I have a solution for my problem but am unable to verify it by simulations. Is there anything wrong with my reasoning?
Assume the following setting:
We have $N$ i.i.d. uniformly distributed continuous R.V., $Y_1, \ldots, Y_N \sim U[0, 1]$ and some order statistic, say the minimum in this case $Z := \min\{Y_1, \ldots, Y_N\}$. We then draw (i.i.d) once again from the same distribution $X \sim U[0, 1]$.
What is the probability that $X$ is smaller than the minimum? In other words, what is $P[X < Z]$?
This is what I have so far: Let $f(x)$ be the PDF of $X$, so $f_X = 1$. Let $g(z)$ be the PDF of $Z$, which, according to this, is Beta-distributed with $Beta(1, n)$.
Since $X$ and $Z$ are independent, we can write: $$P[X < Y] = \int_{z = -\infty}^\infty \int_{x = -\infty}^z f(x)g(z) dx dz \,.$$
Plugging in the support and that $f = 1$:
$$\Rightarrow \int_{z = 0}^1 \int_{x = 0}^z 1 \cdot g(z) dx dz \,.\\ = \int_{z = 0}^1 g(z) \int_{x = 0}^z 1 dx dz \,.\\ = \int_{z = 0}^1 z \cdot g(z) dz = E[Z]. $$
This seems kind of odd to me, how can the probability to be smaller than the minimum value be the expected value of the minimum?
Does this mean that, on average, one needs $\frac{1}{N + 1}$ independent trials to randomly draw a realization $x$ from $X$ that is smaller than the minimum of the other realizations?
Any help would be greatly appreciated, thank you!