Probability of conditional inequalities of random variables

For full disclosure, this problem is drawn from my latest problem set, but at this point, I care more about solving the puzzle than about the credit. The statement is:

Given iid continuous random variables $X_1,X_2,X_3$, compute $P(X_1 > X_2 | X_1 > X_3)$.

Obviously, that's equal to $\frac{P(X_1 > X_2 ,\ X_1>X_3)}{P(X_1>X_3)}$ and that the denominator is $\frac{1}{2}$ is clear enough, but I can't quite grep the numerator. Class discussion and my own work lead me to believe it will be a triple integral on the general joint probability function (separable into the marginals due to independence), but I can't figure out how to set the limits, and then how to get a numeric answer from there. Any hints in the right direction would be greatly appreciated.

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Method 1: The numerator should be $1/3$. HINT: What should be the probability that the maximum of the three numbers $\{ X_1, X_2, X_3 \}$ turns out to be $X_1$?

Method 2: Getting the same answer using integrals: Let $F(\cdot)$ be the c.d.f. of the distribution. Conditioned on $X_1 = x$, the probability that $X_2$ (resp. $X_3$) lies in the interval $(-\infty, x)$ is $F(x)$. Therefore, the desired probability is $$p = \int_{-\infty}^{+\infty} F(x)^2 F^{\ \prime}(x) dx.$$ Using integration by parts, $$p = \left. F(x)^2 F(x) \right|_{-\infty}^{+\infty} - \int_{-\infty}^{\infty} 2 F(x) F^{\ \prime}(x) \times F(x) dx = \left. F(x)^3 \right|_{-\infty}^{+\infty} - 2p.$$ Can you now simplify to get $p = 1/3$?

Method 3: Using triple integrals: Let $f(\cdot)$ denote the p.d.f. of the distribution. Then, the "happy" event whose probability we are computing is that $(X_1, X_2, X_3) \in D$ where $$D := \{ (x, y, z) \mid x \geq y, x \geq z \} \subseteq \mathbb R^3.$$ So the probability is given by $$\iiint_D f_{X_1, X_2, X_3}(x,y,z) \ dx \ dy \ dz = \iiint_D f(x) f(y) f(z) \ dx \ dy \ dz.$$ Now, fixing an $x \in \mathbb R$, we note that $y$ and $z$ lie in the interval $(-\infty, x)$. So the probability is given by $$\int_{-\infty}^\infty f(x) \left( \int_{-\infty}^x f(y) dy \right) \left( \int_{-\infty}^x f(z) dz \right) dx,$$ which can be written equivalently as $$\int_{-\infty}^\infty f(x) F(x)^2 dx,$$ confirming the integral in Method 2.

Another approach. The original problem can also be solved directly as follows. Sort the three numbers $X_1, X_2, X_3$ and rank them in decreasing order, so that $r_i$ is the rank of $X_i$. (We assume that $X_1, X_2, X_3$ are pairwise distinct; this assumption is ok if this fails to occur with zero probability.) Then since $(X_1, X_2, X_3)$ are iid, the random variable $r := (r_1, r_2, r_3)$ has the uniform distribution over the $3! = 6$ permutations of $(1, 2, 3)$.

Conditioned on $X_1 > X_3$, $r$ takes the uniform distribution over $\{(1,2,3), (1,3,2), (2,1,3)\}$. Of these three possibilities, two of them (namely, ${ (1, 2, 3), (1, 3, 2)}$) correspond to the event $X_1 > X_2$. Hence the probability that $X_1 > X_2$ conditioned on $X_1 > X_3$ is $\frac {2}{3}$.

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Let $I=\text{the index }i\text{ for which }X_i=\max\{X_1,X_2,X_3\}$. Then $$\Pr(X_1 > X_2\text{ and }X_1>X_3) = \Pr(I=1) = \frac13$$ simply because all three possible values of $I$ are equally probable, by symmetry.

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