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.
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.
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.
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
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?
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).
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.
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.
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.