Dr Brian Dean The Conventional Wisdom Teams that have earned homefield advantage over the course of a 162game regular season prefer longer sevengame playoff series to fivegame series feeling that the better team is more likely to win in a longer series ID: 279253
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Slide1
Are Seven-game Baseball Playoffs Fairer Than Five-game Series?
Dr. Brian DeanSlide2
The Conventional Wisdom
Teams that have earned home-field advantage over the course of a 162-game regular season prefer longer, seven-game playoff series to five-game series, feeling that the “better” team is more likely to win in a longer series.
Question:
Is the difference between seven-game and five-game series so great that baseball should consider changing the Division Series round to a best-of-seven format?
Goal:
Create a mathematical model to analyze this situation.Slide3
Dr. Lee May (1992):
Seven-game series are
not
significantly fairer than five-game series (where
significantly fairer
is defined to mean that the better team has at least a four percent greater probability of winning a seven-game series than a five-game series.)
May’s model:
Let p denote the probability that the better team will win a given game. Since he’s looking at things from the point of view of the better team, p must lie in the interval [0.5, 1]. (For example, p = 0.7 if the better team has a 70% probability of winning a given game.) Note that May’s model treats each game of the series the same.Slide4
Probability that the better team will win a five-game series
In May’s model, the probability of each
W
for the better team is p, so the probability of each
L
for the better team is 1-p. There are ten total scenarios of victory for the better team in a five-game series. The probability of each scenario is the product of the probabilities of each individual game.
Result
Probability
WWW
p³
L
WWW
p³(1-p)
W
L
WW
p³(1-p)
WW
L
W
p³(1-p)
LL
WWW
p³(1-p)²
L
W
L
WW
p³(1-p)²
L
WW
L
W
p³(1-p)²
W
LL
WW
p³(1-p)²
W
L
W
L
W
p³(1-p)²
WW
LL
W
p³(1-p)²
Adding these, the total probability that the better team would win a five-game series is
6p⁵ - 15p⁴ + 10p³Slide5
Probability that the better team will win a seven-game series
There are a total of 35 different scenarios in which the better team would win a seven-game series. Rather than listing each individually, we will summarize the probabilities of the different scenarios under May’s model:
Series length # of Scenarios Probability of Each
4 games 1 p⁴
5 games 4 p⁴(1-p)
6 games 10 p⁴(1-p)²
7 games 20 p⁴(1-p)³
Adding the probabilities of the 35 scenarios, the total probability that the better team would win a seven-game series is
-20p⁷ + 70p⁶ - 84p⁵ + 35p⁴Slide6
Comparing
five-game and seven-game series
To compare five-game and seven-game series in May’s model, let f(p) denote the probability that the better team would win a seven-game series, minus the probability that it would win a five-game series:
f(p) = (-20p⁷ + 70p⁶ - 84p⁵ + 35p⁴) – (6p⁵ - 15p⁴ + 10p³)
= -20p⁷ + 70p⁶ - 90p⁵ + 50p⁴ - 10p³, 0.5 ≤ p ≤ 1
The maximum value of this function is ≈ 0.0372 (when p ≈ 0.689), and the minimum value is 0 (when p = 0.5). In other words, under May’s model, the better team is at most only about 3.72 % more likely to win a seven-game series than a five-game series. Therefore, a seven-game series is not significantly fairer than a five-game series.Slide7
What are some possible ways to modify May’s model?
My model, the subject of the rest of this talk, will attempt to take home-field advantage into account. That is, the probabilities of victory/defeat in road games will be different from those in home games.
Another possible modification, which we won’t discuss, would be to account for the effects of momentum/morale. That is, would the status of the series after each game affect the probabilities of victory/defeat in the next game? For example, would the probability of victory in game 2 differ depending on whether the team won or lost game 1?Slide8
Model taking home-field advantage into account
Let Team H be the team with home-field advantage in the series, and let p be the probability that Team H will win a given home game. (Since Team H is not necessarily the “better” team, our model does not imply that p ≥ 0.5 like May’s did. Instead, we will allow p to be anything in the interval [0,1], though it seems unlikely that it would ever be much less than 0.5 in practice.)
We will take the probability that Team H will win a given road game to be
rp
, where r is a parameter we will call the
road multiplier
.
Slide9
Road Multiplier
For a given team, we define the road multiplier as the ratio of a team’s road winning percentage to its home winning percentage. If a team’s road multiplier were 0.9, for example, we could say that they would be 90 % as likely to win a given road game as they are a given home game.
For the 112 playoff teams of the first 14 years of the wildcard era (1995-2008), the average road multiplier has been (to three decimal places) 0.883. The three highest and three lowest road multipliers, rounded to three decimal places, have been:
Team Home Road
Road
Multiplier
‘01 Braves 40-41 48-33 1.200
‘97 Orioles 46-35 52-29 1.130
‘01 Astros 44-37 49-32 1.114
‘03 Athletics 57-24 39-42 0.684
‘05 Astros 53-28 36-45 0.679
‘08 White Sox 54-28 35-46 0.656
24 of the 112 road multipliers have been 1.000 or higher, and
17 have been
0.750 or lower.Slide10
Probability that Team H will win a five-game series
In the current five-game series format, Team H plays games one, two, and five at home, and games three and four on the road. Let
W
and
L
denote home wins and losses (with probabilities p and 1-p, respectively) and let
w
and
l
denote road wins and losses (with probabilities
rp
and 1-rp, respectively.) The ten scenarios for victory for Team H are as follows:
Result
Probability
WWw
p²(
rp
)
L
Www
p(
rp
)²(1-p)
W
L
ww
p(
rp
)²(1-p)
WW
l
w
p²(
rp
)(1-rp)
LL
wwW
p(
rp
)²(1-p)²
L
W
l
wW
p²(
rp
)(1-p)(1-rp)
L
Ww
l
W
p²(
rp
)(1-p)(1-rp)
W
Ll
wW
p²(
rp
)(1-p)(1-rp)
W
L
w
l
W
p²(
rp
)(1-p)(1-rp)
WW
ll
W
p³(1-rp)²
The total probability of victory for Team H in a five-game series would therefore be
6r²p⁵ - (9r²+6r)p⁴ + (3r²+6r+1)p³Slide11
Probability that Team H will win a seven-game series
In a seven-game series format, Team H plays games 1, 2, 6, and 7 at home, and 3, 4, 5 on the road.
Series Result # of Scenarios Probability of Each
2
W
, 2
w
1 p²(
rp
)²
1
W
, 3
w
, 1
L
2 p(
rp
)³(1-p)
2
W
, 2
w
, 1
l
2 p²(
rp
)²(1-rp)
1
W
, 3
w
, 2
L
1 p(
rp
)³(1-p)²
2
W
, 2
w
, 1
L
, 1
l
6 p²(
rp
)²(1-p)(1-rp)
3
W
, 1
w
, 2
l
3 p³(
rp
)(1-rp)²
2
W
, 2
w
, 2
L
, 1
l
9 p²(
rp
)²(1-p)²(1-rp)
3
W
, 1
w
, 1
L
, 2
l
9 p³(
rp
)(1-p)(1-rp)²
1
W
, 3
w
, 3
L
1 p(
rp
)³(1-p)³
4
W
, 3
l
1 p⁴(1-rp)³
Adding the probabilities of the 35 scenarios, the total probability that the better team would win a seven-game series is
-20r³p⁷ + (40r³+30r²)p⁶ - (24r³+48r²+12r)p⁵ + (4r³+18r²+12r+1)p⁴Slide12
Comparing five-game and seven-game series
For each fixed value of r, let f(
r,p
) denote the probability that the better team would win a seven-game series, minus the probability that it would win a five-game series:
f(
r,p
) = [-20r³p⁷ + (40r³+30r²)p⁶ - (24r³+48r²+12r)p⁵ + (4r³+18r²+12r+1)p⁴]
- [6r²p⁵ - (9r²+6r)p⁴ + (3r²+6r+1)p³]
= -20r³p⁷ + (40r³+30r²)p⁶ - (24r³+54r²+12r)p⁵ + (4r³+27r²+18r+1)p⁴ - (3r²+6r+1)p³,
0 ≤ p ≤ 1
Note that, if we take r = 1 (that is, treat road games to have the same probability of victory for Team H as home games), then we get
f(1,p) = -20p⁷ + 70p⁶ - 90p⁵ + 50p⁴ - 10p³,
the same function as in May’s model, with the only difference being that we’re no longer requiring p ≥ 0.5.
Let’s first take r = 0.883, the average road multiplier of the 112 playoff teams from 1995-2008. The maximum value of f(0.883,p) is ≈ 0.0339 (when p ≈ 0.728), and the minimum value is ≈ -0.0387 (when p ≈ 0.332). In other words, under our model, using this average road multiplier as our value of r, Team H is at most only about 3.39 % more likely to win a seven-game series than a five-game series (and is actually more likely to win the shorter series if its home-win probability p is low enough).Slide13
Maximum/Minimum Values of f(r,p
) for different values of r
We will consider values of r between 0.650 and 1.200, since the road multipliers of all 112 playoff teams from 1995-2008 have fallen in that interval. All of the max./min. values are rounded off to four decimal places:
r Max. Min. r Max. Min.
0.650 0.0255 -0.0427 0.950 0.0358 -0.0378
0.700 0.0276 -0.0417 1.000 0.0372 -0.0372
0.750 0.0295 -0.0408 1.050 0.0385 -0.0366
0.800 0.0313 -0.0400 1.100 0.0398 -0.0360
0.850 0.0329 -0.0392 1.150 0.0411 -0.0355
0.900 0.0344 -0.0385 1.200 0.0424 -0.0350
In general, the maximum and minimum values of f(
r,p
) are increasing as r is increasing (and, though this is not shown in the table, the values of p at which the max./min. occur are decreasing as r is increasing.)
The value of r for which the maximum value of f(
r,p
) is 0.0400 is r ≈ 1.107. For r below this value, Team H is not significantly likelier to win a seven-game series than a five-game series. Of the 112 playoff teams from 1995-2008, 109 have had road multipliers below 1.107.Slide14
Conclusion and Acknowledgements
Though our model shows that, under certain circumstances, the team with home-field advantage may be significantly likelier to win a seven-game series than a five-game series, a general statement that seven-game series are significantly fairer than five-game series is incorrect.
Thank you to all of the participating schools, students, and teachers in this year’s Eastern Shore High School Mathematics Competition, and to the contest’s co-chairs, Dr. Kurt
Ludwick
and Dr. Barbara Wainwright.