Probability random variable is less or equal to k-th out of two samples when ordered Given the random variable $X$, $\{X_{i}\}_{i=2}^{n}$, $\{Y_{i}\}_{i=2}^{n}$ all iid and lets denote $X_{(k)}$ as the k-th statistic  of $\{X\} \cup \{X_{i}\}_{i=2}^{n}$ and $Y_{(k)}$ for $\{X\} \cup \{Y_{i}\}_{i=2}^{n}$ what is the probability for $X \leq X_{(k)}$ or $X \leq Y_{(k)}$, in other words
$$
\mathbb{P}(X \leq \max\{X_{(k)}, Y_{(k)}\})
$$
If it's easier, assume that they are $U ([0,1])$, in that case it's more or less clear that:
$$
\mathbb{P}(X \leq X_{(k)}) = \frac{k}{n}
$$
due to this other question, the problem I'm facing is that one naturally says:
$
\mathbb{P}(X \leq \max\{X_{(k)}, Y_{(k)}\}) = 1 - \mathbb{P}(X > \max\{X_{(k)}, Y_{(k)}\}) = 1 - \mathbb{P}(X > X_{(k)}, X > Y_{(k)})
$
And it's tempting to say $\mathbb{P}(X > X_ {(k)}, X > Y_ {(k)}) = \left[\mathbb{P}(X > X_{(k)})\right]^{2}$ but I do believe that is not true, because $X_ {(k)} $ and $Y_ {(k)} $ both have $X$ so they should not be independent (I believe), plus numeric simulations suggest that they in fact are not independent. Any help in order to tackle the problem would be welcome
 A: \begin{align}
\mathbb{P}(\forall k\ X > X_ {(k)}, X > Y_ {(k)}) &= \mathbb E\left[\prod_{k=1}^n\mathbb I_{X > X_ {(k)}}\prod_{k=1}^n\mathbb I_{X > Y_ {(k)}}\right]\\
&= \mathbb E^X\left\{\mathbb E\left[\prod_{k=1}^n\mathbb I_{X > X_ {(k)}}\prod_{k=1}^n\mathbb I_{X > Y_ {(k)}}{\Huge|}X\right]\right\}\\
&= \mathbb E^X\left\{\prod_{k=1}^n\mathbb E\left[\mathbb I_{X > X_ {(k)}}\prod_{k=1}^n{\Huge|}X\right]\prod_{k=1}^n\mathbb E\left[\mathbb I_{X > Y_ {(k)}}{\Huge|}X\right]\right\}\\
&= \mathbb E^X\left\{\mathbb{P}(X > X_{(k)}|X)^n\mathbb{P}(X > Y_{(k)}|X)^n\right\}\\
&\ne \left[\mathbb{P}(X > X_{(k)})\right]^{2n}\\
\end{align}
A: If the distribution is continuous then WLOG we can assume that we are dealing with uniform distribution on $(0,1)$.
Then for $u\in(0,1)$:
$$P\left(X=X_{\left(r\right)}\text{ and }X=Y_{\left(s\right)}\mid X=u\right)=\binom{n-1}{r-1}u^{r-1}\left(1-u\right)^{n-r}\binom{n-1}{s-1}u^{s-1}\left(1-u\right)^{n-s}=$$$$\binom{n-1}{r-1}\binom{n-1}{s-1}u^{r+s-2}\left(1-u\right)^{2n-r-s}$$
so that:
$$P\left(X>X_{\left(k\right)}\text{ and }X>Y_{\left(k\right)}\right)=\sum_{r=k+1}^{n}\sum_{s=k+1}^{n}\binom{n-1}{r-1}\binom{n-1}{s-1}\int_{0}^{1}u^{r+s-2}\left(1-u\right)^{2n-r-s}du=$$$$\sum_{r=k+1}^{n}\sum_{s=k+1}^{n}\binom{n-1}{r-1}\binom{n-1}{s-1}\mathsf B(r+s-1,2n-r-s+1)$$
and consequently:
$$P\left(X\leq X_{\left(k\right)}\text{ or }X\leq Y_{\left(k\right)}\right)=$$$$1-\sum_{r=k+1}^{n}\sum_{s=k+1}^{n}\binom{n-1}{r-1}\binom{n-1}{s-1}\mathsf B(r+s-1,2n-r-s+1)$$
Here $\mathsf B$ denotes the beta function and: $$\mathsf B(r+s-1,2n-r-s+1)=\frac{\Gamma(r+s-1)\Gamma(2n-r-s+1)}{\Gamma(2n)}=\frac{(r+s-2)!(2n-r-s)!}{(2n-1)!}$$
I don't know yet whether further simplification is possible.
