I cannot figure out part a) ii) and iii) in the following question "Two students, Karl and Hanna, play a game in which they take it in turns to select a card, with replacement, from a well-shuffled pack of 52 playing cards. The first person to select an ace wins the game. Karl has the first turn.
a. i) Find the probability that Karl wins on his third turn.
ii) Show that the probability that Karl wins prior to his (n + 1)th turn is
$\frac{13}{25}$(1 - ($\frac{12}{13}$)^2n
iii) Hence, find the probability that Karl wins the game."
I got a) i) easily by assuming they both lost 2 rounds each, and Karl then wins the 5th round (which is his turn), so it would be ($\frac{12}{13}$)($\frac{12}{13}$)($\frac{12}{13}$)($\frac{12}{13}$)($\frac{1}{13}$).
However with part ii) I got stuck. I assuming that it since it's Karl's (n + 1)th turn, they both would have failed n times. If he would win on the (n + 1)th turn, it would be ($\frac{12}{13}$)^n($\frac{12}{13}$)^n($\frac{1}{13}$). But seeing that they want prior to that, it would be 1 - ($\frac{12}{13}$)^2n($\frac{1}{13}$). I don't know how to go further from here, and I have no clue how to start iii) without understanding this first.
 A: On iii)
Let $K$ denote the event that Karl wins and let $E$ denote the event
that Karl wins at his first turn. Then:
$P\left(K\right)=P\left(K\mid E\right)P\left(E\right)+P\left(K\mid E^{c}\right)P\left(E^{c}\right)$
It is evident that $P\left(K\mid E\right)=1$ and $P\left(E\right)=\frac{1}{13}$.
Next to that we have: $P\left(K\mid E^{c}\right)=1-P\left(K\right)$. 
This because after the first round - lost by Karl - Hanna will - just like
Karl had in that position - have chance $P\left(K\right)$ to win, hence Karl will have chance $1-P(K)$ to win.
To be solved is the equation:
$P\left(K\right)=1\cdot\frac{1}{13}+\left(1-P\left(K\right)\right)\frac{12}{13}$
A: On $(ii)\;\;and (iii)\;\; following\;\; book\;\; method$
P(karl wins by $n_{th}\;\; turn) = \frac1{13}(1 + (\frac{12}{13})^2 + ... (\frac{12}{13})^{2n})$
which is a G.P. with  sum $= \frac1{13}(\frac{1- (\frac{12}{13})^{2n}}{1 - \frac{144}{169}}) = \frac{13}{25}(1- (\frac{12}{13})^{2n}) $
As $n$ tends to infinity, the sum becomes $\frac{13}{25}$
The book has used summation of a G.P. rather than recursion

Added: another method
Consider the first two rounds:
P(Karl wins on first round) $= \frac1{13}= \frac{13}{13}\frac1{13}$
P(Hannah wins on second round) $= \frac{12}{13}\frac1{13}$
Odds in favor of Karl $=13:12$
P(Karl wins) $=\frac{13}{25}$ 
