The (baseball) World Series consists of seven games, and the first team to win four games wins the title. The first two games are played at home; the next three away, and the last two at home again -- though of course not all games might be played. (As happened this year when the Marlins beat the Yankees 4-2... GRRR!!)

The probabilities involved were mentioned in World Series.

Assuming the teams have the same chance of winning each game, what is the most equitable way to plan the games, so as to minimize the expected difference between home and away games? And, in that case, what are the expected numbers of home games, of away games, and of total games?

According to the solution of the "World Series" problem, the odds of getting a 4 matches series are 1/8; 5 matches, 1/4; 6 or 7 matches, 5/16. So, the four first matches are always played; match #5 is played 7/8 times; match #6 is played 5/8 times, and match #7 is played only 5/16 times.

Using a common denominator, matches #1-#7 are played proportionally to 16, 16, 16, 16, 14, 10, and 5, which add up to 93. We must distribute them among home/away, so both sums are as close to 93/2 as possible. The best distribution is to placing 16, 16, 10 and 5 in a group, and 16, 16, 14 in the other -- and the arrangement that minimizes traveling is the current one: 2 home games, 3 away games, and 2 home games.

The number of home games is then 47/16, away games are 46/16, and the total number of games is 93/16.

Posted by e.g. on 2004-04-01 09:55:32