Counting exercise Three players a,b,c take turns in a game according to the following rules:

At the start A and B play (so C does not play). The winner of the first trial plays against C and so on until one of the players wins two trials in a row.
  Possible outcomes are
  aa,acc,acbb,acbaa,
  bb,bcc,bcaa,bcabb etc.

We have to prove that probability of A winning equals p(A) = 5/14, p(B) = 5/14, p(C) = 2/7 
I have been stuck on this problem for a long time. The only thing I have been able to find out so far is that C can never win on even turns.
Re-edit
Players continue to play until one of them wins two times consecutively and all players are equally good at playing the game.
Apologies for leaving such crucial information. 

I have finally solved this problem with a different approach
Sample space
aa ,acc ,acbb ,acbaa ,acbacc ,acbacbb ,acbabcaa ,...
bb ,bcc ,bcaa ,bcabb ,bcabcc ,bcabcaa ,bcabcabb ,...
Each point in the sample space has a probability (1/2^k) associated with it where k is the number of turns.
For example probability of point (aa) = 1/4
Now let us construct a table enumerating probabilities - 

Let us consider the event that C wins overall. But C can only win if k = 3,6,9,12,...
P(C) = P(C3) + P(C6) + P(C9) ......
where P(Ck) = Probability of C winning overall after k turns.  
P(C) = 1/4 + 1/32  .....
P(C) = a / (1 - r)       ( Sum of an infinite GP )
a = 1/4                  ( First Term )
r = 1/8                  ( Ratio ) 
P(C) = 2/7
P(A) = 5/14 = P(B)
 A: First, let us assume $A$ wins the first game.  Let $a$ be the probability that $A$ wins overall, $b$ the probability that $B$ wins overall, and $c$ the probability that $C$ wins overall.  Then we can write (in this case) $a=\frac 12 + \frac b2$ because $A$ either wins the second game (and wins overall) or loses the second game and is now in $B$'s position.  Similarly, $c=\frac 12a$ because if $C$ wins his first game he is in $A$'s position, while if he loses his first game $A$ wins overall.  Finally $b=\frac 12c$ because $B$ needs $C$ to win and he is now coming in, which is $C$'s position.  This gives 
$$a=\frac 12 + \frac b2\\c=\frac a2\\b=\frac c2\\a=\frac 12+\frac c4==\frac 12+\frac a8=\frac 47\\c=\frac 27\\b=\frac 17$$
This is correct for $C$, but we assumed $A$ won the first one.  Clearly $A$ and $B$ have the same winning probability at the start so we can split their total chances evenly, giving $$P(A)=\frac 5{14}, P(B)=\frac 5{14},P(C)=\frac 27$$
A: Perhaps the problem is supposed to be, 
3 players A, B and C take turns at a game according to the following rules. At the start A and B play while C is out. The loser is replaced by C and at the second trial the winner plays against C while the loser is out. The game continues in this way until a player wins twice in succession, thus becoming the winner of the game. What is the probability that Player A will be the winner? What is the probability that player C will be the winner? (assume in each match, A has .6 chance to beat B, and .4 chance to beat C, B has .5 chance to beat C.) 
which I found at http://www.chegg.com/homework-help/questions-and-answers/3-players-b-c-turns-game-according-following-rules-start-b-play-c--loser-replaced-c-second-q2427231
(and, yes, I know this is not an answer, but I don't think it would fit as a comment). 
A: Here's my crack at the question:
Since both players $A,B$ satisfy:
$$M_k(p) = \{\{1, 1\},\{1,0,1,1\},\{1,0,1,0,1,1\},...\}$$ where $M_k(p)$ is the $k$th way for player $p$ to win the game. $\{1\}$ indicates winning the trial, and $\{0\}$ indicates loss.
As player C, and only $C$, satisfies;
$$N_k(p) = \{\{0,1, 1\},\{0,1,0,1,1\},\{0,1,0,1,0,1\},...\}$$
Trials needed to have elapsed for $C$ to win the game is $\mid N_1(C)\mid = (\mid M_1(A) \mid +\mid M_1(B)\mid)-1 = 3$
Thus, chances of player $p$ winning the game in this case is:

Define players $A,B,C$ as $0,1,2$
$t=$Trials needed to win the game.
$n=$Number of players.
$k=$Number of players with the same chances of winning($A,B$), .
$$
P(p) = \left\{\begin{matrix}
{kt-1 \over k((M_1(p)\;t)+1)} & p < 2\\ 
{t \over (N_1(p)\;t)+1} & p = 2
\end{matrix}\right.
,\quad -1 < p < n
$$
Where $P(p)$ is the probability of player $p$ winning.
Edit:
To clarify: The reason that the chance of player $C$ winning is highest, is because there is a greater chance that player ($A,B$) won't win because they share the same set of combinations needed to win.
Disclaimer:  I have not tested this thoroughly in any way, and drafting was started prior to question being answered.
