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

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.
