Integral rising from difference of chi-squared random variables

Question

Let $X,Y$ be independent random variables such that $X\sim\chi_{n-1}^{2}, Y\sim\chi_{1}^{2}$ are chi-squared distributed (where $n\geq2$ is a natural number). I am trying to evaluate $\mathbb{P}[X\leq Y]$ as a function of $n$, at least asymptotically when $n\to\infty$. Obviously this probability decays to 0, but I'd like to be able to say something more quantative.

I was able to reduce this into a purely analytic question: Denoting $Z=X-Y$, I'm seeking for $F_Z(0)$. Expressing $F_Z$ using $f_X,F_Y$ I found that $$\mathbb{P}[X\leq Y]=\frac{1}{2^{\frac{n-1}{2}}\Gamma(\frac{n-1}{2})}\int_{0}^{\infty}\left(erfc(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}\right)dx $$ where $erfc$ is the complementary error function.

To my suprise, wolfram spits out nice looking algebraic numbers for odd $n$, which gives me hope that this integral can be expressed as a (simpler) function of $n$. Any help, either evaluating the integral or calculating the asymptotic probability in some other way will be much appreciated.

Answer 1

$$F_n=\int_{0}^{\infty}{\rm erfc}\,(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}dx$$ $$\qquad\qquad=2^{2-n} \, _2F_1\left(\tfrac{n-1}{2},\tfrac{n}{2};\tfrac{n+1}{2};-\tfrac{1}{2}\right)\frac{\Gamma \left(n-1\right)}{\Gamma \left(\frac{n}{2}+\frac{1}{2}\right)}.$$ The large-$n$ asymptotics follows by approximating ${\rm erfc}(\sqrt{x})\simeq \frac{e^{-x}}{\sqrt{\pi x } }$, resulting in $$F_n\rightarrow\pi^{-1/2}\left(\tfrac{3}{2}\right)^{1-\frac{n}{2}} \Gamma \left(\tfrac{n}{2}-1\right),\;\;n\gg 1.$$ This plot compares $F_n$ (gold) and the large-$n$ asymptotic (blue), they are nearly indistinguishable for $n$ above 10 or so.

For the probability in the OP I thus find for $n\gg 1$ $$\mathbb{P}[X\leq Y]\rightarrow \frac{3^{1-\frac{n}{2}} \Gamma \left(\frac{n}{2}-1\right)}{\sqrt{2 \pi } \Gamma \left(\frac{n-1}{2}\right)},\;\;n\gg 1.$$