how impossible is this? for math majors

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.

PatsWorldChamps said:

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!

Click to expand...

Good job. Now we know the odds. But what if they repeat it? Hehehe, I didn't say that.

makewayhomer said:

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)

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I think you need to take it a bit further out of the math arena and delve more into the football analysis of this exercise, as mwh alluded to here. For example, looking back on each individual game, what do you feel the probability would be of the Patriots coming out with a win? In some cases (home vs. Jets and Fins) the probability of winning the game would be closer to 95%, imo. At Indy would be 50%, etc. Here's a rough guestimate (again, in my opinion) of their probability going into the game of coming out with a W:

@NYJ - .85
SD - .70
Buff - .85
@Cin - .75
Cle - .75
@Dal - .60
@Mia - .85
Wash - .80
@Ind - .50
@ Buff - .80
Phi - .75
@Balt - .75
Pitt - .70
NYJ - .95
Mia - .95
@NYG - .70
-----------
Jax - .75
SD - .65
NYG - .80

(Note: I don't have time to compare my percentages, so a few might be high or low when looking at where I placed other values, but for this exercise it'll do...maybe we can have a discussion of values that I may have a bit high or low. And remember, this is the % GOING INTO THE GAME, so Baltimore and Philly, teams that simply weren't that good this season, would have high W probabilities despite the outcome of the game)

Results:
(all that mess above multiplied together) = .01126
1/.01126 = 88.8099

Thoughts on this?

hambone1818 said:

Thoughts on this?

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while we can sit here and quibble with the %'s of each individual game, your method is obviously correct.

re: the %'s, a pretty good place to start would be the Vegas moneyline for each game. from that you can extract the winning % of each team in this game. these guys do a better job of predicting results than just about anyone else (and anyone who can do better should be a professional sports better) I'm not gonna bother to do that, but my guess is that it would come out looking pretty similar to your result: if the Pats replayed all of these games, they go 19-0 somewhere around 10% of the time

that basically indicates that it takes a lot of luck to go 19-0. an historically great team + a good amount of luck = 19-0. it's really hard for both of these elements to align, which is why someone (probably) is going to go undefeated for only the 2nd time in 35 years

it's pretty easy to make the argument that if Ryan doesn't call that TO, the Pats go down as the 85 Bears: an unbelievable team who lost only 1 game, but "didn't have what it takes" to go 19-0.

Somehow I don't think this thread is actually for 'math majors'.

makewayhomer said:

while we can sit here and quibble with the %'s of each individual game, your method is obviously correct.

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Actually even the improvement in the methods mentioned in that post aren't really 'obviously correct', they're just better. As rough approximations they work, but they're grossly simplified and not worth much.

The problem is the outcomes aren't independent. This makes calculating probability much more difficult but ...

Game 1 say there's a 60% probability of a win.
Game 2 might have a 70% probability of a win if Game 1 was won and a 55% if Game 1 was a loss.
Game 3 would have four 'histories' etc.

And even thats a simplification. The probability of game 2 would be also partially dependent on what happened in Game 1 for Opponent 2. And a simple binary outcome would be insufficient given the complex nature of the sport.

Plus, based on the original premise, we can't assume a 14-2 team has a .875 probability to win on average. Even if luck was neutral, it could be ~+- .05 and luck is never truly neutral. Its like assuming the probability of flipping a coin heads is .60 just because you happened to get heads 6 times out of 10 or assuming that 1 will come up 1/5 of the time when you roll the die 5 times.

hambone1818 said:

Results:
(all that mess above multiplied together) = .01126
1/.01126 = 88.8099

Thoughts on this?

Click to expand...

i followed you up until you used the reciprocal. I viewed .01126 as a 1.1% chance of winning all the games.

alamo said:

"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.

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This is the most accurate post in the thread.

PantsB said:

The problem is the outcomes aren't independent. This makes calculating probability much more difficult but ...

Game 1 say there's a 60% probability of a win.
Game 2 might have a 70% probability of a win if Game 1 was won and a 55% if Game 1 was a loss.
Game 3 would have four 'histories' etc.

Click to expand...

I think this is v arguable. are you assuming that a team might try harder in week 2 if they lose week 1? or vice versa?

or that momentum will carry a team forward?

I think this stuff is generally overrated. "momentum" can end at any time, and it's unpredictable when it will end. so I'm not sure how if it should factor in, and if it is factored in, the Vegas moneyline would incorporate it

also, the percentages laid out do assume a "W" in every single game. once they lose it's irrelevant since we're talking about the odds of going undefeated.

also, I generally assume professional athletes give it their all every game, especially in an emotional sport like football with unguaranteed contracts. to believe the opposite is to believe that guys sometimes don't try

For any one team,

Game 1 => 8 undefeated teams (100% chance of 1-0)
Game 2 => 4 undefeated teams (100% chance of 2-0)
Game 3 => 2 undefeated teams (100% chance of 3-0)
Game 4 => 1 undefeated team (100% chance of 4-0)
Game 5-19 => (chance of winning game 5)*(chance of winning game 6)*...*(chance of winning game 19)

I think it would be difficult to figure out the chances of winning games 5-19, but it should be greater than .5. For instance, what are the chances of a 4-0 team winning the each of the next 15 games?

At .8 chance of winning each game, this would mean this team would have a 3.5% chance to go 19-0 - or once every 28.5 years.

alamo said:

"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.

Click to expand...

The only major problem with this post is that it is absolutely not true that, for example, a 15-1 team has/had a 93.75% chance to win each game. The teams with the best records in the league are almost by definition a little lucky. Likewise, the teams with the worst records tend to have been unlucky. So a 14-2 team probably had more like a 75-80% chance to win each game, rather than the 87.5% chance that most of the calculations in this thread are assuming.

crafty35a said:

The only major problem with this post is that it is absolutely not true that, for example, a 15-1 team has/had a 93.75% chance to win each game. The teams with the best records in the league are almost by definition a little lucky. Likewise, the teams with the worst records tend to have been unlucky. So a 14-2 team probably had more like a 75-80% chance to win each game, rather than the 87.5% chance that most of the calculations in this thread are assuming.

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this is a good point...I'll just say that I've been using exaggerated/simplified #'s to illustrate a point

crafty35a said:

The only major problem with this post is that it is absolutely not true that, for example, a 15-1 team has/had a 93.75% chance to win each game. The teams with the best records in the league are almost by definition a little lucky. Likewise, the teams with the worst records tend to have been unlucky. So a 14-2 team probably had more like a 75-80% chance to win each game, rather than the 87.5% chance that most of the calculations in this thread are assuming.

Click to expand...

Why are the teams with the best records "lucky"? What about the team that finished 11-5 or 12-4 despite tons of bad luck? Wouldn't they balance off the 14-2 lucky one?

I fail to see where luck factors in at all. Even if you believe in it, it gets more than balanced off in a sample size this large.

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.

this thread makes my head hurt...lets just agree that 19-0 is hard.:rocker:

This probability model is inherently flawed because it assumes that in each football game, the winner is determined at random.

eom said:

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.

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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:

stinkypete said:

This probability model is inherently flawed because it assumes that in each football game, the winner is determined at random.

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A probability is just a relative frequency of events. E.g., if the Pats were to play a team 100 times, what would be their win frequency. E.g., 85 wins means probability of a win is 0.85. While we would never do this experiment, it certainly is a well-formed question about which we can hazard educated guesses.

I don't think those guesses are really very educated based on the sample size you have to work with.

anyway, I hope next year's undefeated season puts all this to bed.

ps

no jinx on the weekend!

neuronet said:

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:

Click to expand...

The methodology seems like it should be correct, but there's something that I cannot articulate about it that seems wrong to me. Maybe it will occur to me.

I don't think I can dispute, however, the underlying premise divorced from the numbers: Even where a team is overwhelmingly likely to win each individual game, the chance of it doing so 19 times in a row is extremely remote.

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.

pheenix11 said:

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?

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