If $X,Y \sim N(0, 1)$, what is $P(X > 3Y \mid Y > 0)$? Given that $X,Y \sim N(0, 1)$ and IID, what is $P(X > 3Y \mid Y > 0)$? I think the answer is $1/12$.
If we look at the $X-Y$ plane of the joint distribution of $X,Y$, we see that conditioning on $Y > 0$, gets rid of the bottom half. The area under the lie $Y = X/3$ is $1/12$ of the area of the top half. However, when I try to use monte carlo, I am seeing about $0.102$ as I increase the number of trials.
Is my solution wrong?
My other question regarding this is, what if $X,Y$ are not independent? How would you compute the conditional probability in this case?

The conditional probability can be rewritten as:
$$P(X > 3Y | Y > 0) = \frac{P(X > 3Y \wedge Y > 0)}{P(Y > 0)} = 2P(X > 3Y \wedge Y > 0),$$
since obviously $P(Y>0)= \frac{1}{2}.$
Now, we need to integrate:
$$P(X > 3Y \wedge Y > 0) = \int_{0}^{+\infty}\int_{3y}^{+\infty} \frac{1}{2\pi}e^{-\frac{1}{2}(x^2 + y^2)}dx dy$$
We can use polar coordinates, where:
$$\begin{cases}
x = r \cos\theta\\
y = r \sin \theta
\end{cases},$$
and
$$dxdy = r d\theta dr.$$
$$P(X > 3Y \wedge Y > 0) = \int_{0}^{+\infty} \int_{0}^{\arctan\frac{1}{3}} \frac{1}{2\pi}e^{-\frac{1}{2}r^2}rd\theta dr = \\
= \frac{1}{2\pi}\left(\int_{0}^{+\infty}e^{-\frac{1}{2}r^2}rdr \right)\left(\int_{0}^{\arctan\frac{1}{3}} d\theta\right) =\\= \frac{1}{2\pi}\cdot 1\cdot \arctan\frac{1}{3} =  \frac{1}{2\pi}\arctan\frac{1}{3} \simeq 0.0512$$
Finally, the probability you are looking for is:
$$P(X > 3Y | Y > 0) = \frac{1}{\pi}\arctan\frac{1}{3} \simeq 0.1024.$$
