Let $X,Y,Z$ be i.i.d. and all $\sim \mathcal{N}(0,1)$.
It can easily be shown that $A = (X-Y) = \sim \mathcal{N}(0,2)$, and thus $\mathbb{P}(X>Y) = 1/2$
But supposing it is already known that $Z > Y$. My instinct is that this changes the distribution of $Y$? In some sense, $Z$ is "probably a bit higher than average" and $Y$ is "probably a bit lower than average".
Thus I would expect that the distribution of $B = (Y_{Z>Y})$ is no longer $\sim \mathcal{N}(0,1)$?
What, then, is it's distribution, and how would I calculate the conditional probability: $\mathbb{P}((X>Y)_{Z>Y})$
By definition, \begin{align} P(X > Y | Z > Y) = P(X > Y, Z > Y) / P(Z > Y) = 2P(X > Y, Z > Y). \end{align}
To calculate $P(X > Y, Z > Y)$, using independence between $X, Z$ and $Y$, we have: \begin{align} & P(X > Y, Z > Y) \\ = & \int_{\mathbb{R}}P(X > y, Z > y)\varphi(y)dy \\ = & \int_{\mathbb{R}}(1 - \Phi(y))^2\varphi(y) dy \\ = & \int_0^1 (1 - t)^2dt = \frac{1}{3}. \end{align}
Therefore the answer is $2/3$.