NCAA

Rosenthal: Some NCAA men's basketball probabilities

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Jeffrey Rosenthal, special to TSN and TSN.ca
3/18/2013 5:14:10 PM
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I was asked by TSN to compute a few fun probability facts about the upcoming 2013 NCAA Men's Basketball "March Madness" tournament, and did so here. Note that this is separate from, and in addition to, my computation of the "Rosenthal Fit" to predict the tournament bracket, which is described elsewhere. Note also that some of the calculations below are somewhat rough or approximate, just using whatever data is easily available, so some of these probabilities are not set in stone.

How many brackets would you have to fill out to track every single possible outcome?

There are a total of 63 games (not counting the First Four), and each one has 2 different possible winners. So, the total number of different ways of filling in a bracket is equal to the product of 63 different 2's all multiplied together, or 2^63, which works out to 9,223,372,036,854,775,808. This is approximately equal to 9 x 10 to the 18th, i.e. a 9 followed by 18 zeros, which equals nine billion billion, or nine million million million. That's a lot of different brackets!

What are the odds of picking a perfect bracket?

If you just fill in your bracket randomly, then your probability of being correct is equal to one divided by the total number of different brackets, which (by the previous question) equals about one chance in nine billion billion - not very high!

Basketball experts tend to call about 70 per cent of the tournament game winners correctly. So, even if you are a true expert and have a 70 per cent chance of getting each game right, the probability of getting the entire bracket right is still only (70 per cent) 63, which works out to about 10^(-10) or one chance in ten billion. This is why perfect brackets are so rare.

What is the probability that one particular #16 seed will beat the #1 seed on the first day?

In the 27 years that the tournament has had the modern 64-team structure, a #16 seed has never beaten a #1 seed. (By comparison, #15 seeds have beaten #2 seeds about 4 per cent of the time.) Specifically, in 27 x 4 = 108 such matchups, the #16 seeds are 0 for 108. So, the "95 per cent Clopper-Pearson Exact Method" statistic concludes from this that the true probability of a #16 seed beating a #1 seed is somewhere between 0 per cent and 3.3 per cent.

So, a #16 seed beating a #1 seed is very unlikely, but not impossible, and indeed the probability could be as high as 3.3 per cent.

What is the probability that AT LEAST ONE #16 seed will beat a #1 seed on the first day?

Since there will be four such matchups, the probability range from the previous question gets multiplied by nearly four: the probability of at least one such victory is somewhere between 0 per cent and 1 - (1 - 0:033)^4 = 12:6 per cent.

So, the probability could be as high as 12.6 per cent. Still not too likely, but it's certainly a possibility.

What is the probability that ALL FOUR #16 seeds will beat #1 seeds on the first day?

This is much less likely. Indeed, even taking the highest possible probability of a single victory, 3.3 per cent, the probability of all four victories is then equal to (3:3 per cent)^4 which equals about 1 / 1,000,000. So, this would be a "one in a million" occurrence!

What is the probability that last year's champion, Kentucky, will repeat as champion this year?

Historically speaking, in the 27-year modern 64-team era, only two teams have repeated as champion, namely Duke (1992) and Florida (2007), for a fraction of 2/27 or about 7.4 per cent.

Now, winning the championship requires winning six games in a row.

Even if Kentucky were the top favourite, they would still have only about a 70 per cent chance of winning each game on average (see above), and therefore a probability of about (70 per cent)^6 or 11.8 per cent of winning the tournament.

In fact, most experts predictions (including the Rosenthal Fit) have Kentucky ranked somewhat lower this year, so it might be more accurate to give them at most a 50 per cent chance of winning each game on average. This would give them a probability of (50 per cent)^6 or about 1.5 per cent of winning the tournament.

LATE-BREAKING NEWS: It seems that Kentucky did not even qualify for the tournament. So, I can now say with confidence that the probability of them repeating as champions is equal to 0 per cent!

What is the probability that one particular #1 seed will make it to the Final Four?

In the 22-year period 1979-2010, a total of 22 x 4 = 88 teams made the Final Four, of which 54 were #1 seeds. This means that 54/88, or 61 per cent of the #1 seeds have advanced to the Final Four.

So, as an estimate, we could say that any one particular #1 seed has about a 61 per cent chance of making it to the Final Four.

What is the probability that ALL FOUR of the #1 seeds will make it to the Final Four?

Based on the previous question, this probability is about equal to four different copies of 61 per cent all multiplied together, i.e. to (61 per cent)^4, which equals about 13.8 per cent. (In fact, this has only happened once in the modern era, in 2008. My calculation suggests that we should expect it to happen a bit more frequently than it has in the past)

What is the probability that this year's Wooden Award winner's team will win the tournament?

In the 36-year period 1997-2012, this has happened six times (Louisville-Darrell Griffith, 1980; Kansas - Danny Manning, 1988; Duke - Christian Laettner, 1992; UCLA - Ed O'Bannon, 1995; Duke - Shane Battier, 2001; Kentucky - Anthony Davis, 2012) for a frequency of 6/36 or about 16.7 per cent.

However, it seems that there is no clear-cut favourite to win the tournament this year, so I am inclined to instead say as above that no team has a higher probability than (70 per cent)^6 or about 11:8 per cent of winning the tournament this year.

So, assuming that the Wooden Award winner is from one of the top teams, then I would estimate that team's probability of winning the tournament at about 11.8 per cent.

Jeffrey Rosenthal is a professor in the Department of Statistics at the University of Toronto, and the author of the bestseller Struck by Lightning: The Curious World of Probabilities. His analysis can seen during TSN's coverage of the 2013 NCAA Men's Basketball tournament.

Anthony Bennett (Photo: The Canadian Press)

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(Photo: The Canadian Press)
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