<s>[INST] Find the probability that "A" wins the game. In a game, A and B play against each other. They throw a die alternatively until they get the face of 6. Once one gets 6, he wins. i.e., there is only one winner and they keep playing until one of them wins. 
Consider that B plays first.
What is the probability of A winning?
I don't even have a clue how to start, but I'm quite sure that it is concerned with the geometric progression. 
Pls do help me, any answers will be greatly appreciated. ( but if possible, pls do explain in detail )
 [/INST] For $A$ to win on their first turn we need $B$ to not get a 6 and $A$ to get a 6, for probability $\frac{5}{6}\cdot\frac{1}{6}$.
For $A$ to win on their second turn, we need both to miss on their first try, $B$ to miss on their second try, and $A$ to get 6 on their second try, for probability $\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}$.
For $A$ to win on their third turn, we need $B$, $A$, $B$, $A$, $B$ to miss in that order and then $A$ to get a 6, for probability $\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}$.
This pattern continues infinitely so the probability $A$ wins the whole thing is $$\frac{5}{6}\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}+\cdots = \dfrac{\frac{5}{6}\cdot\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2} = \dfrac{5}{11}.$$
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