Blog

Let’s everyone take a break from draining the swamp and talk SuperBowl prognostications.

I’m laying mine out here, right now. And further down – for those of you interested, I explain the model I used to obtain the forecast.

Some Preliminaries

In the static calculation the (14-2) Patriots have a 66 percent chance of winning over the (11-5) Falcons; 2 to 1 odds. This win probability is based on Bill James’s Pythagorean Expectation or Log5 Formula.

There is apopular adjustment to the Log5 calculation, however. This alternative adds eleven games of 0.500 ball to the team’s current record to obtain their “true” win percentage (adjusting for variation).

Recalculating with the adjustment gives us the revised “true” win likelihoods.

For the Patriots, the true Wpctg: (14+5.5)/(16+11) = 72.2 percent.

For the Falcons, the true Wpctg: (11+5.5)/(16+11) = 61.1 percent.

And going back to the Log5 formula armed with these “true” Win Percentages – the Patriot’s chances of prevailing in the Super Bowl are now at 62.1 percent; odds to win of 3.1.

This adjusted, “true” win percentage appears to underscore posted results at OddsShark and at FiveThirtyEight. Both sites have the Patriots at 3 to 1 odds and chances of winning of 59 percent (OddsShark) and 61 percent (FiveThirtyEight).

A Simulation

The problem with this static calculation is well ... that it is static; one draw from many possible outcomes for the 2016 season for each of these two teams. One draw.

If we replay the seasons 1000 times- we can come up with a fairly good distribution of win percentage for each team and recalculate the Log5 1000 times. And from the resulting distribution of Win Probabilities we can derive confidence intervals – and have a better understanding of the range of outcomes – and a better estimate of the Pats win chances.

For this I use the football variant of the Pythagorean Expectation: win percentage in terms of points scored and points scored-against (with an exponent of 2.37). Again, I re-run the season for each team 1000 times and via resampling calculate the distribution of possible outcomes for the Patriots. By the way, all data is from The Football Database.

The resulting distribution looks like this:

The expected value (mean) of this distribution is 65 percent with very tight 95% confidence intervals.

So that’s my number. Patriots have a 65 percent chance of winning.

What is the score and the spread?

Score and Spread

I could use the same simulation methodology to calculate the expected points scored by each team. But I like a different model for this. And I like it because it is simple and straightforward – and easy to explain and understand intuitively. (Check out Nate Silver’s methodology – it’s brutal. And who knows what the hell OddShark is using.)

The model is essentially two simple adjustments to the average number of points scored in the league in the 2016 season. That number is 22.8 points per game. There is no need for home-stadium adjustment cause it’s a neutral site.

The first adjustment is to account for offensive prowess: the amount each team outperformed the league average. That is 21 percent for the Patriots and 48 percent for the Falcons. The second adjustment is to account for the defensive weakness of each team. This is the number of points (in percent terms) that were scored against each team relative to the league average. This is -21 percent for the Patriots and + 11 percent for the Falcons.

In other words: the Patriots had 21 percent less points scored against them than the league average; the Falcons had 11 points more than the league average scored against them.

The formula is: Expected Number of Points =

                           (League Average Points per Game)x(Offense Adj)x(Defense Adj)

Here is the tally:

Expected Points, Patriots: (22.775) x (1.21) x (1.11)  = 30.7

Expected Points, Falcons:  (22.775) x (1.48) x (0.69) = 23.2

                                                                               Spread 7.6

The expected number of points and my prediction:

Patriots 31 Falcons 23. The spread is 8.

Here is a graph of the distribution of the possible score.

I use a Poisson distribution using the results in the table above as the expected values even though both Wayne Winston in his book Mathletics and the classic paper by Hal Stern suggest that the distribution is normal.

And the distribution of the Spread – again via a Poisson using 7.6 as the mean.

peace

arod

arodriguez@newhaven.edu

Email me when people comment –

You need to be a member of UNH Economics Collective to add comments!

Join UNH Economics Collective