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Question
A, B and C in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?
Solution
\[P\left( \text { A winning } \right) = P\left( \text{ head in first toss } \right) + P\left( \text{ head in fourth toss }\right) + . . . \]
\[ = \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + . . . \]
\[ = \frac{1}{2}\left[ 1 + \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^6 + . . . \right]\]
\[ = \frac{1}{2}\left[ \frac{1}{1 - \frac{1}{8}} \right] \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{1}{2} \times \frac{8}{7}\]
\[ = \frac{4}{7}\]
\[P\left( \text{ B winning } \right) = P\left( \text{ head in second toss } \right) + P\left( \text{ head in fifth toss } \right) + . . . \]
\[ = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + . . . \]
\[ = \frac{1}{4}\left[ 1 + \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^6 + . . . \right]\]
\[ = \frac{1}{4}\left[ \frac{1}{1 - \frac{1}{8}} \right] \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{1}{4} \times \frac{8}{7}\]
\[ = \frac{2}{7}\]
\[P\left( C \text{ winning } \right) = P\left( \text{ head in third toss } \right) + P\left( \text{ head in sixth toss } \right) + . . . \]
\[ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + . . . \]
\[ = \frac{1}{8}\left[ 1 + \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^6 + . . . \right]\]
\[ = \frac{1}{8}\left[ \frac{1}{1 - \frac{1}{8}} \right] \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{1}{8} \times \frac{8}{7}\]
\[ = \frac{1}{7}\]
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