In this paper we present a generalisation of previously considered Markovian models for Tennis that overcome the assumption that the points played are i.i.d and includes the time into the model. Firstly we postulate that in any game there are two different situations: the first 6 points and the, possible, additional points after the first deuce, with different winning probabilities. Then we assume that the duration of any point is distributed with an exponential random time. We are able to compute the law of the (random) duration of a game in this more general setting

Ferrante, M.

Fonseca, G.

Pontarollo, S.

English

FERRANTE, MARCO

PONTAROLLO, SILVIA

Fonseca, Giovanni

Archivio istituzionale della ricerca - Università di Padova

Original Citation:How long does a tennis game last?Padova University PressPublisher:Published version:DOI:Terms of use:Open Access(Article begins on next page)This article is made available under terms and conditions applicable to Open Access Guidelines, asdescribed at http://www.unipd.it/download/file/fid/55401 (Italian only)Availability:This version is available at: 11577/3240085 since: 2017-08-04T09:52:33ZUniversità degli Studi di PadovaPadua Research Archive - Institutional RepositoryM AT H S P O R T  I N T E R N AT I O N A L  2 0 1 7 C O N F E R E N C E- P r o c e e d i n g s -University of  Padua Botanical Garden 26-28 June 2017UPPADOVA P A D O V A  U N I V E R S I T Y  P R E S SOriginal title Proceedings of MathSport International 2017 ConferenceEditors: Carla De Francesco, Luigi De Giovanni, Marco Ferrante, Giovanni Fonseca, Francesco Lisi, Silvia Pontarollo© 2017 Padova University PressUniversità degli Studi di Padovavia 8 Febbraio 2, Padovawww.padovauniversitypress.itISBN 978-88-6938-058-7All rights reserved.MathSport International 2017 Conference Proceedings       How long does a tennis game last?M. Ferrante*, G. Fonseca** and S. Pontarollo**Dip. di Matematica “Tullio Levi-Civita", Università di Padova, Via Trieste 63, 35121-Padova, Italyemail: ferrante@math.unipd.it, spontaro@math.unipd.it** Dip. di Sc. Econom. e Stat., Univ. di Udine, via Tomadini, 30/A, 33100-Udine, Italyemail: giovanni.fonseca@uniud.itAbstractIn this paper we present a generalisation of previously considered Markovian models forTennis that overcome the assumption that the points played are i.i.d and includes the time intothe model. Firstly we postulate that in any game there are two different situations: the first 6points and the, possible, additional points after the first deuce, with different winning probabil-ities. Then we assume that the duration of any point is distributed with an exponential randomtime. We are able to compute the law of the (random) duration of a game in this more generalsetting.1 IntroductionMarkovian framework is particularly suitable to describe the evolution of a tennis match. The usual assump-tion is that the probability that a player wins one point is independent of the previous points and constantduring the match. Under these hypotheses the score of a game, set and match can be described by a set ofnested homogeneous Markov chains. Hence, theoretical results concerning winning probabilities and meanduration of a game, set and match can be easily obtained. A complete account on this approach can be foundin Klaassen and Magnus (2014).Anyway, some authors criticise the assumption that the point winning probability is constant along thematch and independent of the previous points played, see e.g. Klaassen and Magnus (2001). In particular, itis pointed out that playing decisive points, i.e. points after a deuce score, modifies players attitude and thisreflects heavily on the probability to win these points.In Carrari et al. (2017), we propose a modification of the model at the game’s level. Indeed, we assumethat during any game there are two different situations: the first points and the, possible, additional pointsplayed after the (30,30) score that in our model coincide with the “Deuce”. Under this hypothesis, followingthe approach used in Ferrante and Fonseca (2014), we computed the winning probabilities and the expectednumber of points played in a game.In the present work we include in the model the time needed to play a single point. The aim of such amodification is to obtain the computation of the expected length of a match in terms of actual time and not justas number of points. Indeed there is a concern about the length of tennis matches and several modificationto the game rules are nowadays proposed in order to fix the length of a match or at least to avoid too longmatches. In Section 2. we present the model, in Section 3. we compute the game winning probabilities andin Section 4. we obtain the expected length of a game.122MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo2 The continuous time modelIn this paper we model the tennis game as a continuous-time Markov chain (see Norris (1998) for a completeaccount on this topic). We define the state space S of the chain, which collects all the possible scores in thegame, and the generator matrix Q on S. In order to determine the matrix Q, we define independently thetransition matrix of the associated Jump chain, which is a discrete-time Markov chain, and the exponentialholding times.The transition matrix of the associated Jump chain follows the model defined in Carrari et al. (2017) fora discrete-time Markov chain of the tennis. The classical assumptions previously considered in the literature(see e.g Newton and Keller (2005)) were that the probability to win any point by the player on service wasindependent of the previous points and constant during the game. In Carrari et al. (2017) we assume thatp, the probability to win a point, does not remain the same during the game. As empirical data on thematches confirm, the estimated winning probability of the first 6 points of a game is different from that ofthe additional played points from the “Deuce” on. For this reason, we consider a second parameter p̄, thatdescribe this part of the game and, to avoid trivial cases, we assume that both p and p̄ belong to (0,1).Regarding the holding times, it is not easy to find in the literature data sets to test different scenarios thatbetter describe the true course of a tennis game (see Morante and Brotherhooc (2007) for some statistics onthe duration). For this reason in this paper we assume that all the holding times have the same distribution,i.e. share the same rate l of their Exponential Laws.Let us now define precisely our model: the state space is the set S = {1,2, . . . ,17} which describes thescore of a game as defined in Table 1.Score (0,0) (15,0) (0,15) (30,0) (15,15) (0,30) (40,0) (30,15) (15,30)State 1 2 3 4 5 6 7 8 9Score (0,40) (40,15) (15,40) Deuce AdvA AdvB WinA WinBState 10 11 12 13 14 15 16 17Table 1: Scores and corresponding states used in equationsNote that in the present model the scores (30,30) and Deuce are represented by the single state 13, since theyshare the same mathematical properties, as it happens to the pairs (40,30)–AdvA and (30,40)–AdvB. Thegraph representing the transition probabilities of the Jump process is presented in Fig. 1, where q = 1  pand q̄ = 1  p̄.By our construction, we define the transition rates in the generator matrix Q by l1 = l p and l2 = lq(see Norris (1998)) and in Fig. 2 we present the graph of the continuous time Markov chain describing atennis game. Note that the expected length of a point is equal to 1/l and it does not depend on who is thewinner of the point.From the graph it is immediate to write down the generator matrix Q = (qi j)i, j2S and to prove that the states123MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo1234567891011121314151617pqpqpqpqpqpqqpqpqpqpqqpppp pqqqFigure 1: Graph of the Jump Markov chain16 and 17 are absorbing, while all the other states are transient.Q =26666666666666666666666666666664 l l1 l2 0 0 0 0 0 0 0 0 0 0 0 0 0 00  l 0 l1 l2 0 0 0 0 0 0 0 0 0 0 0 00 0  l 0 l1 l2 0 0 0 0 0 0 0 0 0 0 00 0 0  l 0 0 l1 l2 0 0 0 0 0 0 0 0 00 0 0 0  l 0 0 l1 l2 0 0 0 0 0 0 0 00 0 0 0 0  l 0 0 l1 l2 0 0 0 0 0 0 00 0 0 0 0 0  l 0 0 0 l2 0 0 0 0 l1 00 0 0 0 0 0 0  l 0 0 l1 0 l2 0 0 0 00 0 0 0 0 0 0 0  l 0 0 l2 l1 0 0 0 00 0 0 0 0 0 0 0 0  l 0 l1 0 0 0 0 l20 0 0 0 0 0 0 0 0 0  l 0 0 l2 0 l1 00 0 0 0 0 0 0 0 0 0 0  l 0 0 l1 0 l20 0 0 0 0 0 0 0 0 0 0 0  l̄ l̄1 l̄2 0 00 0 0 0 0 0 0 0 0 0 0 0 l̄2  l̄ 0 l̄1 00 0 0 0 0 0 0 0 0 0 0 0 l̄1 0  l̄ 0 l̄20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 037777777777777777777777777777775In order to compute the winning probabilities, we need to determine the absorption probabilities in states16 and 17 for the transition matrix of the Jump process, while to investigate the distribution of the expectedlength of a game, we need to evaluate the exponential matrix of Q, which is in general not very simple.124MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo1234567891011121314151617l1l2l1l2l1l2l1l2l1l2l1l2l2l1l2l1l2l1l2l̄1l̄2l̄2l̄1l2l1l1l̄1l2l2l̄2Figure 2: Graph of the continuous time Markov chain describing a tennis game, with its transition rates.3 Winning probabilitiesIn this section we recall some of the result proved in Carrari et al. (2017). The transition matrix of the Jumpprocess isP =266666666666666666666666666666640 p q 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 p q 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 p q 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 p q 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 p q 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 p q 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 q 0 0 0 0 p 00 0 0 0 0 0 0 0 0 0 p 0 q 0 0 0 00 0 0 0 0 0 0 0 0 0 0 q p 0 0 0 00 0 0 0 0 0 0 0 0 0 0 p 0 0 0 0 q0 0 0 0 0 0 0 0 0 0 0 0 0 q 0 p 00 0 0 0 0 0 0 0 0 0 0 0 0 0 p 0 q0 0 0 0 0 0 0 0 0 0 0 0 0 p̄ q̄ 0 00 0 0 0 0 0 0 0 0 0 0 0 q̄ 0 0 p̄ 00 0 0 0 0 0 0 0 0 0 0 0 p̄ 0 0 0 q̄0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 137777777777777777777777777777775The winning probability of the game for the player A on service, denoted by h1, coincides with the absorptionprobability in the state 16 of the previous Markov chain starting from 1, which can be obtained (see e.g.125MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloNorris (1998)) as the minimal, non negative solution of the linear systemhi = Âj2Spi jh j for 1  i  15 , h16 = 1 , h17 = 0 .The solution can be easily calculated and we obtain thath1 = p25p2  4p3 +4(p 1)2 pp̄  2(p 1)2 p̄2(p(4 p̄ 2) 2 p̄ 3)2 p̄2  2 p̄+1 .Denoting by G(p, p̄) = h1, by A and B the two players, and by PGxY the probability that the player Y wins agame when X serves, thanks to the symmetry of the model we obtain that:PGaA = G(pA, p̄A)PGaB = G(1  pA,1  p̄A) (1)PGbB = G(pB, p̄B)PGbA = G(1  pB,1  p̄B)Note that, since PGxX + PGxY = 1, G(1   pX ,1   p̄X) = 1   G(pX , p̄X) and that for pX = p̄X , the previousprobabilities coincides with those well known in the literature (see e.g. Newton and Keller (2005)). In Table2 we report the values of G for increasing p and p̄.p̄p 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.1 0.001 0.004 0.011 0.021 0.034 0.049 0.062 0.071 0.078 0.0810.2 0.011 0.022 0.043 0.076 0.119 0.165 0.204 0.233 0.252 0.2630.3 0.040 0.061 0.099 0.158 0.234 0.312 0.378 0.425 0.455 0.4720.4 0.102 0.132 0.185 0.264 0.363 0.464 0.549 0.607 0.643 0.6630.5 0.206 0.242 0.302 0.391 0.500 0.609 0.697 0.758 0.794 0.8120.6 0.357 0.392 0.451 0.535 0.636 0.736 0.815 0.868 0.898 0.9130.7 0.545 0.575 0.622 0.688 0.766 0.842 0.901 0.939 0.960 0.9690.8 0.748 0.767 0.795 0.835 0.881 0.924 0.957 0.978 0.989 0.9930.9 0.922 0.929 0.938 0.951 0.965 0.979 0.989 0.995 0.998 0.9991.0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000Table 2: Winning probabilities of a game4 Duration of a gameIn this section we evaluate the duration of a game and of a game given that the player on serve wins thegame. By the Markovian structure of the present model, the problem we face is equivalent to evaluatethe distribution of the absorption time to one of the states 16 and 17. These distributions are called in theliterature Phase-type distributions (see e.g. Neuts (1981)) and we have explicit formulas for their densitiesand moments. The only drawback is that the computation of the density passes through the evaluation of thematrix exponential of a 16⇥16 matrix, which is usually not feasible. On the contrary, for the moments weonly need to be able to evaluate the inverse of the same matrix and its powers.126MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo4.1 The unconditioned caseDue to the difficulties described above, in this section we consider only the case where p = p. Startingfrom Q, the generator matrix defined above, we can compute the distribution of the duration time of a game.Indeed, let T be equal to the matrix Q where the last two rows and two columns have been erased, that isT =266666666666666666666666664 l l1 l2 0 0 0 0 0 0 0 0 0 0 0 00  l 0 l1 l2 0 0 0 0 0 0 0 0 0 00 0  l 0 l1 l2 0 0 0 0 0 0 0 0 00 0 0  l 0 0 l1 l2 0 0 0 0 0 0 00 0 0 0  l 0 0 l1 l2 0 0 0 0 0 00 0 0 0 0  l 0 0 l1 l2 0 0 0 0 00 0 0 0 0 0  l 0 0 0 l2 0 0 0 00 0 0 0 0 0 0  l 0 0 l1 0 l2 0 00 0 0 0 0 0 0 0  l 0 0 l2 l1 0 00 0 0 0 0 0 0 0 0  l 0 l1 0 0 00 0 0 0 0 0 0 0 0 0  l 0 0 l2 00 0 0 0 0 0 0 0 0 0 0  l 0 0 l10 0 0 0 0 0 0 0 0 0 0 0  l l1 l20 0 0 0 0 0 0 0 0 0 0 0 l2  l 00 0 0 0 0 0 0 0 0 0 0 0 l1 0  l377777777777777777777777775and let T0 be a vector of length 15 where each entries is equal to the jumping time needed for each state(excluding WinA and WinB) to reach one of the two absorbing states, that isT0 = (0,0,0,0,0,0,l1,0,0,l2,l1,l2,0,l1,l2)> .Then, if ↵= (1,0, . . . ,0), the density function of the duration time, denoted by fp,l , can be computed asfp,l =↵eTtT0(see Neuts (1981) for the simple proof). Therefore, if we setAp,l (t)= 3l 2 p4t2(l t 4)+6l 2 p3t2(l t 4)  p2 4l 3t3  21l 2t2 +30 + p l 3t3  9l 2t2 +30  15+l 2t2andBp,l (t) = 15p2 2p2  2p+1 ⇣e2l tp2(1 p)p  1⌘,the time distribution is given byfp,l (t) =l24p(1  p)pe l t⇣1+p2(1 p)p⌘⇣Bp,l (t)+⇣4l tp(1  p)pel tp2(1 p)p⌘Ap,l (t)⌘.In Figure 3. we plot the density fp,l in the cases p = 0.5,0.65 and 0.9, while l = 2.We are also able to evaluate the moments of these random times. Indeed, the expected time µ needed towin a game can be computed as ↵( T ) 1e, where e = (1, . . . ,1)T . Therefore, the average time is given byµ =4  6p6 +18p5  18p4 +6p3 + p2   p+1 l (2p2  2p+1) .127MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloFigure 3: Graph of the distribution of a game duration time for three different values of p, and for l = 2.Finally the moment of order two, computed as µ2 = 2↵( T ) 2e, allows us to obtain the variances2 = 4l 2 (2p2  2p+1)2  144p12 +864p11  2160p10 +2880p9  2232p8 +1152p7  618p6+414p5  197p4 +40p3 +3p2  2p+1 .4.2 The conditioned caseLet us now compute the expected duration of a game given that the player on serve wins the game. In thiscase it is easy to prove that the Jump chain of the conditioned chain is the matrix P0 on the state space{1, . . . ,16} given by:p0i j = pi jh jhiwith i, j 2 {1, . . . ,16},where the hi are the absorption probabilities in 16. Moreover, the conditional generator matrix is the matrixQ0 obtained from P0 and the original exponential holding times of parameter l . The mean and variancecomputed above are nowµ =4 20p5  84p4 +148p3  143p2 +79p 21 l (2p2  2p+1)(8p3  28p2 +34p 15) ,s2 = 4l 2 (2p2  2p+1)2 ( 8p3 +28p2  34p+15)2 320p10  2880p9 +11616p8  27744p7+43608p6  47460p5 +36746p4  20540p3 +8287p2  2288p+336 .128MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloIn Table 3 we show the Mean±StandardDeviation of the actual time of a game when the point winningprobabilities for the serving player are p and p̄ = p (the rate l is set arbitrarily equal to 2 for ease ofexposition). For different values of p (recall that we have set p = p) the first row shows the (mean ±StandardDeviation) time distribution for finishing a game when the player with probability of winning apoint equal to p is serving. The second row shows (mean± StandardDeviation) when the distribution isconditioned to the winning of the player that is serving. Note that, due to symmetry of the problem, whenp < 0.5 and the model is conditioned as above, the results represent both the average time for the player(characterized by p < 0.5) to win a game on his turn of serving, and the average time to win a game for theplayer with same p when the other player is serving. The same empirical quantities are derived in Moranteand Brotherhooc (2007). This is one of the few quantitative studies about the duration of points and gamesin Tennis, although they are more interested in relating playing time with performance indicators. Theycompute the average duration using a sample of Grand Slam matches for both male and female professionalplayers.p = 0.9 p = 0.8 p = 0.7 p = 0.6 p = 0.5 p = 0.4 p = 0.3 p = 0.2 p = 0.12.23±1.13 2.54±1.34 2.92±1.60 3.24±1.82 3.37±1.90 3.24±1.82 2.92±1.60 2.54±1.34 2.23±1.132.23±1.13 2.53±1.33 2.87±1.58 3.18±1.80 3.37±1.90 3.40±1.86 3.30±1.71 3.13±1.52 2.96±1.35Table 3: (mean±StandardDeviation) time for finishing a game (first row) and for winning a game whileserving.5 ConclusionsIn this paper we present a model for Tennis including non i.i.d. point winning probabilities and the time ofplay. Indeed, we allow winning point probabilities to change depending on the score of the game. Moreover,we are interested in describing the time of play since there is some concerns about the excessive lengthof matches, especially in male Grand Slam competitions. In particular, in the present work, we obtain thedistribution of the actual time of a game.References[1] Carrari, A., Ferrante M. and Fonseca G. (2017) A new Markovian model for tennis matches. Electronic Journalof Applied Statistical Analysis, to appear.[2] Ferrante, M. and Fonseca G. (2014) On the winning probabilities and mean durations of volleyball. Journal ofQuantitative Analysis in Sports 10, 91-98.[3] Klaassen, F. and Magnus, J.R. (2001) Are points in tennis independent and identically distributed? Evidencefrom a dynamic binary panel data model. Journal of the American Statistical Association 96, 500-509.[4] Klaassen, F. and Magnus, J.R. (2014) Analyzing Wimbledon. Oxford University Press, Oxford.[5] Morante, S. and Brotherhooc, J. (2007) Match Characteristics of Professional Singles Tennis. http://www.cptennis.com.au/pdf/CooperParkTennisPDF_MatchCharacteristics.pdf.[6] Neuts, M.F. (1981) Matrix-Geometric Solutions in Stochastic Models: An algorithmic approach. Johns HopkinsUniversity Press, Baltimore.129MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo[7] Newton, P.K. and Keller, J.B. (2005) Probability of winning at tennis (I). Theory and data. Studies in AppliedMathematics 114, 241-269.[8] Norris, J.R. (1998) Markov chains. Cambridge University Press, Cambridge.130ISBN  978-88-6938-058-7

How long does a tennis game last?

M. Ferrante

G. Fonseca

S. Pontarollo

Archivio istituzionale della ricerca - Università degli Studi di Udine

MathSport International 2017 Conference Proceedings       How long does a tennis game last?M. Ferrante*, G. Fonseca** and S. Pontarollo**Dip. di Matematica “Tullio Levi-Civita", Università di Padova, Via Trieste 63, 35121-Padova, Italyemail: ferrante@math.unipd.it, spontaro@math.unipd.it** Dip. di Sc. Econom. e Stat., Univ. di Udine, via Tomadini, 30/A, 33100-Udine, Italyemail: giovanni.fonseca@uniud.itAbstractIn this paper we present a generalisation of previously considered Markovian models forTennis that overcome the assumption that the points played are i.i.d and includes the time intothe model. Firstly we postulate that in any game there are two different situations: the first 6points and the, possible, additional points after the first deuce, with different winning probabil-ities. Then we assume that the duration of any point is distributed with an exponential randomtime. We are able to compute the law of the (random) duration of a game in this more generalsetting.1 IntroductionMarkovian framework is particularly suitable to describe the evolution of a tennis match. The usual assump-tion is that the probability that a player wins one point is independent of the previous points and constantduring the match. Under these hypotheses the score of a game, set and match can be described by a set ofnested homogeneous Markov chains. Hence, theoretical results concerning winning probabilities and meanduration of a game, set and match can be easily obtained. A complete account on this approach can be foundin Klaassen and Magnus (2014).Anyway, some authors criticise the assumption that the point winning probability is constant along thematch and independent of the previous points played, see e.g. Klaassen and Magnus (2001). In particular, itis pointed out that playing decisive points, i.e. points after a deuce score, modifies players attitude and thisreflects heavily on the probability to win these points.In Carrari et al. (2017), we propose a modification of the model at the game’s level. Indeed, we assumethat during any game there are two different situations: the first points and the, possible, additional pointsplayed after the (30,30) score that in our model coincide with the “Deuce”. Under this hypothesis, followingthe approach used in Ferrante and Fonseca (2014), we computed the winning probabilities and the expectednumber of points played in a game.In the present work we include in the model the time needed to play a single point. The aim of such amodification is to obtain the computation of the expected length of a match in terms of actual time and not justas number of points. Indeed there is a concern about the length of tennis matches and several modificationto the game rules are nowadays proposed in order to fix the length of a match or at least to avoid too longmatches. In Section 2. we present the model, in Section 3. we compute the game winning probabilities andin Section 4. we obtain the expected length of a game.122MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo2 The continuous time modelIn this paper we model the tennis game as a continuous-time Markov chain (see Norris (1998) for a completeaccount on this topic). We define the state space S of the chain, which collects all the possible scores in thegame, and the generator matrix Q on S. In order to determine the matrix Q, we define independently thetransition matrix of the associated Jump chain, which is a discrete-time Markov chain, and the exponentialholding times.The transition matrix of the associated Jump chain follows the model defined in Carrari et al. (2017) fora discrete-time Markov chain of the tennis. The classical assumptions previously considered in the literature(see e.g Newton and Keller (2005)) were that the probability to win any point by the player on service wasindependent of the previous points and constant during the game. In Carrari et al. (2017) we assume thatp, the probability to win a point, does not remain the same during the game. As empirical data on thematches confirm, the estimated winning probability of the first 6 points of a game is different from that ofthe additional played points from the “Deuce” on. For this reason, we consider a second parameter p̄, thatdescribe this part of the game and, to avoid trivial cases, we assume that both p and p̄ belong to (0,1).Regarding the holding times, it is not easy to find in the literature data sets to test different scenarios thatbetter describe the true course of a tennis game (see Morante and Brotherhooc (2007) for some statistics onthe duration). For this reason in this paper we assume that all the holding times have the same distribution,i.e. share the same rate λ of their Exponential Laws.Let us now define precisely our model: the state space is the set S = {1,2, . . . ,17} which describes thescore of a game as defined in Table 1.Score (0,0) (15,0) (0,15) (30,0) (15,15) (0,30) (40,0) (30,15) (15,30)State 1 2 3 4 5 6 7 8 9Score (0,40) (40,15) (15,40) Deuce AdvA AdvB WinA WinBState 10 11 12 13 14 15 16 17Table 1: Scores and corresponding states used in equationsNote that in the present model the scores (30,30) and Deuce are represented by the single state 13, since theyshare the same mathematical properties, as it happens to the pairs (40,30)–AdvA and (30,40)–AdvB. Thegraph representing the transition probabilities of the Jump process is presented in Fig. 1, where q = 1− pand q̄ = 1− p̄.By our construction, we define the transition rates in the generator matrix Q by λ1 = λ p and λ2 = λq(see Norris (1998)) and in Fig. 2 we present the graph of the continuous time Markov chain describing atennis game. Note that the expected length of a point is equal to 1/λ and it does not depend on who is thewinner of the point.From the graph it is immediate to write down the generator matrix Q = (qi j)i, j∈S and to prove that the states123MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo1234567891011121314151617pqpqpqpqpqpqqpqpqpqpqqpppp pqqqFigure 1: Graph of the Jump Markov chain16 and 17 are absorbing, while all the other states are transient.Q =−λ λ1 λ2 0 0 0 0 0 0 0 0 0 0 0 0 0 00 −λ 0 λ1 λ2 0 0 0 0 0 0 0 0 0 0 0 00 0 −λ 0 λ1 λ2 0 0 0 0 0 0 0 0 0 0 00 0 0 −λ 0 0 λ1 λ2 0 0 0 0 0 0 0 0 00 0 0 0 −λ 0 0 λ1 λ2 0 0 0 0 0 0 0 00 0 0 0 0 −λ 0 0 λ1 λ2 0 0 0 0 0 0 00 0 0 0 0 0 −λ 0 0 0 λ2 0 0 0 0 λ1 00 0 0 0 0 0 0 −λ 0 0 λ1 0 λ2 0 0 0 00 0 0 0 0 0 0 0 −λ 0 0 λ2 λ1 0 0 0 00 0 0 0 0 0 0 0 0 −λ 0 λ1 0 0 0 0 λ20 0 0 0 0 0 0 0 0 0 −λ 0 0 λ2 0 λ1 00 0 0 0 0 0 0 0 0 0 0 −λ 0 0 λ1 0 λ20 0 0 0 0 0 0 0 0 0 0 0 −λ̄ λ̄1 λ̄2 0 00 0 0 0 0 0 0 0 0 0 0 0 λ̄2 −λ̄ 0 λ̄1 00 0 0 0 0 0 0 0 0 0 0 0 λ̄1 0 −λ̄ 0 λ̄20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0In order to compute the winning probabilities, we need to determine the absorption probabilities in states16 and 17 for the transition matrix of the Jump process, while to investigate the distribution of the expectedlength of a game, we need to evaluate the exponential matrix of Q, which is in general not very simple.124MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo1234567891011121314151617λ1λ2λ1λ2λ1λ2λ1λ2λ1λ2λ1λ2λ2λ1λ2λ1λ2λ1λ2λ̄1λ̄2λ̄2λ̄1λ2λ1λ1λ̄1λ2λ2λ̄2Figure 2: Graph of the continuous time Markov chain describing a tennis game, with its transition rates.3 Winning probabilitiesIn this section we recall some of the result proved in Carrari et al. (2017). The transition matrix of the Jumpprocess isP =0 p q 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 p q 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 p q 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 p q 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 p q 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 p q 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 q 0 0 0 0 p 00 0 0 0 0 0 0 0 0 0 p 0 q 0 0 0 00 0 0 0 0 0 0 0 0 0 0 q p 0 0 0 00 0 0 0 0 0 0 0 0 0 0 p 0 0 0 0 q0 0 0 0 0 0 0 0 0 0 0 0 0 q 0 p 00 0 0 0 0 0 0 0 0 0 0 0 0 0 p 0 q0 0 0 0 0 0 0 0 0 0 0 0 0 p̄ q̄ 0 00 0 0 0 0 0 0 0 0 0 0 0 q̄ 0 0 p̄ 00 0 0 0 0 0 0 0 0 0 0 0 p̄ 0 0 0 q̄0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1The winning probability of the game for the player A on service, denoted by h1, coincides with the absorptionprobability in the state 16 of the previous Markov chain starting from 1, which can be obtained (see e.g.125MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloNorris (1998)) as the minimal, non negative solution of the linear systemhi = ∑j∈Spi jh j for 1≤ i≤ 15 , h16 = 1 , h17 = 0 .The solution can be easily calculated and we obtain thath1 = p2[5p2−4p3 +4(p−1)2 pp̄− 2(p−1)2 p̄2(p(4 p̄−2)−2 p̄−3)2 p̄2−2 p̄+1].Denoting by G(p, p̄) = h1, by A and B the two players, and by PGxY the probability that the player Y wins agame when X serves, thanks to the symmetry of the model we obtain that:PGaA = G(pA, p̄A)PGaB = G(1− pA,1− p̄A) (1)PGbB = G(pB, p̄B)PGbA = G(1− pB,1− p̄B)Note that, since PGxX + PGxY = 1, G(1− pX ,1− p̄X) = 1−G(pX , p̄X) and that for pX = p̄X , the previousprobabilities coincides with those well known in the literature (see e.g. Newton and Keller (2005)). In Table2 we report the values of G for increasing p and p̄.p̄p 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.1 0.001 0.004 0.011 0.021 0.034 0.049 0.062 0.071 0.078 0.0810.2 0.011 0.022 0.043 0.076 0.119 0.165 0.204 0.233 0.252 0.2630.3 0.040 0.061 0.099 0.158 0.234 0.312 0.378 0.425 0.455 0.4720.4 0.102 0.132 0.185 0.264 0.363 0.464 0.549 0.607 0.643 0.6630.5 0.206 0.242 0.302 0.391 0.500 0.609 0.697 0.758 0.794 0.8120.6 0.357 0.392 0.451 0.535 0.636 0.736 0.815 0.868 0.898 0.9130.7 0.545 0.575 0.622 0.688 0.766 0.842 0.901 0.939 0.960 0.9690.8 0.748 0.767 0.795 0.835 0.881 0.924 0.957 0.978 0.989 0.9930.9 0.922 0.929 0.938 0.951 0.965 0.979 0.989 0.995 0.998 0.9991.0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000Table 2: Winning probabilities of a game4 Duration of a gameIn this section we evaluate the duration of a game and of a game given that the player on serve wins thegame. By the Markovian structure of the present model, the problem we face is equivalent to evaluatethe distribution of the absorption time to one of the states 16 and 17. These distributions are called in theliterature Phase-type distributions (see e.g. Neuts (1981)) and we have explicit formulas for their densitiesand moments. The only drawback is that the computation of the density passes through the evaluation of thematrix exponential of a 16×16 matrix, which is usually not feasible. On the contrary, for the moments weonly need to be able to evaluate the inverse of the same matrix and its powers.126MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo4.1 The unconditioned caseDue to the difficulties described above, in this section we consider only the case where p = p. Startingfrom Q, the generator matrix defined above, we can compute the distribution of the duration time of a game.Indeed, let T be equal to the matrix Q where the last two rows and two columns have been erased, that isT =−λ λ1 λ2 0 0 0 0 0 0 0 0 0 0 0 00 −λ 0 λ1 λ2 0 0 0 0 0 0 0 0 0 00 0 −λ 0 λ1 λ2 0 0 0 0 0 0 0 0 00 0 0 −λ 0 0 λ1 λ2 0 0 0 0 0 0 00 0 0 0 −λ 0 0 λ1 λ2 0 0 0 0 0 00 0 0 0 0 −λ 0 0 λ1 λ2 0 0 0 0 00 0 0 0 0 0 −λ 0 0 0 λ2 0 0 0 00 0 0 0 0 0 0 −λ 0 0 λ1 0 λ2 0 00 0 0 0 0 0 0 0 −λ 0 0 λ2 λ1 0 00 0 0 0 0 0 0 0 0 −λ 0 λ1 0 0 00 0 0 0 0 0 0 0 0 0 −λ 0 0 λ2 00 0 0 0 0 0 0 0 0 0 0 −λ 0 0 λ10 0 0 0 0 0 0 0 0 0 0 0 −λ λ1 λ20 0 0 0 0 0 0 0 0 0 0 0 λ2 −λ 00 0 0 0 0 0 0 0 0 0 0 0 λ1 0 −λand let T0 be a vector of length 15 where each entries is equal to the jumping time needed for each state(excluding WinA and WinB) to reach one of the two absorbing states, that isT0 = (0,0,0,0,0,0,λ1,0,0,λ2,λ1,λ2,0,λ1,λ2)> .Then, if α= (1,0, . . . ,0), the density function of the duration time, denoted by fp,λ , can be computed asfp,λ =αeTtT0(see Neuts (1981) for the simple proof). Therefore, if we setAp,λ (t)=−3λ 2 p4t2(λ t−4)+6λ 2 p3t2(λ t−4)− p2(4λ 3t3−21λ 2t2 +30)+ p(λ 3t3−9λ 2t2 +30)−15+λ 2t2andBp,λ (t) = 15√2(2p2−2p+1)(e2λ t√2(1−p)p−1),the time distribution is given byfp,λ (t) =λ24√(1− p)pe−λ t(1+√2(1−p)p)(Bp,λ (t)+(4λ t√(1− p)peλ t√2(1−p)p)Ap,λ (t)).In Figure 3. we plot the density fp,λ in the cases p = 0.5,0.65 and 0.9, while λ = 2.We are also able to evaluate the moments of these random times. Indeed, the expected time µ needed towin a game can be computed as α(−T )−1e, where e = (1, . . . ,1)T . Therefore, the average time is given byµ =4(−6p6 +18p5−18p4 +6p3 + p2− p+1)λ (2p2−2p+1) .127MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloFigure 3: Graph of the distribution of a game duration time for three different values of p, and for λ = 2.Finally the moment of order two, computed as µ2 = 2α(−T )−2e, allows us to obtain the varianceσ2 =4λ 2 (2p2−2p+1)2(−144p12 +864p11−2160p10 +2880p9−2232p8 +1152p7−618p6+414p5−197p4 +40p3 +3p2−2p+1).4.2 The conditioned caseLet us now compute the expected duration of a game given that the player on serve wins the game. In thiscase it is easy to prove that the Jump chain of the conditioned chain is the matrix P′ on the state space{1, . . . ,16} given by:p′i j = pi jh jhiwith i, j ∈ {1, . . . ,16},where the hi are the absorption probabilities in 16. Moreover, the conditional generator matrix is the matrixQ′ obtained from P′ and the original exponential holding times of parameter λ . The mean and variancecomputed above are nowµ =4(20p5−84p4 +148p3−143p2 +79p−21)λ (2p2−2p+1)(8p3−28p2 +34p−15) ,σ2 =4λ 2 (2p2−2p+1)2 (−8p3 +28p2−34p+15)2(320p10−2880p9 +11616p8−27744p7+43608p6−47460p5 +36746p4−20540p3 +8287p2−2288p+336).128MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. PontarolloIn Table 3 we show the Mean±StandardDeviation of the actual time of a game when the point winningprobabilities for the serving player are p and p̄ = p (the rate λ is set arbitrarily equal to 2 for ease ofexposition). For different values of p (recall that we have set p = p) the first row shows the (mean±StandardDeviation) time distribution for finishing a game when the player with probability of winning apoint equal to p is serving. The second row shows (mean± StandardDeviation) when the distribution isconditioned to the winning of the player that is serving. Note that, due to symmetry of the problem, whenp < 0.5 and the model is conditioned as above, the results represent both the average time for the player(characterized by p < 0.5) to win a game on his turn of serving, and the average time to win a game for theplayer with same p when the other player is serving. The same empirical quantities are derived in Moranteand Brotherhooc (2007). This is one of the few quantitative studies about the duration of points and gamesin Tennis, although they are more interested in relating playing time with performance indicators. Theycompute the average duration using a sample of Grand Slam matches for both male and female professionalplayers.p = 0.9 p = 0.8 p = 0.7 p = 0.6 p = 0.5 p = 0.4 p = 0.3 p = 0.2 p = 0.12.23±1.13 2.54±1.34 2.92±1.60 3.24±1.82 3.37±1.90 3.24±1.82 2.92±1.60 2.54±1.34 2.23±1.132.23±1.13 2.53±1.33 2.87±1.58 3.18±1.80 3.37±1.90 3.40±1.86 3.30±1.71 3.13±1.52 2.96±1.35Table 3: (mean±StandardDeviation) time for finishing a game (first row) and for winning a game whileserving.5 ConclusionsIn this paper we present a model for Tennis including non i.i.d. point winning probabilities and the time ofplay. Indeed, we allow winning point probabilities to change depending on the score of the game. Moreover,we are interested in describing the time of play since there is some concerns about the excessive lengthof matches, especially in male Grand Slam competitions. In particular, in the present work, we obtain thedistribution of the actual time of a game.References[1] Carrari, A., Ferrante M. and Fonseca G. (2017) A new Markovian model for tennis matches. Electronic Journalof Applied Statistical Analysis, to appear.[2] Ferrante, M. and Fonseca G. (2014) On the winning probabilities and mean durations of volleyball. Journal ofQuantitative Analysis in Sports 10, 91-98.[3] Klaassen, F. and Magnus, J.R. (2001) Are points in tennis independent and identically distributed? Evidencefrom a dynamic binary panel data model. Journal of the American Statistical Association 96, 500-509.[4] Klaassen, F. and Magnus, J.R. (2014) Analyzing Wimbledon. Oxford University Press, Oxford.[5] Morante, S. and Brotherhooc, J. (2007) Match Characteristics of Professional Singles Tennis. http://www.cptennis.com.au/pdf/CooperParkTennisPDF_MatchCharacteristics.pdf.[6] Neuts, M.F. (1981) Matrix-Geometric Solutions in Stochastic Models: An algorithmic approach. Johns HopkinsUniversity Press, Baltimore.129MathSport International 2017 Conference Proceedings       How long does a tennis game last? M. Ferrante, G. Fonseca, S. Pontarollo[7] Newton, P.K. and Keller, J.B. (2005) Probability of winning at tennis (I). Theory and data. Studies in AppliedMathematics 114, 241-269.[8] Norris, J.R. (1998) Markov chains. Cambridge University Press, Cambridge.130

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How long does a tennis game last?

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