Expected Value probability question - answer clarification

Question

this is a relatively simple question but my book's answer is giving a different answer.

Question: Tickets in a game of chance can be purchased for 2 dollars. Each ticket has a 30% chance of winning 2 dollars, a 10% chance of winning $20, and otherwise loses. How much are you expected to win or lose if you play the game 100 times?

My answer: E(X) = -2*60 + 18*10 = $60; the book has 100 dollars instead.

Who is correct?

Thanks

Answer 1

Expectation of one ticket $Y$ is $$ E[Y] = 0.3 \cdot 2 + 0.1 \cdot20 - 2 = 0.6 $$ Then $E[100Y] = 100 E[Y]= 60$

Answer 2

I would read the question the way you did,

$$ -\$2\times 60 +(\$2-\$2)\times 30+ (\$20-\$2)\times 10 = \$60$$

with the possible alternative that you get your original stake back with a winning ticket (e.g. the $30\%$ chance is of a net gain of $\$2$, not just a refund), making the calculation

$$ -\$2\times 60 +\$2\times 30+ \$20\times 10 = \$140$$

There is no reason to take the average of these, but it would give you the book's $\$100$