NCAA Tournament Trends of the Day: Sweet Sixteen and Final Four
We've gotten through the first two rounds and the Final Four . Now, let's get to the Sweet Sixteen and look at the Final Four again.
Like last time, all statistics are since 1989, unless otherwise noted.
Sweet Sixteen
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Which seeds win?
The following table shows the wins and losses of each seed in the Sweet Sixteen. The #/Year column shows the number of wins per year or, in other words, the number of Elite Eight teams per year.
| Seed | W-L | W% | #/Year |
| 1 | 56-14 | 0.800 | 2.8 |
| 2 | 36-13 | 0.735 | 1.8 |
| 3 | 22-21 | 0.512 | 1.1 |
| 4 | 13-25 | 0.342 | 0.65 |
| 5 | 5-21 | 0.192 | 0.25 |
| 6 | 8-21 | 0.276 | 0.4 |
| 7 | 5-9 | 0.357 | 0.25 |
| 8+ | 15-36 | 0.294 | 0.75 |
• About three No. 1 seeds make the Elite Eight every year. This means that one should lose in the Sweet Sixteen.
• About two No. 2 seeds make the Elite Eight. Since we had one No. 2 seed losing in the second round, we should expect one more losing in the Sweet Sixteen. (Note: The exact number of No. 2 seeds losing in the second round was 1.6. If you went "bold" and chose two No. 2 seeds to lose in the second round, you should move your remaining two seeds into the Elite Eight.)
• One No. 3 seed makes the Elite Eight. Two were projected to lose in the second round, so of the remaining two No. 3 seeds, we should advance one.
• Only one No. 4 or 5 seed should reach the Elite Eight. In theory, it's easy to understand: One No. 1 seed will lose. They will most likely play the No. 4 or 5 seed. Of course, that means that only one No. 4 or 5 seed will advance (if a No. 12 or 13 seed faces the top seed, the No. 1 seed will win every time—see below). This reasoning also explains the number of No. 6-and-below seeds in the Elite Eight.
• The table above shows that the winning percentage of the No. 5 through 7 seeds increases as the seed gets worse. It seems unfeasible, but think about the matchups: The No. 5 seed would most likely play the No. 1 seed, No. 6 would play No. 2, and No. 7 would play No. 3. The lower the seed, the easier the matchup.
Which brings us to the next point. How do we quantify the individual matchups for each seed? Is there a clear winner in the No. 2-vs-No. 3 matchup, or any other matchup?
Looking at Individual Matchups
| Matchup | W-L | W% |
| 1 vs 4 | 23-9 | 0.719 |
| 1 vs 5 | 19-5 | 0.792 |
| 1 vs 12/13 | 14-0 | 1.000 |
| 2 vs 3 | 16-10 | 0.615 |
| 2 vs 6 | 16-3 | 0.842 |
| 3 vs 7 | 5-2 | 0.714 |
| 3 vs 10 | 7-3 | 0.700 |
• Here's one way to make use of this table: In my bracket, I have a No. 1 seed facing two No. 4 seeds, a No. 5, and a No. 12. I can move ahead the team playing the No. 12 seed automatically.
For the other three matchups, I find the weighted winning percentage of the No. 1 seed (by weighing the winning percentage against No. 4 seeds twice as high as the winning percentage against No. 5 seeds); this is .739.
Then, with three matchups, they'd be expected to lose 0.784 of them. With that number, combined with the fact that 1.2 No. 1 seeds are expected to bow out in the Sweet Sixteen no matter whom they face, I can be certain that one of my three No. 1s will lose.
Final Four
Which seeds win?
The first article in this series was determining which seeds make the Final Four. Now, I'll look at the results of those matchups to see which seed most often makes the Finals.
This table shows the record of each seed in the Final Four. These exclude matchups where one seed plays the same seed, such as two No. 1 seeds facing each other.
| Seed | W-L | W% |
| 1 | 12-8 | 0.600 |
| 2 | 7-8 | 0.467 |
| 3 | 8-4 | 0.667 |
| 4 | 2-7 | 0.222 |
| 5 | 2-2 | 0.500 |
| 6 | 1-0 | 1.000 |
| 8 | 0-2 | 0.000 |
| 11 | 0-1 | 0.000 |
• The Final Four is unpredictable. No. 1 seeds are only 12-8, No. 2 seeds are under .500, and No. 3 seeds do better than No. 1s.
• By the way, no No. 7 seed has reached the Final Four, let alone the National Championship game.
The table above doesn't show any conclusions to take away from it. How about looking at the resulting matchups in the Finals?
Looking at National Championship matchups
The table below shows the frequency of each matchup in the Finals since 1989, looking at the top seed, the lower seed, and the combined matchup.
| Matchup | Freq. |
| 1, _ | 14 |
| 2, _ | 5 |
| 3, _ | 1 |
| _, 3 | 7 |
| _, 1 | 5 |
| _, 2 | 3 |
| 1, 1 | 5 |
| 2, 3 | 5 |
| 1, 2 | 3 |
| 1, 4 | 2 |
| 1, 5 | 2 |
| 1, 3 | 1 |
| 1, 6 | 1 |
| 3, 3 | 1 |
• Now this is a tough decision. The top seed has been a No. 1 seed 70 percent of the time, while the lower seed has been a No. 3 seed 35 percent of the time. But a matchup of those teams has occurred just once, while a matchup of two No. 1 seeds and a matchup of a No. 2 and No. 3 occurring five times each.
• I'll stick to the latter part of the table, which shows the actual matchups. A No. 1-vs-No. 1 or No. 2-vs-No. 3 matchup is the most probable.
Putting it all together
I won't go through the trends and upsets I've shown in the first two articles; you can read those for the advice.
But I've put all my advice together and made a bracket based on historic trends. You can see my bracket below (click for larger view).
Based on the trends, Syracuse should have beaten Gonzaga, but I think the Bulldogs are a much better team.
The hardest games to pick were definitely UCLA versus Villanova and West Virginia versus Michigan State. I like all the No. 6 seeds this year except Marquette (I could see Arizona State upsetting Syracuse). I think UCLA is slightly better than Villanova, but in Philadelphia, the Wildcats should advance.
As for the second game, West Virginia is rated high by most rating systems, including a No. 8 rankings in Pomeroy's ratings, five spots higher than Michigan State. But the difference in their ratings (.015 in winning percentage) equates to about a two-point difference in an actual game. That difference wasn't enough for me to choose the upset.



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