jmc

We first find the fraction of the time that either Amy or Lily wins by adding $\frac{3}{8}+\frac{3}{10}$. Since the least common multiple of $8$ and $10$ is $40$, we write $\frac{3}{8}\cdot \frac{5}{5}=\frac{15}{40}$ and $\frac{3}{10}\cdot \frac{4}{4}=\frac{12}{40}$, so our sum is: $$\frac{3}{8}+\frac{3}{10}=\frac{15}{40}+\frac{12}{40}=\frac{15+12}{40}=\frac{27}{40}.$$Since they tie the rest of the time, to find the fraction of the time that they tie, we need to subtract this fraction from $1$. We know that $1=\frac{40}{40}$, so we have that $$1-\frac{27}{40}=\frac{40}{40}-\frac{27}{40}=\frac{40-27}{40}=\frac{13}{40}.$$Therefore, Amy and Lily must tie $\boxed{\frac{13}{40}}$ of the time.