Question

two players, a and b, alternately keep rolling a fair dice. the person to get a six first wins the game. given that player a starts the game, the probability that a wins the game is

Answer 1

it is infinite
because in first chance he can also get 6
maybe in the next chance
so there is a infinite probability to get 6 anytime and win the game.

Answer 2

A and B throws a Fair dice one after another. Whoever throws 6 first wins , What is the probability that A wins ?
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Prakash Raj
Prakash Raj, love stats
Answered Feb 2, 2016 · Author has 152 answers and 90.8k answer views
probability of winning is throwing 6. prob of throwing 6 is 1/6.

prob of winning = 1/6

prob of losing is 1-(1/6)= 5/6

assuming that A start the game i.e A throws the die first and then B and the game goes on

suppose if A throws and if 6 comes then the game will stop and A will win. but if A loses then B will throw the die (here B will have to loose because according to question we want A to win the game) and then A will throw . and so on.

it will be like

A+ ~A~BA+ ~A~B~A~BA+..........

HERE A denotes A wins and ~A means A loses i.e 6 does not come . similarly ~B means B loses.

this will form an infinite geometric progession series

converting into probabilities we have

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

prob of losing for A and B is throwing other than 6 i.e 5/6.

here first term is 1/6 and common ration is 25/36.

applying formula for the sum of an infinite geometric progression

= (first term) / (1-common ratio) = (1/6) / (1-(25/36)) = 6/11

prob of A winning is 6/11.