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First time post and I'm not sure if I'm posting in the right place but here goes:

I enjoy Sports and was reading a book over Christmas (Mathletics by Wayne Winston) and to cut a long story short have started ranking sports teams based on least Squares Regression (based on the number of points scored or conceded) to generate Power Rankings. I've also done a little bit of reading around the subject such as on "The Probability of Winning a Football Game (Stern, The American Statistician, Aug 1991).

I've noted that this works out fairly well in predicting results with some leagues (based on margins of error) and has led me to look at occurrences to do with probabilities of scoring goals. Below is a sheet that I have pulled from one of my Excel tables for a given league. The means are based on my least squares data from previous results, used to predict the likely number of goals based on a teams power rankings (offensively and defensively). The probability of scoring a goal I have treated as a Poisson Random Variable and have used the Poisson function in Excel to generate the numbers below:

mean    3.07554771 3.389358782  
goals   team 1          team 2          team 1 wins 
0   0.04616434      0.03373030  
1   0.14198062  0.11432408  0.00478905 
2   0.21833409  0.19374267  0.03232532 
3   0.22383230  0.21888780  0.07650522 
4   0.17210173  0.18547233  0.09649483 
5   0.10586142  0.12572645  0.07898926 
6   0.05426364  0.07102201  0.04731158 
7   0.02384149  0.03438844  0.02248027 
8   0.00916570  0.01456934  0.00895759 
9   0.00313217  0.00548675  0.00310669 
10  0.00096331  0.00185966  0.00096076 
CUMULATIVE  1.00    1.00    
tie 0.159139717     
team 1 wins 0.371920564     
team 2 wins 0.468939719

(sorry for the layout - I haven't got the hang of the formatting on here yet when pasting from Excel)

My query is could I use this to predict the probability of an over under (e.g. over/under 5.5 total goals for instance)? I've got this far and had a total brain fade (high school statistics was a long while ago and the topics we learnt were never that interesting or relevant at the time). For what it's worth I've been able to interpret the means fairly well against totals (64% for this league) and I am just looking at ways of refining this (e.g. putting a probability out there as opposed to in this case 6.45, is over 5.5, I'd go over). I find it is always fun to have an opinion, but it's better when you can back it up with a statistics.

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    $\begingroup$ I've used the code blocking to try to salvage the table, not perfect, but certainly easier to read than with all of those horizontal rules. See the advanced markdown editing help for tips, and welcome to the site. Feel free to edit to improve. $\endgroup$
    – Andy W
    Commented Jan 29, 2012 at 18:22
  • $\begingroup$ Thanks that is spot on - I'll read up on the advanced markdown editing for next time. $\endgroup$
    – user8812
    Commented Jan 29, 2012 at 18:39
  • $\begingroup$ You appear to have started out well. Stern's article is an interesting one and he takes a fairly active interest in this area. There are also some somewhat related questions on this site you might like looking at. For example, there is a fairly active games tag and from there you can find other questions by looking at the "Related" questions listed at the right on each page. Cheers. $\endgroup$
    – cardinal
    Commented Jan 29, 2012 at 23:14

1 Answer 1

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If you believe the scores are well defined by the poisson distribution and that they are independent, you could quickly simulate the probability of certain spreads. Here's some simple r-code that runs 10,000 simulations and finds how often team b wins by 5.5+ points:

mean( (rpois(n=10000,lambda=3.0755) + 5.5) < rpois(n=10000,lambda=3.3894) )

I get about a 2% chance. I know nothing of this domain (sports stats) so you might want to closely visit that independence assumption. You could go deeper and look at alternative discrete distributions like the negative binomial (assuming heterogeneous player contributions?) Or go baysian and treat your lambda as a random variable to capture more of your uncertainty...

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  • $\begingroup$ Hi, I think this could work or it's given me an alternative idea of how to do it. If I could work out the Standard Deviation of my predictions against the actual results (e.g. the deviation or sigma value for US Football is approx. 13.86 - ref: www.stanford.edu/class/stats50/handouts/stern.pdf - the stern text I mentioned) I could just treat this as a normal random variable with the margin of victory and set up a data table in excel based on the expected margin and simulate this). I think I may be able to do this just from the data above though without working out a sigma? $\endgroup$
    – user8812
    Commented Jan 29, 2012 at 21:38
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    $\begingroup$ It appears the OP is asking specifically about over/under rather than the spread. In that case, if the scoring is independent, then the sum of the scores is also Poisson with the mean being the sum of the two means. So finding a confidence interval of some sort is "easy". If looking at the spread, then the difference in the scores follows a Skellam distribution which is known to be unimodal, though explicit tail probabilities are more difficult to calculate. $\endgroup$
    – cardinal
    Commented Jan 29, 2012 at 22:54
  • $\begingroup$ Hi Cardinal - yes I was looking at the over/under in particular (I understand the spread side on certain sports - I just need to understand about calculating the Standard Deviation of my prediction against the actual results). Re: the sum of the scores being Poisson (and the sum of the two means), so hence in the case above we are talking 6.45? How do I calculate the probability that this is greater than or equal to 6 in this case (e.g. over 5.5). Re: my other idea, I'm not sure that will work but am looking into it. $\endgroup$
    – user8812
    Commented Jan 29, 2012 at 23:11
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    $\begingroup$ @user8812: You can calculate it the same general way you have in your table. You now have a Poisson with mean parameter 6.45. So, the probability of scoring five or fewer points is found by summing the probabilities from 0 to 5 (inclusive). Subtract the result from one. (You should get about 0.624 as your final answer.) $\endgroup$
    – cardinal
    Commented Jan 29, 2012 at 23:17
  • $\begingroup$ Thanks - I get it (I think e.g. I do as above but only with one team effectively as it the total score not the team score that is the variable) - I will sleep on it (am in the UK and have work in the morning) have a go and come back. I'm very confident in the figures as I've gone about 60-64% in Handicaps, Totals and Money Line based on my predictions against actual results (200 trials) so I think that is good enough to be statistically significant. P.S. Wasn't sure if I should put these as comments or answers - first day on the site and haven't figured it out yet. $\endgroup$
    – user8812
    Commented Jan 29, 2012 at 23:24

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