Brownian motion positive at time 1 then negative at 2

A standard Brownian motion $X$ starts at $X_0=0$. Given that $X_1>0$, what is the probability that $X_2<0$?

$1/4$. $X_1$ and $X_2$ are jointly normal with correlation $1/\sqrt2$, so $\mathbb{P}[X_1>0,X_2<0]=\tfrac14-\tfrac{\arcsin(1/\sqrt2)}{2\pi}=\tfrac14-\tfrac18=\tfrac18$. Dividing by $\mathbb{P}[X_1>0]=\tfrac12$ gives $\tfrac14$. Geometrically, the increment must exceed $X_1$ in magnitude, a $45^\circ$ wedge out of the $180^\circ$ half-plane.