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 $C$.
Best Answer
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.