Some Foolishly Precise Predictions About the Bracket

March 15, 2009

Last year, I ran some numbers on teams from one year and where they return to the next. Some of the work, updated with last year’s tournament, went into my 5 Things You Should Consider When Filling Out Your Bracket post.

I used some aspects of that in order to make some projections about the tournament. It was based on historic trends, patterns, and a three-year moving average. As it turned out, they were pretty good.

I came up with 16 to 18 teams from the ’07 second round making the ’08 second round, and 17 ended up doing it.

I projected nine teams from the ’07 second round making the ’08 Sweet 16, and 10 did. I projected that from four to six teams (and most likely six) from the ’07 Sweet 16 would make it that far in ’08, and five did.

It didn’t work so well when it came to the Elite Eight, since I had fewer than three Elite Eight teams from ’07 making it back in ’08, and four teams did. In my defense, the back-to-back years of zero returners in ’05 and ’06 messed up the moving average analysis, and it was the first time that the number of repeat Elite Eight teams increased after having been three or more the previous season.

Anyway, I decided to do the same thing again this year. I am also publishing the projections this time just in case it turns out to be good again. I also applied the same analysis to seed in the first round, so we’ll see how this goes.

It could be that this is snake oil and I got lucky last year, or it could be really close again. We’ll have to see. Anyway, here’s what I got for this year:

  • 15 teams from the ’08 second round in the ’09 second round
  • From seven to nine (likely eight) teams from the ’08 second round in the ’09 Sweet 16
  • Five to seven (likely six or seven) teams from the ’08 Sweet 16 in the ’09 Sweet 16
  • Between one and three (likely one) teams from the ’08 Elite Eight in the ’09 Elite Eight
  • One team from the ’08 Elite Eight in the ’09 Final Four
  • Either five or six (likely six) upsets in the ’09 first round

If there are five first round upsets, the second round will have four ones, four twos, four threes, three fours, three fives, three sixes, two sevens, one eight, three nines, two tens, zero elevens, three twelves, one thirteen, and zero each of fourteens, fifteens, and sixteens.

If there are six first round upsets, the second round will have the same as above only with one five, four sixes, three twelves, and zero elevens.

As the post title indicates, I have a feeling that this is probably a fool’s errand. However, I’m hoping when the brackets come out that the combination of the returners and seeds will put together a pretty accurate projection of the field. I think it’s probably more likely that it will be impossible to meet all of the above conditions.

After all, if there’s one thing that I know about the tournament, it changes everything up the moment you think you have it figured out.