Football pools a game for mathematicians

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 square and the square will come up as winners more often than, say, the 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 square or the square, suddenly your odds of winning will have jumped up, whereas if you're stuck with the square, your chances are practically nil.

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?

So, for example, if the final score was , this would correspond to the box at , rather than the box at What about a score like 39? You'll never need to add the digits more than twice, and after this process you'll end up with a digit from , 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 point increase to the score. So, for example, if the score is , and the winning team scores a touchdown and a field goal, the score will likely be - 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 On the other hand, using the digital root, the winning square corresponding to a score of would be , while the winning square for is Here's an example of a football pool which has been tagged in the four squares mentioned above.

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.

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 , to , to , and then to So, while there are four unique scores in the game, with the usual football pool, only two squares will be hit: the square, and the 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 , , , and 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. But how?

I needed to look closer at the lottery. But not just Mark Six… every lottery. The lottery project was so intriguing I got straight to work. Using the undefeated combination of Google and Wikipedia, I began compiling a list of large lotteries around the world.

I would perform cursory research on each of the games, and rank them based on:. On the surface it appeared to be a standard lottery with 6 random numbers being selected from 38 choices 38 choose 6.

It had a major jackpot that snowballed if there were no winners, and a range of smaller secondary prizes. However there was a twist. The selected numbers were not completely random, rather determined based on results of scheduled European football matches.

Taking a deep dive into the game rules, I emerged with the following important information:. If results were identical, the match with a higher match number was selected. Reaction: OK, sure. Are any of these more probable than others, on average?

Need to check data. Again, need high level look at data to observe distribution of match scores. At this point I am jumping out of my boots. A basic familiarity with football scores tells us that the scores are often very low, and therefore frequently the same.

So if 38 matches are played, the chance that many of the results are precisely the same is very high. Which will then trigger this clause, forcing all the highest ranked matches winning lottery numbers to simply be the matches with high match numbers. With the above logic alone, I probably already had a small positive expectancy enough to start beating the game. But it was time to dig deeper….

Specifically, instead of treating all of the football matches according to large sample averages, there was additional information I could use- I knew exactly which teams were playing each other, and could get information about these teams! For example, in the Premier League, Manchester City strong team was likely to beat Huddersfield weak team. This made a draw, a result that ranked highly according to the ranking criteria, unlikely.

Conversely, Manchester City vs Liverpool was a close match that increased the chance of a draw.