Probability of winning a rigged coin-flipping game Betsy and Katie are playing a game with an unfair coin. The coin is rigged to come up heads with probability $\frac35$ and tails with probability $\frac25$. 
Betsy goes first. The two take turns. The first player to flip a tail wins. What is Betsy's probability of winning?
What confuses me is that the game can continue for an unlimited amount of times (at least until Betsy flips a tail). How would I solve this?
 A: Let $p$ be the probability that Betsy wins the game. We check some cases. If Betsy flips tails on her first move, then she wins and the game ends; this happens with probability $2/5$. Otherwise, Katie gets a chance to play; this happens with probability $3/5$. If Katie gets a chance to play, she wins with probability $2/5$, or resets the game with probability $3/5$. Therefore, $$p = \frac{2}{5} \cdot 1 + \frac{3}{5} \left(\frac{2}{5} \cdot 0 + \frac{3}{5} p \right) \implies p = \frac{5}{8}$$
Alternatively, the probability that Betsy wins equals the probability she wins on her first move, plus the probability she wins on her second move, plus ... 
The probability Betsy wins on her $k$th move is $$\left(\frac{3}{5} \right)^{2(k-1)} \cdot \frac{2}{5}$$ since every flip before her winning move must have been a heads. So the answer is $$\frac{2}{5} \cdot \sum_{k=1}^{\infty} \left(\frac{3}{5} \right)^{2(k-1)} = \frac{5}{8}$$ 
Note that both of these solutions are essentially the same.
A: The Betsy wins under these outcomes. $T,\space HHT,\space HHHHT ,\space HHHHHHT$ etc.
Those chances are $(\frac{2}{5}), (\frac{3}{5})(\frac{3}{5})(\frac{2}{5}), (\frac{3}{5})(\frac{3}{5})(\frac{3}{5})(\frac{3}{5})(\frac{2}{5})$ So you can set that up as a summation that is a geometric decreasing summation that is solvable.
That geometric series comes to $$\frac{\text{first term}}{1-\text{ratio}}=\frac{\frac{2}{5}}{1-(\frac{3}{5})(\frac{3}{5})}=\frac{5}{8}$$
A: The probability that Betsy wins at attempt #$1$ is $\frac25$
The probability that Betsy wins at attempt #$2$ is $\frac35\cdot\frac35\cdot\frac25$
The probability that Betsy wins at attempt #$n$ is $(\frac35)^{2n}\cdot\frac25$
The probability that Betsy wins at some point is $\sum\limits_{n=0}^{\infty}(\frac35)^{2n}\cdot\frac25$

$$\sum\limits_{n=0}^{\infty}(\frac35)^{2n}\cdot\frac25=\frac25\cdot\sum\limits_{n=0}^{\infty}(\frac{9}{25})^n=\frac25\cdot\frac{1}{1-\frac{9}{25}}=\frac58$$
