Question

Three persons A, B, C throw a die in succession till one gets a 'six' and wins the game. Find their respective probabilities of winning.

Answer 1

$$\begin{gathered} P\left(\text{six}\right)=\frac{1}{6} \\ P\left(\text{no six}\right)=\frac{5}{6} \\ P\left(A\text{winning}\right)=P\left(\text{6 in first throw}\right)+P\left(\text{6 in fourth throw}\right)+... \\ =\frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+... \\ =\frac{1}{6}\left[1+{\left(\frac{5}{6}\right)}^3+{\left(\frac{5}{6}\right)}^6+...\right] \\ =\frac{1}{6}\left[\frac{1}{1-{\displaystyle \frac{125}{216}}}\right]\left[{\text{1+a+a}}^2{\text{+a}}^3\text{+}...\text{=}\frac{1}{1-a}\right] \\ =\frac{1}{6}\times \frac{216}{91} \\ =\frac{36}{91} \\ P\left(\text{B winning}\right)=P\left(\text{6 in second throw}\right)+P\left(\text{6 in fifth throw}\right)+... \\ =\frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+... \\ =\frac{5}{36}\left[1+{\left(\frac{5}{6}\right)}^3+{\left(\frac{5}{6}\right)}^6+...\right] \\ =\frac{5}{36}\left[\frac{1}{1-{\displaystyle \frac{125}{216}}}\right]\left[{\text{1+a+a}}^2{\text{+a}}^3\text{+}...\text{=}\frac{1}{1-a}\right] \\ =\frac{5}{36}\times \frac{216}{91} \\ =\frac{30}{91} \\ P\left(\text{C winning}\right)=P\left(\text{6 in third throw}\right)+P\left(\text{6 in sixth throw}\right)+... \\ =\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{2}+... \\ =\frac{25}{216}\left[1+{\left(\frac{5}{6}\right)}^3+{\left(\frac{5}{6}\right)}^6+...\right] \\ =\frac{25}{216}\left[\frac{1}{1-{\displaystyle \frac{125}{216}}}\right]\left[{\text{1+a+a}}^2{\text{+a}}^3\text{+}...\text{=}\frac{1}{1-a}\right] \\ =\frac{25}{216}\times \frac{216}{91} \\ =\frac{25}{91} \end{gathered}$$