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Question
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.
Solution
Total of 7 on the dice can be obtained in the following ways:
(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)
Probability of getting a total of 7 = \[\frac{6}{36} = \frac{1}{6}\] Probability of not getting a total of 7 = \[1 - \frac{1}{6} = \frac{5}{6}\] Total of 10 on the dice can be obtained in the following ways: (4, 6), (6, 4), (5, 5)
Probability of getting a total of 10 = \[\frac{3}{36} = \frac{1}{12}\] Probability of not getting a total of 10 = \[1 - \frac{1}{12} = \frac{11}{12}\] Let E and F be the two events, defined as follows:
E = Getting a total of 7 in a single throw of a dice
F = Getting a total of 10 in a single throw of a dice
P(E) = \[\frac{1}{6}\] \[P\left( \bar{E} \right) = \frac{5}{6}\] , P(F) = \[\frac{1}{12}\] \[P\left( \bar{F} \right) = \frac{11}{12}\] A wins if he gets a total of 7 in 1st, 3rd or 5th ... throws.
Probability of A getting a total of 7 in the 1st throw = \[\frac{1}{6}\] A will get the 3rd throw if he fails in the 1st throw and B fails in the 2nd throw.
Probability of A getting a total of 7 in the 3rd throw = \[P\left( \bar{E} \right) P\left( \bar{F} \right) P\left( E \right) = \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\] Similarly, probability of getting a total of 7 in the 5th throw =
\[P\left( \bar{E} \right) P\left( \bar{F} \right) P\left( \bar{E} \right) P\left( \bar{F} \right) P\left( E \right) = \frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\] and so on
Probability of winning of A = \[\frac{1}{6}\] + \[\left( \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6} \right)\] +
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