Gate probability

Question

Two players,$A$ and $B$,alternately keep rolling a fare dice.The person to get six first wins the game.Given that player $A$ starts the game,the probability that $A$ wins the game is

• A. $5/11$
• B. $1/2$
• C. $7/13$
• D. $6/11$

Answer 1

the probability  of throwing $6$ is $1/6$

$P(win) = 1/6$ ,  $P(lose) = 1 - 1/6 = 5/6$

 if A throws and 6 comes then the game will stop and A will win. if A loses then B will throw the die (B has to lose because according to question we want A to win the game) and then A will throw . and so on.

It's like 

$A + A'B'A + A'B'B'A + A'B'B'B'A + .......................................$

$A'$ means A loses the game

$B'$ means B loses the game

Probability to lose the game for both A and 5 is 5/6

. 

$1/6 + (5/6)*(5/6)*(1/6) + (5/6)*(5/6)*(5/6)*(1/6) + .........................$

first term(a) = 1/6

common ratio(r) = 25/36

sum of an infinite geometric progression = $ a / (1 - r ) $

                                                                 = $(1/6) / ( 1 - 25/36 )$

                                                                 = $ (1/6) / (11/36) $

                                                                 = $6/11$