## Poll

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**25 members have voted**

Quote:SOOPOOI disagree. Generally, in the NFL, home field is worth 3 points compared to a neutral field, not 1.5 points.

What is your evidence?

From 1994 to 2012 the average away score was 19.92 and average home score was 22.67. That is a difference of 2.75. This is comparing a team playing on their home field to a team playing on an enemy field. If your theory were true we could expect to see an average six-point difference between home and away team.

If you have 2 evenly matched teams, the home team is installed as a 3 pt favorite (not 1.5 pts). You see this each year with evenly matched division rivals. Team A is favored by 3 at home, then will be a 3 pt dog when playing on the road against the same opponent.

This is common knowledge, and shouldn't be disputed. Just ask any linesmaker.

Quote:Wizard

From 1994 to 2012 the average away score was 19.92 and average home score was 22.67. That is a difference of 2.75. This is comparing a team playing on their home field to a team playing on an enemy field. If your theory were true we could expect to see an average six-point difference between home and away team.

I don't doubt your numbers, but aren't we talking about setting a pointspread? If you routinely set the home team as a 1.5 point favorite against a evenly matched opponent, you will be flooded with money on the home team. And you will be out of business rather quickly because they will cover that number enough to overcome 11/10.

Quote:KeyserSozeThe typical difference from NFL home field and NFL road field is 6 pts.

If you have 2 evenly matched teams, the home team is installed as a 3 pt favorite (not 1.5 pts). You see this each year with evenly matched division rivals. Team A is favored by 3 at home, then will be a 3 pt dog when playing on the road against the same opponent.

This is common knowledge, and shouldn't be disputed. Just ask any linesmaker.

I don't dispute that. What I'm saying is if you calculate the expected number of points each team will score, you add 1.5 points to the team playing on its own field, and subtract 1.5 points from the team playing on an enemy field. The difference is 3 points. Thus your 3-point spread.

Quote:KeyserSozeI don't doubt your numbers, but aren't we talking about setting a pointspread? If you routinely set the home team as a 1.5 point favorite against a evenly matched opponent, you will be flooded with money on the home team. And you will be out of business rather quickly because they will cover that number enough to overcome 11/10.

No, I'm talking about estimating the number of points scored by each individual team. If the teams were evenly matched on a neutral field I would of course make the spread 3 on the underdog, because 2*1.5=3.

Let's look at an example. On a neutral field team 1 and team 2 can each expect to score x points. Then they decide to move the game to team 1's field. Now expected points scored by each team are:

Team 1: x+1.5

Team 2: x-1.5

Difference = point spread = (x+1.5) - (x-1.5) = 3.

My general formula for the probability of an x-point underdog winning is e

^{-0.13176*x}/(1+e

^{-0.13176*x}).

Plugging in x=26.5, we get a probability of the Jaguars winning of 2.96%, which equates to a fair money line of about 33 to 1.

Granted there are zero games with a point spread this big. There are not that many anywhere in the twenties. So this is definitely the kind of bet where there isn't a lot to go on.

Personally, I'll only bet if I can get 40 to 1 or more.

Quote:WizardI see Pinnacle put the spread at 26.5.

My general formula for the probability of an x-point underdog winning is e^{-0.13176*x}/(1+e^{-0.13176*x}).

Plugging in x=26.5, we get a probability of the Jaguars winning of 2.96%, which equates to a fair money line of about 33 to 1.

Granted there are zero games with a point spread this big. There are not that many anywhere in the twenties. So this is definitely the kind of bet where there isn't a lot to go on.

Personally, I'll only bet if I can get 40 to 1 or more.

If nothing out of the ordinary occurs in this game, then I'd argue there's a 0% chance the jaguars win. Anyone who places a bet on the moneyline is basically betting on the chance that Peyton

Manning can be injured and knocked out of the game on theoretically any play. Or, Jax will score an extraordinary number of points on plays where the two teams talent levels are more

even with each other, special teams. So by taking the moneyline at 33-1 you're saying that the odds of one of those events occurring is such that 33-1 payout offers value

As far as I know, The only sporting event in my lifetime where a > 33-1 underdog won outright is Mike Tyson vs Buster Douglas (44-1). And I'd argue that boxing is a bad comparison, because in boxing all it takes is one human being to be unprepared that night (which is what happened) or one instance where he fails to defend himself properly, and it can be over. On Sunday, one, two , or several broncos players can have bad days, and they can still win this game rather easily.

Personally, If I'm looking for a 35-1 payout, I'd rather place my money on a roullette number. I feel that has a much better chance of winning then the jags do

EA SPORTS Madden 25 prediction

"41-10 triumph for the Broncos, with Manning set to go 32-of-39 for 392 yards and four touchdowns. The game simulation has Denver opening up a 38-3 lead, before eventually turning off the gas midway through the third quarter and kindly allowing Jacksonville a consolation touchdown – a 32-yard pass from Chad Henne to Justin Blackmon – while still covering the spread"

Heres the link to whole article

http://sports.yahoo.com/blogs/nfl-shutdown-corner/denver-cover-28-point-spread-against-jacksonville-madden-232840280--nfl.html