How ridiculous is the whole undefeated talk at this point?

[Zeus]

Talk of an undefeated season when the team is barely halfway through the schedule is absurd. And as long as they win enough to secure home field throughout the playoffs, it really doesn't matter all that much to me whether they are 14-2 or 16-0.

What is really important is they continue to improve as the season goes on. The great Belichick teams have played their best football in December and January (also undefeated in February!). As well as this team has played so far, I believe they can - and will - play even better. We haven't seen their best game yet.

 

[Absurdly Metro]

How many points do you ascribe to the officiating in the RCA Dome?

I think it's being generous to say the Colts got 1 TD worth out of the officiating. It was definitely to the point where we threw the ball, and a flag went up. Maybe I'm wrong.

If you call it 7 points for the refs in that game, that "makes us" 11 points better than the closest team we've faced. PFnV

Great point PFnV! You're not wrong about those flags. And I'm not feeling as generous as you are this morning either. So off the top of my head I count 13 points for Indy directly attributed to the officials. They are:

1. 1st Q: Ticky tack 37yd PI call on Samuel = 3 points (would have been 7 if Asante doesn't make up for the penalty with a strip of Gonzalez in the EZ to force the FG)
2. 2nd Q: Absolutely bogus 40 yd PI on Hobbs = 3 points
3. 2nd Q: No call on block in the back on Addai's 73 yd. "dump off" = 7 points

This is not to mention the other penalties (Moss offensive PI in EZ) that aided and abetted the Colts in the attempt to pilfer this game away from NE.

An evenly called game sees the Pats win by at least 10. And this is on the road, with no coach to QB communication, questionable decibal levels, one-sided officiating, vs. the undefeated defending sb champs who are clearly the 2nd best team the league has to offer. If the Pats are 10+ better than all THAT I don't know how you can't like our chances for a run at history.

 

[Rainy-Day Patriot]

I believe it does.

If you assume average win rate is x% (ie 90%)

then the total win rate for two games is (1-x)*(1-y), where x+y = 20%

So (1-x)*(1-.2+x) = win rate = x-.8x-x^2

taking the derivative to find min/maxes
d (win rate)/ dx = .2x -2x = 0
x = .1

So the maximum "total win %" occurs at x=.1

There is probably a more intuitive way to see this too.

I love this explanation. Now, I'm trying to understand it. (I'm taking calculus for the first time at the age of 32, so please bear with me, while this old dog is trying to learning new tricks.)

f(x) = (1-x)*(1-(.2-x)) ?==> Doesn't y =.2-x
= (1-x)*(-.2+x)
= (-.2+x+.2x-x^2)
= (-.2 + 1.2x -x^2)
f'(x) = 1.2 -2x
The root of the derivative is x=.6 which means that f(.6) is a local max.

How do you interpret the slope of the tangent lines to f(x)? Is it the instantaneous rate of change of the total probability of winning both Game A and Game B as the probability for Game A changes, assuming the average probability for Game A and Game B is 80%? In English?

Sean

 

[Rainy-Day Patriot]

I love this explanation. Now, I'm trying to understand it. (I'm taking calculus for the first time at the age of 32, so please bear with me, while this old dog is trying to learning new tricks.)

f(x) = (1-x)*(1-(.2-x)) ?==> Doesn't y =.2-x

Argh! Stupid calculation error!
Of course:
f(x) = (1-x)*(1-(.2-x)) ?==> Doesn't y =.2-x
f(x) = (1-x)*(1-.2+x)
f(x) = 1-.2+x-x+.2x-x^2
f(x) = .8+.2x-x^2

f'(x) = .2-2x

That's right, now, isn't it?

 

[Absurdly Metro]

Sheesh. I don't think we have a 93% chance of winning tonight's game!
Divisionality is a great equalizer in the NFL. I'm not sure if it's the familiarity or the emotions, but when division rivals play, the probability of winning is pulled toward 50% for the proverbially better team on paper.

In other words, I'm looking forward to the game. I'm not planning on humble pie for breakfast, but I'm not counting it out either.

Simple question? Are you looking at this team as a team of the ages? If not then you should be afraid--be very afraid--of messrs JP Losman and Co. tonight. But for those of us drinking the koolaid and buying into the notion that what we're seeing is a run by (what will one day be looked back on as) the greatest team of all time--then it's party time brethren!

Does it make you like our chances any better to realize that the Pats have won 8 straight meetings vs. Buffalo dating back to the "they hate their coach" game in Sept of '03? Eight straight! What were the chances of THAT happening? They have done this with far less talent and drive on the field than this year's edition has demonstrated. Yes, I know, "on any given Sunday" and yada, yada, yada..." But I just can't find it in me to worry about tonight's game. Barring fluke or injury this Pats team walks all over the Bills. If they truly have a date with destiny it can't happen any other way.

 

[Rainy-Day Patriot]

taking the derivative to find min/maxes
d (win rate)/ dx = .2x -2x = 0
x = .1

I see now. The typo threw me off track, but it was obviously just a typo since your root is right.

When we do the averaging method, the max total win percentage occurs when all the probabilities are exactly equally. As they diverge, which they surely do, our estimate becomes an overestimate.
Awesome! Thank you.

 

[Rainy-Day Patriot]

Yes, I know, "on any given Sunday" and yada, yada, yada..." But I just can't find it in me to worry about tonight's game. Barring fluke or injury this Pats team walks all over the Bills. If they truly have a date with destiny it can't happen any other way.

I just think that, especially for divisional games, the any-given-Sunday factor is greater than 7%. It's just a gut feeling. Knowing that there's a 93% chance of rain means that I'm 100% sure that I should bring an umbrella!

"Barring fluke or injury this Pats team walks all over the Bills."
I could not agree more. All I am saying is that there is a +7% chance (in my subjective opinion) of a fluke or injury.

 

[Absurdly Metro]

I just think that, especially for divisional games, the any-given-Sunday factor is greater than 7%. It's just a gut feeling. Knowing that there's a 93% chance of rain means that I'm 100% sure that I should bring an umbrella!

"Barring fluke or injury this Pats team walks all over the Bills."
I could not agree more. All I am saying is that there is a +7% chance (in my subjective opinion) of a fluke or injury.

And all I'm saying is drink more koolaid! Let go of your doubts and believe in this teams destiny; the greatest single season in the history of the NFL. Undisputed, undefeated SB champions. This will mean being willing to abandon your ironclad logic (as above) for what some will deem folly and even madness.

Of course, along with this believe at all costs mentality comes the inherent risk of looking stupid, eating crow, and feeling like you've been gutted through the nether regions with a dull blade, should the impossible happen and the dream season come up short. All of this is but a small price to pay for the opportunity to completely believe, without reservation, that this is THE team that shall conquer and vanquish every obstacle in it's path during its relentless assault on perfection. And I do mean EVERY obstacle... every opponent, every bad call, every scandal, every injury, every fluke--nothing in the whole of the heavens or on the earth can stand between us and history.

So repeat after me... There is no chance of anything but victory. Embrace it my brother and welcome to the lunatic frindge. More koolaids all around!

:eat2:

 

[JoeSixPat]

I just heard a stat on NFL Network that 23 teams in NFL history have been undefeated at this point in the season. If the Pats win tonight, they will be the 19th team to be undefeated after 10 games.

Yes - and and I'm sure most of those 19 teams would routinely beat their opponents by an average score of 24 points - futher illustrating that the Patriots are really nothing special.

C'mon - we all know that there's plenty of teams to have started 10-0...

And we also know that the NFL hasn't seen this sort of dominant team in, well, maybe ever...

And when one looks at the remaining schedule and recognizes that this team is beating opponents by more than 3 touchdowns on average, one can't help but understand the odds of the team remaining undefeated is better than the chance its going to lose.

In other words - which game do YOU think the Patriots are going to be underdogs in? Do you think the Las Vegas oddsmakers are going to have them as underdogs in any of the remainig games?

And if you don't think they'll be underdogs in any games, why do you think its so rediculous to be considering the possibility of a perfect season?

Talking about things like this is what fans and media do - don't try to stop it or chide people because of it. Is it rediculous to be talking about the Dolphins going 0-16? Of course not. Come down off your high horse and join us - it's fun!

 

[Absurdly Metro]

Yes - and and I'm sure most of those 19 teams would routinely beat their opponents by an average score of 24 points - futher illustrating that the Patriots are really nothing special.

C'mon - we all know that there's plenty of teams to have started 10-0...

And we also know that the NFL hasn't seen this sort of dominant team in, well, maybe ever...

Talking about things like this is what fans and media do - don't try to stop it or chide people because of it. Is it rediculous to be talking about the Dolphins going 0-16? Of course not. Come down off your high horse and join us - it's fun!

Amen brother Joe!! Testify!!!

 

[Rainy-Day Patriot]

So repeat after me... There is no chance of anything but victory. Embrace it my brother and welcome to the lunatic frindge. More koolaids all around!

:eat2:

There is no chance of anything but victory.

 

[mcbee]

I love this explanation. Now, I'm trying to understand it. (I'm taking calculus for the first time at the age of 32, so please bear with me, while this old dog is trying to learning new tricks.)

f(x) = (1-x)*(1-(.2-x)) ?==> Doesn't y =.2-x
= (1-x)*(-.2+x)
= (-.2+x+.2x-x^2)
= (-.2 + 1.2x -x^2)
f'(x) = 1.2 -2x
The root of the derivative is x=.6 which means that f(.6) is a local max.

How do you interpret the slope of the tangent lines to f(x)? Is it the instantaneous rate of change of the total probability of winning both Game A and Game B as the probability for Game A changes, assuming the average probability for Game A and Game B is 80%? In English?

Sean

Yes, good catch on the typo.

I think in yours that (1-(.2-x)) you reduced to -.2+x instead of .8+x.

Yes, it would be the rate of change of winning both given that you're changing one (x) (and putting the other loss probability in y, since it's fixed).

It's related to something simpler, which I forgot.

Suppose you have a number Z and x+y=Z. Which two numbers x and y will give you the largest product, xy?

y=Z-x
x *(Z-x) = prod
d(prod)/dx = Z-2x

d(prod)/dx = 0 at min,max
Z-2x = 0
x=Z/2

So the largest product is when both numbers equal each other.

IE if Z = 10 then x=y=5 will give you a bigger product than 8 and 2 for example....or 5.5 and 4.5.