Probability with an unusual pseudo-random generator
The key here is the part of the problem that says "each value, $v$ occurs with a probability that is inversely proportional to $v$". You can do this by assigning $$
P(1)=c, P(2)=\frac{c}{2}, P(3)=\frac{c}{3}, P(4)=\frac{c}{4}
$$
Then, since these values are the only ones possible, we'll have
$$
c\cdot 1+\frac{c}{2}+\frac{c}{3}+\frac{c}{4} = 1
$$
so
$$
1 =c(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}) = c\cdot\frac{25}{12}
$$
and so $c=12/25$, giving us $P(1)=12/25, P(2)=6/25, P(3)=4/25, P(4)=3/25$. It's easy to see that these satisfy the requirements of the problem.