# Answer to Same-City NCAA Sweet 16 Questions

Last week, we posed two questions regarding the NCAA basketball tournament: what are the odds that two teams from the same city would make the Sweet 16? And when was the last time that happened? To the first person to answer correctly, we offered some Freakonomics swag. And you responded, 147 of you to be exact.

Now it’s time to reveal the answers and the winners.

Let’s start with the easy one: When was the last time two teams from the same city advanced to the Sweet 16?

The correct answer is 2007, when USC and UCLA made it. The winner, **bracket**, responded in eight minutes, beating out **Dimitri** by just one minute.

Now onto the probability question.

Here’s how we worded it: The 64-team tournament this year started with three pairs of teams from the same city: U. of Richmond and VCU from Richmond; Vanderbilt and Belmont from Nashville; and Xavier and Cincinnati from Cincinnati. That’s 6 teams out of the original 64. So what are the odds that one of those pairs (U. of R., VCU) makes it to the Round of 16?

It’s a simple question, but the calculation’s tricky. In order to explain it, we turned to an expert, Columbia Business School statistics professor **Nicolas Stier Moses**, who helped us break down the problem into the basics of probability.

The calculation is based on the probability rules of “and,” “or,” and “not”. The simplest way to do the calculation is through Microsoft Excel using the COMBIN function to find the number of combinations that one can have when selecting k elements from n elements. This is used to find the number of ways that teams can advance to the next round. For example, you’ll find in all denominators combin(64,16), which represents the 16 teams that advance to the next round out of the 64 original ones.

Since the wording of the original question could have been interpreted as whether *at least* one pair advances, or that *only* one pair advances, we’ve provided both solutions:

**At least one pair:**

p(at least one team)

= p(1-2 or 3-4 or 5-6)

= p(1-2) + p(3-4) + p(5-6) – p(1-2, 3-4) – p(1-2, 5-6) – p(3-4, 5-6) + p(1-2, 3-4, 5-6)

= 3 p(1-2) – 3 p(1-2, 3-4) + p(1-2, 3-4, 5-6)

= 3 COMBIN(62,14)/COMBIN(64,16) – 3 COMBIN(60,12)/COMBIN(64,16) + COMBIN(58,10)/COMBIN(64,16)

= 3 * 0.0595 – 3 * 0.0028 + 0.0001 =

= **0.1701**

**Only one pair:**

This is harder. You first have to determine the probability that any one given pair advances to the next round, not taking into account the other teams.

p(1-2)=p(1-2 alone)+p(1-2 and another)

=p(1-2 alone)+p(1-2, 3-4) + p(1-2, 5-6) – p(1-2, 3-4, 5-6)

It’s the same figure for all pairs if you take them one by one, but not the same as taking any of the three pairs. This is computed by setting that the pair advances, and from the remaining teams, 14 are to advance from the 62 that remain, hence combin(62,14).

p(1-2 alone)= p(1-2) – p(1-2, 3-4) – p(1-2, 5-6) + p(1-2, 3-4, 5-6)

= p(1-2) – 2 p(1-2, 3-4) + p(1-2, 3-4, 5-6)

= COMBIN(62,14)/COMBIN(64,16) – 2 COMBIN(60,12)/COMBIN(64,16) +COMBIN(58,10)/COMBIN(64,16)

= 0.0595 – 2 * 0.0029 + 0.0001

= **0.0539**

From there, you can get the probability of exactly one pair advancing:

p(exactly one pair)

= p(1-2 alone)+ p(3-4 alone)+ p(5-6 alone)

= 3 p(1-2 alone)

= **0.1617**

For this part of the contest, we’ve selected two winners. Reader **Michael Lugo** was the first to answer the question correctly, and in fact went above and beyond the call of duty, accounting for an additional layer of complexity — the fact that “we can’t have both Nashville teams *and* both Richmond teams making it to the Sweet 16, since Vanderbilt and Richmond competed for the same slot.” But we’ll also be awarding swag to reader **MW**, who exactly matched our answer.

Congratulations to bracket, Michael Lugo, and MW, and thanks to everyone for playing!

## Comments