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In order to have a 4-game series, games 2..4 must end the same way as game 1. p = 1/8
In order to have a 5-game series, games 1..4 must be split 3-1, then game 5 must be won by the leader. The probability of splitting 3-1 is 4 * 1/16 * 2 = 1/2. So overall, for a 5-game series, p = 1/4.
In order to have a 6-game series, games 1..5 must be split 3-2, then game 6 must be won by the leader. The probability of splitting 3-2 is 10 * 1/32 * 2 = 5/8. So overall, for a 6-game series, p = 5/16.
In order to have a 7-game series, games 1..6 must be split 3-3. The probability of this is 20 * 1/64 = 5/16.
The above calculations all rely on the binomial distribution. The factors of 2 are there to account for the possibility of either team winning. (That is, when I calculate the probability of going 3-0, it counts the probability of the NL team leading 3-0 plus the probability of the AL team leading 3-0.) |
