Blogs, Brackets, Entertainment, Modeling

The Perfect Bracket

It is that time of year. College basketball is reaching the end of its season with a huge tournament to cap things off. As such, I am often getting questions this time of the year asking for help in determining the best bracket to go with. While there is no magical trick for predicting the future, I did want to look at what the chances of getting that perfect bracket would be.

As a disclaimer, Stem and Leaf is in no way affiliated with the coming tournament. As such, we will refrain from using any trademarked names or actual data. If you would like to change the results to fit the actual data, I suggest you visit the corresponding websites to find this information.

Random Tournament

In this first case, we will assume that we have a group of 64 teams entered into a tournament randomly. As such, we would have no further information about the teams, so we should assume that each team has an equal chance of winning each game.

For this situation, each outcome of every game would have the same probability. Furthermore, every bracket outcome would be as likely as any other. For reference, this would be equivalent to the situation where you flip a coin for each game. The winner is then determined by whether the result was a heads or tails.

Because each outcome is equally likely, we can determine the probability of a specific bracket being correct by counting the total number of possible outcomes, then dividing 1 by this number. So let’s do that. Note that in each round, if there are n teams left, there will be \(\frac{n}{2}\) games played. Each of these games will have 2 possible outcomes. Therefore, the total number of outcomes will \(2^{\frac{n}{2}}\). Therefore, in each round we have the following number of possible outcomes.

  • First round
    • There are 64 teams.
    • This gives us 32 games, each with 2 outcomes.
    • Therefore, there are \(2^{32} \approx 4.3\) billion possible outcomes for the first round of the bracket.
  • Second Round
    • There are 32 teams left.
    • This gives us 16 games with 2 outcomes each.
    • We then have \(2^{16}=65,536\) total possible outcomes.
  • Third Round
    • There are 16 teams left.
    • This gives us 8 games with 2 outcomes each.
    • We have a total of \(2^{8}=256\)
  • Fourth, Fifth, Sixth Rounds.
    • There will be 8 teams, 4 teams and then 2 teams left.
    • This gives us 4, 2 and 1 games.
    • We then have \(2^{4}=32\), \(2^{2}=4\) and \(2^{1}=2\) total options.

In order to find the total, we now just add these up and find that there will be
2^{32}*2^{16}*2^{8}*2^{4}*2^{2}*2^{1} \approx 9.2 \times 10^{18}=9.2 \text{quintillion}.

Thus, the probability of having a perfect bracket will be \(\frac{1}{9.2 \times 10^{18}} \approx 1 \times 10^{-19}\).

Improving our probability

In the case of many tournaments (including college basketball) the entrants are seeded according to how well they performed during the regular season (or some other period prior to the tournament). With the teams being seeded, the probability that each team wins isn’t equally likely because one team is better than the other. Because of these seeding, we can actually improve the probability of making the correct choice.

For the following, I want to point out that in any given game, the team with a lower seed is more likely to win than the higher seed (or upset team). As such, if we would like to have the highest probability of choosing the correct winner, we should choose the lower seed in each case. I will therefore work under the assumption that we choose the lower seed to win each game.

At this point you may be saying that is a stupid technique because there will not always be an upset. At the end, I will actually talk about the probability of an upset occurring; however, it is more likely in each case that the lower seed wins. Therefore, if you want to have a perfect bracket, your highest probability of getting such a bracket occurs by choosing no upsets. Yes, this is unlikely, but it is more likely than any one given choice of a bracket in which there is an upset.

Probability of lower seed winning.

At this point, I then need to define the probability of the lower seed winning (or conversely the probability of an upset) for each game. Since the seeding gives us an indication of how good a team is, we will work under the assumption that the larger the difference in the seeds, the less likely an upset is. While we could look over the data for this particular tournament and apply empirical probability, I will instead choose to model this probability using a function.

Here we will seed the teams so that there are four teams with each number seed. The 1 seed will face the 16 seed in each round (there will be four such cases), the 2 seed will face the 15 seed, and so on down to the 8 seed verse the 9 seed. The winner will then face each other in the next round, with the winner of the 1 and 16 facing the winner of the 8 and 9, the winner of the 2 and 15 facing the winner of the 7 and 10 and so on.

For my probability function for the lower seed winning, \(l(c)\), I will make the following assumptions.

  • The probability of a 16 seed beating a 1 seed is 1.
    • That is the probability of the lower seed winning with a seed difference of 15 is 1.
    • Therefore \(l(15)=1\).
  • The probability of the lower seed winning with up to a 1 difference in seed is \(\frac{1}{2}\).
    • That is, if there is up to a 1 difference in seeding, then each team is considered equally likely to win.
    • That is, \(l(0)=l(1)=\frac{1}{2}\).
  • For \(1 \leq c \leq 15\), the probability will vary linearly.
    • That is the change in probability over the change in seed will be constant over this interval.
    • This slope will be \(\frac{.5-1}{1-15}=\frac{1}{28}\).

Combining these together, we find that the probability of upset, based on the difference of seeds will be
l(c)=\begin{cases} .5 \text{ if } c=0 \\
\frac{1}{28}(c-15)+1 \text{ if } 1 \leq c \leq 15.

Probabilities for each game

Below we will give a table defining the probabilities of the lower seed winning each game, rounded to the thousandth. However, we focus only on one fourth of the tournament, and will use the symmetry of the tournament in order to determine the total probabilities.

Round 1 Round 2 Round 3 Round 4
1 over 16
1 1 over 8
2 over 15 .714
.964 1 over 4
3 over 14 .571
.857 2 over 7
4 over 13 .643
.786 1 over 2
5 over 12 .5
.714 3 over 6
6 over 11 .571
.643 2 over 3
7 over 10 .5
.571 4 over 5
8 over 9 .5
.082 .131 .286 .5

The totals in the bottom are the probability that there are no upsets in the given round, given that there have been no prior upsets. If we then want to find the probability that there will be no upsets for this portion of the tournament, we would then multiply these together. This gives us a probability of approximately .00154, or 1 in 650. Note that this is much better than the 1 in \(2^{8} \times 2^{4} \times 2^{2} \times 2^{1}=32,768\) we had when we worked with the assumption that each team was equally likely to win.

As a note here, the probability that there will be at least one upset in this portion of the bracket is actually the probability that our bracket will not be perfect. This will be \(1\) minus the probability we are correct, which is \(1-.00154=.99846\). So, yes, it is extremely likely that there will be at least one upset in this portion of the bracket. However, choosing an upset would lower the probability that our bracket would be correct (aside from choosing an upset with a difference of 1 or less in the seeding).

Full bracket

We have now found the probability for getting one fourth of our bracket completely correct up to the choice of last of four teams. In order to have the entire bracket be correct, we would then have to do this same thing 3 more times. This gives us a probability of \(.00154^{4}=5.6 \times 10^{-12}\) that we can correctly predict the outcomes of each game up to the last 4 teams.

Once we get here, we will still have to pick the winner of the 2 games in the fifth round and the 1 game in the sixth. Since we are finding the probability there will be no upsets, each of these teams will be 1 seeds, so the outcome of each team winning is equal. That is, in predicting the last games, we will have a probability of \(2^{2} \times 2^{1}\) different ways to choose outcomes. Therefore, the probability of correctly doing this is \(\frac{1}{8}\).

In order to get the perfect bracket, we would have to get these correct as well, so the total probability of a tournament with no upsets would be \(5.6 \times 10^{-12} \times \frac{1}{8}=7*10^{-13}\). In other words, there is a 1 in 1.4 trillion chance that there will be no upsets in the tournament, thus giving us the best chance of getting a perfect bracket as the same.


We found the probability of correctly choosing the winner of every game in a tournament of 64 teams in both the situation where the teams were randomly assigned and the situation where we looked at the seeding of the teams. While in either case it was extremely unlikely that we would be able to correctly predict the winners, we did improve our chances from 1 in 9.2 quintillion to 1 in 1.4 trillion. This means we improved our probability by a factor of \(6.5 *10^{6}\). That is, we would be 6.5 million times more likely to get the correct answer using the second technique.

I hope you enjoyed the post and learned something along the way. I’m sorry that I can’t predict the future, but hopefully you are better equipped to look at the probabilities of given outcomes. Enjoy the tournament, and make sure to fill out your bracket.  Also, share this post on social media with anyone else you know that may be filling out their own brackets.

1 thought on “The Perfect Bracket”

We'd love to hear your thoughts!

This site uses Akismet to reduce spam. Learn how your comment data is processed.