Prologue: JK commented that it would be interesting to see my analysis for the HRWager sponsored contest involving two NFL team posting identical scores. The following is a write-up of my analysis for attempting to pick two teams with a contest advantage. I've tried to keep the text understandable and not get too verbose so at least a few readers can get through it.

Choose two different NFL teams who will match their scores in a given week's play (and if not match, then will minimize the absolute value of the difference in scores) in a given week. We want to maximize our chances in a contest with an unknown number of contestants each of whom will select one unique pair of teams (denoted as Team A and B). Common sense says we want 2 teams whose scoring patterns are similar. Because football scoring is "lumpy" (majority of scores are 3 and 7), we expect that identical scores by two teams are fairly frequent.

There are two primary variables, denoted TotA and TotB, which are the scores of Teams A and B respectively. We use game theory and postulate the existence of a

Let?s assume we have modeled a discrete probability distribution (DPD) for the scoring patterns of Team A and B in their specific games. We might consider eliminating the case of Team B being Team A?s opponent since the chances of a tie are so rare, however this isn't necessary.

The scoring model may be very elaborate since it could take into account the bettable team total, line, total, offensive stats, defensive stats, team opponent, matchups, injuries, weather, etc. Or it may be quite simple just using a template of the NFL "average" team score distribution and making some adjustments to various key number frequencies to reflect either higher or lower than average scoring.

Let?s compute the probability Pexact (case of TotA = TotB) given that we have a DPD for the scoring patterns (assume the scores are independent events):

Pexact = summation [PtAi * PtBi] from ?i=0 to max score

Now look at the case where Abs (TotA ? TotB) = 1, denoted as ?Off By 1?

Poff_by_1 = summation [PtAi * (PtB(i-1)+PtB(i+1) + PtBi * (PtA(i-1)+PtA(i+1) ] from ? i=0 to max score

? And the general term given by:

Poff_by_n = summation [PtAi * (PtB(i-n)+PtB(i+n) + PtBi * (PtA(i-n)+PtA(i+n) ] from ? i=0 to max score

Let the payoff function be of the form:

Where a1, a2, a3, ? constant coefficients are weighting factors to be determined.

Clearly if we choose the constants appropriately and can maximize F, we have an excellent chance at getting at least a near-optimal pair of teams. As the number of contestants goes up, we expect that we have to reduce the size of the constants a1,a2,a3, etc. Also note that when we look at terms like Poff_by_7 and higher, we might consider making a7 and beyond zero or even negative to penalize cases (we know the probabilities are always nonnegative) that result in large discrepancies.

Let?s consider the average scoring distribution for the NFL over the last 10 years, 2004 through 2013 YTD.

The six highest frequencies are: 17 (7.2%); 20 (6.8%); 24 (6.5%); 27 (5.6%); 10 (5.4%); 13 (5.3%) ... the other values can be obtained from any database of NFL scores. The max score, which we used above, turns out to be 62.

With both teams having the ?standard? DPD we can compute an ?average? Pexact which turns out to be 4.01%. Poff_by_1 turns out to be 9.35%, Poff_by_2 = 6.72% and Poff_by_3 = 13.75%.

Incidentally, if we use only 5 years of data, (2008 ? 2013), these percentages are almost the same, Pexact = 4.02%, Poff_by_1 = 9.40%, Poff_by_2 = 6.60% and Poff_by_3 = 13.70%. However, scoring is up in the last 5 years, and the six highest frequencies are: 17 (7.3%); 24 (6.8%); 20 (6.8%); 27 (5.9%); 31 (5.1%); 13 (5.0%).

Now we need to estimate the number of contestants to see how often someone will hit the exact TotA = TotB and how many times we will go to Off By 1, By 2, etc?

Assume 40 contestants is approximately correct for the contest being run. Using the binomial distribution, we find that the exact score match will be hit by at least one contestant 21.45% of time. ?Off by 1? hits 18.79% if there is no exact match. Off by 2 hits 16.77% of the time if the first two cases aren?t hit, etc? We expect the winner?s difference will be two points (or less) more than half of the time.

We want to choose the terms a1, a2, a3, etc to produce smaller and smaller contributions to the payoff function since the lowest the difference, the more likely it is to win the contest, thus let?s choose a1 = 0.3, a2 =0.2, a3 = 0.05, and a4 = 0, etc, which leads to Faverage = .0401 + 0.3*.0935 + 0.2*.0672 + .05*.1375 = .0885

This is the benchmark value for our payoff function F. F for 5-year data (2008-2013) is the same to 3 significant digits.

If we choose, we can incorporate team total derived by using the line and total for the specific game (using Week 14 of 2013).

The following table is sorted from low to high:

Team Mean Score

Browns,Rams,Vikings 17.5

Colts 18.5

Jets 19.0

Dolphins,Seahawks,Titans 19.3

Bills 19.5

Jaguars,Redskins,Panthers,Raiders 21.5

49ers,Falcons 21.8

Giants,Steelers,Texans,Buccs 23.0

Chiefs 23.8

Cards 24.0

Saints 24.3

Bears,Bengals,Cowboys,Ravens 24.5

Chargers,Packers,Lions 25.8

Eagles 28.3

Patriots 28.5

Broncos 31.3

We can use these team totals and in addition, an NFL scoring model if we have one, to produce all 32 distributions (there are over 900 combinations of two teams). Perform the calculation of the payoff function F with Eqn (1) for each pair of teams, which is straightforward programming or can be done with a spreadsheet. As expected, teams with similar team totals have larger values of F. Teams at the extremes tended to have larger variances and smaller values of F.

There were many good combinations, it turns out that the Cards/Saints pair gave a value of F very near the maximum, at .0928. This selection doesn?t guarantee a significant edge over all other contestants since that depends on their choices as well. It does have about a 5% better payoff function than the ?average? 2-team distribution.

The actual contest ended up with 39 contestants. The result of Week 14 was that the Cards scored 30 and the Saints 31, two ?mid-value? scores (31 is 7[SUP]th [/SUP](4.9%) and 30 is 15[SUP]th[/SUP] (2.6%), both scores went over their projected team total and happened to be almost identical, tying for first place in the contest with 6 others, a higher number of very close guesses than expected (should have been about 5.2 contestants that either had identical or ?Off by 1? results using the ?average? scoring distributions).

The methodology and results appear to give a fairly significant advantage over blindly picking two teams. It does not give a large advantage, but when this method is combined with a good scoring model, it does give good sets of candidate team pairs. It provides an example using a payoff function which is widely used in more complicated optimal control theory and game theoretic problems.

**Coincidental NFL two team score analysis****History:**Recently EOG.com offered a contest with a lone question:*Which two NFL teams will record identical scores in Week 14?***Problem statement**:Choose two different NFL teams who will match their scores in a given week's play (and if not match, then will minimize the absolute value of the difference in scores) in a given week. We want to maximize our chances in a contest with an unknown number of contestants each of whom will select one unique pair of teams (denoted as Team A and B). Common sense says we want 2 teams whose scoring patterns are similar. Because football scoring is "lumpy" (majority of scores are 3 and 7), we expect that identical scores by two teams are fairly frequent.

**Methodology**:There are two primary variables, denoted TotA and TotB, which are the scores of Teams A and B respectively. We use game theory and postulate the existence of a

**payoff function**of these two independent variables that serves as an approximate metric of our chances of winning.**Maximizing that function is one way of determining a solution to this class of problem**.**Nomenclature**: Let PtAn = probability of TotA = n, and PtBn= probability of TotB = n.**Analysis**:Let?s assume we have modeled a discrete probability distribution (DPD) for the scoring patterns of Team A and B in their specific games. We might consider eliminating the case of Team B being Team A?s opponent since the chances of a tie are so rare, however this isn't necessary.

The scoring model may be very elaborate since it could take into account the bettable team total, line, total, offensive stats, defensive stats, team opponent, matchups, injuries, weather, etc. Or it may be quite simple just using a template of the NFL "average" team score distribution and making some adjustments to various key number frequencies to reflect either higher or lower than average scoring.

Let?s compute the probability Pexact (case of TotA = TotB) given that we have a DPD for the scoring patterns (assume the scores are independent events):

Pexact = summation [PtAi * PtBi] from ?i=0 to max score

Now look at the case where Abs (TotA ? TotB) = 1, denoted as ?Off By 1?

Poff_by_1 = summation [PtAi * (PtB(i-1)+PtB(i+1) + PtBi * (PtA(i-1)+PtA(i+1) ] from ? i=0 to max score

? And the general term given by:

Poff_by_n = summation [PtAi * (PtB(i-n)+PtB(i+n) + PtBi * (PtA(i-n)+PtA(i+n) ] from ? i=0 to max score

Let the payoff function be of the form:

**F (TotA, TotB) = Pexact + a1 * Poff_by_1 + a2 * Poff_by_2 + a3 * Poff_by_3 ? Eqn (1)**Where a1, a2, a3, ? constant coefficients are weighting factors to be determined.

Clearly if we choose the constants appropriately and can maximize F, we have an excellent chance at getting at least a near-optimal pair of teams. As the number of contestants goes up, we expect that we have to reduce the size of the constants a1,a2,a3, etc. Also note that when we look at terms like Poff_by_7 and higher, we might consider making a7 and beyond zero or even negative to penalize cases (we know the probabilities are always nonnegative) that result in large discrepancies.

Let?s consider the average scoring distribution for the NFL over the last 10 years, 2004 through 2013 YTD.

The six highest frequencies are: 17 (7.2%); 20 (6.8%); 24 (6.5%); 27 (5.6%); 10 (5.4%); 13 (5.3%) ... the other values can be obtained from any database of NFL scores. The max score, which we used above, turns out to be 62.

With both teams having the ?standard? DPD we can compute an ?average? Pexact which turns out to be 4.01%. Poff_by_1 turns out to be 9.35%, Poff_by_2 = 6.72% and Poff_by_3 = 13.75%.

Incidentally, if we use only 5 years of data, (2008 ? 2013), these percentages are almost the same, Pexact = 4.02%, Poff_by_1 = 9.40%, Poff_by_2 = 6.60% and Poff_by_3 = 13.70%. However, scoring is up in the last 5 years, and the six highest frequencies are: 17 (7.3%); 24 (6.8%); 20 (6.8%); 27 (5.9%); 31 (5.1%); 13 (5.0%).

Now we need to estimate the number of contestants to see how often someone will hit the exact TotA = TotB and how many times we will go to Off By 1, By 2, etc?

Assume 40 contestants is approximately correct for the contest being run. Using the binomial distribution, we find that the exact score match will be hit by at least one contestant 21.45% of time. ?Off by 1? hits 18.79% if there is no exact match. Off by 2 hits 16.77% of the time if the first two cases aren?t hit, etc? We expect the winner?s difference will be two points (or less) more than half of the time.

We want to choose the terms a1, a2, a3, etc to produce smaller and smaller contributions to the payoff function since the lowest the difference, the more likely it is to win the contest, thus let?s choose a1 = 0.3, a2 =0.2, a3 = 0.05, and a4 = 0, etc, which leads to Faverage = .0401 + 0.3*.0935 + 0.2*.0672 + .05*.1375 = .0885

This is the benchmark value for our payoff function F. F for 5-year data (2008-2013) is the same to 3 significant digits.

If we choose, we can incorporate team total derived by using the line and total for the specific game (using Week 14 of 2013).

The following table is sorted from low to high:

Team Mean Score

Browns,Rams,Vikings 17.5

Colts 18.5

Jets 19.0

Dolphins,Seahawks,Titans 19.3

Bills 19.5

Jaguars,Redskins,Panthers,Raiders 21.5

49ers,Falcons 21.8

Giants,Steelers,Texans,Buccs 23.0

Chiefs 23.8

Cards 24.0

Saints 24.3

Bears,Bengals,Cowboys,Ravens 24.5

Chargers,Packers,Lions 25.8

Eagles 28.3

Patriots 28.5

Broncos 31.3

We can use these team totals and in addition, an NFL scoring model if we have one, to produce all 32 distributions (there are over 900 combinations of two teams). Perform the calculation of the payoff function F with Eqn (1) for each pair of teams, which is straightforward programming or can be done with a spreadsheet. As expected, teams with similar team totals have larger values of F. Teams at the extremes tended to have larger variances and smaller values of F.

There were many good combinations, it turns out that the Cards/Saints pair gave a value of F very near the maximum, at .0928. This selection doesn?t guarantee a significant edge over all other contestants since that depends on their choices as well. It does have about a 5% better payoff function than the ?average? 2-team distribution.

**Results:**The actual contest ended up with 39 contestants. The result of Week 14 was that the Cards scored 30 and the Saints 31, two ?mid-value? scores (31 is 7[SUP]th [/SUP](4.9%) and 30 is 15[SUP]th[/SUP] (2.6%), both scores went over their projected team total and happened to be almost identical, tying for first place in the contest with 6 others, a higher number of very close guesses than expected (should have been about 5.2 contestants that either had identical or ?Off by 1? results using the ?average? scoring distributions).

**Conclusion**:The methodology and results appear to give a fairly significant advantage over blindly picking two teams. It does not give a large advantage, but when this method is combined with a good scoring model, it does give good sets of candidate team pairs. It provides an example using a payoff function which is widely used in more complicated optimal control theory and game theoretic problems.

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