In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :

Step-by-step solution

Sum 6 → ( 1 , 5 ) , ( 5 , 1 ) , ( 3 , 3 ) , ( 2 , 4 ) , ( 4 , 2 ) Sum 7 → ( 1 , 6 ) , ( 6 , 1 ) , ( 5 , 2 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 4 , 3 ) P ( A   wins ) = P ( A ) + P ( A ¯ ) · P ( B ¯ ) · P ( A ) + P ( A ¯ ) P ( B ¯ ) · P ( A ¯ ) · P ( B ¯ ) · P ( A ) + … … We see that this is an infinite G.P. with common ratio P ( A ¯ ) × P ( B ¯ ) Thus, probability that A wins = P ( A ) 1 - P ( A ¯ ) P ( B ¯ ) = 5 3