Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear Colleagues,

This is a math question for people who know the rules of (American) football.

Every year my barber runs a “football squares” game. He finds 100 customers, each put in 20 dollars, and each person is assigned a square on a 10 by 10 grid. After all squares are sold he picks numbers out of a hat to label each row and column 0 thru 9. So each contestant winds up being assigned with an ordered pair of numbers – in my case (3,7) this year. Now prizes are awarded by inspecting the last digit of the score of each team in the Super Bowl at the end of the first, second, third quarters and the final game score. Each winner gets $500.

For example, suppose that the scores are AFC 14, NFC 10 at the end of the first quarter. Then the person with (4,0) wins $500.

Now this is a fair bet, in the sense that each square is assigned randomly (and I believe that my barber doesn’t cheat.) However, it would seem that some numbers are “better” than others. Football scores are not random. For instance, at the end of the first quarter it is extremely unlikely that (5,5) will win. On the other hand, ((0,7) would seem like a good number. What is needed is some probabilistic analysis based upon the actual scoring patterns in football together with looking at actual scores of many pro football games. I am unable to find any analysis of this on the internet. I have tagged this as a probability question but since I work in operator algebras for a living this may be mis-tagged, and I ask you to pardon my error.

Let’s make it precise. Let f: {0,1,2,…, 9} x {0,1,2,…, 9} x {1,2,3,final} \to [0,1] be the function that to a point (x,y,z) assigns the probability that the score at the end of the z’th quarter of the Super Bowl will be equal to (x,y) mod 10. Find the function f. Where does f achieve its max?

How about it, colleagues? Inquiring minds want to know!

CS

share|improve this question

3 Answers 3

up vote 2 down vote accepted

Alternative (and rough) idea for a model: Consider a random walk on score vectors $[x,y]$ where x and y represent the scores of the two teams. Set $P_0=[0,0]$. Let $P_1=P_0+v_1$ where $v_1$ is with probability 1 chosen (uniformly distributed or otherwise) from $[0,7]$, $[7,0]$, $[3,0]$, or $[0,3]$. Let $P_2=P_1+v_2$ where $v_2$ is randomly chosen from the same set of vectors but with probability $1/1.1$ (adding the zero vector with probability $.1/1.1$). Repeat: $P_i=P_{i-1}+v_i$ where $v_i$ is chosen from those four vectors with probability $1/(1.1)^i$ or zero otherwise. Letting $i\rightarrow\infty$ and repeating gives you a distribution of possible football scores, avoiding the problem of finding an "expected value of a game", which would be nearly irrelevant to the trailing digit.

Quick notes:

  • The constant $1.1$ is chosen so that the expected total number of scores is 10, which I found online (and thus must be true) as the average number of scores in an NFL game.
  • Truncating after fewer steps would gives you distributions for quarter scores instead of final scores.
  • Probabilities should probably not be uniform, either between the two teams (so this model could build in one team being a favorite) or between field goals and touchdowns.
  • A fun tweak would be to build in 8's for two-point conversions context-dependent on the current score. Safeties are probably negligible.
  • Easily modifiable and poorly written SAGE code below:

def rungame():
    v=vector(ZZ,[0,0])
    p = random()
    enter code here
    def iterate(v,i):
        p = random()
        if (0<p<.25/(1.1^i)):    
            return v+vector(ZZ,[7,0])
        if (.25/(2^i)<=p<.50/(1.1^i)):
            return v+vector(ZZ,[3,0])
        if (.5/(2^i)<=p<.75/(1.1^i)):
            return v+vector(ZZ,[0,3])
        if (.75/(2^i)<=p<=1/(1.1^i)):
            return v+vector(ZZ,[0,7])
        return v;

    for i in range(0,20):
        v=iterate(v,i);

    print v

EDIT: Here's the result of 10,000 trials. The 117 in the top left is the number of times the game ended with (0,0) digits, etc.

[117 91 116 121 117 51 89 118 116 86]
[121 54 91 47 91 110 43 115 117 50]
[ 42 90 111 42 47 52 48 83 118 85]
[119 85 39 99 90 54 122 92 88 120]
[ 89 48 91 44 120 89 92 119 97 42]
[112 109 87 114 86 0 117 121 73 113]
[ 89 99 114 119 47 33 118 121 41 114]
[ 99 88 120 88 88 89 40 120 98 84]
[ 99 120 100 49 91 44 40 43 32 115]
[ 90 100 82 89 98 91 50 115 111 100]

share|improve this answer
    
I would think that the number of scores in a game would be Poisson-distributed, not geometrically distributed. But I don't have data. –  Michael Lugo Jan 27 '10 at 21:22
    
Almost certainly. Good point. –  Cam McLeman Jan 27 '10 at 22:09

See here and here for empirical statistics. These are based on all NFL games since the two-point conversion was added (1994), not just Super Bowls.

Your guesses are right: scores ending in 0, 3, 7 come up a lot; scores ending in 2, 5, 9 come up very rarely.

share|improve this answer

The theorist's answer

It seems natural to model American football as a Markov chain. If you want to be realistic, the state space could be incredibly complicated, with information about things like who's in possession, how many downs, what's the score, etc. In statistical physics, however, ridiculously simple models do a surprisingly passable job of modeling complex processes. I wouldn't be surprised if something as stupid as a normal-distributed random walk, with absorbers at 0 and 100 yards, did an okay job of predicting the gross statistical properties of football scores.

The experimentalist's answer

There have only been 43 Super Bowls, so your sample size is pretty crappy. (For what it's worth, all of the scores are available here.) Super Bowl contenders are drawn from the NFL, though, and there have been hundreds of NFL games. It might be interesting to see whether you can reliably distinguish a Super Bowl game from a regular season game just by looking at the final scores!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.