Of course, the sample size is naturally restricted by the small number of Super Bowls, but if the author (Mark Pesavento) had really been interested in the question of whether or not the coin toss is correlated with the winner in a football game, he could’ve easily collected data over a couple of seasons and obtained an answer to the question.  At the very least, he could’ve owned up to the fact that his analysis is worthless, but instead, to the critics he offers only the following rebuttal: “because of the small sample size, some statisticians argue that the win-loss record of coin-toss winners is statistically insignificant.”

This is completely disingenuous, because it suggests that there would be a debate among statisticians about the significance in the data Pesavento uses, when no such debate exists.  Anyone with even a rudimentary background in statistics would understand that the sample size here would be too small to draw the conclusion he draws.

Moreover, Pesavento falls for one of the most common traps in statistics: mistaking correlation for causation.  Even if the data was much stronger in indicating that the coin toss winner is at a disadvantage, this would not imply that Brees should hope to lose the toss.  A correlation between these two effects does not imply a causal relationship between the two.  I feel like I’ve discussed this before, but just in case, here’s a thorough discussion of this misconception.

Here this point is moot, since we don’t even have a correlation.  I thought no one would need to point out that “No correlation does not imply causation,” but apparently we do.

Thankfully, most of the comments on Pesavento’s post are scathing in regards to his methods.  But that’s cold comfort in light of the fact that the article was deemed fit for posting on the front page of Yahoo.

In the first part of this discussion, we introduced a new way to conduct the pool: rather than looking at the last digit of a team’s score, we looked instead at the digital root of the team’s score.  Recall that the digital root of a team’s score is obtained by adding the digits in their score.  If that sum is between 1 and 9, we stop – if it is larger than 9, we compute the digital root again, until we get a digit between 1 and 9.  For example, the digital root of 14 is 1 + 4 = 5, while the digital root of 38 is 2, since 3 + 8 = 11, and 1 + 1 = 2.  We then analyzed the distribution of scores, and found that the digital root of a team’s score is more evenly distributed between 1 and 9 than the last digit of a team’s score is evenly distributed between 0 and 9 (this is subject to the convention that we assign 0 a digital root of 9, since 0 is the only number with digital root equal to 0).

In the second part of the discussion, we tackled questions of independence.  Namely, we asked whether the last digit in one team’s score is independent of the last digit of the other team’s score, and similarly we asked whether the digital root in one team’s score is independent of the digital root of the other team’s score.  In both cases, we found the answer to be negative.

The subject of this article is based on the following observation: when you have wagered in a traditional football pool, it’s not uncommon for a small number of squares to be hit with high frequency during the course of a game.  For example, suppose you watch a game in which one team scores 7 points, then 3, then 7, then 3, while the opposing team never scores.  This means that the game’s score will go from 0-0, to 7-0, to 10-0, and then to 17-0.  So, while there are four unique scores in the game, with the usual football pool, only two squares will be hit: the 0-0 square, and the 7-0 square.  However, with the digital root pool, four squares will be hit: again using the convention that we assign 0 a digital root of 9, the squares will be 9-9, 7-9, 1-9, and 8-9.

The reason the digital root pool hits more squares in this case is because whenever one team increases its score by 10, the last digit of their score will return to a previous value.  However, with the digital root method, if a team increases its score by 10, the digital root increases by 1.  Because a score increase of 10 is a relatively common occurrence in football (all one needs is a touchdown, extra point, and field goal), one may therefore guess that using the digital root pool, more squares should be hit during the course of the game.

Whether one would want more squares to be hit or not is up for debate, but I see certain benefits.  For example, if more squares are hit during the game, then more people will have something invested in the game as it airs.  If you are sitting on the square that represents the current score, you want the score to remain the same through the end of the quarter so that you can reap the rewards – but if the winning squares keep bouncing around between a small number of people, there may be fewer people actively interested in the score as the game progresses.  This is especially true in Super Bowl parties, when many of the attendees are less interested in the game than they should be.

In other words, I’m of the belief that if more squares are hit, it’s a good thing.  It therefore becomes natural to ask whether or not the digital root pool actually does hit more unique squares than the traditional pool.  Thankfully, we have a wealth of data which we can use to answer this question.

I looked at all the games from this current season, and counted the number of boxes that would have been hit in each game using the traditional pool and the digital root pool.  Averaged over 331 games (this includes preseason and postseason), the number of squares hit using the traditional pool is approximately 6.84.  By comparison, the number of squares hit using the digital root pool is 8.43 – an increase of 1.59 boxes, or an increase of about 23%.  This effect is amplified when one considers the fact that the digital root pool uses only 81 squares, as opposed to the traditional pool’s 100.  This means that as a proportion of the total number of squares, the traditional pool hits about 6.84% of its squares, while the digital root pool hits 10.4% – here we have an increase of over 50%!

This is strong evidence that the digital root pool hits more squares than the tradition pool.  In fact, the data shows that an average game will have a change in the score approximately 8.73 times, which is only a bit higher than the average number of boxes hit by the digital root pool.  This makes sense when we slice the data another way: of the 331 games analyzed, in 252 of them the number of squares hit with the digital root pool was equal to the number of changes of score, meaning that no square got hit more than once.  The same cannot be said of the traditional pool – in this case, the number of games in which no square got hit more than once was only 62.

The data has convinced me that the digital root pool may be better suited for festive gathering, where wagering on football will be but one of many activities designed to induce merriment.  At the very least, it’s hard to argue that the traditional pool will hit as many squares as the digital root pool. Some may balk at a break from football pool tradition, but that’s ok.  I won’t watch football games with them anyway.

This post is a follow-up to an earlier post that looked at betting squares for football scores. In particular, we analyzed the distribution of the second digit of final football scores, and compared that to the digital root of final football scores (recall that the digital root of a number is found by iteratively calculating the sum of the digits in that number until you come up with a single digit number from 1 through 9). We found that on average, the final digits of football scores do not distribute themselves evenly – a score ending in 2 or 5 is much rarer than a score ending in 7 or 0, for example. However, the analysis of the digital root suggested that digital roots may become evenly distributed on average.

We now turn to a related question of independence, which was mentioned at the end of the previous post. More specifically, we address the following question: is the second digit of the home team’s football score independent of the second digit of the away team’s score? What if we replace “second digit” by “digital root” in the above question?

A more basic question that is related to both of these is whether or not the two final scores in a football game are independent. The answer to this question is not entirely obvious. On the one hand, knowing that one team scored 7 points doesn’t tell you anything a priori about the second team’s score. On the other hand, one team’s score during a game will certainly influence the strategy of the opposing team: if one team is ahead by a wide margin, for example, the losing team will likely play riskier in the hopes of making some big plays. Conversely, if the difference in scores is very close, teams are less likely to make plays that are as risky unless it’s late in the fourth quarter.

Alternatively, what if you are told that one team scored 7 points, but that the game went into overtime? Now you know a great deal about the other team’s score: since overtime in the NFL is sudden death, and since the only way to score 7 points is by a touchdown or the (very unlikely) combination of a field goal and two safeties, you know that the other team must have scored 4, 5, 7, 9, 10, or 13 points.

From these examples, it seems that football scores aren’t entirely independent – there is often some relationship between the two. However, it’s natural to think that perhaps on average, we do have independence. Similarly, we may hope that by looking at several games, we have independence of the second digit or the digital root between home and away teams.

One reason to hope that we would have independence is that it allows us to calculate probabilities of winning squares with much less input. If the probability of the home team having a score that ends in a 7 is 17%, for example, and the probability of the away team having a score that ends in an 8 is 7%, we’d like to be able to say that the probability of the (7, 8) square winning in a football pool is .17 x .07. Otherwise, to estimate the probability of any particular square winning we’d have to rely on historical data, and create a running tally of how each square has performed. Needless to say, this is more work.

Unfortunately, it seems that this is necessary work. By looking at data from 4,642 preseason and regular season games from 1994-2008 (the 1994 season was the first in which the two-point conversion was instituted), I tallied the counts for each winning square in the case of both the traditional football pool and the digital root pool. Here is the data:

Recall that for the digital root pool, the digit 0 only occurs if one team doesn’t score – to even things out, I have assigned a score of 0 with a digital root of 9 (I discussed reasons for this in my earlier post).

Using a standard statistical test for independence, we conclude that there is essentially no way that the digital root of one team is independent of another team, or that the last digit of one team’s score is independent of another team’s score.¹ To illustrate one example, 10.8% of the home team’s scores had a digital root of 2, while 9.26% of the away team’s scores had a digital root of 2. Therefore, assuming the digital roots of the away team and home team’s scores were independent, the probability of winding up in the (2,2) square would be 0.108 x 0.0926, which is approximately 1%.

However, as the data shows, only 12 games ended in the (2,2) square this gives a much smaller probability of 12/4,642, which is approximately 0.26%. In other words, if you know that the score of the home team’s digital root is 2, it is suddenly much less likely that the score of the away team’s digital root is also 2.

Some other observations:

1. Using this larger data set, we also find that it’s actually unlikely that the digital roots become equally distributed among the digits from 1 to 9. While the distribution is certainly more uniform than in the case of the second digit, we also see that certain digital roots occur significantly more frequently than others (5 occurs much less frequently than 1, for example).

2. Even though away team and home team 2nd digits and digital roots are not independent, one may expect that the probability of having a score with a given 2nd digit or a given digital root should be independent of whether the team is the home team or the away team. Interestingly, in the case of the 2nd digit, this is not true. In particular, it is more likely for an away team’s score to end in a 3 than a home team’s score, and it is more likely for a home team’s score to end in an 8 than an away team’s score.² In the case of the digital root, however, the data is insufficient for us to reject the hypothesis that the probability that the away team’s score has a certain digital root should be different from the probability that the home team’s score has a certain digital root.³

In considering the relative merits of the traditional football pool versus the digital root method, one question from my earlier post remains unanswered: which method will hit the most squares during a typical football game? It seems reasonable to expect that the digital root method will hit a larger number of squares than the traditional method. For example, if the score is 13-7, and the home team scores a field goal followed by a touchdown, two boxes will be hit by the traditional pool: (3,7), (3,0), followed again by (3,7). However, the digital root method will hit three boxes: (3,7), (3,1), and then (3,8).

This heuristic is certainly reasonable, but does it hold when looking over a large number of games? I’ll return to this question in a later post.

1. In the case of the last digit, we obtain a χ² value of 425.50 with 81 degrees of freedom, which gives a probability that is essentially 0. Similarly, in the case of the digital root, we obtain a χ² value of 326.52 with 64 degrees of freedom, again yielding a probability that is essentially 0.

2. Testing the Null Hypothesis that the probability that a home team’s score ends in a 3 is equal to the probability that the probability that an away team’s score ends in a 3 yields a χ² value of 33.62 with 1 degree of freedom. Replacing 3 with 8 gives a χ² value of 11.86. At a 0.01 significance level and 1 degree of freedom, we reject the null hypothesis if χ² is larger than 6.6348.

3. In other words, the χ² values in the digital root case are never larger than 6.6348. The case of the digital root equal to 5, however, comes close – in this case, the χ² value is 6.6130.

During this month’s Super Bowl, like many of my fellow Americans, I participated in the great tradition of the football pool. This method of betting on a football game is quite simple. For those of you who have never partaken in this activity, here’s how it works:

If you have any knowledge of how football games are scored, it may seem to you that certain squares are more desirable than others. For example, you may guess that the 7-0 square and the 0-7 square will come up as winners more often than, say, the 2-2 square, since football scores that end in 2 are much less common than football scores that end in 7 or 0.

However, it seems reasonable to assume that this effect will diminish in later quarters, because as the scores increase, the possibility for variation in the second digit also increases. In other words, after the first quarter, the number of possible scores for the game will be fewer than the number of possible scores at the end of the last quarter, because the longer time means that there is more opportunity for scoring, and therefore a larger set of possible scores.

This gives rise to a natural question: on average, are the last digits in a football score distributed evenly between 0 and 9? If not, is there a different way we could conduct the pool that would yield more uniform results?

Of course, there is excitement in the fact that once the numbers have been revealed, some squares will be more popular than others. If you have the 0-7 square or the 7-0 square, suddenly your odds of winning will have jumped up, whereas if you’re stuck with the 2-2 square, your chances are practically nil. (In fact, Doug Drinen crunched some numbers back in 2005 by looking at the scores for all football games since 1994, and found that the chance of winning the fourth quarter with a 7-0/0-7 square was about 3.8%, versus a dismal 0.04% for the 2-2 square.) This disparity between squares, however, means that the unlucky people who get the squares less traveled will have less of a stake in the game, and perhaps enjoy it less because of this.

Here’s one simple alternative to the traditional pool described above: instead of looking at the last digit of the score when determining the winning square, what happens if you look at the sum of the digits? With this system, a score of 14 would have the sum of its digits equal to 1 + 4 = 5. So, for example, if the final score was 14-21, this would correspond to the box at 5-3, rather than the box at 4-1.

What about a score like 39? If we add the digits here, we get 3 + 9 = 12. In a case like this, simply add the digits again: 1 + 2 = 3, so 39 would correspond to the digit 3. You’ll never need to add the digits more than twice, and after this process you’ll end up with a digit from 0-9, so the same grid will work. If you want to sound impressive, you can refer to this process as calculating the digital root, and since I want to sound impressive, I will use this terminology throughout. So, for example, the digital root of 39 is 3, the digital root of 68 is 5, the digital root of 17 is 8, and so on.

There is one slight hiccup when using the digital root: namely, the digital root of a number is 0 if and only if that number is itself 0. This means that any box associated to a 0 on the row or column will only be a potential winning candidate if one team remains scoreless.

One way to overcome this slight complication is to simply ignore the digit 0, and make a 9 x 9 grid with digits from 1 to 9. Most of the time, this will be fine, especially later in the game when it is unlikely a team will have no points. In the event that you do encounter a 0, I’ll explain two possible solutions below.

Why might we expect the digital root to give us a more uniform distribution of values than simply looking at the last digit of the score? Well, think about the way football is scored. In the game, a touchdown and a field goal will (usually) result in a 10-point increase to the score. So, for example, if the score is 10-7, and the winning team scores a touchdown and a field goal, the score will likely be 20-7 – this is nice for the winning team, but not so interesting for those in the pool, since this has no effect on which square is winning (in either case, the winner is 0-7).

On the other hand, using the digital root, the winning square corresponding to a score of 10-7 would be 1-7, while the winning square for 20-7 is 2-7.

This is but one example, but it is useful in guiding our intuition. Is this indicative of a larger trend? This may suggest that using the digital root gives a more uniform distribution among digits, but is this really the case?

To answer these questions, we can turn to our trusty friend, mathematics. The Football Database provided me with the final scores for every football game during the 2008 season (including preseason), and this is the data I have used as the basis for the following results. Note that because I only used scores from the end of the game, everything here applies only to the final winning square. Also, there’s no doubt that I could get more accurate data by compiling data over many seasons, but I don’t really have time for that, and one season gives enough data to paint an interesting, if not entirely complete, picture.

Here are the questions the data addresses:

The data gives fairly definitive answers to all of these questions. In the 2008 season, there were 332 games played (including pre- and post-season), yielding a total of 664 final scores. Of those final scores, 120 ended in 0, 63 ended in 1, and so on. Similarly, 9 of those final scores had a digital root of 0 (i.e. 9 games had one team go scoreless), 85 final scores had a digital root of 1, and so on. The results are summarized in the following table:

What can we learn from this table? Well, it’s clear from the second column that the last digit of football scores certainly don’t seem to be evenly distributed. Some digits, such as 7 and 4, have much larger counts than numbers like 2 or 5.

What about for the digital root? In this case, the numbers appear to be much more evenly distributed, with the exception of 0, but this is to be expected. Aside from 0, the digit which appears to deviate the most is 9, with only 63 final scores having this digit as their digital root.

If you’re one for pictures instead of words, this graph can help to visualize the discrepancy between these two methods. Here I’ve divided the counts above by 664 to yield percentages in the graph:

The blue bars, showing the percentage of scores that ended in a particular digit, are visibly less evenly distributed than the red bars, which show the percentages of scores that correspond to a particular digit using the digital root.

This is all well and good, but what if 2008 was just a strange year for football scores? To eliminate all doubt, I tested the probabilities for the traditional pool against the null hypothesis that all the probabilities should be equal (i.e. the probability that a score should end in a certain digit is 10%, regardless of the digit). The result? The odds of getting such uneven percentages if the last digits of the scores truly are evenly distributed is essentially 0%¹. From this, we can safely reject the null hypothesis and conclude that looking at the last digits of football scores will not give you numbers that are evenly distributed among the digits from 0 through 9.

How about with the digital root method? To answer this question, we need to know how to deal with that pesky 0 digit, which clearly skews the data. Here were the results under two different sets of rules governing what to do with 0.

In the first case, I simply ignored the scores that were 0. This amounted to only 9 scores, so I was still left with a sample of 655 scores. To these scores I conducted the same analysis as I did with the traditional pool, but in this case the null hypothesis increases each digit’s probability from 10% to 11.1%, since there are only 9 digits to consider rather than 10. Here, the odds of getting the percentages above if the digital root of the end game football scores are evenly distributed are about 34%². In other words, we can’t reject the null hypothesis in this case – it may be that the digital roots lead to an even distribution between the digits from 1-9.

In the second case, I combined the nine zeros into the count of scores with digital root 9, since of the digits from 1-9, 9 occurred with the lowest frequency. With this method, the odds of getting the percentages above if the digital root of the end game football scores are evenly distributed jumps to 63%³. In either case, we can’t reject the null hypothesis that the digital roots don’t become evenly distributed.

Alternatively, you could simply distribute the zeros randomly between the remaining digits – but this should give a probability that is roughly the same as in the first case.

In summary, using the traditional football pool, the second digits of the scores most certainly do not become evenly distributed between 0 and 9. However, it’s quite possible that the digital roots do become evenly distributed between 1 and 9, regardless of the choice we make in how to deal with the zero scores.

Does this make the digital root method “better” than the traditional method? Well, it depends on what you’re looking for in the football pool. Certainly the data suggests that on average, the digital root method will hit the squares that are rarely hit by the tradition pool with more frequency. However, this is a result that holds true on average, so this analysis can’t say whether or not the digital root method will make individual games more exciting. It does, however, seem to make the less popular squares slightly more popular, and the more popular squares slightly less popular. With the traditional pool, if you get stuck with the 2-2 square, you can basically kiss your money goodbye. But with the digital root method, there is still a glimmer of hope that you will prevail.

Here are some further questions. I won’t address them here, but may tackle them later.

I can’t say yet whether I prefer the digital root method to the traditional pool, but it certainly seems to offer a compelling alternative. In conclusion, here is a video of the 49ers from the 1980s rapping about how awesome they are. If only the 49ers of the 21st century could flow as well as these dudes, maybe they would have a shot at returning to glory.

1. Using a simple Pearson’s χ² test, one obtains here that χ² is approximately 188.7. We have 9 degrees of freedom, so this gives the result (probabilities can be calculated in many places online, e.g. here).
2. In this case, we get χ² is approximately 9.05. Since we ignore zero, this reduced our degrees of freedom to 8.
3. In this case, we get χ² is approximately 6.15, again with 8 degrees of freedom.