$A$ and $B$ are drawing marbles with replacement where there is one white and one black marble. When the individual draws the white ball, they win. Assume $A$ draws first, what is the probability that A wins the game?
Since we only want to look at $P$($A$ wins), we need to look at the (2i + 1)th draw. We also have 1/2 chances of drawing a white marble each trial. From this I have $\frac{1}{2}^{2i+1}$ but I don't understand how to turn this into something I can evaluate as a general probability for the question.
Best Answer
Set up a tree where $A$ has a $\frac{1}{2}$ chance of winning on the first draw and an equal chance of not winning. Along the non-winning branch, there are again two possibilities: $B$ wins or $B$ doesn't in which case the probability of $A$ winning is the same as the original probability.
This should give $$P(A)=\frac{1}{2} +\frac{1}{4}P(A)\implies P(A)=\frac{2}{3}$$