Question 518919
The probability of getting a 7 on two dice is 1/6.


The probability that A wins right away is 1/6. Hence, the probability that B wins right away is (5/6)(1/6) = 5/36. 


The probability that A wins on his second throw is (5/6)(5/6)(1/6) = 25/216. Hence, the probability that B wins on his second throw is (5/6)(5/6)(5/6)(1/6) = 125/1296.


We repeat this pattern to obtain


*[tex \LARGE P(A) = \sum_{i=0}^{\infty} \frac{5^{2k}}{6^{2k+1}}]


*[tex \LARGE = \frac{1}{6}\sum_{i=0}^{\infty} \frac{5^{2k}}{6^{2k}} = \frac{1}{6}\sum_{i=0}^{\infty} (\frac{25}{36})^k]


*[tex \LARGE = (\frac{1}{6})(\frac{1}{1 - \frac{25}{36}}) = \frac{6}{11}]


Therefore, the probability that A wins is 6/11, and the probability that B wins is 1 - (6/11), or 5/11.