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

$P(H) = \frac{1}{3},\qquad P(T) = \frac{2}{3}$

$\text{Stop when two consecutive tosses match}$

$\text{To find: } P(\text{stops with } HH)$

$\text{Plan: track only the last toss } (H \text{ or } T)$

$\text{recursions for } h, t \Rightarrow \text{solve} \Rightarrow \text{combine}$

$h = P(\text{stops with } HH \mid \text{last toss } H)$

$t = P(\text{stops with } HH \mid \text{last toss } T)$

$h = \frac{1}{3}\cdot 1 + \frac{2}{3}\cdot t$

$t = \frac{1}{3}\cdot h + \frac{2}{3}\cdot 0 = \frac{h}{3}$

$h = \frac{1}{3} + \frac{2}{3}\cdot\frac{h}{3} = \frac{1}{3} + \frac{2h}{9}$

$h - \frac{2h}{9} = \frac{1}{3} \Rightarrow \frac{7h}{9} = \frac{1}{3}$

$h = \frac{3}{7}$

$t = \frac{h}{3} = \frac{1}{7}$

$P = \frac{1}{3}h + \frac{2}{3}t$

$= \frac{1}{3}\cdot\frac{3}{7} + \frac{2}{3}\cdot\frac{1}{7}$

$= \frac{1}{7} + \frac{2}{21} = \frac{3}{21} + \frac{2}{21}$

$= \frac{5}{21}$

$P(\text{stops with head}) = \frac{5}{21}$

$\text{Correct option: } (2)$