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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.

Community
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Frank Malenfant
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  • Welcome to math.se! I believe a similar rule of having to win a set/game by at least two points holds for volleyball and badminton (and maybe other sports) ... did you try searching math.se for that? – Bram28 Apr 04 '17 at 19:17
  • I have not searched for these other sports yet. haven't thought of it. Thanks! – Frank Malenfant Apr 04 '17 at 19:28
  • In badminton, the score does not go past 30. If 29-29 is reached then the next point is sudden death. 30-29 is a valid final score. I have experienced it a few times.

    In squash, the normal winning score is 11 but you need a 2 point margin. There is no maximum. The highest I have experienced is 27-25.

     – badjohn Apr 04 '17 at 21:44

2 Answers2

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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.

badjohn
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  • You haven't considered the fact that only the server scores, so from deuce if A wins the next point and B the one after we are at add out with B serving. If B wins the next point and A the one after we are at deuce with A serving. The same technique works, but this needs some work. – Ross Millikan Apr 04 '17 at 22:27
  • Depends on the game and maybe your location. I am unfamiliar with racket ball. In squash and badminton by the current rules, the winner of a rally gets a point regardless of whether he served. In both games, it used to be as you say but it's not at the moment.

    As I said, the argument needs adjusting for badminton due to the 30 point limit. For games with rules as you describe it will also need adjustment.

     – badjohn Apr 04 '17 at 22:40
  • OP stated that only the server could win the point and there is no limit to the points scored, so I think we should just accept that as the rule regardless of the sport. – Ross Millikan Apr 04 '17 at 22:44
  • Sorry, I missed that and slipped into the rules that I was familiar with. I may have been distracted by Bram28's mention of other games. – badjohn Apr 04 '17 at 22:47
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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 $p(0,0)$

Ross Millikan
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