Statistics of a racketball/volleyball/badminton game It's been a while since I've used my knowledge in statistics and I have no idea how to turn that problem into an equation. I wanted to challenge myself but I failed. I thought maybe you too would like this challenge.
Starting with pre-defined score and who currently has the ball, I'd like to estimate the odds of winning of each player/team assuming they are precisely as likely to win any exchange.
A player/team scores a point every time it wins an exchange when it has the ball. If it wins an exchange when the other player/team has the ball, the scores stays the same but it gets the ball and the chance to score some points.
A racketball game, for example, ends when a player reaches 15 points but the game may continue as long as no one leads by 2 points (15-14 is not a valid final score, but 16-14 is).
There is theoretically two possibilities that this game never ends : either you enter a loop where no-one scores any point or no-one can lead by two points, and this is what I find very tricky. Still I am pretty convinced that this can be solved, just not by a newbie like me.
This problem has 4 variables :


*

*The current score of the player/team who has the ball

*The current score of the other player/team

*The score at which a game normally ends

*The minimum number of points by which the winning team/player has to lead


This question looks like mine but starts with the final score. Maybe it could be useful anyway.
 A: Tie breaks requiring a 2 point margin and no maximum are not hard to calculate even if the odds of one player winning a point are not equal.  Suppose the tie break has been reached.  Player $A$ has probability $a$ of winning each point.  Player $B$ has probability $b$.  Of course, $a + b = 1$.  Let $w$ be the probability of $A$ winning the game but we don't know this yet.  Consider the next two points, there are four cases: 
$A$ wins both and the game with probability $a^2$.  
$B$ wins both and the game with probability $b^2$.
$A$ wins the first and $B$ the second with probability $ab$.  
$B$ wins the first and $A$ the second with probability $ab$.  
In the third and fourth cases, we are back to a level score so the probability of $A$ winning is $w$.  The probability of A winning is therefore $$a^2 + 2abw$$
If the players are consistent and the probability of A winning each point is still $a$ then this must be $w$ so $$w = a^2 + 2abw$$
So, $$w = \frac{a^2}{1 - 2ab}$$
If the players have equal probability of winning each point then $a = b = \frac{1}{2}$ and $w = \frac{1}{2}$ which is not very surprising.  
Badminton would be a bit harder due to the maximum of 30.  
A: It is easiest to start at the end, considering cases where the player who leads by $2$ wins.  Let $a$ be the chance that a player up by $1$ and serving wins, $b$ the chance that a player at deuce and serving wins, and $c$ the chance that a player down by $1$ and serving wins.  We can write $a=\frac 12(1)+\frac 12(1-c)$ because with probability $\frac 12$ the server wins the point and the game and with probability $\frac 12$ the server loses the point and must win from up $1$ and receiving.  Similarly $b=\frac 12a+\frac 12(1-b), c=\frac 12b+\frac 12(1-a)$ with solution $a=\frac 45, b=\frac 35, c=\frac 25$  We can also note that the server wins the next point with probability $\frac 12+\frac 1{2^3}+\frac 1{2^5}\ldots=\frac 23$  Now we can define $p(x,y)$ as the probability that a player serving at $x-y$ will win the game.  We will start down from the top and we can write the recurrence $p(x,y)=\frac 23p(x+1,y)=\frac 13(1-p(y+1,x))$  We have $p(14,13), p(14,14), p(13,13), p(13,14)$ already calculated, so you can work backwards to $p(0,0)$
