Given random variables $X_1, X_2..., X_n$ that are iid from uniform $[0,1]$, the order statistics $X_{(1)}, X_{(2)}, ..., X_{(n)}$ are also random variables, defined by sorting the values (realizations) of $X_1, ..., X_n$ in increasing order. Additionally, we have $Y_1, \cdots, Y_n$ that are iid from uniform $[0,1]$ as well, and $Y_{(1)}, Y_{(2)}, ..., Y_{(n)}$ are defined similarly.
I want to compute the probability that $X_{(n)}$ (i.e., largest number) is greater than or equal to $Y_{(n-1)}$ (i.e., second largest number). According to Wikipedia, we know that the PDF for $X_{(n)} \sim \beta(n, 1)$ is $f(x) = n x^{n-1}$; and the PDF for $Y_{(n-1)} \sim \beta(n-1, 2)$ is $f(y) = n(n-1)y^{n-2}(1-y)$. To compute $Pr(X_{(n)} \ge Y_{(n-1)})$. I have the follower two methods, which lead to different answers and I am not sure which part is wrong.
Method 1: $$Pr(X_{(n)} \ge Y_{(n-1)}) = \int_0^1 n x^{n-1} \int_0^x n(n-1)y^{n-2}(1-y) \, dy dx = \frac{3n-1}{4n-2}.$$
Method 2: I define random variable $Z = X_{(n)} - Y_{(n-1)}$, and according to Wikipedia
Using the above formulas, one can derive the distribution of the range of the order statistics, that is the distribution of $X_{(k)} - X_{(j)}$ for $n \ge k > j \ge 1$ also forms a beta distribution $\beta(k-j, n-(k-j)+1)$.
Therefore, when $k=n$ and $j=n-1$, we have $Z \sim \beta(1, n)$ and $Pr(Z \ge 0) = 1$ which means $X_{(n)} \ge Y_{(n-1)}$ with probability $1$?
I am not sure which part goes wrong and which method has the correct answer... One concern I have with Method 1 is that this method (Expression for one random variable greater than another random variable) seems to be used to compute $X_{(n)} > Y_{(n-1)}$ instead of $\ge$, could this be the problem?