We are given a biased coin with the probability of getting heads (H) as $\frac{1}{3}$ and thus, the probability of getting tails (T) is $\frac{2}{3}$ (since the total probability of getting either heads or tails should sum to 1).

The experiment involves tossing this coin repeatedly until we obtain two identical outcomes in a row. We are interested in the case where the experiment ends with two consecutive heads.

Let's consider two different scenarios for the sequences of coin tosses :

1. Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so on).
2. Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and so on).

Both types of sequences will result in the experiment ending, so we need to consider both in our final probability calculation.

Scenario 1 : Sequences beginning with H

The simplest sequence is getting HH right away. The probability of this event is $(\frac{1}{3})^2 = \frac{1}{9}$.

The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The probability of this sequence is $\frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{81}$.

We can see a pattern in these sequences. Each sequence in this scenario can be described as getting one head, followed by some number of "tail, head" pairs, and ending with HH.

Hence, the sequences in this scenario form a geometric series, where the ratio between consecutive terms is $r = P(HT) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$.

The sum of an infinite geometric series is given by the formula $\frac{a}{1 - r}$, where $a$ is the first term of the series and $r$ is the common ratio.

In this scenario, the first term is the probability of getting HH immediately, which is $\frac{1}{9}$. So, the sum of this series is :

$$S_1 = \frac{1/9}{1 - 2/9} = \frac{1/9}{7/9} = \frac{1}{7}.$$

Scenario 2 : Sequences beginning with T

In this scenario, we first get a tail, and then we follow the same pattern as in the first scenario.

The simplest sequence is THH, with a probability of $\frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{27}$.

The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The probability of this sequence is $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{4}{243}$.

As before, these sequences form a geometric series, with the ratio between consecutive terms is $r = P(TH) = \frac{2}{9}$.

In this scenario, the first term is the probability of getting THH, which is $\frac{2}{27}$. So, the sum of this series is :

$$S_2 = \frac{2/27}{1 - 2/9} = \frac{2/27}{7/9} = \frac{2}{21}.$$

Finally, to calculate the total probability of the experiment ending with two consecutive heads, we need to sum up the probabilities calculated in both scenarios:

$$P(\text{HH}) = S_1 + S_2 = \frac{1}{7} + \frac{2}{21} = \frac{5}{21}.$$

So, the answer to the problem is $\frac{5}{21}$ (Option B).