$$\lim_{n\to\infty}n\,\mathbb P(X\gt Y)=2.$$
To show this, note that, for every $x\geqslant2$, the event $[Y\leqslant x,X=x+1]$ corresponds to a sample of size $x$ without duplicate and including $1$ and $2$, and to a duplicate appearing at time $x+1$. Each sample of size $x$ without duplicate and including $1$ and $2$ corresponds bijectively to a sample of size $x-2$ without duplicate and without $1$ and $2$, and to the choice of a position in this sample of size $x-1$ to place $1$, and finally to the choice of a position in this augmented sample to place $2$. And there are $x$ choices for the duplicate at time $x+1$.
Thus, the number of samples corresponding to the event $[Y\leqslant x,X=x+1]$ is
$$
(n-2)_{x-2}\cdot(x-1)\cdot x\cdot x.
$$
The total number of samples of length $x+1$ is $n^{x+1}$ hence
$$
\mathbb P(Y\lt X)=\sum_{x=2}^n\mathbb P(Y\leqslant x,X=x+1)=t_n,
$$
with
$$
t_n=\sum_{x=2}^n\frac{(x-1)x^2(n-2)_{x-2}}{n^{x+1}}=\frac{(n-2)!}{n^{n+1}}\sum_{y=0}^{n-2}\frac{(n-y-1)(n-y)^2n^y}{y!},
$$
where one uses the change of variable $y=n-x$.
Let $N_n$ denote a Poisson random variable with parameter $n$, then
$$
t_n=\frac{(n-2)!}{n^{n+1}}\mathrm e^n\mathbb E((n-N_n-1)(n-N_n)^2:n-N_n\geqslant2).
$$
The central limit theorem indicates that $n-N_n=\sqrt{n}Z_n$ where $Z_n\to Z$ in distribution, with $Z$ standard normal, hence
$$
t_n\sim\frac{(n-2)!}{n^{n+1}}\mathrm e^nn^{3/2}\mathbb E(Z^3:Z\geqslant0).
$$
Stirling's formula and the value $\mathbb E(Z^3:Z\geqslant0)=2/\sqrt{2\pi}$ yield the result stated at the beginning of this answer.
Likewise, for each $x\geqslant2$,
$$
\mathbb P(X\geqslant x\mid X\lt Y)=\frac{\mathbb P(X\geqslant x,X\lt Y)}{\mathbb P(X\lt Y)}=\frac{\mathbb P(X\geqslant x)-\mathbb P(X\geqslant x,X\gt Y)}{\mathbb P(X\lt Y)}.
$$
The denominator is $1-\mathbb P(X\gt Y)=1-t_n$. The numerator involves $\mathbb P(X\geqslant x)$, whose value is known, and
$$
\mathbb P(X\geqslant x,X\gt Y)=\sum_{y=x-1}^{n}\mathbb P(Y\leqslant y,X=y+1),
$$
and one knows from the first part of this post that
$$
\mathbb P(Y\leqslant y,X=y+1)=\frac{(y-1)y^2(n-2)_{y-2}}{n^{y+1}},
$$
hence the value of $\mathbb P(X\geqslant x\mid X\lt Y)$ follows.
