Two persons throw a die alternately till one of them gets a 'three' and wins the game. The ratio of their respective probabilities of winning is

Let the persons be \(\mathrm{A}\) and \(\mathrm{B}\). Suppose that A has the first throw. We know that the probability of getting 'a three' in a single throw \(=\frac{1}{6}\) and that of 'not getting a three' \(=1-\frac{1}{6}=\frac{5}{6} \text {. }\) \(\mathrm{P}(\mathrm{A}\) wins \()=\frac{1}{6}+\left(\frac{5}{6}\right)^2 \frac{1}{6}+\left(\frac{5}{6}\right)^4 \frac{1}{6}+\ldots\). .to \(\infty\) [Since, A can win at the \(1^{\text {st }}, 3^{\text {rd }}, 5^{\text {th }}, \ldots\). mutually exclusive trials with respective probabilities \(\begin{aligned} & \left.\frac{1}{6},\left(\frac{5}{6}\right)^2 \frac{1}{6},\left(\frac{5}{6}\right)^4 \frac{1}{6} \ldots \ldots\right] \\ & =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^2+\ldots \ldots . .\right] \\ & =\frac{1}{6} \times \frac{1}{1-\left(\frac{5}{6}\right)^2}=\frac{1}{6} \cdot \frac{36}{36-25}=\frac{6}{11} \\ & \Rightarrow P(\text { Bwins })=1-\frac{6}{11}=\frac{5}{11} . \end{aligned}\) i.e. the probability of winning of the person having the first throw \(=\frac{6}{11}\) and that of the other \(=\frac{5}{11}\). Hence the ratio of chances of their winning = 6:5.