$$3$$ friends $$A,\,B$$  and $$C$$ play the game “Pahle Hum Pahle Tum” in which they throw a die one after the other and the one who will get a composite number $${1^{st}}$$ will be announced as winner, If $$A$$ started the game followed by $$B$$ and then $$C$$ then what is the ratio of their winning probabilities ?

Solution :
Probability of getting a composite number is $$\frac{2}{6} = \frac{1}{3}$$
$$\eqalign{ & {\text{Probability that }}A{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{1}{3}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{1}{3} \times \frac{{27}}{{19}}} \right) \cr & = \frac{9}{{19}} \cr & {\text{Probability that }}B{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{2}{9}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{2}{9} \times \frac{{27}}{{19}}} \right) \cr & = \frac{6}{{19}} \cr & {\text{Probability that }}C{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{4}{{27}}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{4}{{27}} \times \frac{{27}}{{19}}} \right) \cr & = \frac{4}{{19}} \cr & {\text{So required ratio is}}\,\,9:6:4 \cr} $$