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?

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.
