What are the odds that a best-of-five series will go to five games?

Factual Questions
21

[QUOTE=KarlGauss]
I think the answer is actually 50%

Why?

The probability of the series going 5 games is the chance of it going 3 or 4 games divided by the total number of ways.

Chance of ending in three games

  • 2 possibilities (AAA, BBB)

Chance of ending in four games

  • 4 possibilities (AABA, BABB, ABAA, BBAB)

Remaining possibilities (all leading to a fifth game)

  • 6 (AABB, BBAA, ABBA, BAAB, ABAB, BABA)

So, 2 + 4 + 6 = 12 possible ways for the series to unfold

So we have 6 out of 12 = 50% going to a fifth game.
[/QUOTE]

The mistake made here is to equate the AAA and BBB possibilities with the AABA, BABB, ABAA, BBAB possibilities, with the five game series possibilities. The mistake is that AAA really stands for: AAAAA, AAAAB, AAABA, AAABB. Similarly, BBB stands for four possible five game outcomes, and each of the four-game series stand for two possible five game outcomes. So, of 32 possible five-game combinations, only 12 actually occur, or 3/8.

Or, if you prefer to think of it this way: Out of every 8 series, six will go past the third game (the ones that don’t go AAA or BBB). Of each of those six, they have a 50/50 shot of making it past game four. 50% of 6/8 is 3/8.

22

[QUOTE=Usram]
You have grouped AAAA and AAAB together (and BBBB and BBBA), when they should have been counted separately. So it’s not 6/14 but 6/16, which is 3/8.
[/QUOTE]

Usram: but AAAA and AAAB (and BBBB and BBBA) are not possible series. The series would end after winning 3 games.

23

Yes, but you have to imagine the full sequence of games for the purposes of probability calculation, even if that sequence would lead to an early victory for one team. So sequences like AAAAA and AAAAB should be considered separately, regardless of the fact that they would both be experienced as AAA sequences.

24

[QUOTE=RickJay]
Being that kind of guy I decide to look up all the five-game series ever played in baseball history; the ALCS and NLCS between 1969 and 1984, and the division series played in 1981 and then from 1995 to 2006.

NLCS 5-Gamers : 5 out of 16 series
NLDS 1981: 2 for 2
NLDS 1995+: 4 for 24

ALCS 5-gamers: 5 for 15
ALDS 1981: 1 for 2
ALDS 1995+: 9 for 24

Total: 26 out of 84, or 31 percent

This is a bit lower than you would expect by random chance, which is within a reasonable random distribution but to be honest is what I expected, for the simple reason that teams are NOT evenly matched; in many cases one team is clearly vastly superior to the other, and you would not expect those series to go to 5 games at quite the same rate that perfectly matched serieses would. If all playoff teams were the same, the rate of 5-gamers would always converge on 37.5%, but they aren’t so it will probably always be a bit lower than that.
[/QUOTE]
Since the Modern World Series began in 1903, 98 out of 102 have been played over the best of seven.

35 of these 98 series have gone to the seventh game, giving a percentage of 35.7. Instinctively rather than mathematically this seems high to me, indicating a tendency for the teams to be more evenly matched - in a seven game series there are more possibilities - instead of 3, 4, or 5 games there could be 4, 5, 6,or 7.

My education in probability & chance is lost in the mists of time. If anyone wishes to calculate the chance of a seven game series actually reaching the seventh game it might be interesting to compare the result with the real world.

25

[QUOTE=DSYoungEsq]
The mistake made here is to equate the AAA and BBB possibilities with the AABA, BABB, ABAA, BBAB possibilities, with the five game series possibilities. The mistake is that AAA really stands for: AAAAA, AAAAB, AAABA, AAABB. Similarly, BBB stands for four possible five game outcomes, and each of the four-game series stand for two possible five game outcomes. So, of 32 possible five-game combinations, only 12 actually occur, or 3/8.

Or, if you prefer to think of it this way: Out of every 8 series, six will go past the third game (the ones that don’t go AAA or BBB). Of each of those six, they have a 50/50 shot of making it past game four. 50% of 6/8 is 3/8.
[/QUOTE]

Thank you - this is very helpful. I really appreciate you taking the time to explain things to me.

26

How about the other way? The probability of a sweep?

Heck, considering the way this postseason is going so far, what’re the odds of all four series sweeping?

27

A little late to the thread, but I think that SSG Schwartz has the right answer, but not necessarily the right logic.

The possible combinations of four games is as follows.

AAAA
AAAB
AABB
ABBB
BBBB

The order of wins/losses and whether the fourth game is played is irrelevent, the only combination that leads to the fifth game is AABB, or one chance in five. 20%.

ETA: Apologies SSG Schwartz if this is what you said and I just didn’t pick it up

28

[QUOTE=Projammer]
A little late to the thread, but I think that SSG Schwartz has the right answer, but not necessarily the right logic.

The possible combinations of four games is as follows.

AAAA
AAAB
AABB
ABBB
BBBB

The order of wins/losses and whether the fourth game is played is irrelevent, the only combination that leads to the fifth game is AABB, or one chance in five. 20%.

ETA: Apologies SSG Schwartz if this is what you said and I just didn’t pick it up
[/QUOTE]

Projammer,

If this information is irrelevant, how is it that galt’s javascript code always gives an answer near 37.5% by playing “virtual games” and counting how many of the series get to 5 games?

29

[QUOTE=CJJ*]
I agree that 37.5 is the correct answer assuming 50-50 odds. I’d like to add a wrinkle by taking Home Field Advantage into account.

According to a recent column of Cecil’s, the home team in MLB has a 53% chance of winning any generic game. Assuming Team A plays at home for games 1and 2, while team B has home field for games 3 and 4, the various combination in which the first four games split are as follows:

AABB = .53*.53*.53*.53 = .0789
ABAB = .53*.47*.47*.53 = .06205
BAAB = .47*.53*.47*.53 = .06205
ABBA = .53*.47*.53*.47 = .06205
BABA = .47*.53*.53*.47 = .06205
BBAA = .47*.47*.47*.47 = .0488

All tolled, that’s 37.59%; the home field advantage makes it slightly more likely the series will go five games (not by much–less than 1/10 a percentage point–but still).

Obviously, team A–with homefield in game 5–will win 53% of these series, or there is a 19.9% chance team A will win in five games. Just for fun what are the odds of team A winning in four games? Possible scenarios:

AABA = .53*.53*.53*.47 = .07
ABAA = .53*.47*.47*.47 = .055
BAAA = .47*.53*.47*.47 = .055

Which sums to 18%. Finally, team A will win in a sweep .53*.53*.47 = 13.2 percent of the time. Thus, team A–who has home field advantage for the series–will win 51.1% of the time; the HFA therefore represents just a 2.2% advantage over the other team, versus the single-game advantage of 6%.
[/QUOTE]

That’s really interesting! Good job on the calculations. I wonder if home-field advantage in the post-season has a higher % of winning for the hometeam than your generic regular season game data (since fans will actually sell out the place and make a lot more noise).

30

[QUOTE=Projammer]
A little late to the thread, but I think that SSG Schwartz has the right answer, but not necessarily the right logic.

The possible combinations of four games is as follows.

AAAA
AAAB
AABB
ABBB
BBBB

The order of wins/losses and whether the fourth game is played is irrelevent, the only combination that leads to the fifth game is AABB, or one chance in five. 20%.

ETA: Apologies SSG Schwartz if this is what you said and I just didn’t pick it up
[/QUOTE]

Your calculations are incorrect. The correct answer has already been posted above multiple times.

31

[QUOTE=Chez Guevara]
Since the Modern World Series began in 1903, 98 out of 102 have been played over the best of seven.

35 of these 98 series have gone to the seventh game, giving a percentage of 35.7. Instinctively rather than mathematically this seems high to me, indicating a tendency for the teams to be more evenly matched - in a seven game series there are more possibilities - instead of 3, 4, or 5 games there could be 4, 5, 6,or 7.

My education in probability & chance is lost in the mists of time. If anyone wishes to calculate the chance of a seven game series actually reaching the seventh game it might be interesting to compare the result with the real world.
[/QUOTE]

Ok, seven game series:

Possible outcomes: 2[sup]7[/sup] = 128

Four game Outcomes that we can discard:

All AAAA, all BBBB (2 out of 16, or 1/8). This leaves 7/8 of the possibilities still alive.

Five game Outcomes we can discard:

AAABA, AABAA, ABAAA, ABBBB, BAAAA, BABBB, BBABB, BBBAB (8 out of 28, or 2/7). This leaves 5/7 good series.

Of each of those, 1/2 will continue to game 7, the others will terminate after 6 games.

So we have 7/8 * 5/7 * 1/2 = 5/16 or 31.25%

32

[QUOTE=Projammer]
The order of wins/losses and whether the fourth game is played is irrelevent, the only combination that leads to the fifth game is AABB, or one chance in five. 20%.
[/QUOTE]

As has been pointed out, this is incorrect. The reason is because the order of wins/losses (at least for this calculation) is important.

An easy way to prove this to yourself is to simplify the problem: If you flip a coin exactly twice, what is probability of any possible outcome? You might argue that the outcomes HH, HT, and TT are equally possible; however, this is incorrect. Since each flip is independent, order matters. To sort outcomes that are equally probable, you need to distinguish HT and TH, That means each of the four outcomes HH, HT, TH and TT are equally probable. You can verify this by doing the experimentation yourself.

33

So, is there a definitive answer to the OP’s query?

34

One more vote for 3/8, with an explanation that I didn’t notice above:

There are 32 possible strings of length 5 over the alphabet {A, B}. We’re considering a function that labels these 0 if there are more As, and 1 if there are more Bs. If there are three of any character in the first four symbols, there’s no reason to even look at the fifth character–we can just assign the label and be done with it. So, the only times we have to consider that fifth character is when the first four characters contain exactly two As and two Bs. There are 6 possible four character combinations of As and Bs (four places to put the first A, three to put the second, and divide by two because the first and second As are indistinguishable). There are two choices for the final character, so the total number of strings where the fifth character matters is 12. 12/32 is 3/8 or 37.5%.

Again, this is assuming that the two teams are evenly matched. If not, the counting argument above still holds, but you have to weight each outcome appropriately.

35

I say 1/6.

After three games, its A has won, B has won, or it’s 2-1. So 1/3 for a 4th game.
The 4th game will make it 3-1 or 2-2, so 1/2 for a 5th game.
1/3*1/2=1/6.

36

[QUOTE=Cricket Bat]
I say 1/6.

After three games, its A has won, B has won, or it’s 2-1. So 1/3 for a 4th game.
The 4th game will make it 3-1 or 2-2, so 1/2 for a 5th game.
1/3*1/2=1/6.
[/QUOTE]

Zombie thread.

Look, it’s 37.5%. I don’t know how much more clearly it can be explained.