Three chessplayers compete in tournament by the following scheme:
Players A and B play match, than winner plays with C, then new winner plays with the loser of the previous match. Tournament ends at the event of two consecutive wins of some player.
It's given, that players are equal in skills, hence outcome of the match should be considered as coin flip.
We have to find winning probabilities of chessplayers A, B and C, given that first match was between A and B, and A scored it.
Any ideas?;
Edit I misread task terribly. It's two wins required, not three. Already made an edit. Now it's easier significantly.
[Math] Probability of winning the chess tournament
probability
Best Answer
There are four states to consider, namely $AC$, $CB$, $BA$, and the ending state $E$. Here $XY$ encodes that the next match is between $X$ and $Y$, whereby $X$ has scored immediately before. At the start we are in state $AC$. Denote the probability that $Z$ will win when the game is in state $XY$ by $p_Z(XY)$.
From a figure one easily reads off that one has $$p_A(AC)={1\over2} +{1\over2}p_A(CB),\quad p_A(CB)={1\over2}p_A(BA),\quad p_A(BA)={1\over2}p_A(AC)\ .$$ Solving this system gives $$p_A(AC)={4\over7},\quad p_A(BA)={2\over7},\quad p_A(CB)={1\over7}\ .$$ Using the inherent circular symmetry of the problem we then obtain $$p_B(AC)=p_A(CB)={1\over7} ,\quad p_C(AC)=p_A(BA)={2\over7}\ .$$