Forecasting a Tennis Match at the Australian Open

Tristan BarnettStephen ClarkeAlan Brown

Introduction Match Predictions

Markov Chain Model

Collecting Data

Exponential Smoothing

Combining Player Statistics

Real Time Predictions Combining Sheets from Markov Chain Model

Bayesian Updating Rule

Excel Computer Demonstration

Markov Chain Model Modelling a game of tennis

Recurrence Formula: P(a,b) = pP(a+1,b) + (1-p)P(a,b+1)

Boundary Conditions: P(a,b) = 1 if a=4, b ≤ 2

P(a,b) = 0 if b=4, a ≤ 2

where for player A:

p = probability of winning a point on serve

P(a,b) = conditional probability of winning the game when the score is (a,b)

22

2

)-1(+=)3,3(

ppp

P

Markov Chain Model

Table 1: The conditional probabilities of player A winning the game from various score lines for p = 0.6

Similarly

sheet for player B serving

sheets for a set (from sheets of a game)

sheet for a match (from sheets of a set)

B score 0 15 30 40 Game

0 15 30 40

A score

Game

0.74 0.84 0.93 0.98

1

0.58 0.71 0.85 0.95

1

0.37 0.52 0.69 0.88

1

0.15 0.25 0.42 0.69

0 0 0

Collecting Data The ATP tour matchfacts: http://www.atptennis.com/en/media/rankings/matchfacts.pdf

Collecting Data fi = ai bi + ( 1 - ai ) ci

gi = aav di + ( 1 - aav ) ei

where the percentage for player i :

fi = points won on serve

gi = points won on return

ai = 1st serves in play

bi = points won on 1st serve

ci = points won on 2nd serve

di = points won on return of 1st serve

ei = points won on return of 2nd serve

where the percentage for average player on the ATP tour:

aav = 1st serves in play = 58.7%

Exponential Smoothing Fi

t = Fit-1 + [ 1 - ( 1 – α )n ] [ fi

t - Fit-1 ]

Git = Gi

t-1 + [ 1 - ( 1 – α )n ] [ git - Gi

t-1 ]

where:

For player i at period t

Fit = smoothed average of the percentage of points won on serve after observing fi

t Gi

t = smoothed average of the percentage of points won on return of serve after observing gi

t

Initialised for average ATP tour player

Fi0 = the ATP average of percentage of points won on serve

Gi0 = the ATP average of percentage of points won on return of serve

n = number of matches played since period t-1

α =smoothing constant

When n=1, [1-(1-α)n] = α, as expected When n becomes large, [1-(1-α)n] → 1, as expected

Combining Player Statistics fij = ft + ( fi - fav ) - ( gj - gav )

gji = gt + ( gj - gav ) - ( fi - fav )

where:For the combined player statistics

fij = percentage of points won on serve for player i against player jgji =percentage points won on return for player j against player I

For the tournament averagesft = percentage of points won on servegt = percentage of points won on return of serve

For the ATP tour averagesfav = percentage of points won on servegav = percentage of points won on return of serve

Since ft + gt = 1, fij + gji = 1 for all i,j as required

Combining Sheets The equation for the probability of player A winning a best-of-5 set match

from (e,f) in sets, (c,d) in games, (a,b) in points, player A serving.

P''(a,b:c,d:e,f ) = P(a,b) P'B(c+1,d) P''(e+1,f ) +

P(a,b) [1-P'B(c+1,d)] P''(e,f+1) +

[1-P(a,b)] P'B(c,d+1) P''(e+1,f ) +

[1-P(a,b)] [1-P'B(c,d+1)] P''(e,f+1)

where for player A :

P''(a,b:c,d:e,f ) = probability of winning the match from (a,b:c,d:e,f )

P'B(c,d) = probability of winning the set from (c,d) when player B is serving

P''(e,f ) = probability of winning the match from (e,f )

Bayesian Updating Rule

where:

θ ti = updated percentage of points won on serve at time t for player i

μi = initial percentage of points won on serve for player i

φ ti = actual percentage of points won on serve at time t for player i

n = number of points played

M = expected points to be played

When n=0, θ 0i= μi as expected

When M → 0, θ ti → φ t

i

tii

ti φ

nMn

μnM

Mθ

++

+=

Computer Demonstration ISF3.XLS

2003 Australian Open Quarter Final

El Aynaoui versus Roddick

Computer Demonstration ISF4.XLS

Chance of winning current Point Game Set MatchEl Aynaoui 30% 10% 50% 68%Roddick 70% 90% 50% 32%

End of 1st set

where: = game to El Aynaoui

= game to Roddick

= set to El Aynaoui

Chances of winning match

0%

25%

50%

75%

100%

0 10 20 30 40 50 60Number of points played

El Aynaoui0%

25%

50%

75%

100%

Roddick

Computer Demonstration End of match

where: = game to El Aynaoui by breaking serve

= game to Roddick by breaking serve

= set to El Aynaoui

= set to Roddick

Chances of winning match

0%

25%

50%

75%

100%

0 100 200 300 400Number of points played

El Aynaoui0%

25%

50%

75%

100%

Roddick