A counterintuitive problem in conditional probability/combinatorics Please help me to finish/validate my solution for the following problem:

Three players $A,B,C$ play a game. A game is played between two
  players. After a game the winner plays against player who did not play
  last time. Everyone has equal winning chance of $\frac12$. To become a
  champion a player have to win 2 games in a row. Find the probabilities
  of being a champion for each player, if the first game is played
  between $A$ and $B$.

Ok, my attempt:
Let ${PL}_n$ is the probability of an event "player $PL$ became a champion after game $n$". Then it is easy to get that
$A_1=B_1=C_1 =C_2=0,A_2=B_2=\frac14$, $C_3=\frac14$.
With the probability $\frac14$ $4$-th game is played between $A$ and $B$, so the game tree with winning leaves cut off will repeat itself with a period $3$.
So $A_{3n}=B_{3n}= {\frac14(\frac14)}^{n-1}={(\frac14)}^{n}$,$C_{3n+1}= {\frac14(\frac14)}^{n-1}={(\frac14)}^{n}$. Then $A_{\infty}=B_{\infty}=$ sum of infinite sequence with first term $\frac14$ and factor $\frac14$, so  $A_{\infty}=B_{\infty}=\frac13$. Since the probability the no one will will win the game after game $3n+1$ is equal to ${(\frac14)}^{n}\to0$ for large $n$, we get $C_{\infty}=1-A_{\infty}-B_{\infty}=\frac13$.
I do not see any errors in my solution and I like it and believe it is correct, but the result looks counterintuitive - it looks like players $A$ and $B$ should have better chance of being a champion. 
Is my solution correct? Thanks a lot for your help!
 A: The game tree won't repeat itself after 3 games because even though the 4th game is between $A$ and $B$, one of them will be coming in with one win under her belt and will now win with probability $\frac{1}{2}$, not $\frac{1}{4}$.
However, your solution idea works with a little modification, now that we understand that the first game is special. (Spoiler alert: don't read on yet if you don't want to try it out yourself!)
Suppose $A$ wins the first game. Now $A$ wins with probability  $\frac{1}{2}$ (by winning the second game), $C$ wins with probability $\frac{1}{4}$ (by winning the second and third games), $B$ wins with probability $\frac{1}{8}$ (by $A$ losing the second game, and then $B$ winning the next two), and with probability $\frac{1}{8}$ ($C$ wins, $B$ wins, then $A$ wins) the fifth game is between $A$ and $C$ and is in exactly the same situation as the second game. So, given that $A$ wins the first game, $A$ wins with probability $\frac{1}{2}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{2} \cdot \frac{8}{7} = \frac{4}{7}$, $C$ with probability $\frac{1}{4}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{4} \cdot \frac{8}{7} = \frac{2}{7}$, and $B$ with probability $\frac{1}{8}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{8} \cdot \frac{8}{7} = \frac{1}{7}$.
Given that $B$ wins the first game, the roles of $A$ and $B$ are reversed: $B$ wins with probability $\frac{4}{7}$, $C$ with probability $\frac{2}{7}$, and $A$ with probability $\frac{1}{7}$.
So $A$ wins the competition with probability $\frac{1}{2}\cdot\frac{4}{7}+\frac{1}{2}\cdot\frac{1}{7} = \frac{5}{14}$, $B$ with probability $\frac{1}{2}\cdot\frac{1}{7}+\frac{1}{2}\cdot\frac{4}{7} = \frac{5}{14}$, and $C$ with probability $\frac{1}{2}\cdot\frac{2}{7}+\frac{1}{2}\cdot\frac{2}{7} = \frac{2}{7}$. This confirms your intuititon: $A$ and $B$ each have a slightly better chance than $C$.
A: Since $A$ and $B$ obviously have equal chances, we can focus on $C$.
A little thought will show that $C$ can only win in rounds numbered in multiples of $3$: e.g. by $ACC$ or $BCC$, with probability $\frac14$,
or in round $6$, by $ACBACC$ or $BCABCC$, with probability $\frac1{32}$, and so on.
So P($C$ wins) is the sum of an infinite G.P with $a = \frac14, r = \frac18,\text{ and }s_\infty = \frac{1/4}{1 - 1/8} = \frac27$. The balance $\frac57$ will be equally divided between $A$ and $B$, giving each a probability of $\frac5{14}.$
A: Concentrate on the player $C$. After the initial round there are the following three non-terminal states of the game:
$N:\quad$ $C$ has not taken part in the foregoing round.
$W:\quad$ $C$ has won the foregoing round.
$L:\quad$ $C$ has lost the foregoing round.
Denote by $p_N$, $p_W$, $p_L$ the probability that $C$ will become champion, given that the game is in state $N$, $W$, $L$ respectively. The rules of the game then imply
$$p_N={1\over2}\cdot 0+{1\over2}p_W,\quad p_W={1\over2}p_L+{1\over2}\cdot 1,\quad p_L={1\over2}\cdot0+{1\over2}\cdot p_N\ .$$
The solution of this system is $p_N={2\over7}$, $\>p_W={4\over7}$, $\>p_L={1\over7}$. After the initial round the game is in state $N$, hence $C$ becomes champion with probability $p_N={2\over7}$. Since the probability of becoming champion is the same for $A$ and $B$ these probabilities then are ${5\over14}$ each.
A: The problem can be analyzed in terms of a Markov chain. The chain has 3 absorbing states $A_W$ (= player $A$ won 2 games in a row), $B_W$ and $C_W$, and 7 more states $A_0B_0$ (initial state, never returned), $A_0B_1$, $A_1B_0$, $A_0C_1$, $A_1C_0$, $B_1C_0$, $B_0C_1$. Here for example $A_0B_1$ means that the current game is played by $A$ and $B$ and the previous game was won by $B$.

We can come to $C_W$ state only from the states $B_0C_1$ and $A_0C_1$, so $$ P(C_W)=\frac{1}{2}\left(P(B_0C_1) + P(A_0C_1)\right)$$ and by $A-B$ symmetry $$ P(C_W)=P(B_0C_1)$$
We can come to $B_0C_1$ state only from the state $A_1C_0$, so$$ P(B_0C_1)=\frac{1}{2}P(A_1C_0)$$
We can come to $A_1C_0$ state only from the states $A_0B_0$ and $B_1A_0$, so$$ P(A_1C_0)=\frac{1}{2}+\frac{1}{2}P(B_1A_0)$$
We can come to $B_1A_0$ state only from the state $B_0C_1$, so$$ P(B_1A_0)=\frac{1}{2}P(B_0C_1)$$
And we already know that $P(C_W)=P(B_0C_1)$
So putting it all together$$P(C_W)=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}P(C_W)\right)$$ and
$$P(C_W)=\frac{2}{7}$$
