A and B throw with a pair of dice.A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6.if A begins, what is his change of winning---------------

The sum 6 can be got in 5 ways. [i.e., (1,5) , (2,4) , (3,3) , (4,2) , (5,1) ] . The probability of A throwing 6 is 5/36 . Thus , the probabilityof A not throwing 6 is 1 – 5/36 = 31/36 . similarly the probability of B throwing 7 is 6/36 and the probability of B not throwing 7 = 1 – 6/36 = 30/36 = 5/6 . Now A can win if he throws 6 in the first , third , fifth , …throws. Hence , the chance of A winning = 0.1

 The probability of A throwing ‘6’ = 5/36 . So , the probability of A not throwing ‘6’ = 1 – 5/36 = 31/36 .The probability of B throwing ‘7’ = 6/36 .So, the probability of B not throwing ‘7’ = 1 – 6/36 = 30/36 .The probability that somebody wins =[1-{(31/36)(30/36)}].So, The probability of winning of A provided that somebody wins = (5/36)/[1-{(31/36)(30/36)}]=30/61.