MATHS POINT: Finding Tennis’ Winning Formula

From a purely mathematical point of view, tennis is an extraordinary game. Being such a complex game with unique, rigid and repetitive scoring structures throws up a huge range of possibilities in terms of analysing the game from both a predictive and performance-driven perspective. Given the nature of this blog I obviously find it fascinating in a statistical sense.

There are countless ways that one could go about predicting the outcome of a tennis match. Regression-based models and Elo models are popular ways of going about it but the one for me that I find most interesting is points-based modelling.

Points-based modelling operates on the principle that in order to predict the outcome of a tennis match between two players (say Player A and Player B), you simply need to determine the probability of Player A winning any given point versus Player B. From a point-by-point level, we can extrapolate the probability we find upward from a point-level to find the probability of Player A winning a game, set or the whole match.

So essentially the name of the game here is to predict the outcome of an entire match based on the probability of a player winning a point.

The one underlying fundamental principle upon which this approach relies is that we assume that the probability of a player winning a point is independently and identically distributed – aka the i.i.d. assumption. Essentially this means that we are assuming that the probability of a player winning a point does not change between points or across the duration of the match. Whilst this isn’t necessarily exactly true, there is plenty of tennis literature out there that suggests it isn’t far enough from the reality of the situation to be of major immediate concern here.

From what I have seen in this space, most points-based models use a Monte-Carlo approach which simulates an entire tennis match a point at a time. Rather than going down that route, I want to explore whether I can find a mathematical equation/formula/expression/rule which takes an input at a point-by-point level and can then determine the probability of a player winning the whole match.

Okay that’s enough background info, I’m keen to get into having a crack at building a basic points-based tennis model.

Winning a point on-serve versus returning

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While it would probably be much more straight-forward to find the probability of a player winning any given point to use in a points-based model, the power of the serve in tennis is far too important to neglect, so adding a layer of (slight) complexity to incorporate who is serving is a no-brainer to me. This is confirmed below.

The ATP Tour sees around 63.4% points go the way of the server while the WTA Tour is slightly less lopsided but still very much in favour of the server at a win percentage of 56.2%. A points-based model which does not factor in who is on-serve clearly wouldn’t be likely to be capable of generating accurate predictions for match outcomes.

So rather than trying to use the probability of a player winning any given point, we should be modelling with two probabilities in mind:

• The probability that a player wins a point on-serve
• The probability that a player wins a point when returning

Lets start by focusing on when a player is on-serve.

Winning a service game

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Now is where things start getting a bit more mathematical.

The ways that a player on serve can win a game are:

• game-love – i.e. winning 4 points to 0.
• game-15 – i.e. winning 4 points to 1.
• game-30 – i.e. winning 4 points to 2.
• game-40 – i.e. a deuce game where the server wins 2 more points than his/her opponent and at least 5 points overall.

Let’s denote the probability of winning a point on-serve as \(p_s\) and the probability of winning a service game as \(g_s\).

Game-Love

There is only one way a player can win a game game-love and that is by winning all four points consecutively, i.e.:

There is only one way a game can be won game-love (i.e. 4 points won in a row):

Game-15

Since $p$ denotes the probability of winning a point of serve, the probability of losing a point on serve therefore intuitively must be $1-p$.

Winning game-15 can be achieved in multiple ways (i.e. drop the first point v drop the second point etc.) so we’ll need to introduce a little bit of combinatorial math. The last point of the game needs to be won by the server, but aside from that the one point that was lost can occur at any time within the game. For a game-15 scoreline, there are a total of 4 points played before the last point, one of which was lost by the server.

Game-30

Pretty similar to above, this time though the server has dropped two points. Again we know that the last point must be won by the server, but before this there are a total of 5 points, any two of which can be lost by the server.

Game-40 (Deuce)

It would’ve been too easy without there being a twist in the tail, right? Deuces really throw a spanner in the works here so let’s break it down into two parts:

1. The probability of reaching deuce (40-40), and
2. The probability of winning from deuce

Reaching Deuce:

To reach deuce, the two players in our hypothetical game need to split the points at 3 apiece. This time, however, it does not matter who wins the last point.

Winning from Deuce:

If you’ve read this far without knowing how tennis scoring works I’d be shocked and slightly concerned, but just on the off chance – to win from deuce, a player needs to win two consecutive points from the deuce state to take out the game.

Let’s add another variable to the mix to make things easier to understand – let d denote the probability of winning from the deuce state (40-40).

The $p^2$ here represents the scenario in which a player serving wins two consecutive points from deuce to take out the game. Everything after that illustrates the situation in which the two players trade points, thereby returning to the deuce state (hence the d at the end of the equation).

Time to do a bit of math to get the d out on its own.

We now have a formula for the probability of a player on serve winning a game from deuce.

Combining (multiplying) our expression for d with the previously mentioned probability of reaching deuce, we get:

Bring it all together

We can now create an expression for g by simply summing all of the previous expressions and simplifying.

The probability that a player winning a service game therefore is:

Absolute poetry in motion for a stats nerd like me. Totally indecipherable garbage to everyone else (sane people).

We can explore the relationship between the probability of winning a point on serve with the probability of winning a service game visually too.

The relationship is defined by a nice s-shaped curve, which (as one would expect) passes through the point (0.5,0.5) which essentially means when a player is at equal odds to win or lose a point in serve, he/she is also at equal odds to win or lose the service game. Makes sense.

Winning a Return Game

Having found an expression for winning a service game, we can adapt it simply by replacing point/game on serve (s) probabilities with point/game on return (r) probabilities which allows us to have a separate (albeit incredibly similar) equation for winning a return game as that will definitely take on a different value to the probability of winning a service game, as already discussed.

Additionally, we can rewrite the formula so that it represents the probability that a player wins a return game. We will replace probabilities for winning points/games on-serve (s) with probabilities for winning points/games on return (r) in order to create a similar but separate expression for winning a game when returning serve.

This will come in handy when looking at predicting sets where the serve rotates after each game.

What About Tie-Breakers?

Whilst the rotation of serve has an influence on how tie breakers are played, they are designed to be as balanced as possible in that the two players will end up serving the exact same number of points, or in some situations they may differ but only by one. For the sake of simplicity I’ll consider the probability of winning any point in a tie break as the average of winning a point on serve and on return, i.e.:

This of course does not factor in for any extraneous situational variables such as player fatigue at the end of a set or a player’s “clutch” ability, but that is something that could definitely be explored if this model were to be improved upon in the future.

Calculating the probability of winning a tie break is the exact same process as for a service/return game, except this time the player needs 7 points to win rather than 4. Skipping over some repeated math, this leaves us with:

We’ll be able to use this to determine an equation for winning sets which are decided by a tiebreaker.

Winning a set

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Time to move up a level and look at the probability of winning a set.

A complicating factor that needs to be considered here is the rotation of serve. Unlike in the previous section where we addressed individual points within a game in a uniform manner, we can’t treat all games within a set as equal as the serve rotates after each game. I’ve seen plenty of similar articles and models out there which don’t consider the rotation of serve, but as touched on earlier its too major of an influence on a tennis match to not be considered.

Taking into account rotation of serve, each game in a set can be categorised as having one of four “states”:

1. Player is on-serve and wins
2. Player is on-serve and loses
3. Player is returning and wins
4. Player is returning and loses

Whilst we acknowledge that the probability of winning each game will not be identical, we will retain the independence assumption from our previous i.i.d. assumption (i.e. the outcome of one game does not influence the outcome of the next game).

This next part is a bit gritty so I’m going to try to keep it as light and breezy as possible by giving a high-level overview of how to find an expression for the probability that a player wins a set while including for rotation of serve.

There are two initial conditions we must consider due to the rotation of serve:

• Our hypothetical player serves first, and
• Our hypothetical player returns first

The reason we need to consider this is that if there are an odd number of games played within a set (e.g. 6-1 = 7, 6-3 = 9) one player will have served an extra game compared to their opponent, which could have an impact on their overall probability of winning the set.

Set Outcomes When Serving First

Fun Fact: There are 966 unique permutations by which a player can win a game when serving first in a set. Isn’t that fun! Okay, just me then.

Each of these permutations is unique in terms of the order in which games are won and/or lost. As stated earlier, there are four possible states for a service game. The 966 permutations can be categorised into 1 of 28 different combinations of the four states, i.e. the exact number of service and return games are won and/or lost by a player within the set.

Take a 6-0 scoreline for example. There is only one permutation by which achieving that is possible: win all six games (i.e. WWWWWW). Of our four possible individual game states:

1. There were 3 service games
2. 0 service games were lost
3. There are 3 return games

# Service Game
service_game <- function(p_s) {
  return(p_s^4 + 4*p_s^4*(1-p_s) + 10*p_s^4*(1-p_s)^2 + (20*p_s^5*(1-p_s)^3)/(1-2*p_s+2*p_s^2))
}

# Return Game
return_game <- function(p_r) {
  return(p_r^4 + 4*p_r^4*(1-p_r) + 10*p_r^4*(1-p_r)^2 + (20*p_r^5*(1-p_r)^3)/(1-2*p_r+2*p_r^2))
}

# Tie Breaker
tie_breaker <- function(p_s,p_r) {
  p_t = 0.5*p_s + 0.5*p_r
  return(p_t^7 + 7*p_t^7*(1-p_t) + 28*p_t^7*(1-p_t)^2 + 84*p_t^7*(1-p_t)^3 + 210*p_t^7*(1-p_t)^4 + 462*p_t^7*(1-p_t)^5 + (924*p_t^8*(1-p_t)^6)/(1-2*p_t+2*p_t^2))
}

# Set
set <- function(p_s,p_r) {
  s = service_game(p_s)
  r = return_game(p_r)
  t = tie_breaker(p_s,p_r)
  s_firstserve = 
    1* s^3 * r^3 + 3* s^3 * r^3 * (1-s)^1 + 3* s^4 * r^2 * (1-r)^1 + 12* s^3 * r^3 * (1-s)^1 * (1-r)^1 + 6* s^2 * r^4 * (1-s)^2 + 3* s^4 * r^2 * (1-r)^2 + 24* s^3 * r^3 * (1-s)^2 * (1-r)^1 + 24* s^4 * r^2 * (1-s)^1 * (1-r)^2 + 4* s^2 * r^4 * (1-s)^3 + 4* s^5 * r^1 * (1-r)^3 + 60* s^3 * r^3 * (1-s)^2 * (1-r)^2 + 40* s^2 * r^4 * (1-s)^3 * (1-r)^1 + 20* s^4 * r^2 * (1-s)^1 * (1-r)^3 + 5* s^1 * r^5 * (1-s)^4 + 1* s^5 * r^1 * (1-r)^4 + 100* s^3 * r^4 * (1-s)^3 * (1-r)^2 + 100* s^4 * r^3 * (1-s)^2 * (1-r)^3 + 25* s^2 * r^5 * (1-s)^4 * (1-r)^1 + 25* s^5 * r^2 * (1-s)^1 * (1-r)^4 + 1* s^1 * r^6 * (1-s)^5 + 1* s^6 * r^1 * (1-r)^5 + 200* s^3 * r^3 * (1-s)^3 * (1-r)^3 * t + 125* s^4 * r^2 * (1-s)^2 * (1-r)^4 * t + 125* s^2 * r^4 * (1-s)^4 * (1-r)^2 * t + 26* s^5 * r^1 * (1-s)^1 * (1-r)^5 * t + 26* s^1 * r^5 * (1-s)^5 * (1-r)^1 * t + 1* s^6 * (1-r)^6 * t + 1 * r^6 * (1-s)^6 * t
  s_firstreturn = 
    1* s^3 * r^3 + 3* s^3 * r^3 * (1-r)^1 + 3* s^2 * r^4 * (1-s)^1 + 12* s^3 * r^3 * (1-s)^1 * (1-r)^1 + 6* s^4 * r^2 * (1-r)^2 + 3* s^2 * r^4 * (1-s)^2 + 24* s^3 * r^3 * (1-s)^1 * (1-r)^2 + 24* s^2 * r^4 * (1-s)^2 * (1-r)^1 + 4* s^4 * r^2 * (1-r)^3 + 4* s^1 * r^5 * (1-s)^3 + 60* s^3 * r^3 * (1-s)^2 * (1-r)^2 + 40* s^4 * r^2 * (1-s)^1 * (1-r)^3 + 20* s^2 * r^4 * (1-s)^3 * (1-r)^1 + 5* s^5 * r^1 * (1-r)^4 + 1* s^1 * r^5 * (1-s)^4 + 100* s^4 * r^3 * (1-s)^2 * (1-r)^3 + 100* s^3 * r^4 * (1-s)^3 * (1-r)^2 + 25* s^5 * r^2 * (1-s)^1 * (1-r)^4 + 25* s^2 * r^5 * (1-s)^4 * (1-r)^1 + 1* s^6 * r^1 * (1-r)^5 + 1* s^1 * r^6 * (1-s)^5 + 200* s^3 * r^3 * (1-s)^3 * (1-r)^3 * t + 125* s^2 * r^4 * (1-s)^4 * (1-r)^2 * t + 125* s^4 * r^2 * (1-s)^2 * (1-r)^4 * t + 26* s^1 * r^5 * (1-s)^5 * (1-r)^1 * t + 26* s^5 * r^1 * (1-s)^1 * (1-r)^5 * t + 1 * r^6 * (1-s)^6 * t + 1* s^6 * (1-r)^6 * t
  return(0.5*s_firstserve + 0.5*s_firstreturn)
}

# Match
match <- function(p_s, p_r, best_of = 3) {
  x = set(p_s,p_r)
  if (best_of == 5) {
    return(10*x^3 - 15*x^4 + 6*x^5)    
  } else {
    return(3*x^2 - 2*x^3)
  }
}

# Test
match(best_of = 5, p_s = 0.63, p_r = 0.41)