Blogs, Brackets, Entertainment, Modeling

The Average Bracket

In the post last Tuesday, The Perfect Bracket, we found that probability that you would be able to find the perfectly predict the outcome of every game in the upcoming college basketball tournament. We then decided to improve our odds by looking at the seeding of the teams.

Here, we will determine how many points you would expect to get if you filled out your bracket randomly and if you used the technique of choosing the highest seed. By looking at the expected values, we will have a nice benchmark to compare our final results to, so that we can see how we did.

If you’d like to join the fun and make your own picks, we’ve created a bracket for STEM and leaf through CBS sports. I hope you join us, so that we can compare results and look at more math along the way. If you would like to join the fun, head to STEM and leaf Bracket.

Random choices

Here we will work under the assumption that the choices you make for your bracket are completely random. That is, you didn’t look at the seeds, or do any research and you just chose a team to win.

In this case, since we have no information about what teams we choose, we will have that each team will have a 50& chance of winning. That is, either choice will have the same probability of winning. In order to find the expected number of points we will earn through the bracket, we will need to determine the probability of getting each point. We will therefore look, round by round at the expected outcome of each game.

First Round

In the first round, each game will consist of two teams playing each other. Since we assumed each team is equally likely to win, we would then get that the probability that each of your choices is correct in the first round is \(\frac{1}{2}\). Since each correct chose is worth 1 point in the first round, we find that the expected value for each game will be \(\frac{1}{2}*1+\frac{1}{2}*0=\frac{1}{2}\). That is, we multiply the probability by the value and add these together.

There are 64 teams in the tournament at this point, so there will be a total of 32 games. Since each game has an expected value of \(\frac{1}{2}\), we will get the total expected value to be \(32*\frac{1}{2}=16\) points. Therefore, we would expect to earn 16 points in the first round.

Second Round

In order to get points in the second round, you will need to predict the winner in the second round. In order for the team you choose for the second round to win, they would have to first win the first round game and then win the second round game. With the probability of winning each game being \(\frac{1}{2}\), the probability of winning both games would be \(\frac{1]{2}*\frac{1}{2}=\frac{1}{4}\).

Now, each correct choice in the second round is worth twice as many points as the previous round. Therefore, each choice will be worth 2 points. Therefore, the expected value for each game will be \(\frac{1}{4}*2+\frac{3}{4}*0=\frac{1}{2}\). Since there are 32 teams in the tournament at this point, there will be 16 games. Hence, the total expected outcome for the second round will be \(16*\frac{1}{2}=8\). That is, we expect to get 8 points out of the second round.

Third through sixth rounds

For the subsequent rounds, we note that the probability of winning will be half of what it was the previous round, and the point value will be doubled. Therefore, the expected of each game in the nth round will be \((\frac{1}{2})^{n}*2^{n-1}=\frac{1}{2}\). In each round there will be \(2^{6-n}\) games, therefore, the expected number of points from each round will be \(\frac{1}{2}2^{6-n}=2^{5-n}\) points. That is, we will have the following expected number of points per round.

  • First Round-16 points
  • Second Round-8 points
  • Third Round-4 Points
  • Fourth Round-2 points
  • Fifth Round-1 point
  • Sixth Round-\(\frac{1}{2}\) point

Full Tournament

If we now look over the entire tournament, we would find that the total number of points earned would be the sum of points earned in each round. Therefore, if you were to randomly choose teams for each game, you would expect to have 31.5 points through the tournament. At the end of the tournament make sure to come back and look at this. Did you do better than you would expect by randomly guessing?

Lowest Seed

If we refer back to The Perfect Bracket, we found a function modeling the probability that a team would win a given game based on the difference in seeding between the teams. This probability function, \(l(c)\), was given by
\[
l(c)=\begin{cases} .5 \text{ if } c=0 \\
\frac{1}{28}(c-15)+1 \text{ if } 1 \leq c \leq 15.
\end{cases}
\]

As we did in the case when working with random choices, we can find our expected value for each game by multiplying the probability of a correct choice by the points that a correct choice will be worth. We made a table of probabilities already, so we will fill in the corresponding points to determine expected outcome based on our choices.

Round One

For the first round we will have the following expected values for each quarter of the tournament.

Totals

Round 1 Probability Value Expected outcome
1 over 16 1 1 1
2 over 15 .964 1 .964
3 over 14 .857 1 .857
4 over 13 .786 1 .786
5 over 12 .714 1 .714
6 over 11 .643 1 .643
7 over 10 .571 1 .571
8 over 9 .5 1 .5
6.035

Therefore, you expected number of points for each quarter of the bracket will be 6.035. This will give you a total of 24.14 expected points in the first round. For comparison, this would be an 8.14 point improvement over randomly guessing for the first round.

Round Two

For the second round, we will again look have to find the probability that we made the correct choice in each case so that we can find the expected number of points. Note that for each choice, we will have a different probability of a team winning depending on who won the opposing game leading into this game.

For example, we can have the 1 seed winning the second round by beating the 8 seed or the 9 seed. In order to have the 1 seed beat the 8 seed, we must have had both win their first games. Therefore, the probability of having a 1 over 8 win would be \(1*.5 * .714=.357\). That is, we would multiply the probabilities of the 1 seed winning the first round by the probability the 8 seed won the first round by the probability the 1 seed beats the 8 seed. In a similar manner, the probability we have the 1 seed over the 9 seed in the second round would be \(1*.5*.75=.375\). Therefore, the total probability of the 1 seed winning the second round game would be \(.357+.375=.732\).

In the following table we list the ways we would have chosen the correct answer for the second round and the probabilities each of the occurred. We will then multiply by the point value for this round, (2), and add these all up in order to get the expected outcome for the second round.

Totals

Round 2 Probability Value Expected outcome
1 over 8 1*.5*.714 2 .714
1 over 9 1*.5*.75 2 .75
2 over 7 .964*.571*.643 2 .708
2 over 10 .964*.429*.75 2 .620
3 over 6 .857*.643*.571 2 .629
3 over 11 .857*.357*.75 2 .459
4 over 5 .786*.714*.5 2 .561
4 over 12 .786*.286*.75 2 .337
4.730

The table gives us the probabilities and expected outcome of one-fourth of the total second round. Therefore, we would expect to earn \(4*4.730=18.921\) total points in the second round. Again for comparison, this is 10.664 more points than we would expect to get if we randomly guessed.

Rounds 3 and 4

We can continue to do the same process and find the probabilities that our chosen teams wins in the given round. Instead of listing all the match up possibilities, we will instead give the probabilities that our chosen team wins the following rounds.

Round Team Probability Value Expected Value
3 1 .463 4 1.852
3 2 .383 4 1.532
4 1 .268 8 2.144

Therefore, the total expected value for round three will be \((1.852+1.532)*4=13.536\). Also the expected value for the fourth round will be \(4*2.144=8.576\).

Rounds 5 and 6

In order to simplify the calculations, we will assume that if a team makes it to the last two rounds, that it will be as likely to win the game as the other team, regardless of seed. That is, if a 4 seed make it to face a 1 seed in the fifth round, then we will now assume each team has a probability of winning of .5.

We then find that the probability that our pick for the 5th round will is now \(.268*.5\) and we have a winning value of \(16\), therefore, the expected value of each game to be \(.268*.5*16=2.144\). Since there are 2 games in this round, the total expected value for the round will be \(4.288\).

In the sixth round, we have the probability that our given team will win will be \(.268*.5*.5\) with a value of 32 points. Therefore, the expected value for each game, and therefore the round, will be \(.268*.5*.5*32=2.144\).

Expected Outcome

In order to find the total expected outcome for our technique of choosing the lower seed for each game, then randomly choosing the winner for the last two rounds, we just need to add up the expected outcomes of each of the rounds. We therefore get \(24.14+18.921+13.536+8.576+4.288+2.144=71.605\).

Conclusion

We found the expected number of points that we would earn under using two techniques to choose the teams for our brackets. In the case we chose teams randomly, we found that we would expect to get 31.5 points. On the other hand, if we choose the lower seed to win each game, we would expect to earn 71.605 points. Note that our technique of basing our choices off the seed resulted in more than doubling our projected score. This would therefore give us a much better chance of winning our bracket group.

I hope you enjoyed today’s post and learned something along the way. We also hope you fill out a bracket for the STEM and leaf group at STEM and leaf Bracket, so that we can compare our scores to the each other and to the values we would expect based on the technique used.

We will continue to look at these brackets by determining the amount of variation we would expect in our choices. We will then use this to predict how many points will be needed in order to win a bracket. If we get enough people filling out brackets, we will also compare our results to the expected results.

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