Step 1: Rewrite $X$ in terms of a standard Bernoulli variable.
Let $Z$ be a standard Bernoulli variable that takes the value $1$ for heads (probability $p=1/3$) and $0$ for tails (probability $q=2/3$). The variable in this question, $X$, can be written as $X = 2Z - 1$: when $Z=1$ (heads), $X = 2(1)-1 = 1$; when $Z=0$ (tails), $X=2(0)-1=-1$. This matches the definition of $X$ given in the question.
Step 2: Recall the variance of a Bernoulli variable.
For a Bernoulli variable $Z$ with success probability $p$, the variance is a standard result:
\[ \text{Var}(Z) = p(1-p) = pq \]
Here $p = \frac{1}{3}$ and $q = \frac{2}{3}$, so:
\[ \text{Var}(Z) = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} \]
Step 3: Use the scaling property of variance for $X = 2Z - 1$.
For any linear transformation $X = aZ + b$, variance scales as $\text{Var}(X) = a^2 \text{Var}(Z)$; the shift $b=-1$ does not affect the variance. Here $a=2$:
\[ \text{Var}(X) = 2^2 \times \text{Var}(Z) = 4 \times \frac{2}{9} = \frac{8}{9} \]
Step 4: Match with the given options.
This value, $\frac{8}{9}$, corresponds to option (C).
Final Answer:
The variance of $X$ is $\frac{8}{9}$.
\[ \boxed{\frac{8}{9}} \]