The recent proliferation of post-season playoff series means that most career post-season records are held by recent players. The question looked at here is “How many more post-season games does the average player play in?”<\/p>\n
Let’s begin with the “expected length” of a post-season series. Assume that each team is equally likely to win each game. Then the expected length of the series is easy to compute. This is not entirely accurrate, but the probabilities rarely deviate much from 50%, so the actual values should not have a large impact on expected series length. The expected lengths are as follows:
\n5 game series<\/b>: 4.125 \u00a0 \u00a0 \u00a0 \u00a0 7 game series<\/b>: 5.8125 \u00a0 \u00a0 \u00a0 \u00a0 9 game series<\/b>: 7.5390625<\/p>\n
The table lists the seasons, followed by the expected number of post-season games for a “random” team, which is then followed by the expected number of World Series games.<\/p>\n
Years PS WS \r\n1903 0.942 0.942\r\n1905-1918 0.727 0.727\r\n1919-1921 0.942 0.942\r\n1922-1960 0.727 0.727\r\n1961NL 0.727 0.727\r\n1961AL 0.581 0.581\r\n1962-1968 0.581 0.581\r\n1969-1976 1.172 0.484\r\n1976-1984NL 1.172 0.484\r\n1977-1984AL 1.004 0.415\r\n1985-1992NL 1.453 0.484\r\n1993NL 1.246 0.415\r\n1985-1993AL 1.246 0.415\r\n1995-1997 \r\nin 5 team div. 2.333 0.400\r\nin 4 team div. 2.653 0.454\r\n1998-2002\r\nAL in 5 team div. 2.333 0.400\r\nAL in 4 team div. 2.653 0.454\r\nNL in 6 team div. 1.989 0.341\r\nNL in 5 team div. 2.200 0.377\r\n<\/pre>\n
\nExpected length of series<\/h3>\n
In a seven game series, the probability that the series last exactly six games is (5 choose 3)\/25<\/span><\/sup>. One team wins the sixth game to finish the series, their other three wins are distributed among the other 5 games in (5 choose 3) ways. Each arrangement of wins and losses occurs with probability 1\/25<\/span><\/sup>. In this way, the probablity that the series last k<\/i> games can be determined for any value of k<\/i>.
\nLength of a 3 game<\/b> series: 2G<\/b>: 1\/2, 3G<\/b>: 1\/2. Expected length: 2(1\/2)+3(1\/2) = 5\/2 = 2.500
\nLength of a 5 game<\/b> series: 3G<\/b>: 1\/4, 4G<\/b>: 3\/8, 5G<\/b>: 3.8. Expected length: 3(1\/4)+4(3\/8)+5(3\/8)=33\/8 = 4.125
\nLength of a 7 game<\/b> series: 4G<\/b>: 1\/8, 5G<\/b>: 1\/4, 6G<\/b>:5\/16, 7G<\/b>:5\/16. Expected length: 4(1\/8)+5(1\/4)+6(5\/16)+7(5\/16) = 93\/16=5.8125
\nLength of a 9 game<\/b> series: 5G<\/b>: 1\/16, 6G<\/b>: 5\/32, 7G<\/b>: 15\/64, 8G<\/b>: 35\/128, 9G<\/b>: 35\/128 Expected length: 5(1\/6)+6(5\/32)+7(15\/64)+8(35\/128)+9(35\/128) = 965\/128 = 7.5390625<\/p>\n
\nWild card teams<\/h3>\n
Letting a wild-card team into the playoffs complicates the expected value computation. Making the simplifying assumption that teams are equally distributed allows for an approximate expected value to be computed.
\nThe equal distribution assumption means that it is assumed that the team with the best record in a 14 team league has probability of 5\/14 of being in a particular 5 team division, and probability 4\/14 of being in a four team division.<\/p>\n14 team league<\/h4>\n
The team with the second best record will be the wild card team if they are in the same division as the team with the best record.
\nThey will be in a five team division with probability (10\/14)*(4\/13).
\nThey will be in the four team division with probability (4\/14)*(3\/13).
\nThis somputation can be repeated for teams with the third best and fourth best records.<\/p>\nWild-card 5 team division\r\n2nd best (10\/14)*(4\/13) \r\n3rd best (10\/14)*(5\/13)*(8\/12)+(10\/14)*(4\/13)*(7\/12)+(4\/14)*(10\/13)*(4\/12) \r\n4th best (10\/14)*(5\/13)*(4\/12)*(8\/11)+(10\/14)*(4\/13)*(5\/12)*(8\/11)+(4\/14)*(10\/13)*(5\/12)*(8\/11)\r\n\r\nWild-card 4 team division\r\n2nd best (4\/14)*(3\/13) \r\n3rd best (10\/14)*(4\/13)*(3\/12) + (4\/14)*(10\/13)*(3\/12)\r\n4th best (10\/14)*(5\/13)*(4\/12)*(3\/11)+(10\/14)*(4\/13)*(5\/12)*(3\/11) + (4\/14)*(10\/13)*(5\/12)*(3\/11)\r\n<\/pre>\nSimplifying these sums;<\/p>\n
wild-card 5 team div. 4 team div.\r\n2nd best 20\/91 6\/91\r\n3rd best 30\/91 10\/91\r\n4th best 200\/1001 75\/1001\r\nTotal 750\/1001 251\/1001\r\n<\/pre>\nSo, the expected number of playoff teams from 5 team divisions is 2+750\/1001=2752\/1001.
\nThe expected number of playoff appearances for each of the teams is (2752\/1001)\/10=1376\/5005.
\nThe expected number of playoff teams from a 4 team division is 1+251\/1001=1252\/1001.
\nThe expected number of playoff appearance for each of the teams is (1252\/1001)\/4=313\/1001.<\/p>\nThese teams will play in the 5 game division series, expected number of games: 4.125 games.
\nHalf of them will play in the LCS, expected number of games: (5.8125)\/2.
\nOne-quarter of the teams will play in the World Series, expected number of games: (5.8125)\/4.<\/p>\n16 team league<\/h4>\n
Wild-card 5 team division\r\n2nd best (10\/16)*(4\/15) \r\n3rd best (10\/16)*(5\/15)*(8\/14)+(10\/16)*(6\/15)*(9\/14)+(6\/16)*(10\/15)*(9\/14) \r\n4th best (10\/16)*(5\/15)*(6\/14)*(8\/13)+(10\/16)*(6\/15)*(5\/14)*(8\/13)+(6\/16)*(10\/15)*(5\/14)*(8\/13)\r\n\r\nWild-card 6 team division\r\n2nd best (6\/16)*(5\/15) \r\n3rd best (10\/16)*(6\/15)*(5\/14)+(6\/16)*(10\/15)*(5\/14) \r\n4th best (10\/16)*(5\/15)*(6\/14)*(5\/13)+(10\/16)*(6\/15)*(5\/14)*(5\/13)+(6\/16)*(10\/15)*(5\/14)*(5\/13)\r\n<\/pre>\nSimplifying these sums;<\/p>\n
wild-card 5 team div. 6 team div.\r\n2nd best 1\/6 1\/8\r\n3rd best 11\/42 5\/28\r\n4th best 15\/91 75\/728\r\nTotal 54\/91 37\/91\r\n<\/pre>\nThe expected number of playoff teams from 5 team divisions is 2+54\/91=236\/91.
\nThe expected number of playoff appearances for each of the teams is (236\/91)\/10 = 118\/455.
\nThe expected number of playoff teams from the 6 team division is 1+37\/91=128\/91.
\nThe expected number of playoff appearances for each of the teams is (128\/91)\/6 = 64\/273.<\/p>\nThese teams will play in the 5 game division series, expected number of games: 4.125 games.
\nHalf of them will play in the LCS, expected number of games: (5.8125)\/2.
\nOne-quarter of the teams will play in the World Series, expected number of games: (5.8125)\/4.<\/p>\n
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