The 19 plays yield outcomes which are independent of one another.

Let $Y_i$ be a random variable representing the payoff for game $i$, where $i \in \{1,2,...,19\}$.

From the problem description, we can write,

$$P(Y=y) = \begin{cases} \frac{18}{38}, &y = $ 1\\ \frac{20}{38}, &y = $ -1 \\ 0, &otherwise\end{cases}$$

The Expected Value notated as $E[Y]=$

$$1*\frac{18}{38} + -1*\frac{20}{38} = $ -0.053$$

Let $Z$ be a random variable defined as

$$Z = \sum_{i=1}^{19} Y_i$$

Then

$$E[Z] = E[\sum_{i=1}^{19} Y(i)]$$

By the linear operator property of expected value, we have

$$E[Z(19)] = \sum_{i=1}^{19} E[Y(i)]$$

Substituting the expectation we computed above,

$$E[Z] = \sum_{i=1}^{19} -0.053 = 19*(-0.053) = $ -1$$

Which is the desired expected value.