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

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