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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.
this is a good point...I'll just say that I've been using exaggerated/simplified #'s to illustrate a point
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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.
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.
__________________ "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]."
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.
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.
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:
__________________ Ice_Ice_Brady writes:
The difference is that Brady calmly calls audibles while Manning flaps like a chicken, barks 11 code words, and makes sure every camera in the stadium has documented his once-in-a-generation (and patented, I believe) ability to see a defensive formation and change the play. Both have the same effect, but Manning transcends measurable human intellect while Brady merely chooses a different play.
This probability model is inherently flawed because it assumes that in each football game, the winner is determined at random.
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.
__________________ Ice_Ice_Brady writes:
The difference is that Brady calmly calls audibles while Manning flaps like a chicken, barks 11 code words, and makes sure every camera in the stadium has documented his once-in-a-generation (and patented, I believe) ability to see a defensive formation and change the play. Both have the same effect, but Manning transcends measurable human intellect while Brady merely chooses a different play.
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:
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.
Re: how impossible is this? for math majors
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