Given,

$X$ gets the bigger part than $Y$

So, the bigger part could be any number from $10000$ to $20000$.

Now, if the bigger part is to be at least twice as much as the smaller part, we have

$X\ge 2Y$

$\Rightarrow X\ge 2\left(20000-X\right)$

$\Rightarrow X\ge \frac{40000}{3}$
So, the probability is equal to $\frac{20000-40000/3}{20000-10000}=\frac{2}{3}$

Hence the answer is  $\frac{2}{3}$