Question

In each of a set of games it is $2$ to $1$in favour of the winner of the previous game $:$ what is the chance that the player who wins the first game shall win three at least of the next four?

Answer 1

He has to either win all the games or he can loose one and win the rest out of four

Probability of winning all four $={\left(\frac{2}{3}\right)}^4=\frac{16}{81}$

loose firsrt and win all three $=\frac{1}{3}.\frac{1}{3}.\frac{2}{3}.\frac{2}{3}=\frac{4}{81}$

Loose second and win rest $=\frac{2}{3}.\frac{1}{3}.\frac{1}{3}.\frac{2}{3}=\frac{4}{81}$

Loose third and win rest $=\frac{2}{3}.\frac{2}{3}.\frac{1}{3}.\frac{1}{3}=\frac{4}{81}$

Loose fourth and win rest $=\frac{2}{3}.\frac{2}{3}.\frac{2}{3}.\frac{1}{3}=\frac{8}{81}$

$\Rightarrow$ probability of winning atlest three $=\frac{16}{81}+\frac{4}{81}+\frac{4}{81}+\frac{4}{81}+\frac{8}{81}=\frac{36}{81}=\frac{4}{9}$