smc

Suppose Laurent's number is in the interval $[0,2017]$. Then, by symmetry, the probability of Laurent's number being greater is $\frac{1}{2}$. Next, suppose Laurent's number is in the interval $[2017,4034]$. Then Laurent's number will be greater with a probability of $1$. Since each case is equally likely, the probability of Laurent's number being greater is $\frac{1+\frac{1}{2}}{2}=\frac{3}{4}$, so the answer is $\boxed{\frac{3}{4}}$.