Conditional probability that one of three random variables is the largest

Question

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

$\Pr[X_3 > X_2 \mid X_3 > X_1] = 0.5 * 1 + 0.5\frac{0.5}{0.5} = 1 \neq \frac{2}{3}$.

Answer 1

The easiest thing is to look at it combinatorically. If you rank the numbers $X_i$ by 1,2 and 3 according to their magnitude (3 is the largest), then you have 6 ranked combinations of ($X_1,X_2,X_3$) in total: (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2) and (3,2,1).

The ones with $X_3>X_1$: (1,2,3),(1,3,2) and (2,1,3).

Out of these two combinations also have $X_3>X_2$: (1,2,3) and (2,1,3).

Hence your probability $P(X_3>X_2\mid X_3>X_1)$: 2/3