I’ve been meaning to write about perfect brackets for a while but haven’t been able to figure out what I wanted to say until today. This morning I woke up to find my son playing a game he calls “dice racing.” The game involves rolling two 6-sided dice and keeping track of the totals. The first total you get 42 times “wins.” You also win since you get a nice little binomial distribution picture when you finish. This game seemed like a nice lead into a discussion of brackets. The initial focus for the bracket discussion was exploring some simple 2 and 4 team tournaments:

Next was a 16 team tournament. The main point here was simply to try to understand how many games you’d have in general. From there we moved on to talk about the probability of getting selecting a perfect bracket in a 64 team tournament (assuming that all games were 50/50 coin flips):

Next, with the camera off, we talked through the probabilities of picking a perfect bracket if your chance of picking each game was equal but different from 50%. In addition to the 9 quintillion number that comes up when you have a 50% chance of picking each game, we looked at what happens when you have a 60%, 70%, and 80% chance of picking the outcome of each game correctly. The numbers you deal with in this problem are all so large that they are a little difficult for kids to understand, that’s why I wanted to spend time on this part of the perfect bracket problem. The other thing that I think kids will find surprising is how much the numbers change when you change the chance of picking each game right. Raising numbers to the 63rd power is not intuitive!

Next we looked at a few other probabilities – 40%, 90%, 69% and 71%. The first two numbers came from the boys, the second two came from me. The size of the numbers caused my older son to wonder what would happen if you picked a different bracket every second. It turns out that if you had a 52% chance of picking each game right and picked one bracket per second it would take about 14 billion years (or about the age of the universe) to pick a perfect bracket!

Finally, some fun numbers from this year’s tournament. ESPN has close to 11 million people selecting brackets on their website. They also publish stats at the end of every round. That allowed us to watch and see what happened to the number of perfect brackets at every stage. There were a couple of really interesting surprises. The surprising math behind these specific stats is a little over their heads, but still fun to talk about.

The Mercer upset over Duke was a surprise for two reasons. First, obviously, any time a 15 seed beats a 2 seed it is a surprise. Second, only about 2.5% of the 11 million people who entered brackets in the ESPN contest had selected Mercer, but more than 4 times the percent of the people who had perfect brackets at the end of the first day had picked that upset. Pretty amazing. But the bigger surprise came next.

Two games later #10 seed Stanford beat #7 seed New Mexico (the game in between was a 1 vs 16 seed game). At the time of this game there were about 2,000 remaining perfect brackets in the ESPN contest and when Stanford won the number of perfect brackets dropped to about 60. Roughly 97% of the perfect brackets were eliminated even though about 40% of the 11 million people in the overall contest had picked Stanford. Based on the overall numbers you would have expected the number of perfect brackets to go from 2000 to 800, not 2000 to 60. I have no explanation for why the people with perfect brackets up to that point did not like Stanford, but I’d love to know!

For me, studying perfect brackets is a great way to show kids some fun math. With so many people enjoying the basketball tournament, I’m sure that many kids will find it interesting to talk about this math. For me personally, the math behind the perfect brackets is one of the most interesting “real world” math problems I’ve ever worked on. I hope to be able to write more about it soon.

Hey Tross, Alisha here, from smite. great set of videos… I got a nice refresher by watching them. 🙂 One thing that made me think twice was the computation for being 40% accurate at predicting. With a binary outcome, a 40% probability of getting something right is the same as a 60% probability of getting it wrong, so does it really make sense to compute those odds? (“Make sense” having a very subjective meaning, of course.) I guess you’re making some assumptions on whether or not you know how accurate you are, but if you somehow did know, you could just always make the “wrong” pick and be 60% accurate. So if you’re trying to make some sort of risk assessment or something, it doesn’t really make sense to compute odds below 50%. Is that how I should be looking at it? Or is 40% not the complementary outcome to 60%?

the 40% correct outcome was the result of a question from my 10 year old. You are right, it is a strange case – if you only have two choices and you know your choice will be right only 40% of the time, then the one you aren’t picking will be right 60% of the time. You wouldn’t consider probabilities below 50% for picking a game right if you were really looking at analyzing the odds of picking a perfect bracket. Still, it is a fun case for kids to think through.

Funny enough, there is a famous probability problem where a similar point comes up. Consider this situation:

I hand you two envelopes. All you know is that one has an amount of money – call it X, and the other envelope has an amount of money equal to 2X. I let you open one of the envelopes and look at the amount of money inside and then offer you this choice: (i) keep what you have, or (ii) switch to the other envelope and keep what it has.

A common way of thinking through this problem is assuming that there’s a 50% chance that the other envelope has 1/2 as much and a 50% chance that the other envelope has twice as much, so you get a positive expected value by switching ( 1/2 * 1/2 + 1/2 * 2 = 1.25). Thus, you’ll choose to switch to the other envelope no matter what amount of money you see when you open the first one. The problem with that reasoning is this -> if you always believe that the other envelope has more money in it, why not just open that one first?

Ah! The envelope problem, yes. Wait, isn’t it three doors, a car, and a goat? Well, in any case, I know the answer is always switch, but when I think about it too much… gah math is confusing. I’m going to go build something.

ha ha, no, that’s the Monty Hall problem. The set of kooky probability problems is pretty small 🙂