Two wins in a row in a game involving three players Three players, let's call them $A,B$ and $C$, play a game of chess. The first match is between $A$ and $B$. The winner will go on to play the third player (who is $C$ in the second match). The game continues until a player win $2$ matches in a row, who will be the eventual winner.
The chance for each player to win a match is one half.
Find the chance of winning the game for $A,B$ and $C$.
 A: Let's deal with some notation first. Let $KL.M$ correspond to the probability of a situation that the current match is between $K$ and $L$, $K$ has won the previous game, and $M$ is the ultimate winner of the game. Here, the understanding of the probability is the conditional one with the condition being we already have arrived at the state where the $K$ has won the previous game and currently playing against $L$. 
Now, in the first game, if $A$ wins, then the next state will have $AC.X$. If $A$ wins it is over, if $C$ wins we jump to a situation with $CB.Y$, and here if $C$ loses we jump to $BA.Z$ type of a situation. If $A$ wins we get back to our $AC.X$ situation again. Therefore, there are three relevant situation, $AC.X$, $CB.Y$ and $BA.Z$. 
Here, we would have several equalities. For instance $AC.A$ would be the probability that $A$ wins conditional on the fact that $A$ won the previous game and would be playing with $C$. Now, the chances that $A$ wins right away is equal to $1/2$. The chances that $A$ wins after a loss to $C$ would be equal to $CB.A\times 1/2$ where we multiply by 1/2 because we jump to the state $CB$ with $1/2$ probability. Then you would observe that $AC.A=1/2+CB.A$. The chances for $AC.B$ would correspond to a situation where $A$ necessarily loses to $C$ and $B$ wins after the $CB$ state, which would equal to $1/2 \times CB.B$. We can make these calculations for $9$ different situations which would yield
$$
\begin{align}
AC.A &=& 1/2+CB.A\\
AC.B&=&1/2 CB.B\\
AC.C&=&1/2 CB.C \\
CB.A &=& 1/2 BA.A\\
CB.B&=&1/2 BA.B\\
CB.C&=&1/2+1/2 BA.C \\
BA.A &=& 1/2 AC.A\\
BA.B&=&1/2+1/2 AC.B\\
BA.C&=&1/2 AC.C \\
\end{align}
$$
So you have $9$ equations in $9$ unknowns here. 
Next, you would look at the situations with $B$ winning in the first round. Then your states would look like $BC.X, CA.Y, AB.Z$ which again would yield similar $9$ equations in $9$ unknowns. 
Now, the probability that $A$ wins is equal to $1/2\times AC.A + 1/2 \times BC.A$, for $B$ this would equal $1/2\times AC.B + 1/2 \times BC.B$, and for $C$ we would have $1/2 \times AC.C+1/2\times BC.C$. The reasoning is quite straightforward. For $A$, there are two cases in the beginning, with $1/2$ probability we would get to the state $AC.X$ and with $1/2$ chance we would end up with $BC.Y$. Similar arguments would imply the other equalities.
A: Clearly the chance that $A$ and $B$ win is the same by symmetry.  Let the chance that $C$ wins be $c$.  When $C$ plays the first game, if he loses the match is over.  If he wins twice in a row he wins the match.  If he wins one and loses the next, there is $\frac 12$ chance the match will be over before he plays again. If not, he is in the same position as in his first game.  So $c=\frac 14$ (that he wins his first two games) $+\frac 18c$ (that he wins the first, loses the second, then wins eventually).  This gives the result that $c=\frac 27=\frac 4{14}$, while the chance for each of $A,B$ is $\frac 5{14}$.
A: Since $A$ and $B$ must have equal chances, we shall focus on $C$.
Using the symbols $W,L,N$ to indicate Win, Loss, Not-in-match, we can chart $C's$ course.
Note that if a player is not in a match:


*

*she has to necessarily win her next match else someone else will  

*she can win only $2$ matches later at the earliest.

*it is immaterial (for $C$) who wins the first match 
So $C$ can win like $NWW,\;\; NWLNWW,\;\; NWLNWLNWW$ etc
This is an infinite $G.P.$ with $a = 1/4, r = 1/8, S(\infty) = \frac{1/4}{1-1/8}= \frac27$
The rest gets equally divided between $A$ and $B$
$P(A) = 5/14,\;\; P(B) = 5/14,\;\;P(C) = 2/7$ 
A: After the first match (between $A$ and $B$), there will be a latest winner, a next opponent, and a latest loser.  Let's let $w$, $n$, and $\ell$ denote the probability of the game eventually being won by the latest winner, next opponent, and latest loser, respectively.  Note that our main interest is in $n$, since the game begins with $C$ as the next opponent, regardless of the outcome of the first match.
Now if the game does not end with the latest winner beating the next opponent, which happens with probability $1/2$, then the next opponent becomes the latest winner, the latest loser becomes the next opponent, and the latest winner becomes the latest loser.  This implies $n={1\over2}w$ and $\ell={1\over2}n$.  But we must have $w+n+\ell=1$, and thus
$$2n+n+{1\over2}n=1$$
which solves to $n=2/7$.
