The probability of rolling a 1 is 0.5, so the remaining five possible outcomes (2-6) each have probability 0.1.
If [tex]W[/tex] represents your winnings from the playing the game, then
[tex]P(W=w)=\begin{cases}0.6&\text{for }w=2\\0.4&\text{for }w=-1\end{cases}[/tex]
because the probability of rolling a 1 or 2 is
[tex]P(1\text{ or }2)=P(1)+P(2)=0.5+0.1=0.6[/tex]
and rolling any of the other values is complementary to this event.
The expected value of the game is then
[tex]E[W]=\displaystyle\sum_wwP(W=w)=2\cdot0.6+(-1)\cdot0.4=\boxed{\$0.80}[/tex]