Question

A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12 : 11.

Answer 1

$$\begin{gathered} \text{There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is 10, i.e.}\left[\left(4,6\right)\left(5,5\right)\left(6,4\right)\right]\text{.} \\ \therefore P\left(\text{sum of the numbers is 10}\right)=\frac{3}{36}=\frac{1}{12} \\ P\left(\text{sum of the numbers is not 10}\right)=1-\frac{1}{12}=\frac{11}{12} \\ P\left(\text{any numner other than six}\right)=\frac{5}{6} \\ P\left(A\text{winning}\right)=P\left(\text{10 in first throw}\right)+P\left(\text{10 in third throw}\right)+... \\ =\frac{1}{12}+\frac{11}{12}\times \frac{11}{12}\times \frac{1}{12}+... \\ =\frac{1}{12}\left[1+{\left(\frac{11}{12}\right)}^2+{\left(\frac{11}{12}\right)}^4+...\right] \\ =\frac{1}{12}\left[\frac{1}{1-{\displaystyle \frac{121}{144}}}\right]\left[{\text{1+a+a}}^2{\text{+a}}^3\text{+}...\text{=}\frac{1}{1-a}\right] \\ =\frac{1}{12}\times \frac{144}{23} \\ =\frac{12}{23} \\ P\left(B\text{winning}\right)=1-P\left(A\text{winning}\right)=\frac{11}{23} \\ \mathrm{Now}, \\ \frac{P\left(\text{A winning}\right)}{P\left(\text{B winning}\right)}=\frac{{\displaystyle \frac{12}{23}}}{{\displaystyle \frac{11}{23}}}=\frac{12}{11} \\ \text{Hence proved.} \end{gathered}$$