
