## Expected number of post-season games

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?"

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:
5 game series: 4.125         7 game series: 5.8125         9 game series: 7.5390625

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.

```Years              PS     WS
1903              0.942  0.942
1905-1918         0.727  0.727
1919-1921         0.942  0.942
1922-1960         0.727  0.727
1961NL            0.727  0.727
1961AL            0.581  0.581
1962-1968         0.581  0.581
1969-1976         1.172  0.484
1976-1984NL       1.172  0.484
1977-1984AL       1.004  0.415
1985-1992NL       1.453  0.484
1993NL            1.246  0.415
1985-1993AL       1.246  0.415
1995-1997
in 5 team div.    2.333  0.400
in 4 team div.    2.653  0.454
1998-2002
AL in 5 team div. 2.333  0.400
AL in 4 team div. 2.653  0.454
NL in 6 team div. 1.989  0.341
NL in 5 team div. 2.200  0.377
```

### Expected length of series

In a seven game series, the probability that the series last exactly six games is (5 choose 3)/25. 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. In this way, the probablity that the series last k games can be determined for any value of k.
Length of a 3 game series: 2G: 1/2, 3G: 1/2. Expected length: 2(1/2)+3(1/2) = 5/2 = 2.500
Length of a 5 game series: 3G: 1/4, 4G: 3/8, 5G: 3.8. Expected length: 3(1/4)+4(3/8)+5(3/8)=33/8 = 4.125
Length of a 7 game series: 4G: 1/8, 5G: 1/4, 6G:5/16, 7G:5/16. Expected length: 4(1/8)+5(1/4)+6(5/16)+7(5/16) = 93/16=5.8125
Length of a 9 game series: 5G: 1/16, 6G: 5/32, 7G: 15/64, 8G: 35/128, 9G: 35/128 Expected length: 5(1/6)+6(5/32)+7(15/64)+8(35/128)+9(35/128) = 965/128 = 7.5390625

### Wild card teams

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.
The 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.

#### 14 team league

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.
They will be in a five team division with probability (10/14)*(4/13).
They will be in the four team division with probability (4/14)*(3/13).
This somputation can be repeated for teams with the third best and fourth best records.

```Wild-card            5 team division
2nd best  (10/14)*(4/13)
3rd best  (10/14)*(5/13)*(8/12)+(10/14)*(4/13)*(7/12)+(4/14)*(10/13)*(4/12)
4th 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)

Wild-card            4 team division
2nd best  (4/14)*(3/13)
3rd best  (10/14)*(4/13)*(3/12) + (4/14)*(10/13)*(3/12)
4th 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)
```
Simplifying these sums;
```wild-card  5 team div.  4 team div.
2nd best      20/91        6/91
3rd best      30/91       10/91
4th best    200/1001      75/1001
Total       750/1001     251/1001
```
So, the expected number of playoff teams from 5 team divisions is 2+750/1001=2752/1001.
The expected number of playoff appearances for each of the teams is (2752/1001)/10=1376/5005.
The expected number of playoff teams from a 4 team division is 1+251/1001=1252/1001.
The expected number of playoff appearance for each of the teams is (1252/1001)/4=313/1001.

These teams will play in the 5 game division series, expected number of games: 4.125 games.
Half of them will play in the LCS, expected number of games: (5.8125)/2.
One-quarter of the teams will play in the World Series, expected number of games: (5.8125)/4.

#### 16 team league

```Wild-card            5 team division
2nd best  (10/16)*(4/15)
3rd best  (10/16)*(5/15)*(8/14)+(10/16)*(6/15)*(9/14)+(6/16)*(10/15)*(9/14)
4th 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)

Wild-card            6 team division
2nd best  (6/16)*(5/15)
3rd best  (10/16)*(6/15)*(5/14)+(6/16)*(10/15)*(5/14)
4th 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)
```
Simplifying these sums;
```wild-card  5 team div.  6 team div.
2nd best      1/6         1/8
3rd best     11/42        5/28
4th best     15/91       75/728
Total        54/91       37/91
```
The expected number of playoff teams from 5 team divisions is 2+54/91=236/91.
The expected number of playoff appearances for each of the teams is (236/91)/10 = 118/455.
The expected number of playoff teams from the 6 team division is 1+37/91=128/91.
The expected number of playoff appearances for each of the teams is (128/91)/6 = 64/273.

These teams will play in the 5 game division series, expected number of games: 4.125 games.
Half of them will play in the LCS, expected number of games: (5.8125)/2.
One-quarter of the teams will play in the World Series, expected number of games: (5.8125)/4.

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