Question

In each of a set of games it is 2 to 1 in favour of the winner of the previous game. The chance that the player who wins the first game shall win three at least of the next four is $k/9$. Find the value of $k$.

Answer 1

Let w stand for the winning of a game and I for losing it.
Then there are 4 mutually exclusive possibilities
(i)w,w,w (ii)w,w,l,w (iii)w,l,w,w (iv)l,w,w,w
[Note that case(i)includes both the cases whether he loses or wins the fourth game].
According to the conditions of the question, the probabilities for (i),(ii),(iii)and (iv) are respectively
$\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{2}{3},\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot ,\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{2}{3}\cdot and\frac{1}{3}\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{2}{3}\cdot$
Hence the required probability =$\frac{8}{27}+\frac{4}{81}+\frac{4}{81}+\frac{4}{8}=\frac{36}{81}=\frac{4}{9}$