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 At the World Series (Posted on 2004-04-01)
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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 solution | Comment 1 of 7

First, the definition of "played at home" in this case apparently means the home team of the first game plays at home, as each game has one team playing at home and one playing away.  So we can assume the first game is played "at home".

The solution for World Series shows that the probabilities of going given numbers of games are:

`4  1/85  1/46  5/167  5/16`

The first four games of course each have probability 1 of being played.  Game 5 has probability 1-1/8 = 7/8 of being played.  Game 6 has probability 1-1/8-1/4 = 5/8 of being played (that's also 5/16+5/16). Game 7 of course has probability 5/16 of being played.

If we multiply each of these probabilities by 1 if the game is a home game and 0 if it is an away and add the seven products, we get the expected number of home games.  If the 1's and 0's are reversed we get the expected number of away games.

The following table lists, first, for every set of 7 games where the number of home games is 3 or 4, what the expected number of home games, away games and the difference is:

`1  0  0  0  0  1  1          1.9375    3.875    -1.93751  0  0  0  1  0  1          2.1875    3.625    -1.43751  0  0  0  1  1  0          2.5       3.3125   -.81251  0  0  0  1  1  1          2.8125    3        -.18751  0  0  1  0  0  1          2.3125    3.5      -1.18751  0  0  1  0  1  0          2.625     3.1875   -.56251  0  0  1  0  1  1          2.9375    2.875     .06251  0  0  1  1  0  0          2.875     2.9375   -.06251  0  0  1  1  0  1          3.1875    2.625     .56251  0  0  1  1  1  0          3.5       2.3125    1.18751  0  1  0  0  0  1          2.3125    3.5      -1.18751  0  1  0  0  1  0          2.625     3.1875   -.56251  0  1  0  0  1  1          2.9375    2.875     .06251  0  1  0  1  0  0          2.875     2.9375   -.06251  0  1  0  1  0  1          3.1875    2.625     .56251  0  1  0  1  1  0          3.5       2.3125    1.18751  0  1  1  0  0  0          3         2.8125    .18751  0  1  1  0  0  1          3.3125    2.5       .81251  0  1  1  0  1  0          3.625     2.1875    1.43751  0  1  1  1  0  0          3.875     1.9375    1.93751  1  0  0  0  0  1          2.3125    3.5      -1.18751  1  0  0  0  1  0          2.625     3.1875   -.56251  1  0  0  0  1  1          2.9375    2.875     .06251  1  0  0  1  0  0          2.875     2.9375   -.06251  1  0  0  1  0  1          3.1875    2.625     .56251  1  0  0  1  1  0          3.5       2.3125    1.18751  1  0  1  0  0  0          3         2.8125    .18751  1  0  1  0  0  1          3.3125    2.5       .81251  1  0  1  0  1  0          3.625     2.1875    1.43751  1  0  1  1  0  0          3.875     1.9375    1.93751  1  1  0  0  0  0          3         2.8125    .18751  1  1  0  0  0  1          3.3125    2.5       .81251  1  1  0  0  1  0          3.625     2.1875    1.43751  1  1  0  1  0  0          3.875     1.9375    1.93751  1  1  1  0  0  0          4         1.8125    2.1875`

The smallest difference, in absolute value, is .0625, shared by the following:

`1  0  0  1  0  1  1          2.9375    2.875     .06251  0  0  1  1  0  0          2.875     2.9375   -.06251  0  1  0  0  1  1          2.9375    2.875     .06251  0  1  0  1  0  0          2.875     2.9375   -.06251  1  0  0  0  1  1          2.9375    2.875     .06251  1  0  0  1  0  0          2.875     2.9375   -.0625`

The next-to-last row is the current scheme, which has the further advantages of reducing travel between cities.

The following program produced the above, as well as verified that no set of home/away other than a 4-3 or 3-4 split will produce even as good an equitability:

`CLSminDiff = 99`
`FOR g2 = 0 TO 1FOR g3 = 0 TO 1FOR g4 = 0 TO 1FOR g5 = 0 TO 1FOR g6 = 0 TO 1FOR g7 = 0 TO 1`
`h = 1 + g2 + g3 + g4 + g5 * 7 / 8 + g6 * 5 / 8 + g7 * 5 / 16away = 3 - g2 - g3 - g4 + (1 - g5) * 7 / 8 + (1 - g6) * 5 / 8 + (1 - g7) * 5 / 16diff = h - awayIF ABS(diff) < minDiff THEN minDiff = ABS(diff)IF g2 + g3 + g4 + g5 + g6 + g7 = 2 OR g2 + g3 + g4 + g5 + g6 + g7 = 3 THEN PRINT 1; g2; g3; g4; g5; g6; g7; TAB(30); h; TAB(40); away; TAB(50); diffEND IF`
`NEXTNEXTNEXTNEXTNEXTNEXT`
`PRINT`
`FOR g2 = 0 TO 1FOR g3 = 0 TO 1FOR g4 = 0 TO 1FOR g5 = 0 TO 1FOR g6 = 0 TO 1FOR g7 = 0 TO 1`
`h = 1 + g2 + g3 + g4 + g5 * 7 / 8 + g6 * 5 / 8 + g7 * 5 / 16away = 3 - g2 - g3 - g4 + (1 - g5) * 7 / 8 + (1 - g6) * 5 / 8 + (1 - g7) * 5 / 16diff = h - awayIF ABS(diff) = minDiff THEN PRINT 1; g2; g3; g4; g5; g6; g7; TAB(30); h; TAB(40); away; TAB(50); diffEND IF`
`NEXTNEXTNEXTNEXTNEXTNEXT`

The expected number of games is of course the total of the expected number of home + expected number of away games, or 5.8125 games.  It's just the total of the probabilities of playing each game, or 1 + 1 + 1 + 1 + 7/8 + 5/8 + 5/16.

 Posted by Charlie on 2004-04-01 09:39:31

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