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
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
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
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
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
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
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
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:
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 Level
0-0 .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