Mikie2times Posted December 20, 2004 Share Posted December 20, 2004 I am afraid of math, but was wondering this: If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? or, in other words... It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once? Can anyone solve that equation? I would love to know. Thanks! 168163[/snapback] I just made a post about this with my own calculations. Here is a simple way to figure out the odds of those teams winning out. Take your odds for each game. For example in my post I have the Jaguars at 6/10 vs Houston. In other words for every 10 games they play against Houston in Jacksonville they will win 6 of them. Then you take the odds of your second game, in Jacksonville's case I have them at 8/10 vs Oakland. Turn those fractions into percentages and multiply them together. So its . 60*.80= 48%. So within my estimates the Jaguars have a 48% chance at winning out. Link to comment Share on other sites More sharing options...
eSJayDee Posted December 20, 2004 Share Posted December 20, 2004 f each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? I think you asked 2 questions so, I'll answer this one. The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%. If a team fails to lose at least one game, that means they win both. The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%. You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%. I was a math major, BTW (concentrating in calculus, though). Also, give SDS the gold star. He got the answer 1st. I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong. Link to comment Share on other sites More sharing options...
Guest cal t Posted December 20, 2004 Share Posted December 20, 2004 I am afraid of math, but was wondering this: If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? or, in other words... It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once? Can anyone solve that equation? I would love to know. Thanks! 168163[/snapback] Let me correct the Math here.If each game is a 50/ 50 chance.The each individual team has a 25% chance of losing or winning both games. Each team has a 50% chance of winning one/losing one.If this is the case...you multiple .50 x .50 x .50 and you get .125 or 12.5 percent.-BUT--there is also a .25x.25 x.25 chance of allthese teams losing both games(if they dont play eachother) or about 2 percent(less actually) BUTthere is a chance that one team lose both & the others lose 1--.25 x .5 x .5=.625 or approx 6 percent. BUT--2 teams losing both plus one winning 1---.25x.25 x.5=3.125..So add 12.5 plus 1.5625plus6.25plus3.125...soooo There is a 23.4375 % chance that the Bills will make the playoffs ASSUMING 2 Bills wins--using the 50 percent scenario that you chose. Link to comment Share on other sites More sharing options...
Fan in San Diego Posted December 20, 2004 Share Posted December 20, 2004 Here is my entry in the stats derby 4 events of 50/50 2 * 2 * 2 *2 =16 1 in 16 or 6.25 % chance of 4 heads in a row. Link to comment Share on other sites More sharing options...
Guest cal t Posted December 20, 2004 Share Posted December 20, 2004 I think you asked 2 questions so, I'll answer this one.The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%. If a team fails to lose at least one game, that means they win both. The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%. You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%. I was a math major, BTW (concentrating in calculus, though). Also, give SDS the gold star. He got the answer 1st. I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong. 168319[/snapback] Sorry you were both wrong. sincerely, cal t ...statistics expert. You have to get the odds of all possibilities putting the bills in the playoffs.Those scenarios are all 3 teams losing both. all 3 teams losing once, 1 losing both 2 winning 1,2 losing both 1 winning 1.then you add these......and you ASSUME the Bills win their 2(for true odds--you'd need to add that in--but we are ASSUMING 2 Bills wins)...and you get the 24 + % figure. Thank you, You've been a lovely audience. Link to comment Share on other sites More sharing options...
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