Are Seven-game Baseball Playoffs Fairer Than Five-game Series?

1 / 14

# Are Seven-game Baseball Playoffs Fairer Than Five-game Series? - PowerPoint PPT Presentation

Are Seven-game Baseball Playoffs Fairer Than Five-game Series?. Dr. Brian Dean. The Conventional Wisdom.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## Are Seven-game Baseball Playoffs Fairer Than Five-game Series?

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Are Seven-game Baseball Playoffs Fairer Than Five-game Series?

Dr. Brian Dean

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.

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.

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.

ResultProbability

WWW p³

LWWW p³(1-p)

WLWW p³(1-p)

WWLW p³(1-p)

LLWWW p³(1-p)²

LWLWW p³(1-p)²

LWWLW p³(1-p)²

WLLWW p³(1-p)²

WLWLW p³(1-p)²

WWLLW p³(1-p)²

Adding these, the total probability that the better team would win a five-game series is

6p⁵ - 15p⁴ + 10p³

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 + 35p⁴

Comparingfive-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.

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?

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.

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:

‘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.

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:

ResultProbability

WWw p²(rp)

LWww p(rp)²(1-p)

WLww p(rp)²(1-p)

WWlw p²(rp)(1-rp)

LLwwW p(rp)²(1-p)²

LWlwW p²(rp)(1-p)(1-rp)

LWwlW p²(rp)(1-p)(1-rp)

WLlwW p²(rp)(1-p)(1-rp)

WLwlW p²(rp)(1-p)(1-rp)

WWllW 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)p³

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, 1L, 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)p⁴

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).

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.

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 Ludwickand Dr. Barbara Wainwright.