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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.
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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!
Good job. Now we know the odds. But what if they repeat it? Hehehe, I didn't say that.
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)
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:
(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)
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.
Last edited by makewayhomer; 01-30-2008 at 08:40 AM..
Somehow I don't think this thread is actually for 'math majors'.
Quote:
Originally Posted by makewayhomer
while we can sit here and quibble with the %'s of each individual game, your method is obviously correct.
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.
__________________
"All we do is win Superbowls." BB to Urban Meyer
"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.
This is the most accurate post in the thread.
__________________ "They have one objective," said Washington cornerback Shawn Springs. "They whoop people's [backsides]. If you understand that, it answers all your questions. They might not lose [this year]."
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.
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
Last edited by makewayhomer; 01-30-2008 at 09:54 AM..
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.
"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.
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.
Re: how impossible is this? for math majors
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