Here is a spreadsheet of all 4,096 scenarios for men’s and women’s Ivy Madness seeding.
To answer your question – not quite. There is the possibility where
Penn > Dartmouth and Harvard > Princeton
Penn loses other 2 games (against Harvard/Brown)
Columbia wins out (Brown, Yale Harvard)
Cornell wins out (Yale, Brown, Dartmouth)
In that case, Cornell finishes at 8-6, Penn finishes 7-7, Columbia at 7-7, Dartmouth at either 5-9 or 6-8. Cornell not in any ties so they get 3 seed. 4 seed goes to Columbia (split H2H w/ Penn, but have better win [Harvard]) in this case).
A lot would have to go right though for it not to be good enough.
Isn’t it mathematically possible, in the world of Lotto, that Brown could still make the playoffs, or does it lose in a tiebreaker among all 6-8 teams?
If Brown and Yale sweep next weekend, and Princeton and Penn beat Dartmouth, then there are 4 teams with 5-8 records (Cornell, Princeton, Dartmouth, Brown) and one team (Columbia) with a 4-9 record. Then, if Brown beats Penn, and Princeton loses its final game against Yale, can Brown beat Cornell or Dartmouth in a tiebreaker?
Brown wouldn’t win those 2 way tiebreakers because both Cornell and Dartmouth have victories against one the highest ranked Ivy teams (Yale or Harvard), which is the next tiebreaker after head to head (which would be tied at 1-1).
Yes Brown is still technically alive. There are some scenarios they make it but happened < 0.1% of the time in the sims (hence the 0.0%). And all require a > 2 team tie at 6-8.
At first I was bit surprised at first to see Brown as having a higher probability to be the 3 seed than Harvard. But perhaps I am overlaying projected win probabilities vs. each team having an equal chance of winning each future game.
Yes these scenarios are just “if this combination of games happens X, here is the ordering”, certainly not all equally likely. I’m not quite sure where you are seeing Brown has a higher chance at 3 seed? Harvard is 3 seed in 1,824 of 4,096 scenarios compare to 1,088 for Brown. Harvard also has 712 of the 4,096 scenarios where there are the 2 seed compared to 112 for Brown.
Are the seeding percentages based on each of the 4,096 possible outcomes being equally likely? Or are the win probabilities (of the favorites) of each game taken into account? Say, using ESPN Matchup Predictor (or something similar)?
The sheet of 4,096 doesn’t wait them or anything. Obviously they aren’t all equally likely. That’s accounted for in the simulations where each game is simulated w/ model-based odds (similar to KenPom or ESPN or your favorite site). So flip a weighted coin for each game and see which row you end up in. And then repeat a bunch of time. If helpful, I could add a column to the sheet sorting by probability of that scenario.
For any Brown fans, taking a quick look at Brown’s scenarios, in addition to winning out (possible), I think all require Yale to beat Cornell and Princeton (possible), most require Harvard to beat Princeton and Penn on the road (tough but possible), most require Dartmouth to beat Penn and Cornell (challenging but possible?), and most require Columbia to beat Yale and Harvard (improbable). I see one scenario in which Columbia can lose to Yale, but still has to beat Harvard while Yale, Harvard and Dartmouth otherwise win out. We’ve still got a chance!
For future versions of the chart, wouldn’t it make sense to distinguish between (a) elimination from playoff (or one or more of the seeding options) possibility and (b) vanishingly small? Maybe a 0.1% would make that point? But more generally, Luke, thank you for this; it’s great.
I understand your point but there is only a chart to verify every case close to seasons end. There are 4,096 scenarios now but before last Saturday there was 16 times as many (> 65,000) and so it’s not super easy to verify is 0 times in simulation actually means 0 or vanishingly small until there are 3 or 4 games left in the season.
Anyway glad you like this stuff makes it worth doing
Though I could be mistaken, I had thought that @Eli1143770312 may be noting that the final tabulated probabilities could be displayed with a “<0.1%” in place of zero for situations without technical impossibility. As I roughly understand it, this would only require modifying the final tabulated probabilities by ad hoc adjustment based on whether there is any remaining scenario for possibility X, rather than inferring from the simulations whether there is any remaining scenario for possibility X, if that makes sense. Even if the simulations resulting in a zero could come from either vanishing probability or technical impossibility, technical impossibility might be derived, I took it, from just whether there is a scenario. I’m not even an amateur modeler and that too might be not straightforward or raise other issues.
I understand. My point is that determining if there is a scenario is actually extremely challenging. If it was not super challenging I would just list it < 0.1%. If there are X games left per team in the season, the number of scenarios is 2^(4X) = 16^X. There are currently 3 games left per team, which is where the 4,096 scenarios come from. Before last week there would be 65,336, and with 5 games left per team (each team had played 9 games) the total number of scenarios is 1,048,576.
Now you can simulate it 5,000 or 10,000 or however many times but the fact of the matter is at 1-8, even if not mathematically eliminated, Brown making the playoffs (after correctly applying tiebreakers) is probably so small it may not show up in any one of those.
But to definitively proclaim they are mathematically eliminated, you’d need to apply the tiebreakers to over 1 million scenarios, which just isn’t worthwhile at that point given the scale.
Certainly w/ 4,096 case or the 65,000ish it could be feasible but before that (early enough in the season to where it would really still matter) it’s too tough to do.
Anyway, not trying to be defensive or knock a good suggestion – believe me I would if I easily could. I am just trying to share some of the details of how the sausage is made.
