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

Let the probability of getting Head on a single toss be $$p=\frac13$$. Hence, the probability of getting Tail is $$q=1-p=\frac23$$.

When the experiment ends, it must end either with the pair $$HH$$ or with the pair $$TT$$. We want the probability that it ends with $$HH$$.

Define the following conditional probabilities:
  • $$P_H$$ : probability that the experiment finally stops with $$HH$$, given that the most recent toss is a Head.
  • $$P_T$$ : probability that the experiment finally stops with $$HH$$, given that the most recent toss is a Tail.

1. Recurrence for $$P_H$$ If the last toss is H, then on the next toss:
  • With probability $$p$$ we get another H → the pair $$HH$$ appears and we stop; this contributes $$p\times1$$.
  • With probability $$q$$ we get T → no stop; the new “last toss” is now T, so the chance of ending with $$HH$$ becomes $$P_T$$. Hence $$P_H = p + qP_T \quad -(1)$$

2. Recurrence for $$P_T$$ If the last toss is T, then on the next toss:
  • With probability $$p$$ we get H → no stop; the new “last toss” is H, so the chance becomes $$P_H$$.
  • With probability $$q$$ we get another T → the pair $$TT$$ appears; we stop, but this is a failure for our requirement, so contributes 0. Thus $$P_T = pP_H \quad -(2)$$

3. Solve the two equations Substitute $$P_T$$ from $$(2)$$ into $$(1)$$:
$$P_H = p + q(pP_H) = p + pqP_H$$ $$P_H(1 - pq) = p$$ $$P_H = \dfrac{p}{1-pq}$$

With $$p=\frac13,\; q=\frac23$$ we get $$pq = \frac13\cdot\frac23 = \frac29,\quad 1-pq = 1-\frac29 = \frac79$$ $$P_H = \frac{\tfrac13}{\tfrac79} = \frac13 \times \frac97 = \frac37$$

4. Overall probability of stopping with $$HH$$ The very first toss can be H or T:
  • First toss H (probability $$p$$): we are in the “last toss = H” state, so eventual success chance is $$P_H$$.
  • First toss T (probability $$q$$): we are in the “last toss = T” state, so eventual success chance is $$P_T = pP_H$$ (from $$(2)$$).
Therefore $$P=\;pP_H + qP_T = pP_H + q(pP_H) = pP_H(1+q)$$

Substituting the values: $$1+q = 1+\frac23 = \frac53$$ $$P = \frac13 \times \frac37 \times \frac53 = \frac{15}{63} = \frac{5}{21}$$

Hence the probability that the experiment stops with Head is $$\dfrac{5}{21}$$.

Option B which is: $$\frac{5}{21}$$