Value of a Yard?

[Chaperone's Runner]

One of the real difficult things I encounter while betting SB props is with O/U bets. How do I determine the value of an extra yard or sack etc. The example I am looking at is the total number of yards in the game. My number is 23 yards off of the line of 666.5. How much of a value is that?
I'm certain that the initial number is betable, I am more looking at the second or third pop on my bet after the line adjustments. As you know one pop is never enough:)

Thank ya

[Prop Man]

This isn't exact, but should be close (I think), using all games in the database I have

Over 600 net yards: 62.2%
Over 601 net yards: 61.8%
Over 602 net yards: 61.3%
Over 603 net yards: 61.1%
Over 604 net yards: 60.8%
Over 605 net yards: 60.5%
Over 623 net yards: 54.6%

So roughly 0.35% per yard

Hope that helps.

[custer]

This isn't exact, but should be close (I think), using all games in the database I have

Over 600 net yards: 62.2%
Over 601 net yards: 61.8%
Over 602 net yards: 61.3%
Over 603 net yards: 61.1%
Over 604 net yards: 60.8%
Over 605 net yards: 60.5%
Over 623 net yards: 54.6%

So roughly 0.35% per yard

Hope that helps.

The problem with looking at it this way IMO is that two important factors in a push rate on something where every point is valued the same are the range and the standard deviation. The standard deviation and range for game yards is going to be different than for a running back's yards, for instance. Take a RB like Hightower. There's almost no way he gets more than 100 yards, or less than zero. So each of his yards have to be at least 1%.

Maybe someone who knows more about this can pick it up from here.

[Prop Man]

The problem with looking at it this way IMO is that two important factors in a push rate on something where every point is valued the same are the range and the standard deviation. The standard deviation and range for game yards is going to be different than for a running back's yards, for instance. Take a RB like Hightower. There's almost no way he gets more than 100 yards, or less than zero. So each of his yards have to be at least 1%.

Maybe someone who knows more about this can pick it up from here.

You're right. Sorry, I didn't think anyone would confuse the value of a yard in the net yards of a full game to the value of a yard for a RB that is expected to gain 75 yards. There is a difference, you're right. My post was only an approximation on full-game net-yardage situations for both teams.

[custer]

I didn't read the OP well enough. Don't know why I ever thought you'd give a non great answer anyways.

[Prop Man]

I'll try to answer the question in a general way that would apply for every yardage situation. The user will have to know all the variables for the answers to make sense, as usual, garbage in means garbage out, so the key isn't the formula, but rather the inputs.

Use the NORMDIST function on Excel.

You need the following information (stolen from a teaching site found on google, not my original work on most of this stuff)

NORMDIST gives the probability that a number falls at or below a given value of a normal distribution.
x - The value you want to test
mean - the average value of the distribution
standard_dev - the standard deviation
cumulative -- If FALSE or zero, returns the probability that x will occur; if TRUE or non-zero, returns the probability that the value will be less than or equal to x.

So let's use the following numbers

x = Under 660.5 yards (this is the line, you'll have to adjust for the vig later)
mean = 640 yards (this is what you think the mean is for the particular game)
standard_dev = 122 yards
cumulative = TRUE

Plugging it into Excel, the answer is .5667. So this means 56.7% of the time, with these assumptions, Under 660 is the winner.

To get the value of a yard, we just need to change X from 660.5 to 659.5. If we do that, we get .5635.

So the value of a yard with these assumptions is .0032.

Now let's look at rushing yards for the home team in my database:

x = 115.5 yards
mean = 117.2 yards
sd = 51.5 (this looks really high, remember this is for all games for the home team. the standard deviation for any particular team is gotta be significantly lower than this)

using the normdist function:
Under 115.5 = .4868
Under 116.5 = .4945

So the value of a rushing yard for a team is roughly .0077. As mentioned above, I do think the sd is too high if we know the team that is playing. Just looking through some individual teams, 45, but that's stab in the dark.

using 45 as the standard deviation for rushing yards for a particular team, I get:

Under 115.5 = .4849
Under 116.5 = .4938

so the value of a yard has gone up to 0.0088


We can also apply this to individual player yardage, but the great unknown is the stdev. Here's an example with Edge:

x = 42.5 rushing yards
mean = 41
stdev = 20 (stab in the dark)

Under 42.5 = .5299
Under 43.5 = .5497
value of a yard = .0198

[Prop Man]

Applying it to QB passing yardage, I used a stdev of 75 yards and a mean of 210 yards. with those numbers, the value of a yard is 0.53%, very close to what mikevegas posted.

One other thing I forgot to mention, the further away you get from the mean, the lesser value of a yard. With the normdist function in Excel, it is easy to look for the difference in value of a range of yards.

[Prop Man]

Adding: this excel function is useful if you think the distribution is close to a normal distribution. but if it isn't, then I'd be weary of it. stuff like Arrington total rushing yards is probably not a good fit because there is a good chance he has 0, and his expected rushing yards is probably not normally distributed.

[The Chaperone]

Arrington not gonna get the ball much imo. Brings back fond memories of my Heath Evans UN 2.5 yards from last year :-)