The Mathematics Behind The Probability of a Perfect Bracket
Everyone loves filling out brackets. It’s a way to flex your knowledge and dunk on your friends. It gives you rooting interests in games you might otherwise be a totally neutral fan. But what is the probability of picking a perfect bracket?
Without a doubt, articles about upsets in March Madness talk at length about how unlikely a perfect bracket is. They’ll claim that the probability of a perfect bracket is 2^{-63} (or 2^{-67} depending if the play-in games are counted). I claim that this number is wrong. And it’s not just a little wrong, it is dramatically wrong.
It turns out that the key tool is the law of total probability – a very simple tool learned even in high school statistics.
More importantly, though, we’ll need to explore the question from a higher level. The “probability of a perfect bracket” can be interpreted in a few different ways, each leading to a different answer. This serves as a great example of an important concept in mathematics and sports analytics – precision in asking and answering the correct questions.
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The Naive Approach
Overwhelmingly, when people talk about “the probability of the perfect bracket”, they take the same perspective. They assume each game is a coin flip and that to get a perfect bracket, you need to get every coin flip right.
If you take this perspective, then computing the probability of a perfect bracket is quite straightforward! If there are N teams in the tournament, then N-1 teams are eliminated one-by-one over the course of N-1 games. This means that we need to get N-1 consecutive coin flips correct in order to have a perfect bracket. This leads to a probability of a perfect bracket of \frac{1}{2^{N-1}} .
Here is an example. In the NFL there are 14 teams. Thus, the probability of a perfect NFL playoff bracket using this naive calculation method is 2^{-13} . This is a 1 in over 8000 chance of picking a perfect bracket, or about 0.01%.
I claim that the probability of a perfect bracket is actually much, much larger.
Why This Doesn’t Work
The problem with the previous approach to finding the probability of a perfect bracket is the assumption that every game is a coin flip. This is certainly not the case. Miami did not have a 50% chance of winning at Buffalo in the wild card round of the 2022-23 NFL playoffs.
Rather, Miami’s win probability was much closer to 5-10%. That means if you picked Buffalo, you had something like a 90-95% chance of being right! That makes a perfect bracket much easier. Each game is decidedly not a 50/50 coin flip.
Similarly, San Francisco had about a 90-95% chance to beat Seattle in the first round. Because these games are considerably easier to pick the winner of, it is easier to pick a perfect bracket.
In fact, because these two games are almost ‘gimmes’, picking a perfect bracket requires making roughly 11 correct picks, not 13. Thus, the probability of a perfect bracket should be closer to 2^{-11} instead of 2^{-13} .
Though not every game is a ‘gimme’, there will almost always be a favorite or a consensus pick. Every time this happens, the probability of a perfect bracket becomes that much larger.
The Probability of a Perfect Bracket for a Specific Bracket
This is the point at which things get tricky. We want to start by computing the probability that a submitted bracket is perfect. Let’s simplify things by assuming a college football-like playoff with only four teams. The following table contains the probability of any one team (A, B, C, D) beating another in a hypothetical playoff.
A | B | C | D | |
|---|---|---|---|---|
Beats A | – | 20% | 35% | 30% |
Beats B | 80% | – | 75% | 60% |
Beats C | 65% | 25% | – | 40% |
Beats D | 70% | 40% | 60% | – |
If you look at this table, it is easy to see A is the best team, followed by C. The worst team is B and D is the second worst. What if somebody gives me the following bracket and asks me what is the probability it comes true?
This is pretty easy to compute! For this bracket to be correct, we need team A to beat team B, team C to beat team D and team A to beat team C. The probability of each of these three things happening is given by the product of their probabilities because each event is independent!
More specifically, the probability of a perfect bracket (that is, this specific bracket being perfect) is 80% \times 60% \times 65% = 31% . Notice that if you use the coin flip technique, the probability is 2^{-3} = 12.5% . The true probability of this bracket being perfect is much larger than the number you get with the naive approach.
This methodology is trivial to extend to larger brackets. We simply keep multiplying by the probabilities that the chosen team wins in each matchup. All we need is
- The selected winners in each matchup, and
- The probability of any one team beating any other team in the tournament (the table we provided above, for example).
Perhaps this is what somebody means by “the probability of a perfect bracket”. However, I think a better name for what we’ve just computed is “what is the probability my bracket is perfect”. I think we want something closer to “what is the probability an arbitrary bracket is perfect?”
Probability of a Perfect Bracket and the Law of Total Probability
Computing the probability of an arbitrary bracket being perfect is easily done by using the law of total probability. Suppose the different brackets are labeled x_1, x_2,\dots, x_N . Then, the law of total probability tells us how to compute the probability of a perfect bracket! The formula is relatively straightforward:
Computing the probability of a perfect bracket requires:
- Knowing how likely it is for an individual bracket to be correct. We did this in the previous section with the table of probabilities.
- Knowing how likely it is for an individual bracket to be selected. How do we do this?
The Probability of A Specific Bracket Being Picked
The only problem remaining is putting a probability distribution on how a bracket is picked or selected. Mathematics is – to a large degree – an exercise in precision in question asking and answering. For us, we need to decide what the word arbitrary means when talking about an arbitrary bracket. Different answers actually lead to different answers for the probability of a perfect bracket.
Consider the following ways of defining P(Bracket xi is Selected).
- A bracket is selected randomly according to the probability of each team winning. For example, if team A beats team B 80% of the time, then if these two teams match up in our bracket there is an 80% chance we pick team A to beat team B.
- A bracket is selected uniformly at random. This means each bracket has an equal chance to be selected. This is equivalent to coin-flipping each matchup.
- The bracket is chosen by picking the favorite in each matchup. This is the “chalk” bracket.
Each of these interpretations lead to different probabilities of perfect brackets.
Realistic Brackets
In the first case, brackets are chosen according to the probability of each team winning. The resulting brackets tend to “look like” real brackets. There are upsets, but the favorites tend to win more often than not.
I call these brackets realistic because, at least in my experience, this is roughly how people pick brackets. Everyone picks some upsets, but most people rely on favorites. It’s just good business to pick games this way.
In this setting, the probability of a perfect bracket is quite simple to compute. Notice that the probability of selecting a bracket is the exact same as the probability of that bracket being the truth. Recall our formula for the probability of a perfect bracket.
In the “realistic brackets” scenario, the probability of a perfect bracket simplifies to \sum_i P(x_i)^2 where P(x_i) is the probability that bracket x_i comes true.
Chalk Bracket
The next reasonable way to pick a bracket is to pick only the favorites. I’ll be entirely honest with you. When I enter small bracket pools for money, this is typically the approach I take. Every time you pick an upset, you decrease your expected number of correct picks. In small bracket pools, it doesn’t matter that your bracket is chalk because you don’t need your bracket to differentiate itself from the others as much.
In this setting, the probability of selecting bracket x_i is 1 if i is the index corresponding to the most probable bracket and 0 otherwise.
It is straightforward to show that the probability of a perfect bracket when picking chalk is larger than the probability of a perfect bracket when picking realistic brackets.
Let k be the index corresponding to the most probable bracket and P(x_i) the probability of bracket i coming true. The proof of this fact arises from noticing that \sum_i P(x_i)^2 \leq \sum_i P(x_i) P(x_k) because this is the most probable bracket! Then, factoring and remembering that the sum of the probabilities is 1 yields \sum_i P(x_i) P(x_k)=P(x_k) \sum_i P(x_i) = P(x_k).
Random Brackets
Lastly, if we pick a random bracket, going through the math it becomes easy to see that the probability of a perfect bracket is 1/N where N is the total number of possible brackets!
This is precisely the number that most articles and journalists cite when talking about the probability of a perfect bracket. However, by using the Cuachy-Schwarz inequality and the inequality relating the 1 and 2 p-norms in finite dimensions, it can be shown that the probability of a perfect bracket using either the chalk or the perfect brackets method is larger than this number.
To put that another way, the analysts are always underestimating the probability of a perfect bracket.
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