Total number of the possibilites are $\binom{20}{4}$.
Now we calculate the possibilities where Adam can go, and John can't. Thus, 18 students remain, and 3 of them gets a ticket. They are $\binom{18}{3}$.
The chance to get one of such combinations, are: $\frac{\binom{18}{3}}{\binom{20}{4}}$, which is $\frac{\frac{18!}{3! \cdot 15!}}{\frac{20!}{4! \cdot 16!}}$.
If you don't like calculators because you want to train your mind, you can simplify the result:
$\frac{\frac{18!}{3! \cdot 15!}}{\frac{20!}{4! \cdot 16!}}$ = $\frac{4! \cdot 16!}{20!} \cdot \frac{18!}{3! \cdot 15!} = \frac{4 \cdot 16}{20 \cdot 19} = \frac{16}{5 \cdot 19} = \underline{\underline{\frac{16}{95}}}$.