Answer to Same-City NCAA Sweet 16 Questions

Photo: iStockphoto

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!


Nah. Still reckon I'm right. 0.176 & 0.165. But good try Prof Moses!


Seems to me the question wasn't worded as a probability question, so much as an ODDS question, since that was the word actually used in the question. Wouldn't the methodology then be to take the pre-tournament betting-line odds of team A from city X and multiply them by the same odds for team B from city X?

It doesn't seem right to calculate that the odds of two teams making the Sweet 16 are the same as any other two teams.

Ben D

You also left out the fact that teams are grouped into fours by the tournament structure, so only 1 team from each group of four can make the sweet 16. Many of the 64 comb 16 combinations aren't possible. There are actually only (4 comb 1)^16 possibilities.

Andrew Bressler

Am I missing something? Are you assuming each team has an equal chance of winning? That makes no sense- don't you have to factor in those odds? Wouldn't Kentucky have a greater chance of winning any individual game than Richmond?

Michael Lugo

Andrew: I admit that my method doesn't take into account the fact that some teams are better than others. In fact, I know nothing about college basketball. But the problem didn't seem to call for that sort of knowledge. For one thing, there wouldn't be a "correct" answer if we actually had to take that into account, since we can't really get probabilities of each team winning each individual game. (Even if we assume that betting odds would work, we'd need odds for games that never actually took place.)


Most people's calculations failed to account for the fact that events are not independent -- if one team gets into the sweet 16, it reduces the chances of other teams getting in as there are now fewer available slots. Another error was to calculate the chance of one pair getting to sweet 16, then multiplying by three as there were three possible pairs. As we're calculating the chance of at least one pair getting in, the 'multiply by three' strategy fails to account for the fact that sometimes one of these pairs getting in will be 'wasted' because another pair has already got in.

In this case these errors make only a small difference to the odds - if one of a pair gets in, only one of 16 slots is removed for the second team. The odds of one pair getting in are fairly small, so the odds of 'wasting' a pair getting in because another has already done so are low. Hence the naive calculation of 3*(1/4*1/4) = 3/16=0.1875 comes pretty close.

Also, I'd like to reflect a little more glory on Michael Lugo - I'd forgotten the inclusion/exclusion method, so his post hugely simplified my calculation.

Mike M: This is a matter of interpretation of the question. The answer above is correct *if* we have no additional information beyond what was posed in the question (no knowledge of the draw, no knowledge of the relative strengths of the teams.)

Ben D: No, the calculation above does not depend on the structure of the tournament *IF* we have information on the tournament structure *BUT* no information on which team starts where or what their relative strengths are. E.g. if 15 of the teams automatically got through to sweet 16 and the remaining 49 held a knockout tournament for the final spot, the calculation would be exactly the same, given that we don't know where the teams from same-city-pairs start off in this structure.

I'll illustrate this using the actual tournament structure. The simple calculation for a single pair both getting through is 16/64 odds for the first team (16 available slots, 64 teams that could fill them) and then 15/63 for the second team (15 available slots, 63 teams which could fill them.) so odds = 16*15/(64*63) = 15/(4*63)=15/252. Now lets calculate that accounting for the structure. Once team 1 is assigned its starting slot, there are 63 open slots for team 2, 3 of which will put it in direct competition with team one for a sweet-16 slot. So there is a 3/63 chance that it will be impossible for both teams 1 and 2 to advance. If this does not happen (60/63 chance), then teams 1 and 2 independently have a 1/4 chance to advance. So the odds are (60/63)*(1/4)*(1/4)=15/252, the same as before.


Erik of Dallas

Siding with Anthony and Ben D. Of each set of 4 teams paired in a mini tournament A against B and C against D then the winners against each other one only one of these 4 teams can make it into the sweet 16. There are no possible sweet 16 combination with none of ABCD in it, and there are no sweet 16 combination with more than one of ABCD making it...

The 1/16 (first pair make it) + 15/16 * 1/16 (first pair does not make it but second pair make it)+ 225/265 * 1/16 (3rd pair makes it but first & second do not) =721 / 4096 = 17.6025391% is the simple answer for at least one pair making it. This can also be view as 1 - probability that none of the pairs make it 1-(15/16)^3 = 17.6025391%. The 1/16 in this derivation comes from the 0.5*0.5 =1/4 that the first paired team make it time the same 1/4 that the second paired team makes it.

However if we consider both Nashville teams and both Richmond can't both make it due to the brackets than I still stand by my: 23 / 128 (17.96875%) of the time at least one (and sometimes two) paired teams make the sweet 16. Or two pairs make it 1/128 (0.78125%) of the time and exactly on pair makes it 11/64 (17.1875%) of the time.


Erik of Dallas

MW we are told this is the "NCAA basketball tournament" asking us to comp the odds of paired teams to "make the Sweet 16" form the "The 64-team tournament" and it even goes not to name the teams. We have to presume that we know something about the "NCAA basketball tournament" and that we not to ignore this publically available knowledge.

Erik of Dallas

Nicolas Stier Moses answer presumes that there are Factorial(64)/Factorial(64-16) = 1.02213464591443x10^28 possible Sweet 16 combination. My computation is that there are 4^16 = 2^32 = 4,294,967,296 possible Sweet 16 combination, most of the 64!/48! are not possible and will skew the probability results if they are used.


I'm still looking to see if the use of a combination from a field of 64 accounts for the fact that certain combination are impossible given the tournament structure. E.g all the # 1,2,15 and 16 seeds can't advance to the 3rd round even though this is one way of choosing 16 teams of 64 entrants.
However I'd further ask if this detail affects the overall probability

Erik of Dallas

A) OK. If we knew nothing about the seeding in the tournament (the seeding was random and was done after we calculated the probability) then the 64!/48! possibilities is correct. For instance each pair of cities could be seeded in the same bracket of 4 so that it would be impossible for both of them to get to the same sweet 16 spot. There are lots of other different combination that contribute to this set of 1.02213464591443×10^28 outcomes. This results in Professor Nicolas Stier Moses answer.

B) If we know the seeding or for that matter just know the seeding of the 6 team we care about then the tournament become a problem with just 4^16 possibilities that we care about. The problem did not provide this seeding (a vote for case A) but it did name the teams, so the enterprising problem solver could have looked up the seeding and arrived at case B. As was noted with the relationship between the Nashville and Richmond pairs, it is important to know if any of the 6 teams we care about share a bracket of 4 with any of the other teams we care about… Then based on this knowledge of what if any team share a bracket, then the probabilistic outcomes become specific to that case of seedings / pairings of the 6 teams we care about. Our specific Nashville Richmond case would result in 23 / 128 (17.96875%) of the time at least one pair being in the sweet 16.

C) A vote against the 1-(15/16)^3 = 17.6025391% hypothesis. We were not provided the seeding (case A); if we looked up the seeding we arrive at the Nashville Richmond case B; we should not hypothesize (adding invented data that changes the answer) that all 6 teams were seeded in separate sweet 16 brackets of 4 and would never face each other. This would be the correct answer were we given these facts or were these facts consistent with looking up the seeding.

Some peace of mind, at least for me.