Geometric Distribution in Card Game Probability 1) A and B are playing a game where they each write their name on a card. A goes
first and chooses one of the two cards with equal probability. If A chooses her own card
then she wins. If she chooses B’s card, then it is B’s turn to play. He chooses one of the
two cards randomly with equal probability, and he wins if gets his own card. Otherwise, the
turn passes back to A and she tries again. This pattern continues until one or the other
has won. Calculate the probability that A will win this game. (it is not 1/2.)

I'm assuming the way to do this is geometric distribution, but I'm still a but confused on this. I was trying to use sigma notation to denote how to calculate it, but I don't know how to go about this. I'm getting 1/2 which obviously isn't the answer. 

2) C walks up and wants to join the game so she adds a card with her name into the pile.
Now, if A draws B’s card then he goes next and if she draws C’s card then C
goes next. At each turn, the person drawing a card will win if they draw their own card,
and otherwise the turn will pass to whoever is named on the card. Every card is drawn
independently with all three cards having equal probability. They continue to take turns
until somebody wins. Calculate the probability of A winning, the probability of B
winning, and the probability of C winning.

Also assuming this is the same approach as the last problem, but instead of using 1/2, we use 1/3.
 A: The two questions can be very easily solved using recursion.
$Q1$
Let P(A ultimately wins) $=p$
P(A wins on first turn) $= \frac12$
P(A and B both lose on their turn) $= \frac12\cdot\frac12 = \frac14$
Now this brings us back to square $1$, so A's probability from here is again $p$
Thus $p = \frac12 + \frac14\cdot p\;\;$ which yields $ p = \frac23$
$Q2$
Let P(A ultimately wins) $=p$
If A doesn't win on the first toss, the two others have a probability $\frac{1-p}2$ each of ultimately winning.
Either $A$ wins on first turn with probability $p$, or, so to say, becomes one of the "others".
Thus $p = \frac13 + \frac23\cdot\frac{1-p}2,$ which yields $p = \frac12$
P(A wins) $= \frac12\;\;$ P(B wins) $=$ P(C wins) $= \frac14$  
A: Regarding the first question: Let $W$ be the event that $A$ wins. This event can be "A draws A's card on the first draw"; "A draws B's card on the first draw, then B draws A's card, then A draws A's card"....
If we let the string of $C$ and $I$s denote the outcomes, where the $ith$ $C$ means the $i$th play was the player drawing his/her own card, and the $i$th $I$ means the $i$th play was the player drawing the other player's card (e.g. $IIC$ means A drew B's card, then B drew A's card, then A drew A's card), $W=\{C\}\cup \{IIC\} \cup \{IIIIC\} \cup ...$. Note these events are mutualy disjoint, so by countable additivity of probability measures, $P(W)=P(C)+P(IIC)+P(IIIIC)+...=\frac{1}{2}+\frac{1}{8}+...=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{2^{2n}}$.

For the second question, you could describe the situation as a Markov chain, and then find the probability of absorption. This method is explained carefully in Karlin and Pinsky's An Introduction to Stochastic Modeling. I'll try to think of a way that doesn't use Markov chains as well.
