October 20, 2011
The BP Wayback Machine
The Importance of Being 1-0
While looking toward the future with our comprehensive slate of current content, we'd also like to recognize our rich past by drawing upon our extensive (and mostly free) online archive of work dating back to 1997. In an effort to highlight the best of what's gone before, we'll be bringing you a weekly blast from BP's past, introducing or re-introducing you to some of the most informative and entertaining authors who have passed through our virtual halls. If you have fond recollections of a BP piece that you'd like to nominate for re-exposure to a wider audience, send us your suggestion.
In the wake of the Cardinals' Game 1 victory, revisit Rany's investigation of what it means to go down 1-0, which originally ran as a "Doctoring the Numbers" column on October 28, 2001.
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
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
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
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:
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.