College Teaching Methods & Styles Journal First Quarter 2007 Volume 3, Number 1

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1 The Economics Of The Duration Of The Baseball World Series Alexander E. Cassuto, ( California State University, Hayward Franklin Lowenthal, ( California State University, Hayward ABSTRACT This note examines some statistical features of the major league baseball World Series. We show that, based upon actual historical data, we cannot reject the hypothesis that the two World Series teams are evenly matched. Yet, we can also calculate the relative strengths of the teams that would best match the actual outcomes, and we find that those relative strengths are not equal. Including the home field advantage in the calculations indicates that the differential in relative strength between the competing teams can be explained by this advantage. We present the relative team strengths that would maximize the probability of four, five, six and seven game series. We find that a six or seven game series is most likely when the two teams are evenly matched, a four game series is most likely when the probability of the stronger team winning is one, while the probability of a five game series is maximized if one team has a relative strength of We also show that, on average, the expected number of World Series games will be between 4 and 5.81, depending upon the relative strengths of the teams and the home field advantage. Contracts that do not consider the likelihood of less than seven game Series will create windfall gains to MLB and marginal economic losses to broadcasters. INTRODUCTION R elative team strength is a factor determining the number of games played in baseball s World Series. This paper examines some statistical aspects of expected outcomes using the binomial probability distribution and data from the initial World Series in 1903 to the 2003 Marlins-Yankees Series. By extending the analysis we can also determine the revealed relative strength of World Series teams based upon the actual outcome of the ninety-three World Series played as best of seven series. The four World Series played as best of nine (1903, ) were not included in the results and games played to ties (1907, 1912, 1922) were not considered although the actual World Series results were included. DATA AND ANALYSIS The formula: P n! r!( n r)! r r r p q n ( ), where r is the number of successes (in this case 4), n the number of trials (games played) and p the probability of one team winning a game (q=1-p), describes the probabilities of all possible outcomes of a World Series. There are a total of seventy possible outcomes, not 2 7 = 128. To see this outcome recall that for team A (American League) to win in M games they must win exactly 3 of the first M-1 games and the M th game. There is only one way that team A can win in 4 games while there are 4!/(3!1!) = 4 ways to win in 5 games, 5!/(3!2!)=10 ways to win in 6 games and 6!/(3!3!)=20 ways to win in 7 games. The number of ways that team N (National League) can win is the same (although one of the authors believes that team Y (Yankees) should always win). The probabilities for each outcome are presented below assuming each team has an equal probability of winning each game (p=q). 1

2 Table 1 Number of Total Games Probability of Outcome Examining the actual results from all the World Series that have been played we find that seventeen series were decided in four games, twenty-one were decided in five games, twenty-two were decided six games while thirty five went the full seven games. The theoretical outcomes, based upon the probabilities in Table 1 are for a four game series, for a five game series and for both six and seven game series. Using the null hypothesis that the World Series teams are evenly matched the computed value of Chipermissible type one error and three degrees of freedom, we do not reject this hypothesis. One of the more interesting aspects of this analysis eme in order to find the revealed relative strengths of the teams given the actual results. In this sense we are finding the expected probabilities that best match the actual outcome over all World Series played. Although we cannot reject the null at a value of 5.08 the two teams entering the series do not have the same probabilities of emerging victorious in term of the revealed team strengths. The minimum Chi-Square can be found by calculus; the detailed calculations are extremely lengthy so we instead used the SOLVER feature in EXCEL. A word of caution is in order. SOLVER finds only a local minimum, one that depends upon the initial value of the independent variable, here the probability of the superior team winning a given game. However, we believe that we have made a sufficient number of different initial values to justify our claim that this is a global minimum. One of the most surprising aspects to the actual outcome data is that while Table 1 indicates that if the two teams are equally matched, six and seven game series are equally probable, there have been only 22 six game series but 35 seven game series. Under the assumption that the teams are evenly matched we compute the probability that of the 57 series that lasted more than 5 games, 35 or more would last the full 7 games to be only Here we used the normal distribution as an approximation to the binomial distribution. However, if we assume that a two-tail test is appropriate the probability doubles to For a one-tail test we can reject the null hypothesis at the 95% confidence level while for a two-tail test we cannot reject it at the 95% confidence level, but do reject it at the 90% confidence level. Clearly the probability of a series that is not decided in 5 games ending with the sixth game should increase if the teams are not evenly matched i.e. if p is greater than.50. In fact, the probability that the series will end in 6 games given that the series stands at 3 to 2 is given by p 2 + (1 p) 2. This is because the probability that the better team is ahead is p (it won the extra game) and their probability of winning the 6 th game is p. Likewise, the worse team has a probability of leading of (1-p) and the same probability of (1-p) of winning the 6 th game. The function p 2 + (1 p) 2 has a minimum at p =.5 and is, of course, maximum at the end points. If we use the value of p =.569 which minimizes the chi-square statistic, we find the probability of the series ending with the 6 th game when one team leads 3 to 2 is.510. Under this assumption, we compute the probability that of the 57 series that lasted more than 5 games, 35 or more would last the full 7 games to be only However, if we assume that a two-tail test is appropriate the probability doubles to For a one-tail test we can reject the null hypothesis at the 95% confidence level while for a two-tail test we can reject it at the 90% confidence level, but not at the 95% confidence level. We offer a non-statistical explanation for this statistical anomaly. The team that is behind must play to win game at all costs; thus it may change its rotation to start its best pitcher, use its best reliever for more innings than usual etc., while the team that is ahead will formulate its strategy so as to win one more game, not necessarily the sixth game. 2

3 Finally we can use the theoretical outcomes to see what relative team strengths are most likely to bring about each possible outcome in terms of games played. In other words, for what value of p are the probabilities in Table 1 above maximized? Our results are summarized in Table 2 below. Table 2 Number of Games in Series Probability that Maximizes Maximum Probability 4 games p = 1 or p = games p =.789 or p = games p = games p = The four game outcome is obvious; a relative team strength of 1 and 0 will bring about a probability of 100% that the series will end in four games. We could also verify this by examining the function f(p) = p 4 + (1-p) 4 that represents the probability that one of the two teams wins all 4 games (p is the probability that a given team wins one game-we assume throughout that this probability does not change from game to game). The maximum occurs at the endpoints 0 and 1; to find the minimum, compute the first derivative df/dp = 4p 3 4(1-p) 3 = 8p 3 12p p 4 = 4(2p-1)(p 2 p +1) This is zero if and only if p = ½ since the quadratic factor has no real root. The minimum value for the probability of a sweep is the value 1/8 =.125 given in Table 1. According to Table 2 if the relative team strength is and the probability of a five game series is maximized at 33.3%. To verify this we must examine the function f(p) = 4(p 4 )(1-p) + 4p(1-p) 4 which represents the probability that one of the two teams wins a 5 game series (p is the probability that a given team wins one game). The minimum of zero occurs at the endpoints 0 and 1 (a 5 game series is impossible because there is a sweep); to find the maximum, compute the first derivative df/dp = 16(p 3 )(1-p) 4p 4 + 4(1-p) 4 16p(1-p) 3 = 4(2p-1)(-6p 2 +6p 1) This is zero if and only if either p = ½ or the quadratic factor is zero. These roots are just ½ + sqrt(12) /12 = and ½ - sqrt(12)/12 = It is easy to see that p = ½ leads to the relative minimum value of.25 listed in Table 1 while the solutions of the quadratic lead to the absolute maximum value of 1/3. The relative minimum at p = ½ is what we would expect since for evenly matched teams it is likely that the series will last at least 6 games. To summarize, at the endpoints p = 0 and p = 1, a sweep is certain; as the probability of a sweep decrease the probability of a 5 game series increases, reaching a maximum of 1/3 at the symmetric points p =.789 and p =.211. But then as the teams become still more evenly matched the probability steadily decrease to the relative minimum value of ¼ achieved at p = ½. In the recent World Series between the New York Yankees and the Arizona Diamondbacks a sports columnist predicted a five game series with the Yankees as winners. When is the probability of a specified team winning in 5 games maximized? This is a slightly different problem than the one we addressed above. The probability of such an event is represented by only the first term of f(p), i.e. 4(p 4 )(1-p). The derivative of this function is routinely found to be zero at p = 4/5 =.80. In that case we obtain the maximum probability of a 5 game Yankee win (which sadly did not occur) to be (4/5) 5 =.32768; interestingly this is only slightly less than the maximum probability of 1/3 of either the Yankees or the Snakes winning in 5 games. The explanation of why there is so little difference between 3

7 REFERENCES 1. CNNSI.com, Fox Lands Exclusive TV Rights to Postseason Baseball, 2. CNNSI.com, Fox Posts Lowest Rated Series Ever, 3. The Sporting News, The Vault, 4. Staudohar, Paul D., The Symbiosis Between Baseball and Broadcasting, in Alvin L. Hall. editor, The Cooperstown Symposium on Baseball and American Culture, 2001, McFarland & Company, Summer 1997: 75 Years of National Baseball Broadcasts, 6. Out of Control Yet? 7

8 NOTES 8

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