Sixty-sided die beating a forty-sided die

You roll one fair die with faces $1$–$40$ and another fair die with faces $1$–$60$. What is the probability that the $60$-sided die shows the strictly larger number?

$79/120$. With probability $1/3$ the big die lands above $40$ and wins outright. Otherwise both dice are uniform on $1$–$40$: they tie with chance $1/40$ and otherwise each is equally likely to be higher, giving $\tfrac{39}{80}$. Total $\tfrac13+\tfrac23\cdot\tfrac{39}{80}=\tfrac{79}{120}$.