Suppose I have three random variables $X_1$, $X_2$, and $X_3$ that are i.i.d $N(0,1)$ distributed. The chance that $X_3$ is larger than $X_2$, given that $X_3$ is also larger than $X_1$ should equal $\frac{2}{3}$ in theory. However, when trying to calculate this conditional probability, I always end up with an probability equal to 1, which obviously is not correct. Perhaps you could point out where my mistake is?
$\Pr[X_3 > X_2 \mid X_3 > X_1] = \Pr[X_2 < X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 < X_1]+\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1]$.
I then rewrite the last term as
$\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1] = \Pr[X_2 \geq X_1]\frac{\Pr[X_3 > X_2]}{\Pr[X_3 > X_1]}$.
This gives me
$\Pr[X_3 > X_2 \mid X_3 > X_1] = 0.5 * 1 + 0.5\frac{0.5}{0.5} = 1 \neq \frac{2}{3}$.