I wonder when the last time was......

[stcjones]

When ALL 4 teams in the AFC east LOST their first preseason games......anyone?

 

[patsfan13]

I'm sick but not sick enough to keep track of preseason recors across multiple seasons!

 

[PromisedLand]

When ALL 4 teams in the AFC east LOST their first preseason games......anyone?

Statistically it should happen every 16 (2 to the 4th power) seasons, on average.

 

[Patsfanin Philly]

Statistically it should happen every 16 (2 to the 4th power) seasons, on average.

but only if the teams have a 50/50 chance of winning their games (i.e. flip of a coin). If one of those teams wins 60% (.6)of their games, another one wins 40% (.4) and the other two win 50% (.5) of their games, IIRC it would be .4 times .6 times .5 times .5 which is equal to .06 (which is actually less than 2 to the 4th which is .0625.) or once very 16.7 years........

Just my $0.02,

 

[PromisedLand]

but only if the teams have a 50/50 chance of winning their games (i.e. flip of a coin). If one of those teams wins 60% (.6)of their games, another one wins 40% (.4) and the other two win 50% (.5) of their games, IIRC it would be .4 times .6 times .5 times .5 which is equal to .06 (which is actually less than 2 to the 4th which is .0625.) or once very 16.7 years........

Just my $0.02,

Congratulations, I've been out-geeked!

 

[stcjones]

Congratulations, I've been out-geeked!

Thanks for the work guys....god knows I can't create that kind of mathematical analysis......just curious!!! thanks again!!!

 

[Patsfanin Philly]

Congratulations, I've been out-geeked!

I resemble that comment.............
I take that as a compliment.

 

[rabthepat]

Well...with the state of the AFC East it probably will happen more often until the Bill and Jets regroup.

 

[patchick]

but only if the teams have a 50/50 chance of winning their games (i.e. flip of a coin). If one of those teams wins 60% (.6)of their games, another one wins 40% (.4) and the other two win 50% (.5) of their games, IIRC it would be .4 times .6 times .5 times .5 which is equal to .06 (which is actually less than 2 to the 4th which is .0625.) or once very 16.7 years........

Just my $0.02,

Outstanding.

A question...is it NFL policy that division rivals don't meet in the preseason? A non-zero probablility of two AFCE opponents meeting would certainly throw the numbers off.

 

[BruschiOnTap]

Outstanding.

A question...is it NFL policy that division rivals don't meet in the preseason? A non-zero probablility of two AFCE opponents meeting would certainly throw the numbers off.

I used to wonder if teams played the same opponents each preseason. The pats played Cincy two years in a row and the Panthers (local team round here) had the Skins the same two years. Not sure about whether or not they'd play division rivals though. Would it be bad for your rivals to get extra experience/game film before playing you for real?

 

[zoostation]

Statistically it should happen every 16 (2 to the 4th power) seasons, on average.

The division hasn't had only 4 teams in it long enough to use a 16 year measure. Besides when has a preseason loss ever meant anything?

 

[PromisedLand]

The division hasn't had only 4 teams in it long enough to use a 16 team measure. Besides when has a preseason loss ever meant anything?

Who said anything about 16 teams?

Without any a priori knowledge of the teams probability of winning, each team has a 50% chance of winning each game, or a probability of 0.5. The probability of each team winning two games out of four preseason games is therefore also 50% or 0.5. Assuming independent "events" (which means two teams in the same division cannot play each other in this scenario, which is normally true), the probability of N teams each winning 2 games out of 4 is therefore (0.5) to the Nth power. If there are 4 teams this works out to 1/16th.

Patsfanin Philly, how'd I do?

PS It was meant as a compliment.

 

[zoostation]

Who said anything about 16 teams?

Without any a priori knowledge of the teams probability of winning, each team has a 50% chance of winning each game, or a probability of 0.5. The probability of each team winning two games out of four preseason games is therefore also 50% or 0.5. Assuming independent "events" (which means two teams in the same division cannot play each other in this scenario, which is normally true), the probability of N teams each winning 2 games out of 4 is therefore (0.5) to the Nth power. If there are 4 teams this works out to 1/16th.

Patsfanin Philly, how'd I do?

PS It was meant as a compliment.

16 years not teams... .

 

[PromisedLand]

16 years not teams... .

Doesn't make sense that way either. Are you just jerking my chain?

 

[spacecrime]

but only if the teams have a 50/50 chance of winning their games (i.e. flip of a coin).

So if TAILS = Loss, and you flip three tails in a row. what are the odds of flipping another tail

 

[Patsfanin Philly]

So if TAILS = Loss, and you flip three tails in a row. what are the odds of flipping another tail

1 in 2 or 50% becasue each event is independent. If you want the chances of flipping 4 tails in a row you multiply the chances of each event by itself, so it would be .5 x .5 x .5 x .5 or .0625 or 1/16 for 4 in a row of tails.

 

[Patsfanin Philly]

Who said anything about 16 teams?



Patsfanin Philly, how'd I do?

PS It was meant as a compliment.

You done good . Either that or we're both wrong........