Two teams of equal ability meet in the World Series. What is the chance that the team winning game 5 wins the series?
(“Equal ability” implies the chance of either team winning a given game is 50%. For non-American and/or non-baseball fans, World Series competitors play single games until one team has won 4.)
Well, let us consider. If the teams are perfectly evenly matched in all games, there are three scenarios:
The Game 5 winner:
Leads 3-1, in which case they win it all
Trails 1-3, in which case they are now down 2-3
Is tied 2-2, so they’re now up 3-2
There are sixteen possible combinations of wins and losses for the first four games, two of which make a Game 5 impossible (either team winning 4 straight) so there are fourteen possible combinations that will lead to a Game 5: A 2-2 tie can happen six different ways (WWLL, LLWW, LWLW, WLWL, WLLW, LWWL) while a 3-1 lead either way can happen 8 different ways, 4 ways for one team and 4 for the other (LLLW, LLWL, LWLL, WLLL, WWWL, WWLW, WLWW, LWWW.)
Now, if the team was leading 3-1 (4 chances out of 14) then they’ve already won the series by winning Game 5, so there’s a 1.00 chance in 14 tries the Game 5 winner wins. If a Game 5 win means they lead 3-2 they have a .75 chance of winning (the odds of winning just one of the remaining two games) so there’s a .75 chance of winning if they’re tied 2-2. If they go in trailing 1-3 and win, they have a .25 chance of winning (the odds of winning both the last two games) and so that’s .25 in 4 tries. Adding all that up, the Game 5 winner should have a(14)+(.756)+.25*4)/14=9.5/14=.679 67.9% chance of winning the World Series.
The interesting question would be whether or not this holds up compared to reality. There have been a lot of best-of-7 championship series over the years that have gone to at least five games.
I’m reading the OP a bit differently from RickJay. Do you mean that the series is over after game 5, or that the winner of game 5 goes on to win the series?
OK, I think it’s a bit easier if you use the second meaning, that whichever team wins game 5 will win the series eventually.
In that case, you’ve got a sequence of wins and losses of the form XXXXWXX, where X is either a win or a loss for team 1. So you’re asking what the probability that three of those X’s are wins is. In this case, we’ve got a binomial (6, .5) distribution, so the probability is 5/16, which is .3125. Why is this so much lower than RickJay’s answer? Well, I think that he actually computed the probability that the series ends after 5 games.
FWIW, that answer doesn’t depend on which game you fix.
So let’s see what happens if we restrict the series to end after five games. In that case, you’re looking at the sequence XXXXW, and you’re asking for the probability that 3 of those X’s are wins. When I calculate this (from a binomial (4, .5) distribution), I get .25. However, this isn’t the final answer, as I have to divide by the probability that the series only goes to 5 games. That’s .25/.679, which is approximately .3681.
Later today I might write a simulation of this, to see how closely my answer holds up.
ultrafilter, I’m pretty sure RickJay is correct. I’m not quite sure what you’ve done, but let me ask this reality-check question: Given that Team A has won one game of a seven-game series, AND the probability that Team A wins any other single game is 50-50, how is it that the probability of Team A winning the entire series is LESS than 50%?
Ultrafilter
RickJay’s approach is correct. Your error is in calculating the chance of the team winning exactly 3 out of 6 games to be played. RickJay computed the chance of the team winning 3 other games before the other team wins 4, which is how the WS is played.
Also, the OP clearly does not mean that the series is over after game five. When the series ends at five games the winner of game 5 is always the winner of the series.
In fact, there is a 100% chance that the winner of the last game of the World Series is the winner of the series, no matter if it ends after 4, 5, 6, or 7 games.
As to the question posed by the OP, I’m pretty sure RickJay is dead on.
I’m not sure that I follow exactly what you’re saying, but I agree with zut that there is a problem with my answer. Guess I need to think about it some more. Care to expound?
There are two problems with your approach:
The binomial distribution assumes that the sample size is a constant - in this case that there will definitely be six games in addition to the fifth one, so that one might focus on the likelihood of a given team winning a given number of those games. In actuality, the number of games played is itself a function of the results of previous games - as soon as one team wins 4 games the series is over.
The reason you got such a counterintuitive result is because you calculated the results for the likelihood that the team would win exactly 3 additional games, no more and no less. Even supposing that every series consisted of seven games - that the additional games were played after the winner had clinched - this would be an incorrect approach. If you were to calculate the chances that the winner of the fifth game would win most of the games in the series, you would have to include the chances of that team winning more than 3 games as well, which would get you over 50% (21/32).
Statistics ain’t my thing and I agree that RickJay’s solution is more right than ultrafilter’s, but I think there’s a problem in your logic on this point. The winner of the World Series would always win exactly four games. Period. That’s how best-of-seven tournaments work. Maybe I’m missing something in your argument.
Well yeah, you’re missing the fact that this was in fact the first point I made to ultrafilter (reread my post). My second point was that
My point was to explain why he got a counterintuitive result of less than 50%.
RickJay is indeed correct.
The probability rises, by the way, after Game 4. A team that wins Game 1 under the circumstances posed by the OP will win the series 21/32nds of the time, or about a 65.6% chance. That’s also true for Game 2, Game 3, and Game 4. (I’ll spare you the calculations.)
As RickJay’s math demonstrates, the Game 5 winner has a slightly higher chance of taking the series. The Game 6 winner is higher still–a 75% shot, all things being equal. Either you win right away or you have a 50% chance the next day; .5 + (.5 x .5) = .5 + .25 = .75. And as Lance Turbo suggested, the probability that the team that wins the seventh game of the a seven-game series is 100%.
One very, very minor nitpick. There is a home field advantage in baseball, though it’s not much; so the team that wins Game 5 is therefore a little more likely to be the home team. Because of the way the World Series works, that means that the winner of Game 5 is likely to be the road team during Game 6 and Game 7, thereby slightly lowering the chances of winning those games and depressing the overall average a tad. (But only a tad. Go ahead and place your bets.)
I’m going to try (and most likely fail) to clear this up.
The reason a simple binomial expansion doesn’t work is because it includes cases in which there would be no game 5. If you do a binomial expansion and eliminate those cases you get the correct answer.
Using the ultrafilter setup. (XXXXWXX) and a binomial expansion.
There is one case in which all X’s are W’s
There are 6 cases in which 5 X’s are W’s
There are 15 cases in which 4 X’s are W’s
There are 20 cases in which 3 X’s are W’s
There are 15 cases in which 2 X’s are W’s
There are 6 cases in which 1 X’s are W’s
And there is one case in which zero X’s are W’s
64 cases in all. 20/64 = .3125 = The team that wins game 5 wins the series 4-3, but no one asked about that.
Ok, which cases can we elimate.
The one case in which all X’s are W’s would have meant a 4 game sweep with no game five. Eliminated.
Of the six cases in which 5 X’s are W’s two would mean a 4 game sweep. So they are gone. Leaving 4 cases.
Of the 15 in which 4 X’s are W’s. One would be a sweep. Elimiate. Leaves 14 cases.
All 20 cases in which three X’s are W’s fit the problem. Keep all 20.
Of the 15 in which 2 X’s are W’s we can get rid of 1. Leaving 14.
Of the 6 in which 1 X is a W we can get rid of 2, leaving 4.
And we can get rid of the one case in which no X’s are W’s.
That leaves us with 4+14+20+14+4 = 56 cases in which 4+14+20 = 38 cases the team that wins game five wins the series. 38/56 = 67.9%
Did you expect anything else?
OK, I got it. Thanks to everyone for their corrections.
Sorry for any ambiguity in the OP.
RickJay, I agree it would be interesting to see how the theoretical answer jibes with the historical record. Any takers?
I agree with your answer LT and your approach.
Note how useful it is to pretend that all 7 games will be played, even if a team has already won 4 games. The impact is to make each possible scenario equally likely, which greatly simplifies the calculations.
OK,
Here’s what I added up fairly quickly.
First of all, I ignored the Best of 8 series (1903, 1919-1921)
Second of all, I threw out the two 5-game series that finished 4-0-1 (1907, 1922)
Third, I counted the result of Game 6 in the 1912 World Series as a “Game 5” since Game 2 was a tie.
So, 21 series have ended in five games.
Of ones that 6 or 7 games, the winner of Game 5 has won 29 times in the World Series (most recently the Marlins in 1997). The winner of Game 5 has lost the World Series 25 times (Most recently the 1995 Clevelands).
So, if I read BobT’s stats correctly, we have
21 + 29 = 50 series in which the game 5 winner went on to win the Series (or won it by virtue of winning game 5)
and 25 in which it did not.
50/75 = 66.7 – or [n this B] just a little bit less than the “expected” 67.9%. (No doubt due to home field advantage, of course :D))
Art so rarely imitates life…
The WS ends after four wins, so there are no cases in which 5 X’s would be W’s.
Let A = winner
Let B = Loser
@ = A wins game 5
B has no wins (0/1):
AAAA
B has 1 win (4/4):
BAAAA @
ABAAA @
AABAA @
AAABA @ (remember that B can never lose win the last game)
B has 2 wins (6/10):
BBAAAA @
BABAAA @
BAABAA @
BAAABA
ABBAAA @
ABABAA @
ABAABA
AABBAA @
AABABA
AAABBA
B has 3 wins (10/20):
BBBAAAA @
BBABAAA @
BBAABAA
BBAAABA @
BABBAAA @
BABABAA
BABAABA @
BAABBAA
BAABABA @
BAAABBA
ABBBAAA @
ABBABAA
ABBAABA @
ABABBAA
ABABABA @
ABAABBA
AABBBAA
AABBABA @
AABABBA
AAABBBA
Total: 20/35 = 57.1%
zigaretten, you left out the times when B has 2 wins and A wins it in a thrilling game 7. But you did manage to introduce the first approach that touches on what I was going to say.
I’m going to say that the question is ambiguous as worded. Most people here seem to concur with RickJay that 4-game series don’t count. But when I read the question, I took it to mean that a four-game series is in the space of ‘failing’ events. In other words, ‘the team winning game 5’ in the context of an unplayed World Series does not, to me, necessarily imply that there will even be a game 5, since the Series could end with game 4. I have no objection to reading it the other way (i.e. mention of game 5 implies game 5 exists, even in the future).
So my rephrasing would be, “What is the probability that both a game 5 is played and the team that wins it wins (goes on to win) the Series?”. Now, in this situation, a four-game series is not a success, even if the team wouldn’t really care. The answer follows RickJay’s reasoning, only one divides by 16 instead of 14.
The result is then a slightly lower 59.4 %
Now some might mention that the historical record fits closely with the higher answer of 67.9 % . But this statistic leaves out 4-game series as well. So both answer the same question. To put it in Bayesian terms : “Given that a game 5 is played, what is the probability that the team winning it wins (goes on to win) the Series?”
This might seem a bit of an academic distinction, but I just wanted to make sure no one goes running out making bets like “the team that wins game 5 wins the Series” at 2/3 odds.