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

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