Let’s say we have two teams, and the first team to win three games wins the series. Not taking into account anything like player or team statistics, what are the odds that the series will go all the way to 5 games?
My dad emailed me, my sister, and my sister’s boyfriend this question today, and we’ve all taken different approaches and reached different answers. Any thoughts? It’s been driving me crazy all day.
Thanks!
(go tribe!)
[QUOTE=Grasshopper]
Let’s say we have two teams, and the first team to win three games wins the series. Not taking into account anything like player or team statistics, what are the odds that the series will go all the way to 5 games?
My dad emailed me, my sister, and my sister’s boyfriend this question today, and we’ve all taken different approaches and reached different answers. Any thoughts? It’s been driving me crazy all day.
Thanks!
(go tribe!)
[/QUOTE]
Simple answer 20%.
Out of 5 games there is a 20% chance that a team will win any given game, so 20% to win game one, also 20% to win games 2 and 3.
SSG Schwartz
There are two ways the series can end in three games:
AAA
BBB
There are six ways the series can end in four games:
ABAA
AABA
ABBB
BABB
BBAB
BAAA
There are 12 ways the series can end in five games:
AABBA
ABABA
ABBAA
BBAAA
BABAA
BAABA
BBAAB
BABAB
BAABB
AABBB
ABABB
ABBAB
Since 12 of the 18 possible series go to 5 games, the odds are 2 in 3.
Short answer: assuming random chances between which team wins and loses each game. There’s a 25% chnace of a three game series, a 37.5% chance of a four game series, and a 37.5% chance of a five game series.
Long answer:
Consider two teams which we’ll call A and B. There are eight possible outcomes for the first three games, each of which have a 12.5% chance of occurring:
AAA - A wins - 12.5%
AAB - no winner, go to fourth game - 12.5%
ABA - no winner, go to fourth game - 12.5%
ABB - no winner, go to fourth game - 12.5%
BAA - no winner, go to fourth game - 12.5%
BAB - no winner, go to fourth game - 12.5%
BBA - no winner, go to fourth game - 12.5%
BBB - B wins - 12.5%
Now consider the results of the possible four game series. There’s a 75% chance of a fourth game and there are twelve possible outcomes, each with a 6.25% chance of occurring.
AABA - A wins - 6.25%
AABB - no winner, go to fifth game - 6.25%
ABAA - A wins - 6.25%
ABAB - no winner, go to fifth game - 6.25%
ABBA - no winner, go to fifth game - 6.25%
ABBB - B wins - 6.25%
BAAA - A wins - 6.25%
BAAB - no winner, go to fifth game - 6.25%
BABA - no winner, go to fifth game - 6.25%
BABB - B wins - 6.25%
BBAA - no winner, go to fifth game - 6.25%
BBAB - B wins - 6.25%
Now consider the outcomes of the possible five game series. There’s a 37.5% chance of a five game series and there are twelve possible outcomes, each with a 3.125% chance of occurring and each of which will give one team a total of three wins.
AABBA - A wins - 3.125%
AABBB - B wins - 3.125%
ABABA - A wins - 3.125%
ABABB - B wins - 3.125%
ABBAA - A wins - 3.125%
ABBAB - B wins - 3.125%
BAABA - A wins - 3.125%
BAABB - B wins - 3.125%
BABAA - A wins - 3.125%
BABAB - B wins - 3.125%
BBAAA - A wins - 3.125%
BBAAB - B wins - 3.125%
If you add up all of the winning series, you’ll get the numbers I gave above.
[QUOTE=SSG Schwartz]
Simple answer 20%.
Out of 5 games there is a 20% chance that a team will win any given game, so 20% to win game one, also 20% to win games 2 and 3.
[/QUOTE]
No.
Assuming the teams are evenly matched, the likelihood of a series going to five games is the likelihood that the first four games will include two wins, no more and no less, for one team. There are fourteen different ways that a 4-game series that ends after three wins for one team can unfold:
Win-Win-Win (no game 4)
Lose-Lose-Lose (no game 4)
Win-Win-Lose-Win
Win-Lose-Win-Win
Lose-Win-Win-Win
Win-Lose-Lose-Lose
Lose-Win-Lose-Lose
Lose-Lose-Win-Lose
Win-Win-Lose-Lose
Win-Lose-Win-Lose
Lose-Lose-Win-Win
Lose-Win-Lose-Win
Win-Lose-Lose-Win
Lose-Win-Win-Lose
Possibilities 1 and 2 are twice as likely to happen as any of the others, since they don’t rely on the outcome of a fourth game, so in effect we have the equivalent of sixteen equally likely outcomes, if the teams are assumed to be evenly matched. Since there are six possible outcomes that result in a Game 5, the likelihood of a series goes five games shnould be about 37.5%, or three in eight.
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=SSG Schwartz]
Simple answer 20%.
Out of 5 games there is a 20% chance that a team will win any given game, so 20% to win game one, also 20% to win games 2 and 3.
SSG Schwartz
[/QUOTE]
Uh no.
The OP wants a 50% chance of winning each game by each team. (So the chances are the same as flipping two heads and two tails in four flips of a coin.) The simplest way to show the answer is by enumeration. Let’s arbitrarily call the team that wins the first game Heads
HHHH series ends in 3 games
HHHT series ends in 3 games
HHTH series ends in 4 games
HHTT SERIES GOES 5 GAMES
HTHH series ends in 4 games
HTHT SERIES GOES 5 GAMES
HTTH SERIES GOES 5 GAMES
HTTT series ends in 4 games
so the chances are
3/8 or 37.5% that a best of five game series goes 5 games
3/8 or 37.5% that a best of five game series goes 4 games
2/8 or 25% that a best of five game series ends in 3 games
Another intuitive way to see this is to note that the chance one team sweeps
is 1/2 * 1/2 * 1/2 = 1/8. The chance that either team sweeps is twice this or 1/4. If neither team sweeps in three, then the score must be 2-1 in games one way or the other after 3 games. Since each team is equally likely to win game 4, it it now equally likely that the series will end after one more game (the 2-1 team wins) or the series will go 5 games (the 1-2 team wins). So the probablity the series goes five games is 1/2 of 1 - 1/4 which is 3/8.
Sorry for my post. I should have added I failed statistics. :smack:
SSG Schwartz
My probability is a little rusty, but lemme see if I can work this out.
Assuming each team has a 50% chance to win any game, a series would have a 37.5% chance of going to 5.
Let C be the event that Cleveland wins a game and Y be the event that the Yankees win. There are six different scenarios that could take the series to 5 games:
CCYY (Cleveland wins first two, Yanks win next two)
CYYC
YYCC
YCCY
CYCY
YCYC
The likelihood that the first one of those scenarios would occur is (0.5 x 0.5 x 0.5 x 0.5) = 0.0625 = 6.25%. Likewise with the remaining 5 scenarios. Summing these percentages gives us 37.5%.
On preview: I see I was beaten to the punch…several times over, but I typed this up, darnit. I never answer GQs, so I’m posting it.
Of course, now that the Tribe is up 2-0, the odds shift a little. Now the only thing that can get it to 5 is a YY. So at this point the Yanks have a 25% chance of pushing it to 5.
I read the question differently. My examples agree with the rest that 37.5% of the series go to a fifth game. I went on to give the results of all possible best of five series. I think that’s wrong after reading the other answers, but if it isn’t what the OP wanted it does answer another possible question that might arise.
I have a short solution. Look at it from A’s perspective. A can win 0, 1, 2, or 3 games in the series. These have 1, 3, 6, and 6 equally likely ways, respectively, of occurring. Of these 16 ways in which the series can end for A, 6 involve 5 games. So the probability of a 5 game series is 6/16 = 3/8 = .375.
[QUOTE=Exapno Mapcase]
There are two ways the series can end in three games:
[…]
There are six ways the series can end in four games:
[…]
There are 12 ways the series can end in five games:
[…]
Since 12 of the 18 possible series go to 5 games, the odds are 2 in 3.
[/QUOTE]
The mistake here is that those 18 possible outcomes are not equally likely.
It’s 3/7.
First (for some of you), the question is what are the chances of getting to game 5, not the chances of winning game 5.
2 series end at game 3:
AAA
BBB
6 series end at game 4:
ABAA
AABA
BAAA
BABB
BBAB
ABBB
6 series get to game 5:
ABBA
ABAB
AABB
BAAB
BABA
BBAA
Out of the (6+6+2) = 14 series that can occur before game 5 occurs (if it can occur), only 6 get there. 6/14 = 3/7.
I guess my 3/7 answers the question “What fraction of possible series can lead to game 5”, but not the question “What fraction of possible series will lead to game 5”, which seems to be 3/8.
Does that sound right?
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=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]
I generated some random coin-toss data, and of the 126 series, 38 had 5 games. That gives 30.1%. Maybe someone can write a script to generate data better.
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
Chance of ending in four games
Remaining possibilities (all leading to a fifth game)
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.
Respectfully, is there a definitive answer?
That can be agreed upon!
[QUOTE=bonitahi]
Respectfully, is there a definitive answer?
That can be agreed upon!
[/QUOTE]
It’s been answered. It’s 3/8 or .375. Other answers are wrong for one reason or another.
The tricky thing with these types of problems is that it’s so easy to fool yourself with bogus logic that sounds solid.
I’d be inclined to believe 37.5% myself. Here’s a test I ran which corroborates that:
<html>
<head>
<script language="javascript">
function series()
{
var a = 0;
var b = 0;
while(a < 3 && b < 3)
{
if(Math.random() > 0.5)
{
a++;
}
else
{
b++;
}
}
return a+b;
}
var fivecount = 0;
var numtrials = 1000000;
for(var i=0; i<numtrials; i++)
{
if(series() == 5) fivecount++;
}
alert("percentage of five-gamers: " + (fivecount / (numtrials/100)) + " (" + fivecount + " out of " + numtrials + ")");
</script>
</head>
</html>
Note: I’m not actually sure if “Math.random() > 0.5” is a perfectly fair 50%, so this might be slightly skewed.
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%.