how impossible is this? for math majors

[PatsWorldChamps]

a team that is EXPECTED to win 75% of its games; a very, rare team (12 and 4 teams are super rare)
would have to stay together and play for 236.5 years to go 19-0.
here is the math...
1/(.75^19)

if you think you will be alive the next time, so did sam adams and ben franklin.

a team that is EXPECTED to win 50% of its games...

would have to stay together and play for 524 thousand years to go 19-0... yeah fred flinstone and barney had season tix to that one... before the scalpers screwed up the ticket prices!

 

[PonyExpress]

In NFL history, roughly 2000 teams have had the opportunity to go undefeated and untied. Only 1 has done it, the 1972 Dolphins, pending the outcome of this SB.

 

[pheenix11]

I think your math is right. The odds of winning 19 games straight for any team is:

.5^19

(1/2 *1/2 *1/2 etc)

or

.000001907

or

1 in 524,288

Is that right?

 

[PatsWorldChamps]

In NFL history, roughly 2000 teams have had the opportunity to go undefeated and untied. Only 1 has done it, the 1972 Dolphins, pending the outcome of this SB.

not at 19 and 0

 

[PatsWorldChamps]

I think your math is right. The odds of winning 19 games straight for any team is:

.5^19

(1/2 *1/2 *1/2 etc)

or

.000001907

or

1 in 524,288

Is that right?

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!

 

[Wildo7]

I think your math is right. The odds of winning 19 games straight for any team is:

.5^19

(1/2 *1/2 *1/2 etc)

or

.000001907

or

1 in 524,288

Is that right?

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.

 

[PatsWorldChamps]

In NFL history, roughly 2000 teams have had the opportunity to go undefeated and untied. Only 1 has done it, the 1972 Dolphins, pending the outcome of this SB.

the 72 dolphins never sniffed the neighborhood of 19-0... they will admit that... it's a new development, and if any dolphins want to buy in, they can ring our doorbells!

 

[PatsWorldChamps]

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.

what would you like? a perennial 12-4 team? (how many of those were there ever?)... anyways, i'll go with your idea... a .750 team (every year, for 236 years) would be expected to go 19-0 ONCE in 236 years!

 

[Marko]

Exactly, it's not like flipping a coin. Theoretically, that would be true if you were talking about beating the spread(which really doesn't hold either); there's no way the Pats only had a 50% chance of beating the fins and the Jets etc.

 

[Wildo7]

what would you like? a perennial 12-4 team? (how many of those were there ever?)... anyways, i'll go with your idea... a .750 team (every year, for 236 years) would be expected to go 19-0 ONCE in 236 years!

What? I like the OP's original analysis if you are focusing on only the projected record and then rolling the dice. I was just responding to a post that described the odds somewhat inaccurately.

Anyway, I think that most knowledgeable estimates had the Pats at around 14-2 this year, which would improve the odds somewhat. It's great and all to do this mathematically, but there's too many factors involved to look at this in a mathematical vacuum. It's special not because of luck but because of an accumulation of mostly controllable factors.

 

[Hardboiled]

I think your math is right. The odds of winning 19 games straight for any team is:

.5^19

(1/2 *1/2 *1/2 etc)

or

.000001907

or

1 in 524,288

Is that right?

Looks good asuming the 50%. It would be the same as flipping a fair coin and getting heads 19 straight times or tails 19 straight times.

Alas, the 19th flip or game would still be 50%.

I like your 75% figure better as a team capable of going 19-0 is expected to win more than 50% of its games.

 

[PatsWorldChamps]

Looks good asuming the 50%. It would be the same as flipping a fair coin and getting heads 19 straight times or tails 19 straight times.

Alas, the 19th flip or game would still be 50%.

I like your 75% figure better as a team capable of going 19-0 is expected to win more than 50% of its games.

i agree, but there have been very few 12-4 teams... ever... and a 12-4 team would theoretically have to stay that good for 236 years to go 19-0. amazing.

 

[alamo]

i agree, but there have been very few 12-4 teams... ever... and a 12-4 team would theoretically have to stay that good for 236 years to go 19-0. amazing.

"very few 12-4 teams" ??? "super rare" ???
Not in the NFL... there have been 36 12-4 or better teams this decade, that's an average of 4.5 per year. That's not rare at all.

And your 236 year figure is meaningless, because it gives the odds of one given team consistently winning 3/4 of its games to go 19-0. But since on average 4.5 teams do it every year, we could logically project that 19-0 is a once in every 236/4.5 year event, or once every 52 years.

But figure that many of those 12-4 teams I counted were actually even better (13-3, 14-2, 15-1) their odds are even better. So let's estimate that every-52-years figure should be more like every-35-years. Hmmm, it's been 35 years since 1972... sounds about right.

I believe we will see another perfect season in the next 35 years. It's not as impossible as you portray it. And the Pats will (I hope) have shown it's indeed possible. Teams which never considered it possible will now set it as a goal.

 

[moosekill]

Hopefully we will see a team go 19-0 this year, and that same team go 19-0 next year, but lets not count our chickens before they hatch.

 

[Truck]

I think you need to change your math to use 14-2 as the expected baseline. Every year there's 1 to 2 teams that are expected to go 14-2 (I belive 14-2 would've been a good expectation going into this season with the pats as they were the heavy favorites).
The the probability is; given there's one 14-2 team every year, whats the probability that team will go 16-0 instead

 

[BelichickFan]

I agree with Truck on the 14-2 thing.

Also, I think you're off because there's multiple chances each year - certainly at 12-4 you'd have to divide by 3 or 4 as there's always a few of them. Even at 14-2 you probably need to divide by two.

 

[makewayhomer]

the odds are even low that this Patriots team, forced to replay this seasons schedule, would go 16-0.

even if you give the Pats a a 60% chance to win @ the Colts and a 95% chance to beat every other team, they only win 16 games 28% of the time. also consider that the Pats were very fortunate wrt to injuries.

it obviously takes a great team to even have a chance, but it also takes luck. you only have to go back to the Baltimore game, where we survived only b/c Ryan called that TO. (and please, don't gimme the stuff about how we stopped the play, Brady was kidding)

 

[Patsfanin Philly]

I agree with Truck on the 14-2 thing.

Also, I think you're off because there's multiple chances each year - certainly at 12-4 you'd have to divide by 3 or 4 as there's always a few of them. Even at 14-2 you probably need to divide by two.

14-2 is a .875 winning percentage....
.875^19 is .079

 

[makewayhomer]

I think you need to change your math to use 14-2 as the expected baseline. Every year there's 1 to 2 teams that are expected to go 14-2 (I belive 14-2 would've been a good expectation going into this season with the pats as they were the heavy favorites).
The the probability is; given there's one 14-2 team every year, whats the probability that team will go 16-0 instead

the over/under in Vegas on the Pats this year was 11.5 wins, and they were the highest in the league. (remember their schedule was projected as brutal)

I don't recall any team ever getting to 14 for an o/u on wins, around 13.5 is the highest I can recall. if the Pats bring back Moss and a couple other guys (but not Asante) the o/u on the Pats might still not be 14...)

 

[BelichickFan]

I think to do this really right you'd need to get a typical distribution of records, do the probability of each of the 32 records and add them up for the probability someone does it in a given year. Of course, the records that have a less than .500 chance each game would make a negligible contribution to your end result. Fencer, or Andromeda Pats Fan (whatever he's calling himself these days) can probably set us straight.