<s>[INST] 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!
 [/INST] 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 $C$.
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