how impossible is this? for math majors

[neuronet]

Here's a question I'd be interested in the answer to. Let's assume a prior win proability of .92, which I think results in about a 20 percent chance of 19-0. What is the likelihood that team will finish 18-1?

A few teams have finished 18-1. I'm trying to get a mathematical idea of how much more rare it should be for a team to finish 19-0.

Replace one of the P(WIN) for an individual game with 1-P(WIN), so 18 times P(WIN) times 1-P(WIN). But there are 19 places in the schedule they could have lost, so it becomes more complicated as you have to start to do combinatorics.

There are 19 ways to lose one game, so you have to add the probabilities of each of those seasons (e.g., loss in first game OR loss in second game, etc). So 19*P(loss in game 1, win in remaining games) . You can add them because the outcomes are independent (like the probability of rolling a six or a one: you can add the individual probabilities because there is no overlap in the sample space).

Anyway, doing the calculations, the probability of 1 loss, assuming .92 prob of win per game, is:
0.34.

Probably of winning all 19, assuming same prior is 0.205.

I think that's it anyway. Someone let me know if there is a mistake.

It would be cool to make a 3-d plot: probability of win on x axis, number of games lost on y axis, and z axis is plot of the probability of that combination. It would look sweet. It would take me an hour to put it together though...so not unless I really need to procrastinate... It will get even more complicated combinatorically when you have more than one loss (there are lots of ways to have two losses in a season: first and second game, first and third, second and third, etc etc). It would become a mess.

 

[Urgent]

The critical assumption, obviously, is the chance of success.

Since each year there are often 13-3 and 14-2 teams, then there are commonly teams with a greater than 75% chance of winning each game.

If you take a team that 'should be' a 14-2 team, and check the odds of them going 16-0, those odds drop. Say each season, the best team goes 14-2. Each year, the best team, whichever it is, has an 87.5% chance of winning each game.

Using those odds, each year the best team in the league has a 12% chance of going 16-0. Or, you would expect the best team in the league to go undefeated about every eight years.

At this point, it hasn't happened for 35 years, so maybe the odds are a little worse.

But it should happen again.

 

[neuronet]

Using those odds, each year the best team in the league has a 12% chance of going 16-0. Or, you would expect the best team in the league to go undefeated about every eight years.

I think one weakness in the simple probability models we've been using is the assumption that each game is independent. But the probability of winning, given that you have won, say 15, is probably lower than the probability of winning (for the exact same team), if they have won 13.

Playoff concerns, pressure on the players in the undefeated team, etc., serve to counteract perfection. This makes the Pats' performance this year all the more impressive. They could have sat on their butts in New York week 16, but they dug deep and pulled the win quite impressively.

The Pats ability to maintain poise and fight for 60 minutes is incredible.

 

[sdaniels7114]

How do all these calculations account for Baltimore and Philly being super fired up just to beat the Patriots? The Ravens lost to Miami the week after they played us. It may have been the same bunch of players; but it wasn't the same team. Which of those two Baltimores went into the calculations?

The Pats were obviously capable of not talking about 19-0 until the time was right. Perhaps not even among themselves; I just can't believe none of them thought of it. How did that affect things?

Seems like these sorts of calculations are best left to inhuman things like coin flips.

 

[sbeausol]

exactly why i bit the bullet and i am making the trip to AZ!

 

[godef]

yes, that is right... if you want, you can divide it by the number of teams in the league (then it will be ANY team doing it) and still get a number of years that will probably have some continents merged... you will never see this again!

Actually, you would MULTIPLY it by the number of teams, because you're allowing for the possibility of any 32 teams to do it, not just any one specific team. You want to end up with a higher odds.

 

[NEPatriot]

anything is possible if you have the right combination. When you have somebody like BB and TB and RM, your odd of achieving your objective increases tremendously.

 

[godef]

the .5 is assuming all things being equal, which they aren't. That would be the odds of a team (8-8) with a 50% chance of winning each game.

No, he had it right, he said any team (presuming he literally meant any specific team). Since there is a loss for every win, the average record among all the teams in a league will be 8-8 (presuming no ties). So the odds of any one team going undefeated is 0.5**19

 

[pheenix11]

I think you guys are putting too much into expectations. This should be calculated for any team and the probability of any team winning a game is 1 in 2.

I don't think the actual probability of the Pats winning a game can be calculated with any certainty. There are too many variables.

 

[godef]

These numbers are complete hogwash when applied to real performance. Sure, you can say "this team was 12-4 and therefore had a 75% winning percentage and therefore a .75 chance to win every game" but it's not like that. There are many more factors, none of which can be fully realized before a game. If it were actually possible to predict the % chance of things happening in football, they would be.

Take this counter-example: The Patriots finished the season 16-0 with a 1.000 winning percentage. By your logic, they had a 100% chance of winning every game. The winning percentage is correct looking backwards, but a poor predictor.

Actually, you'd have iy nailed if you say simply: if a team is expected to go 16-0, then their odds to go 16-0 are 1.000. In other words, this extreme example with its obvious results shows the meaningless of such reasoning.

 

[pheenix11]

We need that guy from Numbers, he could nail this LOL

 

[billikens]

Of course random theory dictates that, no matter the odds, the next time of occurrence can be next year. I think we'll probably see this again in our lifetimes - maybe more than once.

 

[PatsWorldChamps]

Of course random theory dictates that, no matter the odds, the next time of occurrence can be next year. I think we'll probably see this again in our lifetimes - maybe more than once.

lol... yes, it could be next year... but the odds would be over 200:1

 

[PatsWorldChamps]

Actually, you would MULTIPLY it by the number of teams, because you're allowing for the possibility of any 32 teams to do it, not just any one specific team. You want to end up with a higher odds.

yeah, you multiply the odds of it happening, but you divide the number of years it would take... that is what i meant.

 

[PatsWorldChamps]

besides all the rest of the nonsense, I would like to know why I keep seeing the percentages of 12-4, 14-2, etc teams extrapolated out to 19-0.
as a math major I can tell you that 12+4, and 14+2 = 16 games.

it's just a little goofy to peg a team at 12-4 and then ask the question of how often they're 19-0.
12-4 teams are never 19-0.

lol. you are a math major? that's funny... the point is, if you took, for example, the 15-1 vikes team, and let that team play 100 seasons, would they go perfect once? no, you are right... the same team cant play 100 straight years. great analysis.

if a 12-4 team had the same players, at the same age, with the same coach, they would rarely go 12-4, even though you seem to think they'd always go 12-4.

 

[PatsWorldChamps]

A few teams have finished 18-1. I'm trying to get a mathematical idea of how much more rare it should be for a team to finish 19-0.

unless im mistaken, arent the 85 bears the only team to go 18-1 ever?

 

[PatsWorldChamps]

It isn't silly at all. The prior probability of winning a game is not one, and that is the relevant quantity. Of course the posterior probability of winning, giving that you won, is 1. But that's not the right value to use.

The product of the 19 individual probabilities seems the right calculation (though it assumes independence, but whatever).

The probability of going 19-0 is less than 0.5 when the prior probability of an individual win is less than .96. They are amazing.

I attached a plot of the probability of winning 19 games as a function of the prior probability for winning an individual game:

neuro... just curious... are you an actuary?

 

[Keegs]

the probable ability of winning 19 games is related to spectative quantification of the numerator.

That is directly related to the pontification of the denomination divided by the square root of the cosign.

Then you take the remainder and multiply it by the 4th prime number and subtract that from the average law of theoretic probably and pi. 3.1414689074

The answer to this theorem is 1.21 gigawats.

 

[PatsWorldChamps]

the probable ability of winning 19 games is related to spectative quantification of the numerator.

That is directly related to the pontification of the denomination divided by the square root of the cosign.

Then you take the remainder and multiply it by the 4th prime number and subtract that from the average law of theoretic probably and pi. 3.1414689074

The answer to this theorem is 1.21 gigawats.

lol... actually, keegs, a more accurate interpretation of pi would be 3.14159265... go play some games; kevin misses you!

 

[psychoPat]

Patriots' ENORMOUS "prior probability"

It isn't silly at all. The prior probability of winning a game is not one, and that is the relevant quantity. Of course the posterior probability of winning, giving that you won, is 1. But that's not the right value to use.

The product of the 19 individual probabilities seems the right calculation (though it assumes independence, but whatever).

The probability of going 19-0 is less than 0.5 when the prior probability of an individual win is less than .96. They are amazing.

I attached a plot of the probability of winning 19 games as a function of the prior probability for winning an individual game:

Several weeks ago i began rather seriously trying to answer this question for myself:
"How good does a team have to be
to have a 50% probability of winning 19 straight?"

Although i made several false starts, i couldn't even
frame the question in a way that mathematics could address.

Now i see that "prior probability" is the key.
Thanks for showing that
... and for showing how truly exceptional and "against all odds"
this team is !