Probability that $x>y$ and $x>z$ I have three independent normally distributed variables:
$$x\sim N(0,1) $$
$$y\sim N(0,1) $$
$$z\sim N(0,1) $$
What it the probability that $x$ is greater than both $y$ and $z$ ?
$$P(x>y , x>z)$$
Analytically, I approached the problem this way:
$$P(x>y , x>z)=P(x>y) P(x>z) = 0.5 \times 0.5 = 0.25$$
But when I run simulations I get:
$$P(x>y , x>z)=0.33$$
What is wrong?
 A: Because the events $x>y$ and $x>z$ are not independent, thus you can't write
$$P(x>y , x>z)=P(x>y) P(x>z)$$
It's $1/3$ because these continuous iid RVs have equal probability of being the maximum, and $x$ is just one of them.
A: One way to look at this is to note that the vector $(x,y,z)$ is exchangeable, and since they are continuous random variables, the probability of a tie is zero.  Consequently, each possible order for the values is equally likely.  Here are all the possible orders and their probabilities, with a star next to the orders that satisfy your statement.
$$\begin{matrix}
\text{Order} & & & \text{Probability} \\[6pt]
x<y<z & & & \tfrac{1}{6} & \\
x<z<y & & & \tfrac{1}{6} & \\
y<x<z & & & \tfrac{1}{6} & \\
y<z<x & & & \tfrac{1}{6} & * \\
z<x<y & & & \tfrac{1}{6} & \\
z<y<x & & & \tfrac{1}{6} & * \\
\end{matrix}$$
(Note that orderings with ties have zero probability when the random variables have a continuous distribution.)  Consequently, regardless of the particular (continuous) distribution for the variables, you have:
$$\mathbb{P}(x > y, x > z) = \frac{1}{6}+\frac{1}{6} = \frac{1}{3}.$$
