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