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The Perfect March Madness Bracket, Part II

23 March 2013

In the last post, we saw that the odds of picking the two play-in games and the Final Four perfectly are a straightforward 1/32.  Three out of every 100 people will pick those five games perfectly, on average.  But there are 60 more games to pick, and the simplest method of estimating those odds is to assume a monkey is making them, so that the odds of predicting all 60 games perfectly are (1/2)^60, or one chance in 1.15E18 (1.15 quintillion).  If everyone in the U. S. (315 million of us) picked these 60 games using that method each year, it would take almost 3.7 billion years for a perfect 60-game bracket to be repeated.  Since the Big Bang, then, there would have been four occurrences of a perfect 60-game bracket, randomly selected.

But we don’t have to pick these games randomly, as we said last time.  The odds of a 16-seed beating a 1-seed in a regional tourney, historically speaking, are zero; the odds of a 9-seed beating an 8-seed are 53%, and all of the other matchups in the first full round (32 games) have intermediate odds.

So let’s assume, again, that the probability of the higher-seeded team winning can be described by this graph of historical NCAA tournament data during the 60 games between the first full round and the Final Four.


And let’s adopt a rational, intelligent strategy:  In any given matchup between an X seed and a Y seed, we’re going to pick the statistically most likely winner, and we’re going to assign a probability to that outcome from the chart to left.  To simplify the process, we’ll assume that all four regional tourneys have an equivalent distribution of probabilities, so we can compute the probability of a single 15-game regional tourney, and then raise that probability to the power of four to get the 60-game estimate.

First Round

In the first round, 1 plays 16, 2 plays 15…8 plays 9, and so forth.  When 1 plays 16, we’re going to pick 1, and we’re going to assume that the probability of that outcome being correct is 1.0.  When 2 plays 15, we’ll pick 2, with a corresponding probability of 0.95.  When 3 plays 14, we’ll pick 3 (seems like this one shows an upset frequently, but we’re playing the odds), with a corresponding probability of 0.83.  For 4 vs. 13, we pick 4 (0.78); for 5 vs. 12, we pick 5 (0.66); for 6 vs. 11, we pick 6 (0.67); and for 7 vs. 10, we pick 7 (0.60).

The 8 vs. 9 game is an oddity; we pick 9 (0.53) instead of 8 (0.47).

The combined probability of correctly picking the eight games in the first round of a single region is therefore the product (1.0)*(0.95)*(0.83)*(0.78)*(0.66)*(0.67)*(0.60)*(0.53) = 0.0864853, or about 9%.

Second Round

In the second round, seed #1 plays seed #9, which requires us to interpolate on the chart.  Let’s use the regression line y = (3.631 * x) + 45.702, with the value of x being 9-1 or 8.  Thus y = (3.631 * 8) + 45.702 = 74.748%, or 0.75 expressed as a fraction, and we pick the top seed.

Seed #2 plays #7, which is a seed difference of 5, which implies a probability of 0.67.  Subsequently, for 3 vs. 6, we pick 3 (0.6); and for 4 vs. 5, we pick 5 (0.53).

The combined probability of picking the second round perfectly, assuming we picked the first round perfectly, is therefore (0.75)*(0.67)*(0.60)*(0.53) = 0.159258, or about 16%.  Most importantly, the odds of picking BOTH the first and second rounds correctly in a single region is 0.0864853 * 0.159258 = 0.0137735, or slightly better than 1%.

At this point, we need to combine all four regions by raising 0.0137735 to the power of 4, which yields odds of 0.0000000359894, or 3.6E-08.  Let’s put that in perspective by noting that it’s equivalent to one person out of 28 million picking the first and second full rounds of the tournament exactly right.  If every person in Mexico City picked a bracket, only one of them would be likely to get these 48 games exactly right.  As a point of reference, the odds of our proverbial monkey picking those games correctly would be one in 2.814E+14, or one in 281 trillion monkeys.  The intelligent human, using the seedings as her guide, is 10 million times more likely to pick the first two rounds perfectly than the monkey.

Third Round, or “Sweet Sixteen”

We’re starting to see a weakness in our approach now.  You may have noted that in the second round, 4 played 5, and the seeding difference of 1 always gives us a 47% chance of 4 beating 5, which means we pick 5 (0.53).  From here on out, we’re going to assign even odds (0.5) to these games pitting the X seed against the X-1 seed, and we’re going to pick the X seed in two regions and the X-1 seed in the other two regions.  Everything else remains the same as before.

In the third round, then, 1 plays 5, and we pick 1 (0.60); 2 plays 3, and we pick 2 in two regions (0.5 each), and we pick 3 in the two other regions (0.5 each).  So the combined probability across all four regions in the third round is (0.6^4)*(0.5^4) = 0.0081, or 0.8%.  The cumulative probability of picking the first three rounds perfectly across all four regions is therefore 0.0000000359894*0.0081 = 2.91514E-10, equivalent to one person in 3 billion getting them right…that’s two persons on the whole globe in any given year, if every single human on the planet picked a bracket.

Fourth Round, or “Great Eight” – the Quarterfinals

Because we’ve already calculated the odds for the Final Four, this is our last round-by-round task before putting it all together.

We have two games pitting #1 vs. #2.  We’ll pick one #1 and one #2, both with probabilities of 0.5.  We also have two games pitting #1 vs. #3, which ever so slightly favors #1 (0.53) in each case.  Our combined probability is therefore (0.5^2)*(0.53^2) = 0.07, and the combined probability for rounds 1-4 across all four regions is therefore 2.91514E-10*0.07 = 2.04453E-11.

Putting it All Together

The last step is to multiply all of the rounds’ individual combined probabilities together.  The play-in games yielded 0.25; the first four full rounds yielded 2.04453E-11; and the Final Four yielded 0.125.  The combined probability, therefore, is 6.3892E-13, or one person in 1.565 trillion.  If all citizens of today’s United States had been picking the brackets since the time of Noah, we’d have seen one person get it exactly right.

Still, those odds are a lot better than the monkey’s, which would be 1 monkey in 37 quintillion (3.68935E+19) picking all 65 games exactly right.  For tasks like this, the human reasoning ability stacks things in our favor compared to our closest genetic relatives, the chimps and bonobos, by a factor of 23.6 million!

More Sophisticated Methods

You might be saying, “qb’s approach sounds OK, but surely there’s more human intelligence than just the seedings.”  And you’d be right.  Take this guy’s approach, which was designed for March Madness pools (i. e., $$$) rather than simply computing the probability of picking a perfect bracket.  But still, there’s a lot of heuristic merit in what he brings to the table.  The problem is, of course, that his approach is mathematically very complicated, and the question for our purposes is, “how much better are the odds of picking a perfect 66-team bracket using Nate Silver’s method than qb’s method?”

To do that, you’d need a BIG computer, maybe even a supercomputer, because you’re going to need to run what’s known as a Monte Carlo simulation, with all of the data that Silver’s method requires, for the specific 66 teams that actually get picked for the big dance.  And then you’re going to need to run that simulation at least twice as many times as you expect the probability to be, and compare its results to the actual results.  You could run the Monte Carlo simulation a couple billion times and see if one of the predicted brackets reproduces the actual tournament outcome perfectly, game by game.

The big difference, of course, is that qb’s method is simple and can be used even before you know who’s in the tournament.  Silver’s method, by contrast, requires mountains of data for the 66 teams actually picked, and then it takes a supercomputer a while to run all of the simulations.  Note that sports networks on TV like to use Monte Carlo simulations of individual games, and they typically run 10,000 simulations of that one game to come up with a probability for that matchup.  It may be more accurate to do that, but can you imagine running 10,000 simulations for each of the possible matchups in a 66-team bracket?

So we’re starting to see how the Law of Diminishing Returns comes into play.  It may be – it probably is, for the sake of argument – that Silver’s sophisticated method increases the odds of picking a perfect bracket.  But by how much?  Hard to say.  But if Silver’s method conferred an advantage over qb’s method that is comparable to the advantage qb’s method has over the chimpanzee’s random-selection method, then the odds of Silver’s method producing a perfect bracket would be 1 in 68,000, roughly one person among the population of Missouri City (near Houston) or Temple (home of Scott & White).  Sorry, but qb don’t buy it!

Still, Silver’s method is fascinating, and I’d love to have access to the code that does the analysis.

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