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 total a 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 :

Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)} <br><br>$P(6) = {5 \over 36}$ <br><br>Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)} <br><br>$\,P(7) = {6 \over {36}}$ = ${1 \over 6}$ <br><br>Game ends and A wins if A throws 6 in 1^st throw or A don’t throw 6 in 1^st throw, B don’t throw 7 in 1^st throw and then A throw 6 in his 2 ^nd chance and so on. <br><br>P(A) = A + $\overline A \overline B A$ + $\overline A \overline B \overline A \overline B A$ <br><br>= $${5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + $$ ..... $\infty$ <br><br>= $${{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}$$ <br><br>= ${{5 \times 36} \over {36 \times 36 - 31 \times 30}}$ <br><br>= ${30 \over {61}}$