Ask the Wizard Column 158 Addendum
In the NBA, there are 3 divisions per conference, and 8 teams per conference make the playoffs. The top 3 seeds in each conference are the respective division winners, and the #4 through #8 seeds are the non-division winners with the best records. This year, 2 teams from the same division in the Western Conference, the Spurs and Mavericks, have the top record in the West. If this keeps up, it means that the 2nd best team in the Western Conference will end up with the #4 seed and will have to face the best team in the 2nd round if they both win. Many people are pointing to this as a problem with the system, while the NBA considers it an anomaly. In an effort to explain that its not an anomaly because it could happen frequently, an analyst from ESPN recently made the following statement in a blog: "There are 15 teams in each conference, and five teams in each of the three divisions. That means that there's a 4-in-14 chance that the team with the second-best record will be from the same division as the team with the best record." Is he correct that there is a 4/14 chance of it happening in a particular conference? How would you figure this out? If he is correct, than it would happen in at least 1 of the conferences 57% of the time, right? — AJ from Huntington Woods, MI
Yes, he is right. There are combin(15,2)=105 ways to choose the two best teams out of 15. There are 3*combin(5,3)=30 ways to choose them from the save division. So the probability the two best teams are from the same division is 30/105 = 4/14. The probability of this happening in at least one conference is 1-(10/14)2 = 48.98%.
A reader wrote to me with criticisms of my assumptions. Here is his e-mail, in its entirety.
Hi Wizard,
I'm a big fan of your site - tons of great information there. In going through some of your older 'Ask the Wizard' pages, I came across a question that I think is actually solved in too simplistic terms. There is more to look at than just the information used to solve the problem. I think the answer you give is actually slightly higher than the true answer should be.
The question is in 'Ask the Wizard' # 158, and is regarding the seeding of NBA teams. The claim is that the top two teams in a conference will be from the same division 4 times in 14. The answer that you give seems to make the assumption that each team has equal likelihood of finishing in any given spot on the final conference standings. While this may be true, and can be a safe assumption for the problem, it then seems that you make the assumption that each team from that list has an equal likelihood of being in any of the 3 divisions. I don't believe that this is exactly the case, and I believe that the reason is because of the way in which the standings are arrived at. Namely, games are played, points are awarded, and each team's ranking is not independent of the other team's rankings, and also not independent on the division that it is in.
I'm not often good at explaining things, but I'll try to explain why here. I believe that there is an effect on the final standings that comes from the fact that teams play a slightly unbalanced schedule. Each team will play 16 games against teams in it's own division - 4 against each team. But each team will play only 36 games against the other teams in it's conference - 4 against 6 teams and only 3 against 4 teams. This has the effect that teams are fighting against teams in their own division for points more often than taking them from other divisions. As a result, the team that finishes first overall in the conference will often have gotten there by taking more points from it's own division than the other 2 division winners. This means that the second place team the conference winner's division has had a harder time accumulating points, and is less likely to end up with the second most points in the conference as a result.
I realize that this discrepancy would be miniscule, and I have no idea what it could be quantified at, but I believe that a computer simulation could be written (well beyond my capabilities) to figure out what effect this might have.
To make it even more apparent, the NHL uses an even more unbalanced schedule. Each team plays the same 82 games as in the NBA. But each team plays 32 games in their own division - 8 against each of the other 4 teams. Then they play 40 games against inter-conference opponents - 4 against each of the other 10 teams in the conference. So here, the top team in a given division may be taking a lot more points away from the second place team in that division, thus making it harder for that second place team in the division to finish with the second most points in the conference overall.
Finally, to give an example that makes it much clearer, let's consider a conference that has only 4 teams in it - 2 divisions with 2 teams each. Using the logic you used to come up with the answer to the original NBA question, the probability that the two best teams come from the same division should be 1 in 3.
If we use a balanced schedule, let's say that each team plays each other team once. If you properly weight ties in the standings, then a computer simulation shows that the top two teams will come from the same division 33.33% of the time, as expected.
However, if we stick in an unbalanced schedule, this will change things. Let's say that each team plays the other team in it's own division 2 times, but plays each team in the other division only 1 time. And let's suppose that the 2 games against your same division opponent happen first. So after each team has played 2 games, it should be clear that the second place team in either division is going to be tied or behind the first place team in the opposite division. So it will have a much harder time making up those wins in the final 2 games to pass the teams that it needs to so that it can climb into second place overall in the conference. A quick computer simulation of these conditions using this schedule shows me that the top two teams will come from the same division only 27.07% of the time. A decrease of about 6.26%.
Now, this is only for a 4 team conference over a 4 game schedule, where the ratio of in-division to cross-division games is 2:1. If you extend it out to a 15 team conference over an 82 game schedule, where the ratio of in-division to cross-division games is a much closer 4:3.6, as I said before, I would expect the deviation from that 4/14 number that you came up with to be miniscule. It would probably be well below 1% by a very large margin, I would guess. But it would still be there, so the answer can't strictly be 4/14.
Anyway, I just thought I would point it out, and also thank you for providing me with an enjoyable problem to think about and stretch my mind.
Regards,
Jonathan Bradford