I break down the mathematics of creating a perfect March Madness bracket, with the power of statistics.

*Who doesn’t love alliteration?*

**Introduction**

Before we get started, I’m sorry this post is a day late. I got busy and didn’t realize it wouldn’t be ready on time^{1}. At this point, you’ve probably already read the title and know what this post is about. In case you didn’t, this is a post about mathematics relating to the March Madness basketball tournament. If you’re not that interested, you can go back to the homepage or watch some cat videos. I promise this isn’t going to become a sports blog. However, since March Madness ~~ends in a couple of days~~ ended a couple of days ago, I thought it might be fun [*air quotes*] to do something March Madness-related^{2} for the blog. One quick warning before we begin: I’ve only filled out one bracket before, and based on the fact that I know almost nothing about basketball, you can imagine it didn’t work out that well. Some people can get pretty serious about their brackets, though. There’s even a website (probably more than one) dedicated to selling paper brackets online (among other things), and even a Wikipedia page for the process of “bracketology“. With that out of the way, let’s get started.

**Time for Some Math**

I wanted to learn a little more about how these figures might be calculated, so I decided to try to work it out for myself. There are different ways of filling out a bracket, but I’m going to assume we’re starting with 64 teams, and only 1 of those teams will win. Given this, we’ll try a couple different scenarios^{3}.

**Random Guessing**

First, let’s start with a completely random bracket, where winning teams are selected arbitrarily each game. The process for calculating the probability of filling out a perfect bracket in this case is outlined in this video, but I’ll break it down here:

We’ll start with the first round of games. There are 64 teams, so there will be 32 games, each with 2 teams. The total number of possible outcomes for the first game is 2. The total number of possible outcomes for the second game is also 2. The total number of possible outcomes for the third game is 2, as well. Let’s skip to the end. . . the total number of possible outcomes for the thirty-second game is *drumroll* you guessed it. . . 2. Unfortunately, to get a perfect bracket, you have to guess *every single game* correctly. So what’s the number of outcomes for any 2 games together? – 2 times 2 is 4. Three games? 2 • 2 • 2 = 8. Four? 2 • 2 • 2 • 2 = 16. Let’s use exponents to make our lives a little easier: 2 • 2 • 2 • 2 = 2^{4}. For all 32 games the number of possibilities is 2^{32}, which works out to 4,294,967,296, a huge number. This means that you only have a 1 in 4,294,967,296 chance of correctly guessing who wins the entire first round of games.

**Let’s Write Some Equations**

Chance of correctly predicting the outcome of one game:

Numerator: 1 prediction

Denominator: 2 possible outcomes

Pretty simple, right?

Chance of correctly predicting the outcome of *n* games:

1 prediction out of 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2. . . possible outcomes.

**We Need More Math**

But why stop there? Even correctly filling out the first round is certainly [very] impressive, but this is a blog post, so we’re going to be as excessive and unnecessary as possible. Excluding this “First Four” thing (whatever that’s all about), there are 63 rounds^{4}. For each of these games, there is (again) a 1 in 2 chance of guessing that the correct team wins. We can write this number as 2^{63} (63 games with 2 possible outcomes for each). Evaluating this leaves us with a very large number of possible brackets: **9,223,372,036,854,775,808**^{5}. Going step by step, we can represent this as 2, 2 • 2, 2 • 2 • 2 • 2, 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2, and so on. Starting with the first round, there are 32 games, then 16 games, then 8, then 4, 2, and 1. 2 must be raised to the power of each of these to find out how many ways that round could go: 2^{n}. Then, to find the total number of possible brackets, we can multiply each of these quantities together 2^{32} • 2^{16} • 2^{8} • 2^{4} • 2^{2} • 2^{1} – the *exponent *is halving each time ^{6}, until we reach the last round, with only 1 game. This means that for each consecutive round, the number of possible outcomes of all games in that round combined works out to 4294967296, then 65536, then 256, then 16, then 4, then finally 2. Multiplying these together leaves us with the same answer: 9,223,372,036,854,775,808^{7}.

The chance of you guessing correctly, then, would be 1 over 9,223,372,036,854,775,808.

**Even More Equations**

The chance of correctly predicting the entire bracket, by random chance:

The big symbol on the left is an enlarged Pi letter from the Greek alphabet, representing “capital Pi product notation” (a form of a summation operation). It indicates that the equation on the right of the symbol should be multiplied by itself a certain number of times, with a changing variable *n*, increasing by 1 with each iteration^{8}.

In this case, on the right we have the probability of correctly guessing all the games in a round with *n* games^{9}. The 2^{(2n)} is the number of possible outcomes On the bottom, the *n=0* shows that we start with a value of *n* equal to zero, which represents the number of the round (minus 1). On the top, the 5 represents the maximum value of *n*, the total number of rounds in the bracket (again, minus 1).

Expanded version:

We can expand the product notation into 6 fractions representing the 6 rounds of basketball, each with more games than the last (going backwards from the final round to the first round). Each set of parentheses represents the chance of guessing the winner of that round correctly. Solving this will give us 1 over 9,223,372,036,854,775,808.

Let’s take a look at a visual to try to show just how small of a chance this is.

In the image below, there are three 500 pixel by 500 pixel boxes with 250,000 pixels each (inside the black outline). Try to find the red pixel in the first box (hint: it’s near the upper left corner). Now we’ll compress the first box down to the size of a pixel, and fill the second box with 250,000 of these, for a total of 62,500,000,000 (6.2 • 10^{10}) pixels.

The third box has 15,625,000,000,000,000 (1.5625 • 10^{16}) pixels, 250,000 times that of the second box. The final box contains 590.2 copies of box 3, for a total of 9,221,875,000,000,000,000 pixels. This roughly represents the number of possible bracket configurations (9,223,372,036,854,775,808, in case you forgot). The one red pixel represents your chance of creating a perfect bracket. Good luck.

Here’s another cool example: let’s say that you’re trying to find a specific grain of sand. I’ll give you a clue: it’s somewhere in the United States. The chance of you finding that one grain of sand by looking at a random 1 mm (millimeter) by 1 mm area inside the U. S. is roughly equal to the chance of you creating a perfect March Madness bracket by random guessing^{10}. Don’t believe me?

**(9.631 × 10 ^{12} meters^{2}) / 9,223,372,036,854,775,808 • 1000^{2} ≈ 1.044 square millimeters**

9.631 × 10^{12} meters^{2} – Area of the United States (in square meters)

9,223,372,036,854,775,808 – Approximate chance of creating a perfect March Madness bracket by randomly guessing

1000^{2} – Number of square millimeters in 1 square meter

1 square millimeter – Approximate size of a grain of sand

I also wrote this short section about how the outcome of one round affects another, and published it as a separate post, if you’re interested.

**Guessing Scores (oh no)**

Let’s try another example: say that you wanted to also correctly predict the score of each team in every game, again randomly choosing winning teams and scores. This is *significantly* more difficult to do, because of the number of possible score combinations for each game. We’ll assume, for simplicity, that the score will not be any higher than 100 in any game, and that all scores have an equal probability of occurring.

**Time for More Math**

First, we’ll return to our fraction that represents the chance of correctly predicting which team wins 1 game:

However, since we’re guessing the score and not just who wins, we’ll use 1 in 100 instead (since we limited our score to 100, and also excluding 0).

But since we have two teams that we need to guess the score for, the number is actually around 1 in 100 times 100 . . .

. . . or 1 over 100 squared.

This is equal to 1 in 10,000. Not great odds.

Then it should make sense that we would multiply the 2 fractions together to find the correct probability:

However, it turns out that we wouldn’t, because if we know both the scores, we know who wins the game anyway.

There’s one more thing to consider before we create our final equation, though. If the two teams tie, the game will go into overtime until one team scores^{11}. This means that we should exclude the 101 (including 0) possible score combinations where both teams have the same score^{12}.

Leaving us with about 1 in 9,899 as our [very] rough probability of correctly predicting both teams’ scores in one game.

And to find the chance of correctly guessing the scores in *n* rounds:

And finally, here’s our revised equation to determine the probability of correctly guessing the scores in every game of a tournament with 6 rounds:

. . . which works out to exactly **1 over**

**527,537,610,550,586,627,685,911,576,197,805,037,160,211,976,528,295,311,400,761,430,697,367,010,561,429,534,468,685,420,561,956,445,274,785,485,806,305,707,269,790,374,694,944,031,194,885,382,968,620,802,367,263,373,255,134,599,340,206,150,593,916,973,354,603,735,705,696,369,118,457,286,923,096,481,646,367,068,181,233,915,676,093,699**

. . . or about 1 in 5.2753761 • 10^{251}

This is well over a googol googols^{13}. If you didn’t enjoy finding that grain of sand earlier, have fun looking for a specific Planck volume somewhere in the *observable universe*. And be ready to look through 16 unvigintillion universes before finding it^{14}.

**Conclusion**

With the creation of more powerful AI and machine learning-based tools, it may be possible that we do see extremely accurate or even perfect brackets being created. If you do happen to do this, contact me; I’m sure we can work out a deal.

If you did make it to the end of this post, congratulations, and thanks for reading. A new post will be out sometime in the future. Bye.

**Credits**

Wolfram|Alpha: Computational Knowledge Engine – Helped with many, many calculations and other information

Microsoft Word – Equation formatting

**Resources**

en.wikipedia.org/wiki/March_Madness_pools#Brackets

www.businessinsider.com/perfect-march-madness-bracket-2013-4

www.si.com/college-basketball/2017/03/12/perfect-ncaa-bracket-odds

If you’re interested in this kind of thing, you may also enjoy my “Math With Dice” miniseries of posts:

Okay, bye (for real this time)^{15}.

- Although I doubt anyone would have realized had I not just told you.
- Gotta love those compound adjectives.
- I’m going to exclude “educated guessing” from my calculations because and simply assigning an arbitrary percentage chance of correctly guessing the outcome of a match would be a massive oversimplification. And I ran out of time.
- 32 in the first round + 16 in the second + 8 in the the fourth + 4 in the fifth + 2 in the semi-final + 1 final
- Also written as 9 quintillion 223 quadrillion 372 trillion 36 billion 854 million 775 thousand 808.
- It is very convenient that the bracket works this way.
- Also note that 2
^{63}is equal to 2^{32}• 2^{16}• 2^{8}• 2^{4}• 2^{2}• 2^{1} - To be honest, I’m not
*entirely*sure that I did this right, but hopefully you get the point. - For simplicity, this equation starts with the final round and increases the number of games with each iteration, although you
*could*write it to start with the first round. - But of course, you must create a whole new bracket for every square millimeter of land that you want to check for that grain of sand. Enjoy.
- I think.
- I know we didn’t count a score of 0 before, but we’ll just exclude it here anyway.
- I recently learned that this is about 8 orders of magnitude above a number called an Octogintillion. Cool.
- You would not believe how long it took to find an even remotely suitable physical comparison, after trying the areas and volumes of the United States, the Earth, the Sun, atoms, grains of sand, electrons, and our solar system to try and find something that worked. Thank the universe for Wolfram Alpha.
~~25~~~~26~~~~27~~28 revisions . . . I’m probably going to stop updating this, because every time I do, I add another revision.