Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are

Q: Wait, why can't we just calculate the probability of getting two heads in a row immediately? Why is this so complicated?

That is a classic trap! If you only calculate $P(HH)$, you are ignoring the fact that the game continues if you get $HT$ or $TH$. The experiment doesn't stop at the second toss unless it's $HH$ or $TT$. You must account for all the 'waiting' time where the sequence alternates. The game is dynamic, not static!

Q: Why do we have to split this into two cases? Can't we just find one big series?

You absolutely could! But splitting into cases based on the first toss is a powerful strategy to reduce cognitive load. It turns one intimidating problem into two manageable ones. It is the 'Divide and Conquer' philosophy of JEE problem solving. It prevents you from getting lost in the infinite series.

Q: Is there a way to solve this without infinite series?

Yes, you can use recursive probability equations. Let $x$ be the probability of success starting from a fresh state. You can write $x = P(H) \cdot P(Success|H) + P(T) \cdot P(Success|T)$. It is elegant, but the series method is often more intuitive for visualizing the sequence structure. Both lead to the same beautiful result!

Question

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

Options:

(A) $\frac{1}{3}$

(B) $\frac{5}{21}$

(C) $\frac{4}{21}$

(D) $\frac{2}{7}$

Sigma Percentile · Accepted Answer

### The Dance of Randomness: Decoding the Infinite Coin Toss

Imagine you are standing in a quiet room, a single coin in your hand. You are tasked with a simple yet profound experiment: toss this coin repeatedly until you see two consecutive outcomes that are the same.

It sounds simple, almost trivial. But as you begin to toss, you realize that the game has a life of its own. It could end in two tosses, or it could drag on for ten, twenty, or even a hundred tosses. Today, we are going to master this chaos.

### Phase 1: The Anatomy of the Game

First, let us ground ourselves in the physics of the problem. We are given the probability of a Head as $P(H) = p = \frac{1}{3}$.

Consequently, the probability of a Tail is $q = 1 - \frac{1}{3} = \frac{2}{3}$. The experiment stops when we hit $HH$ or $TT$. We are only interested in the universe where the experiment concludes with $HH$.

### Phase 2: The Fork in the Road

To solve this, we must embrace the 'Divide and Conquer' mindset. Let us split our reality into two mutually exclusive cases based on the very first toss.

**Case 1: The First Toss is a Head.**
If our first toss is a Head, we are already halfway to our goal. If the second toss is also a Head, the game ends immediately with $HH$. Success!

But what if the second toss is a Tail? We are now in a state of $HT$. We cannot stop here because the outcomes are different. To avoid the failure of $HTT$ (two tails), we must get a Head.

This forces us into the sequence $HTH$. From here, another Head gives us $HTHH$ (Success!). If we get a Tail, we are back to an alternating pattern. This creates an infinite chain of successful sequences: $HH, HTHH, HTHTHH, \dots$.

The probability of this case, $P_1$, is the sum of these probabilities:
$$P_1 = p^2 + p^3q + p^4q^2 + \dots$$

Factoring out $p^2$, we see a beautiful infinite Geometric Progression:
$$P_1 = p^2(1 + pq + (pq)^2 + \dots)$$

### Phase 3: The Infinite Dance

Now, let us solve this series. The sum of an infinite G.P. is given by $S = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio. Here, $a = p^2$ and $r = pq$.

Substituting our values, we get:
$$P_1 = \frac{p^2}{1 - pq}$$

With $p = \frac{1}{3}$ and $q = \frac{2}{3}$, we calculate:
$$P_1 = \frac{1/9}{1 - 2/9} = \frac{1/9}{7/9} = \frac{1}{7}$$

### Phase 4: The Mirror Image

Now, consider **Case 2: The First Toss is a Tail.**
If we start with a Tail, we must avoid $TT$ at all costs. The only way to survive is to get a Head next, leading to $TH$. From $TH$, we need another Head to reach $THH$ (Success!).

If we get a Tail, we are forced into an alternating pattern again. The successful sequences here are $THH, THTHH, THTHTHH, \dots$.

The probability $P_2$ is:
$$P_2 = qp^2 + q^2p^3 + \dots = qp^2(1 + pq + (pq)^2 + \dots)$$

This is the same G.P. structure! Substituting our values:
$$P_2 = \frac{qp^2}{1 - pq} = \frac{(2/3)(1/9)}{7/9} = \frac{2/27}{7/9} = \frac{2}{21}$$

### The Grand Finale

We have conquered both universes. The total probability of the experiment stopping with a Head is simply the sum of our two cases:
$$P = P_1 + P_2 = \frac{1}{7} + \frac{2}{21}$$

Converting to a common denominator, we get:
$$P = \frac{3}{21} + \frac{2}{21} = \frac{5}{21}$$

There it is. The complexity of an infinite process distilled into a single, elegant fraction. Remember, in JEE Advanced, the math is never just about the numbers; it is about the structure you build to contain the randomness. You have done well. The final answer is **$\frac{5}{21}$**.

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

Select Answer:

Visualized Solution

Defining the Target Event

Probabilities of a Single Toss

Case 1: Starting with Head

Case 1: Avoiding Failure

Case 1: The Alternating Pattern

Probability Expression for Case 1

Summing the Infinite GP (Case 1)

Evaluating $P_{1}$

Case 2: Starting with Tail

Case 2: The Alternating Pattern

Probability Expression for Case 2

Summing the Infinite GP (Case 2)

Evaluating $P_{2}$

Total Probability Setup

Conclusion

The Sigma Insight: Addition and Multiplication Theorems

The Dance of Randomness

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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