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Three friends A, B and C are playing with a pair of dice. They throw two dice alternately. Coming of a doublet on two dice leads to a success and the game stops. If A starts the game, then the probability of his winning, is :

Solution

✅ Correct Option: 2

P(doublet)=636=16P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}, q=56q = \frac{5}{6}. A wins on turns 1, 4, 7, ...: P(A)=p+q3p+q6p+…=p1−q3=1/61−125/216=1/691/216=2166⋅91=3691P(A) = p + q^3 p + q^6 p + \ldots = \frac{p}{1-q^3} = \frac{1/6}{1-125/216} = \frac{1/6}{91/216} = \frac{216}{6\cdot 91} = \frac{36}{91}.

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