How important is Game 1? Statistically speaking, what are the odds that a

team that lost Game 1 of a best-of-seven–in this case, the Yankees–can

come back to take the series?

To examine the issue, let’s simplify the question. Instead of looking at a

best-of-seven, let’s start with a best-of-three series, like the one played

by the Dodgers and Giants in 1962 to decide the NL pennant. (Major League

Baseball’s current setup essentially rules out a best-of-three series as a

tiebreaker, meaning that the 1962 series is probably the last one we’ll ever

see.)

Assuming both teams are equal in quality–an assumption we will make–each

team has a 50% chance of winning at the start of a best-of-three series. The

team that loses Game One, then, will have to win Games Two and Three to take

the series. Since each team has exactly a 50/50 chance to win each game, the

chance that the Game One loser can win two straight is 50% x 50%, or 25%.

The winner of Game One has a 75% chance of taking a best-of-three series.

Moving on to a five-game series…the loser of Game One has a variety of

paths it can take to win the series. It can win the four remaining games, or

can it win the next three games and lose the fifth game. (From a practical

standpoint, of course, the fifth game would never be played in either of

these scenarios, but we have to account for both of them in a statistical

sense.) It can lose the next game, and win the next three–a.k.a., "the

Yankee approach"–it can win Game Two, lose Game Three, and win Games

Four and Five; or it can win Games Two and Three, lose Game Four, and win

Game Five.

All told, there are five different ways in which the first-game loser can

come back to take series. Since there are two different outcomes to every

game, the total number of different outcomes over four consecutive games is

2^4, or 16. Hence, the Game One loser has a 5/16, or 31.3% chance, of

bouncing back to win.

As we look at a seven-game series, the number of possible permutations

becomes a little intimidating. Fortunately, we have Binomial Theory to make

our work a little easier. More precisely, we have Pascal to thank for his

Triangle, which is a simple but powerful way to calculate the probabilities

for us.

Pascal’s Triangle is a pyramid of numbers, starting with the number

"1" all alone at the apex, with each row containing one more

number than the one above it. Each number is calculated by adding together

the two numbers immediately adjacent to it in the row above. It starts like

this:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

Pascal’s Triangle has many uses, one of which is directly pertinent to the

question at hand. Take a look at the fourth row. This row represents the

possibilities of what can happen in a three-game series. The digits in this

row (1, 3, 3, and 1) add up to 8, and indeed there are exactly eight

different permutations to a three-game series.(The eight possibilities are

WWW, WWL, WLW, LWW, WLL, LWL, LLW, and LLL.) There are four digits in the

row, and in fact the eight different permutations can be categorized into

four different outcomes (Team A wins three and loses none, wins two and

loses one, wins one and loses two, and wins none and loses three.)

Finally, the probability of a specific outcome (like Team A sweeping the

series) can be calculated by dividing the number corresponding to this

outcome (1) by the total sum for the entire row (8).

Team 1 wins 3: 1/8

Team 1 wins 2: 3/8

Team 1 wins 1: 3/8

Team 1 wins 0: 1/8

Total: 8/8

This makes intuitive sense. If Team A wins two or three games, then they win

the series, and the chance they will do this is (1/8 + 3/8) = 1/2. The

chance that they win one or no games–and Team B wins the series–is also

1/2.

Now look at the third row, with three digits (1, 2, and 1). This row

represents the final two games of a three-game series. In this instance,

there are a total of 1+2+1=4 possibilities. For the loser of Game One to win

the series, they have to win both of the remaining games. This is

represented by the first digit, 1, meaning the chance that this will happen

is 1/4, or 25%, as we already showed.

The fifth row, then, represents the final four games of a five-game series.

The sum of the numbers in the row is 16, corresponding to the number of

permutations in the four games. There are five digits in the row, which can

be listed as:

Team goes 4-0: 1/16

Team goes 3-1: 4/16

Team goes 2-2: 6/16

Team goes 1-3: 4/16

Team goes 0-4: 1/16

In order to win the series, the loser of Game 1 would have to go 3-1 or 4-0,

meaning their chances are (1/16 + 4/16) = 5/16, exactly as shown above.

So for a seven game series, we merely need to extend Pascal’s Triangle:

1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

The final row represents the final six games of a seven-game series. Let’s

convert this row into a column of possible outcomes:

Team goes 6-0: 1/64

Team goes 5-1: 6/64

Team goes 4-2: 15/64

Team goes 3-3: 20/64

Team goes 2-4: 15/64

Team goes 1-5: 6/64

Team goes 0-6: 1/64

The team that loses Game One of a seven-game series needs to win at least

four of the next six games. The probability of that happening is (1/64 +

6/64 + 15/64) = 22/64, or 34.4%

The series winning percentage of a Game One loser goes up the longer the

series:

Best-of-three: .250

Best-of-five: .313

Best-of-seven: .344

This is intuitive; the longer a series, the more time a team has to make up

a one-game deficit. Just for kicks, let’s look at that winning percentage in

series of greater length:

Best-of-nine: .363

Best-of-11: .377

Best-of-13: .387

Best-of-15: .395

Best-of-17: .402

Having established the theoretical winning percentage of a Game One loser,

we can determine the importance of Game One by comparing a team’s chances of

winning the series when they win or they lose the first game.

In a three-game series, the team that wins Game One has a 75% chance of

winning the series, compared to a 25% chance if they lose. Therefore, the

importance of Game One = 75% – 25%, or 50%. We can say that the

"Importance Level" of Game 1 is .500. In a five-game series, it’s

.375; in a seven-game series, it’s .313.

Not only that, but we can apply the same method to determine the winning

percentage in a variety of different situations. For example, we know that

in a best-of-three series, if the team that loses Game One wins Game Two,

then both teams are level; they both have a 50% chance of winning.

(Obviously, if the Game One loser also loses Game Two, they have a 0% shot

at winning the series.) Therefore, for the loser of Game 1 in a three-game

series, the 7quot;Importance Level" of Game Two is 50% – 0% = 50%, or

.500, exactly the same as it was for Game 1. This is not expected; you would

expect the importance level to increase as a series progresses.

Without going into the math here, take my word that the Importance Level for

any particular game is the same for both sides; obviously, if one team’s

chances increase by winning a particular game, their opponent’s chances go

down by the same amount.

Let’s run the numbers for every possible situation in a five-game series:

Situation Importance Level0-0 .375 1-0 .375 2-0 .250 1-1 .500 2-1 .500 2-2 1.000

Interestingly, when the teams split the first two games of a series, Game

Three is twice as important as when the same team wins Games One and Two.

The same numbers for a seven-game series:

Situation Importance Level

0-0 .313 1-0 .313 2-0 .250 1-1 .375 2-1 .375 3-0 .125 3-1 .250 2-2 .500 3-2 .500 3-3 1.000

The numbers generally trend up as a series gets closer to its conclusion,

but a runaway sweep by one team can render the series a foregone conclusion.

Game One is two-and-a-half times more important than a Game 4 when one team

has swept the first three games of a series.

So when you sit down to watch Game Two tonight, remember that the

Diamondbacks are nearly twice as likely (.656 to .344) to win a World

Championship as the Yankees. Then temper that knowledge with the fact that

the Yankees are a win away from evening those probabilities at .500 apiece.

*Rany Jazayerli is an author of Baseball Prospectus. You can contact him by
clicking here.*