Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are the same. If the probability of a random toss resulting in a head is $\frac{1}{3}$, then the probability that the experiment stops with a head is:

The correct option is $\frac{5}{21}$.

Let us analyze the problem step-by-step to calculate the probability that the experiment stops with a head.

1. Define the given probabilities for a single toss:
Let H denote getting a Head and T denote getting a Tail.
Probability of getting a Head, P(H) = $\frac{1}{3}$.
Probability of getting a Tail, P(T) = 1 - $\frac{1}{3}$ = $\frac{2}{3}$.

2. Understand the stopping condition:
The experiment stops as soon as two consecutive tosses yield the same outcome (i.e., HH or TT).
Therefore, for the experiment to stop with a head, the sequence of tosses must end with HH, and no two consecutive outcomes prior to the last toss can be identical.

3. Identify the valid sequences ending in HH:
Since consecutive tosses before the final pair must alternate (to avoid stopping earlier), there are two possible types of sequences depending on the initial toss:

Case A: Sequence starts with Head (H)
The sequence of outcomes must alternate until the final toss: H, T, H, T, ..., H, H.
• 2 tosses: H H (Not alternating, just two heads)
Probability = P(H) × P(H) = $\left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{1}{9}$
• 4 tosses: H T H H
Probability = P(H) × P(T) × P(H) × P(H) = $\left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{2}{81}$
• 6 tosses: H T H T H H
Probability = $\left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{4}{729}$
In general, for a sequence of length 2n starting with H:
Probability P_A,n = ${\left(\frac{1}{3}\right)}^n\times {\left(\frac{2}{3}\right)}^{n-1}\times \left(\frac{1}{3}\right)$

Summing this infinite geometric series for Case A:
$P_A=\frac{1}{9}+\frac{2}{81}+\frac{4}{729}+\cdots$
This is an infinite geometric series with first term $a=\frac{1}{9}$ and common ratio $r=\frac{2}{9}$.
$P_A=\frac{a}{1-r}=\frac{\frac{1}{9}}{1-\frac{2}{9}}=\frac{\frac{1}{9}}{\frac{7}{9}}=\frac{1}{7}$

Case B: Sequence starts with Tail (T)
The sequence of outcomes alternates starting with T: T, H, T, H, ..., H, H.
• 3 tosses: T H H
Probability = P(T) × P(H) × P(H) = $\left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{2}{27}$
• 5 tosses: T H T H H
Probability = P(T) × P(H) × P(T) × P(H) × P(H) = $\left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{4}{243}$
Summing this infinite geometric series for Case B:
$P_B=\frac{2}{27}+\frac{4}{243}+\cdots$
This is an infinite geometric series with first term $a=\frac{2}{27}$ and common ratio $r=\frac{2}{9}$.
$P_B=\frac{a}{1-r}=\frac{\frac{2}{27}}{1-\frac{2}{9}}=\frac{\frac{2}{27}}{\frac{7}{9}}=\frac{2}{27}\times \frac{9}{7}=\frac{2}{21}$

4. Total Probability:
The total probability that the experiment stops with a head is the sum of probabilities of Case A and Case B:
$P=P_A+P_B=\frac{1}{7}+\frac{2}{21}=\frac{3}{21}+\frac{2}{21}=\frac{5}{21}$

Thus, the probability that the experiment stops with a head is $\frac{5}{21}$.