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(a)
Singles
It is assumed that player A has a probability PA of winning a game
on service when the games' scores are equal (ie 0-0, 1-1, 2-2,
),
that it is PA+D when he/she is ahead in the games' score, and that
it is PA-D when he/she is behind. Correspondingly, it is assumed
that player B has a probability PB of winning a game on service
when the games' scores are equal, that it is PB+D when he/she is
ahead in the games' scores, and PB-D when behind. Thus, the psychological
advantage of being ahead might be called the 'front-runner' effect,
and is represented by D.
(b)
A set of singles
Firstly we consider a tiebreak set of tennis between two equal players
(PA=PB) with equal psychological factors, D. For simplicity, it
is assumed throughout this paper that the two equal players have
an equal chance of winning the tiebreak game if it is played (at
6-6). Assuming player A serves in the first game of the set, the
probability player A wins the set can be evaluated using a branching
diagram or using recurrence methods. For example, when PA=PB=0.6
and D=0.1, the probability that the games' score reaches 2-0, 1-1
and 0-2 is (0.6)(0.5) = 0.30, (0.6)(0.5) + (0.4)(0.3) = 0.42 and
(0.4)(0.7) = 0.28 respectively. Further, the probability the games'
score reaches 3-0, 2-1, 1-2 and 0-3 is (0.30)(0.7) = 0.210, (0.30)(0.3)
+ (0.42)(0.6) = 0.342, (0. 42)(0.4) + (0.28)(0.5) = 0.308 and (0.28)(0.5)
= 0.140 respectively. Continuing in this manner, and adding the
probabilities that player A wins 6-0, 6-1, 6-2, 6-3, 6-4, 7-5 or
7-6, it follows that the probability player A wins a tiebreak set
is equal to 0.5164 (see Table 1).
(c)
A match of the present best-of-three tiebreak sets
We now consider a match of the best-of-three tiebreak sets between
two such equal players. We note that if player A serves in the first
game of a set and the set lasts an even number of games (ie the
set score is 6-0, 6-2, 6-4 or 7-5), then, under the present service
exchange rules, player A also serves first in the next set (otherwise
player B serves first). Thus, if player A has an advantage of serving
first in the first set, he/she also has that advantage in the second
set when an even number of games is played in the first set. The
probability that player A wins a tiebreak set in an even number
of games, and the probability he/she wins the set in an odd number
of games are given in Table 1.
Corresponding probabilities are also given for player B.
Table 2 lists ten mutually
exclusive outcomes for a best-of-three sets match won by player
A, given player A serves in the first game of the first set. Note
that in Table 2 a prefix is
used to denote the server in the first game of the set, a capital
letter is used to denote the winner of the set, and odd/even classifies
the number of games played in the set. The fourth column in Table
2, headed probability, is obtained by multiplying the probabilities
of the events in columns 1 and 2, or columns 1, 2 and 3 for the
case in which PA=PB=0.6 and D=0.1. These probabilities can in turn
be obtained from Table 1. Note that to obtain probabilities for sets with player
B serving first, we can just reverse the roles of player A and player
B (as they are equal players). Thus,
for example, the probability of the event bBeven equals the probability
of the event aAeven.
It can be seen from Table 2
that when PA=PB=0.6 and D=0.1, the probability that player A wins
a best-of-three tiebreak sets match given he/she serves first in
the match, is equal to 0.5091. Column (c) in Table 3 gives corresponding results for the
other values of PA, PB and D in Table
1. Thus, it is clear that player A gains a match advantage by
serving first in the first set.
(d)
An alternative best-of-three tiebreak sets system
We now consider the effect of modifying the service exchange mechanism
'across-sets'. The case in which service alternates at the beginning
of each set is considered. It can be shown using values from Table
1 that, given player A serves first in the first set, player
B first in the second set and player A first in the third set (if
necessary), the probability that player A wins the match is equal
to 0.5082 when PA=PB=0.6 and D=0.1. This is a slight improvement
on the present situation analysed in (c) above. The corresponding
probability values for other values of PA, PB and D are given in
column (d) of Table 3. Also, it can be seen that if we modify
this service exchange mechanism so that player B serves first in
both the second and third sets (if necessary), player A's probability
of winning the match is now 1-0.5082 when PA=PB=0.6 and D=0.1. Thus,
this modification to the third set server leads to no overall difference
in fairness on simply alternating service at the beginning of each
set. Also, note that when a set lasts an even number of games under
this system, the same person serves the last game in that set and
the first game of the following set (if played). This should not
be a problem as the players have a two minute rest between sets.
(e)
An alternative 'across- sets' service exchange mechanism
We now consider a slight variation in the third set to the service
exchange mechanism considered in (d) above. We suppose the server
in the third set is determined as at present. That is, given player
B served first in the second set, player A serves first in the third
set if there is an odd number of games in the second set, and player
B serves first if there is an even number of games. It can be shown
using values from Table 1 that,
with this variation to (d) above, the probability player A wins
the match is equal to 0.4993 when PA=PB=0.6 and D=0.1. This represents
a considerable improvement on the situation analysed in (d) above.
The corresponding probability values for other values of PA and
PB are given in column (e) of Table
3. (Another variation in the third set to this service exchange
mechanism could
be where the server in the third set is the player who won the most
number of games in the first two sets. If this countback procedure
leads to a tie in the number of games won, we use the present service
exchange mechanism as in (c) above. This two stage countback mechanism
in fact leads to a very small increase in fairness, but this is
not considered to be worthy of further discussion. Other countback
methods have been considered in another context (Pollard and Noble,
2006).)
(f)
A 'within- set' service exchange mechanism
A service exchange mechanism similar to that used in the tiebreak
game is considered. Player A serves in the first game, player B
serves in the next two games, player A serves in the following two
games,
(ie A,B,B,A,A,B,B,A,A,B,B,A). The present stopping rules
(6-0, 6-1, 6-2,
7-5) are used and the tiebreak game is played if
the games' score reaches 6-6. Under this service exchange mechanism,
assuming that the two equal players have an equal chance of winning
the tiebreak game if played, the probability player A wins the set
is equal to 0.5046 when PA=PB=0.6 and D=0.1. It can be seen by comparing
Table 4 column 4 with Table
1 that this service exchange mechanism within a set considerably
reduces the advantage that player A obtains by serving first. A
major disadvantage of this mechanism is that player B is required
to serve two games in a row on (up to) three occasions, and player
A is required to do the same on (up to) two occasions. This mechanism
is not considered to be of particular practical relevance. However,
if this mechanism was used, change-of-ends might occur after an
even number of games is played, so that, when a player serves two
games in a row, they are from different ends of the court with a
time-break between those service games.
(g)
Another 'within-set' service exchange mechanism
A variation of the 'tiebreak-like' service exchange mechanism in
the above paragraph is the following-A,B,B,A; B,A,A,B; B,A,A,B.
Using the present stopping rules and this mechanism, it can be shown
that the probability player A wins a set is equal to 0.5032 when
PA=PB=0. 6 and D=0.1 (see Table
4 column5). It can be seen from Table 4 that this mechanism gives a slight improvement
on that in the previous paragraph. Players A and B would each have
to serve two games in a row on (up to) two occasions, and change-of-ends
could again be 'on-the-even'. This mechanism is also considered
to be of little practical relevance.
(h)
A third 'within-set' service exchange mechanism
A further 'within-set' mechanism is now considered. Suppose player
B, the server in the second game of the set, is allowed to serve
two games in a row on (up to) one occasion in the set (whilst player
A never serves two games in a row). The possibilities for the (maximum
of) twelve games in a set (up to 6-6) are
(i) A,B;B,A,B,A,B,A,B,A,B,A
(ii) A,B,A,B;B,A,B,A,B,A,B,A
(iii) A,B,A,B,A,B;B,A,B,A,B,A
(iv) A,B,A,B,A,B,A,B;B,A,B,A
(v) A,B,A,B,A,B,A,B,A,B;B,A
For
these five alternatives it can be shown that the probability player
A wins the set when PA=PB=0.6 and D=0.1 is
(i) 0. 4976
(ii) 0.5037
(iii) 0.5078
(iv) 0.5111 and
(v) 0.5140 (see Table 5)
Thus,
the mathematics suggests that it would be in player B's interest
to elect to serve the two games in a row early (rather than later)
in the set. However, he/she might prefer to elect to do it later
in the set when the games are more important, or alternatively just
after having played an 'easy' service game. It would seem that such
a system would
increase the 'excitement' of the set. 'Change- of-ends' might again
be 'on-the- even'.
The player who serves first in a match against an equal opponent
has been shown to have an overall advantage in the situation in
which each player has the same psychological advantage when ahead
in games' score within the set. Several methods of decreasing this
advantage have been considered, and two of them would seem appropriate
for consideration. Firstly, if service alternates at the beginning
of each set (except the final third or fifth set), the benefit a
player receives from serving first in the match is reduced. Secondly,
if the player who serves second in a set is allowed to serve on
two consecutive occasions within that set, the benefit the player
receives from serving first in the set is reduced.
(i)
Another psychological factor
It has been argued by some players, commentators, spectators,
that
some players possess a different psychological factor called the
'back-to-the-wall' effect. In this case the player is assumed to
have a higher probability of winning a game when behind. We firstly
consider the case in which both players possess this factor. Thus,
player A is assumed to have a probability PA of winning a game on
service when the games' scores are equal, and that it is PA+D when
he/she is behind and that it is PA-D when ahead. Correspondingly,
it is assumed that player B has a probability PB of winning a game
on service when the games' scores are equal, and that it is PB+D
when he/she is behind and that it is PB-D when ahead. Thus, the
psychological advantage of being behind is represented by D. Similarly
to (b) above, and assuming player A serves first in the set, the
probability player A wins a tiebreak set is 0.4806 (refer to Table
6) and the probability player A wins a best-of- three tiebreak
sets match is 0.4896 (refer to Table
7 column (c) ) when PA=PB=0.6 and D=0.1. Using the 'across-sets'
service exchange mechanism described in (e) above, player A's probability
of winning such a modified best-of-three tiebreak sets match is
0.5007 when PA=PB=0.6 and D=0.1 (refer to Table
7 column (d)). This represents a considerable improvement on
the number immediately above.
The 'within- set' modifications considered in (f), (g) and (h) all
decrease player B's probability of winning a set from (1-0.4806)=0.5194.
The reason for this is that under these modifications player B is
required to serve (on average) earlier in the match so he/she is
less often behind (when his/her p- values are higher). The mechanisms
in (f) and (g) were considered to be of little practical relevance.
With respect to the mechanism in (h), as player B's probability
of winning the set is decreased for all cases (i) to (v), player
B would presumably not elect to serve two games in a row as he/she
would only decrease his/her probability of winning the set.
(j)
The combination of the two psychological factors
We now assume that player A possesses a 'front- runner' factor D1,
and player B possesses a 'back- to-the-wall' factor D2. Player A's
probability of winning
a service game is equal to PA when the games' scores are equal,
PA+D1-D2 when he /she is ahead in games' scores, and PA-D1+D2 when
he/she is behind. Correspondingly, player B's probabilities on service
are PB when equal, PB+D1-D2 when B is ahead and PB-D1+D2 when behind.
It can be seen that player A's probability of winning a game on
service is always PA when D1=D2, and player B's is always PB when
D1=D2. Thus, the present scoring system is fair for this situation,
as are the two recommendations in (h) above.
(k)
Doubles
The situations for doubles are very similar. As an example, we consider
section (b) above for the case in which PA1=PB1=0.65 and PA2=PB2=0.55
(PA and PB both average 0.6), the only psychological factor being
the 'front- runner' effect for every player and it is assumed to
be D=0.1, and the teams' chances at the tiebreak game are assumed
to be equal.
When the service order is A1, B1, A2, B2,
( the typical case in
which each team uses their more effective server first) the probability
team A wins the tiebreak set is 0.5189. When the service order is
A2, B2, A1, B1,
the probability team A wins the set is 0.5139 (
a fairer outcome than for the order in the previous sentence). This
suggests a minor adjustment to the 'across-sets' modification in
(e) above. Namely, if team A serves first in the first set (with
service order A1, B1, A2, B2) and team B serves first in the second
set (with service order B1, A1, B2, A2), then if team A serves first
in the third set, the first two servers should be reversed (ie A2,
B2, A1, B1,
), and if team B serves first in the third set, the
order should be B2,A2,B1,A1.
Looking
at the 'within-set' changes considered in section (h) (cases (i)
and (ii) in particular), when the order is A1,B1;B2,A2,B1,A1,
,
the probability team A wins the set is 0.4820, and when the order
is A1,B1,A2,B2;B1,A1,B2,A2,
, the probability team A wins the set
is 0.5054. This suggests that after four games have been played
within a set, team B be allowed to play two service games in a row.
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