Biggest among three uniform random variables Question: Given three random variables $X, Y, Z$ of independent uniform distribution in range [0,1], what's the probability for $X$ to be the biggest one?
I've come up with two solutions but they give totally different answers. 


*

*Sol1: Since $X, Y, Z$ are independent and symmetric, anyone of them has equal probability to be the biggest one, so answer is $\frac{1}{3}$.

*Sol2: $\Pr[X\text{ is biggest}]=\Pr[X > Y, X > Z] = \Pr[X > Y]\Pr[X > Z] = \frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.
I guess the first answer is correct because 
$$\Pr[X\text{ is biggest}] + \Pr[Y\text{ is biggest}] + \Pr[Z\text{ is biggest}] $$
should better be 1. But I didn't see why sol2 is incorrect...Thanks in advance for answering!
 A: The logic for the first solution is sound, and all premises are clearly justified.   It is valid.
The second solution would only be valid if the event $X>Y$ were independent of the event $X>Z$.   You have just demonstrated that that premise is not justified.

Also:
$$\begin{align}
\mathsf P(X>Z\mid X>Y) & = \frac{\mathsf P(X>Z\cap X>Y)}{\mathsf P(X>Y)}
\\ & = \frac{\int_0^1 \int_0^x \int_0^x \operatorname d z\operatorname d y \operatorname d x}{\int_0^1 \int_0^x \operatorname d y \operatorname d x}
\\ & = \frac 2 3
\end{align}$$
A: The two events $X>Y$ and $X>Z$ are not independent.  They both depend on $X$, so that shouldn't be surprising.  Knowing that $X>Y$ makes it more probable than it would otherwise be that $X$ is big; hence more probable than it would otherwise be that $X>Z$.  We have $\Pr(X>Y)=1/2)$ and
$$
\Pr(X>Z\mid X>Y) = \frac{\Pr(X>Z\ \&\ X>Y)}{\Pr(X>Y)} = \frac{1/3}{1/2} = \frac 2 3.
$$
The two probabilities in the numerator and the denominator are $1/3$ and $1/2$ is seen to be true be just the method that appears in the posted question.
