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To answer these questions, we'll need a basic fact: the odds of a team winning 4 to N are (1/2)(3+N) * C(4+N,N). Thus, the odds of a 4-0 result are 1/8; 4-1, 1/4; 4-2, 5/16; and 4-3, 5/16. (This results also appear in the solution to the original "World Series" problem.) The expected number of games, however played, is 4*1/8 + 5*1/4 + 6*5/16 + 7*5/16 = 93/16, just a little under 6.
The odds of the first 4 matches being played is 1, for they will always have to be played. The odds for needing a 5th game are 1-1/8=7/8. The odds for a 6th game are 7/8-1/4=5/8. Finally, the odds for a 7th game are 5/8-5/16=5/16.
The most equitable way to play the games is: split the first 4 games 2-2, and give a team the 5th match, and the 6th and 7th to the other: this gives the first team 2+7/8 games, and the second 2+15/16 games -- just a 1/16 game difference.
If you want to minimize travel, the optimum solution is the one actually used: start with two home matches, then three away matches, and finish with two home matches, but there are eleven more possibilities. |
