$\Pr(X>Y|Y>Z)$ for independent random variables $X, Y$ and $Z$

Question

Suppose we have three independent random variables, $X, Y$ and $Z$, that are all uniformly distributed over the unit interval (I'm not sure the specific distribution actually makes a difference). I'm interested in the conditional probability $\Pr(X>Y|Y>Z)$. I can think of a few ways of calculating this but, unfortunately, each approach gives a different answer.

1. $\Pr(X>Y|Y>Z)=\Pr(X>Y>Z)=\frac{1}{3!}=\frac{1}{6}$;
2. $\Pr(X>Y|Y>Z)=\Pr(X>\max\{Y, Z\})=\frac{1}{3}$; and,
3. $\Pr(X>Y|Y>Z)=\Pr(X>Y)\Pr(Y>Z)=\frac{1}{4}$.

My intuition suggests that approach 3 is incorrect, but I couldn't explain why. My hunch is that the events $X>Y$ and $Y>Z$ are not independent. The difference between approach 1 and approach 2 seems to be that in 1 I have fixed the order of $Y>Z$, but for 2 it could also be $Z>Y$, and hence the equivalent expression would be $2/3!=1/3$. However, maybe none of these approaches are correct!

Answer 1

By the definition of conditional probability, $$ P(X > Y \vert Y > Z) = \dfrac{P(X > Y, Y > Z)}{P(Y > Z)} $$ Now, since $X, Y, Z$ are independent random variables, we have $$ P(X > Y, Y > Z) = \int_0^1 \int_0^x \int_0^y dzdydx = \dfrac{1}{6} $$ Similarly, $$ P(Y > Z) = \int_0^1 \int_0^y dzdy = \int_0^1 ydy = \dfrac{1}{2} $$ Thus, $$ P(X > Y \vert Y > Z) = \dfrac{1}{3} $$

Answer 2

To elaborate on the comments:

By symmetry, the six possible orderings of $X,Y,Z$ are all equally likely, each has probability $\frac 16$. Those with $Y>Z$ are $$X>Y>Z\quad \quad Y>X>Z\quad \quad Y>Z>X$$

Of those, only $X>Y>Z$ satisfies $X>Y$, so the answer is $\frac 13$.

I note that your method ($2$) claims to yield $\frac 13$, but I do not understand the method. Knowing that $X>\max (Y,Z)$ does not tell you that $X>Y>Z$. Further argument is needed in order to establish that this answers the question.