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how impossible is this? for math majors

Discussion in 'PatsFans.com - Patriots Fan Forum' started by PatsWorldChamps, Jan 30, 2008.

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  1. PatsWorldChamps

    PatsWorldChamps Rookie

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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!
  2. PonyExpress

    PonyExpress Rookie

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    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.
  3. pheenix11

    pheenix11 Rookie

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    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?
    Last edited: Jan 30, 2008
  4. PatsWorldChamps

    PatsWorldChamps Rookie

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    not at 19 and 0
  5. PatsWorldChamps

    PatsWorldChamps Rookie

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    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!
    Last edited: Jan 30, 2008
  6. Wildo7

    Wildo7 Totally Full of It

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    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.
    Last edited: Jan 30, 2008
  7. PatsWorldChamps

    PatsWorldChamps Rookie

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    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!
  8. PatsWorldChamps

    PatsWorldChamps Rookie

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    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!
  9. Marko

    Marko Rookie

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    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.
  10. Wildo7

    Wildo7 Totally Full of It

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    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.
  11. Hardboiled

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    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.
  12. PatsWorldChamps

    PatsWorldChamps Rookie

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    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.
    Last edited: Jan 30, 2008
  13. alamo

    alamo praedica numerum! PatsFans.com Supporter

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    "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.
  14. moosekill

    moosekill Rookie

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    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.
  15. Truck

    Truck Rookie

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    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
  16. BelichickFan

    BelichickFan B.O. = Fugazi PatsFans.com Supporter

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    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.
  17. makewayhomer

    makewayhomer Rookie

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    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)
    Last edited: Jan 30, 2008
  18. Patsfanin Philly

    Patsfanin Philly Rookie

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

    makewayhomer Rookie

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    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...)
    Last edited: Jan 30, 2008
  20. BelichickFan

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    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.
  21. xmarkd400x

    xmarkd400x Rookie

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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.
  22. FrontSeven

    FrontSeven Rookie

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    Good job. Now we know the odds. But what if they repeat it? Hehehe, I didn't say that.
  23. hambone1818

    hambone1818 Rookie

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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?
  24. makewayhomer

    makewayhomer Rookie

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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.
    Last edited: Jan 30, 2008
  25. PantsB

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

    Patsfanin Philly Rookie

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

    Oswlek Rookie

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    This is the most accurate post in the thread.
  28. makewayhomer

    makewayhomer Rookie

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    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: Jan 30, 2008
  29. patsrgr8t

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    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.
  30. crafty35a

    crafty35a Rookie

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    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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