Let $X$ be the number of heads. Since the coin is flipped $6$ times with a probability of $1/3$ for a head, $X$ follows a binomial distribution with $n = 6$ and $p = 1/3$. The probability of getting a tail is $q = 1 - 1/3 = 2/3$.
The formula for the probability of getting exactly $k$ heads is:
$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$
We are looking for the conditional probability $P(X = 4 \mid X \text{ is even})$.
By definition, $P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$.
Since getting exactly 4 heads is an even number of heads, this simplifies to:
$$P(X = 4 \mid X \text{ is even}) = \frac{P(X = 4)}{P(X \text{ is even})}$$
Step 1: Calculate the probability of getting an even number of heads, $P(X \text{ is even})$
An even number of heads means getting $0, 2, 4,$ or $6$ heads. Let's calculate the probability for each:
$\mathbf{0}$ heads: $P(X = 0) = \binom{6}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^6 = 1 \cdot 1 \cdot \frac{64}{729} = \frac{64}{729}$
$\mathbf{2}$ heads: $P(X = 2) = \binom{6}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^4 = 15 \cdot \frac{1}{9} \cdot \frac{16}{81} = \frac{240}{729}$
$\mathbf{4}$ heads: $P(X = 4) = \binom{6}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^2 = 15 \cdot \frac{1}{81} \cdot \frac{4}{9} = \frac{60}{729}$
$\mathbf{6}$ heads: $P(X = 6) = \binom{6}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^0 = 1 \cdot \frac{1}{729} \cdot 1 = \frac{1}{729}$
Now, sum these probabilities to find $P(X \text{ is even})$:
$$P(X \text{ is even}) = \frac{64}{729} + \frac{240}{729} + \frac{60}{729} + \frac{1}{729} = \frac{365}{729}$$
Step 2: Use the conditional probability formula
We already calculated $P(X = 4) = \frac{60}{729}$.
Now, divide this by the total probability of an even number of heads:
$$P(X = 4 \mid X \text{ is even}) = \frac{\frac{60}{729}}{\frac{365}{729}} = \frac{60}{365}=\boxed{\frac{12}{73}}$$
