Assuming the odds are fair, the odds paid are the odds against the bettor. When a draw with money returned is a possibility these odds are given by $$\frac{Pr(Bettor LosesIfNotDrawn)}{Pr(Bettor Wins If Not Drawn)}:1=\frac{Pr(Bettor Loses)}{Pr(NotDrawn)}\times\frac{Pr(NotDrawn)}{Pr(Bettor Wins)}:1=\frac{Pr(Bettor Loses)}{Pr(Bettor Wins)}:1$$
Suppose the odds against a goal difference of more than $0.5$ are $a:1$ and the odds against a goal difference of more than $1.5$ are $b:1$, where $b>a$. Then the probability of a score of $0$ or less is $\frac{a}{a+1}$ and the probability of a score of $2$ or more is $\frac{1}{b+1}$.
As a check, the probability of a score of exactly $1$ is $$1-\frac{a}{a+1}-\frac{1}{b+1}>\frac{a}{a+1}+\frac{1}{a+1}=1-\frac{a+1}{a+1}=0$$ and so the probability that the score is exactly $1$ is positive, as it needs to be.
The the bettor wins with a goal difference of more than $1.0$ and loses with a goal difference of less than $1.0$, so the odds against him winning will be $$\frac{Pr(Bettor Loses)}{Pr(Bettor Wins)}:1=\frac{Pr(GD<1.0)}{Pr(GD>1.0)}:1=\frac{\frac{a}{a+1}}{\frac{1}{b+1}}:1=\frac{a(b+1)}{(a+1)}:1$$