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At the World Series (Posted on 2004-04-01) Difficulty: 3 of 5
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?

See The Solution Submitted by Federico Kereki    
Rating: 3.2000 (5 votes)

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Some Thoughts Most equitable | Comment 3 of 7 |

The most equitable way to plan the World Series (maybe not the most practical way) would be to play the first game at the stadium of the weaker of the two teams (as determined by a survey of leading sportswriters), and then to play every succeeding game at the stadium of the team that lost the previous game. That would put an end to sweeps, since to sweep, you would have to defeat your opponents at their stadium in at least three straight away games. That never happens at the championship level.  


  Posted by Penny on 2004-04-01 13:41:57
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