Sunday, September 30, 2012

Honnold-Mudd Library and The Odds of Winning the World Series

Greetings everyone!

Honnold Mudd Library

Yesterday I visited the Honnold Mudd Library, which is part of the libraries of the Claremont Colleges. The library has open for most of the day and has room for the public. Amazingly, this is the first collegiate library that I have ever been to where the mathematics section is located on the first floor. The bookshelves for this section are on rollers - which allowed one aisle to be open at a time.

I spent the better part of four hours there, reading about game theory, probability, and the Zeta function.

Here is the link to the Claremont College Libraries' web page.

The Odds of Winning the World Series

Every October in the United States (and Canada should the Toronto Blue Jays be one of the contenders), Major League Baseball (a.k.a. MLB) has an annual tournament of 10 teams to determine the World Series winner. The 10 teams consists of six division winners and four "wild-card" teams. MLB started the current format this year.

Out of the 10 teams, 2 final teams, the National League Champion and American League Champion meet for the World Series. The series is a best of seven series, where the first team to win four games wins the Series and is the Championship team for that year.

We can calculate the odds of a team winning that series by using the Binomial distribution, with the following formula:

nCr × p^r × (1-p)^(n-r)

Where:
n = number of trials
r = number of successes
p = probability of success of one trail
1 - p = q = probability of failure of one trail. Note that p + q = 1.
nCr = combination of n objects taken r objects at a time = n! ÷ ((n - r)! r!)

Even Odds

Let's say two teams are evenly matched. That is, p = 0.50, and q = 1 - p = 0.50. What will be the odds whether one team wins the World Series in four games? Five games? Six games? Seven games?

Win the Series in Four Games

If a team wins a game in four games, it is said that the victorious team swept the losing team. To achieve the probability of a sweep, find the combination of winning all four games and multiply by the odds. That is:

4C4 × 0.50^4 × 0.50^0 = 1 × 0.50^4 = 0.0625

Winning the Series in Five Games

In order to win the series in five games, you must win three games and lose one in the first four games. Obviously, you have to win the fifth game. In calculating probability for this situation, it really does not matter which order you win the first three in four, but that it happens. The "lack of order" requirement allows us to use combinations.

Multiply the result of winning 3 out of 4 games by the odds of winning the fifth game to get the total probability. The total probability of winning the World Series in five games is:

P(winning 3 games out of 4) × P(winning the fifth game)
(4C3 × 0.50^3 × 0.50^1) × (0.50)
= 0.25 × 0.50
= 0.125

Winning the Series in Six Games

In order to win the series in six games, you must win three games and lose two in the first five games. Using a similar approach to the 5 game situation, the total probability of winning the World Series in six games is:

P(winning 3 games out of 5) × P(winning the sixth game)
(5C3 × 0.50^3 × 0.50^2) × 0.50
= 0.3125 × 0.50
= 0.15625

Winning the Series in Seven Games

Similarly, in order to win the series in seven games, you and your opponent win three games apiece in the first six games. Using a similar approach to the 6. game situation, the total probability of winning the World Series in seven games is:

P(winning 3 games out of 6) × P(winning the seventh game)
(6C3 × 0.50^3 × 0.50^3) × 0.50
= 0.3125 × 0.50
= 0.15625

Not surprisingly, the odds of your team winning the World Series in an evenly matched series is 50%. (0.0625 + 0.125 + 0.15625 + 0.15625 = 0.50.)

Underdogs and Favorites

What if the teams are not evenly matched? Instead, we have an underdog, whose odds of winning each game is less than 50%; and a favorite, whose odds of winning each game is greater than 50%. We can use the analysis presented above to calculate the odds of winning the World Series.

For example, say the National League Champion San Francisco Giants, the underdog, plays against the American League Champion New York Yankees. Expert odds makers estimate that the Yankees win about 55% of their games against the Giants. What are the odds of each team winning the World Series?

From the Giants' Point of View, the odds of winning a game is 45%, or p = 0.45. (q = 0.55) Remember each team can win the World Series in one of four ways: sweep (in four games), in 5 games, in 6 games, and in 7 games.

Sweep:
P(win 4 games of 4) = 4C4 × 0.45^4 × 0.55^0 ≈ 0.0410063

Win in 5 Games:
P(win 3 games of 4) × P(winning the 5th game)
= (4C3 × 0.45^3 × 0.55^1) × 0.45 ≈ 0.0902138

Win in 6 Games:
P(win 3 of 5 games) × P(winning the 6th game)
= (5C3 × 0.45^3 × 0.55^2) × 0.45 ≈ 0.1240439

Win in 7 Games:
P(win 3 of 6 games) × P(winning the 7th game)
= (6C3 × 0.45^3 × 0.55^3) × 0.45 ≈ 0.1364483

Total Chance of the Giants winning the World Series (with p = 0.45):
0.0410063 + 0.0902138 + 0.1240439 + 0.1364483 = 0.3917122 (approximately 39.17%)

Going to the Yankees, the favorites in this example. Now the probabilities are reversed, with p = 0.55 (q = 0.45).

Sweep:
P(win 4 games of 4) = 4C4 × 0.55^4 × 0.45^0 ≈ 0.0915063

Win in 5 Games:
P(win 3 games of 4) × P(winning the 5th game)
= (4C3 × 0.55^3 × 0.45^1) × 0.55 ≈ 0.1647113

Win in 6 Games:
P(win 3 of 5 games) × P(winning the 6th game)
= (5C3 × 0.55^3 × 0.45^2) × 0.55 ≈ 0.1853002

Win in 7 Games:
P(win 3 of 6 games) × P(winning the 7th game)
= (6C3 × 0.55^3 × 0.45^3) × 0.55 ≈ 0.1667701

Total Chance of the Yankees winning the World Series, with p = 0.55,
0.0915063 + 0.1647113 + 0.1853002 + 0.1667701 = 0.6082878 (approximately 60.83%)

Note: 0.3917122 + 0.6082878 = 1. As a check, see if the total probabilities add to 1. Being a favorite can significantly increase your chance of winning the World Series!

A RPN Program (HP 15C)

Here is a RPN (Reverse Polish Notation) keystroke program to calculate

nCr × p^r × (1-p)^(n-r)

I used a HP 15C. On the 15C (and 32S, 35S), recall arithmetic can be used to save space. Adjust accordingly.


(Key: Key Code)
LBL A: 42 21 11
RCL 3: 45 3
RCL 2: 45 2
y^x: 14
1: 1
RCL- 3: 45 30 3
RCL 1: 45 1
RCL- 2: 45 30 2
y^x: 14
×: 20
RCL 1: 45 1
RCL 2: 45 2
Cy,x: 43 40
×: 20
RTN: 43 32
(15 steps)


Input:
Preload the following memory registers: R1 = n, R2 = r, and R3 = p

Output:
Probability on the x stack level (display), with the previous contents of the x stack level on the y, z, and t stack levels.


Source of the formula: Packel, Edward. "The Mathematics of Games and Gambling". MAA 1981. New York

Before I go for today, I leave you with several fun facts:

Let z be any complex number, where z = a + bi, with a the real component and b the imaginary part

The Riemann Zeta Function has no zeros (roots) when a is greater than 1.

The Gamma Function has no zeros when a is greater than 0.

Source: Patterson, S.J. "An Introduction to the Theory of the Riemann Zeta Function" Cambridge University Press 1988.


This blog is property of Edward Shore, 2012.

2 comments:

  1. As a note, you can do the same thing by saying that the team that wins 4 *or more* games is the winner. (It wouldn't happen in real life that a team would win 6-1, but you can work out that all the different things that start "WLWWW" (for instance) get covered once and only once.)

    This way, if you happen to have the cumulative binomial distribution programmed into your calculator, as I do, you can use that as a ready-to-go way to get all the probability. (I don't have a 15C, just an 11C, but the program's basically the same--I can send you it if you want, but it's not too hard to work out.)

    ReplyDelete
  2. Tabstop:

    Great insight. As we know with combinations, the order doesn't matter, as long as we get the required number of wins.

    Eddie

    ReplyDelete

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