Question

A and B take turns in throwing two dice, the first to throw 9 being awarded the prize. Show that their chance of winning are in the ratio 9:8.

Answer 1

\[\text{ Total number of events = 36 } \]
\[P\left( \text{ getting } 9 \right) = \frac{4}{36} = \frac{1}{9}\]
\[P\left( \text{ A winning }  \right) = P\left( \text{ getting 9 in first throw }  \right) + P\left( \text{ getting 9 in third throw } \right) + . . . \]
\[ = \frac{1}{9} + \left( 1 - \frac{1}{9} \right)\left( 1 - \frac{1}{9} \right) \times \frac{1}{9} + . . . \]
\[ = \frac{1}{9}\left[ 1 + \frac{64}{81} + \left( \frac{64}{81} \right)^2 + . . . \right]\]
\[ = \frac{1}{9}\left[ \frac{1}{1 - \frac{64}{81}} \right] \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{1}{9} \times \frac{81}{17}\]
\[ = \frac{9}{17}\]
\[P\left( \text{ B  winning } \right) = P\left( \text{ getting 9 in second throw } \right) + P\left( \text{ getting 9 in fourth throw }  \right) + . . . \]
\[ = \left( 1 - \frac{1}{9} \right)\frac{1}{9} + \left( 1 - \frac{1}{9} \right)\left( 1 - \frac{1}{9} \right)\left( 1 - \frac{1}{9} \right) \times \frac{1}{9} + . . . \]
\[ = \frac{8}{81}\left[ 1 + \frac{64}{81} + \left( \frac{64}{81} \right)^2 + . . . \right]\]
\[ = \frac{8}{81}\left[ \frac{1}{1 - \frac{64}{81}} \right] \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{8}{81} \times \frac{81}{17}\]
\[ = \frac{8}{17}\]
\[ \therefore \text{ Winning ratio of A to B } = \frac{\frac{9}{17}}{\frac{8}{17}} = \frac{9}{8}\]