Question

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 : 

(1) 31/61 

(2) 5/6 

(3) 5/31 

(4) 30/61

Answer 1

Answer is (4) \(\frac{31}{61}\)

P(A) = \(\frac{5}{36}\), P(B) = \(\frac{1}{6}\)

P(A wins) = W + FFW + FFFFW + .........

= \(\frac{5}{36} + \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}\)+ \(\frac{31}{36} \times \frac{5}{6} \times \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36} + ........\)

= \(\frac{\frac{5}{36}}{1-\frac{31}{36} \times \frac{5}{6}}\)

= \(\frac{\frac{5}{36}}{\frac{36 \times 6 - 31 \times 5}{36 \times 6}}\)

= \(\frac{30}{216-155}\)

= \(\frac{31}{61}\)

Answer 2

P(6)=5/36

P(7)=6/36=1/6

P(A wins)=1/6 + 31/36 * 5/6 * 1/6 ..........................

                 =1/6[1/(1-155/216)]

                  = 30/61