f.5 probability

This is a geometric series

(1 / 6) + [(5 / 6) x (5 / 6)] x (1 / 6) + {[(5 / 6) x (5 / 6)] ^ 2} x (1 / 6) + {[(5 / 6) x (5 / 6)] ^ 3} x (1 / 6) + ... ... + {[(5 / 6) x (5 / 6)] ^ n} x (1 / 6) + ... ...

= (1 / 6) x {1 + (25 / 36) + [(25 / 36) ^ 2] + [(25 / 36) ^ 3] + ... ... + [(25 / 36) ^ n] + ... ...

= (1 / 6) / [1 - (25 / 36)]

= 6 / 11

The answer is 6/11 and I would like to give you the whole vision of this question
In a game, two players A and B take turns to throw a dice. The one who gets a 6 first wins the game. If A starts the game, find the probability of each of the following events.

(a)A wins the game

(b)B wins the game

(c)Determine whether the first player is more likely to win or not.
(a)
A wins the game when
A throw a dice and get 6. probability=1/6
A throw a dice and not get 6, B throw a dice and not get 6, A throw a dice and get 6
probability=(5/6)(5/6)(1/6)
So the probability of each of A wins the game
=1/6+(5/6)(5/6)(1/6)+(5/6)(5/6)(5/6)(5/6)(1/6)+...
=(1/6)/(1-25/36)
=6/11
(b) So the probability of each of B wins the game
=1-6/11
=5/11
(c) yes, the first player is more likely to win by part (a) and (b)

2008-11-24 19:19:10 補充：
it shoule be version