CosmicBills 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! Link to comment Share on other sites More sharing options...
daquixers Posted December 20, 2004 Share Posted December 20, 2004 I failed Geometry --- Link to comment Share on other sites More sharing options...
Guest Guest Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? Link to comment Share on other sites More sharing options...
daquixers Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] You roll them on ?? Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 Its 1/8th. If you flip a coin 3 times, the likelyhood it will turn up heads (or tails) 3 times in a row is 1 in 8 Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] That can't be the answer...can it? Because wouldn't you have to factor in the chances that only 2 of the 3 teams get a Heads, or the chance that only 1 of the 3 teams get a Heads... It's not 50/50. Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 Its 1/8th. If you flip a coin 3 times, the likelyhood it will turn up heads (or tails) 3 times in a row is 1 in 8 168178[/snapback] Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] I don't know the right answer, but I know that you're wrong.... flipping a penny is an independent event too, but the prob. that it is flipped tails 3x in a row is not 50%. Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! And its sure alot better than I would have thought we had back in october... Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 And its sure alot better than I would have thought we had back in october... 168189[/snapback] Amen! GO BILLS! Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! 168182[/snapback] I doubt that is correct either because you only need one loss out of 2 possible tries for each team. If it was the final game - yes, but not now. I think it is 0.75^3 = 42% WW, LW, WL, LL = 75% success for each team.... Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 I doubt that is correct either because you only need one loss out of 2 possible tries for each team. If it wa sthe final game - yes, but not now. 168193[/snapback] Oh, great point! See, this is why math and I don't get along too well. Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Oh, great point! See, this is why math and I don't get along too well. 168196[/snapback] I think I gave the right answer in my edited post. Link to comment Share on other sites More sharing options...
crackur Posted December 20, 2004 Share Posted December 20, 2004 GOOD DEAL SDS Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 I think I gave the right answer in my edited post. 168199[/snapback] Thanks so much. I know it's meaningless, and even though I hate math, I like numbers. I am a walking contradiction I guess. Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 I have posed the question to some people who know much more than I do regarding statistics...I will post the results as I get them.... Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Thanks so much. I know it's meaningless, and even though I hate math, I like numbers. I am a walking contradiction I guess. 168202[/snapback] not to bum you out, but assuming I'm right - I believe I am - then if you include OUR odds of WW then all the scenarios play out to a 10.5% chance of occuring... Link to comment Share on other sites More sharing options...
Guest TSquareBoston Posted December 20, 2004 Share Posted December 20, 2004 42.2% Assuming equal (50% probability) in each game: Each independent team has a 75% chance of losing at least one game: 4 possible combinations, of which 3 consist of a loss: W/W,W/L,L/W and L/L. 3/4=75% The chance of all three having a positive outcome (at least one loss) is 75% x 75% x 75%. Dated a math major way back. Link to comment Share on other sites More sharing options...
ATBNG Posted December 20, 2004 Share Posted December 20, 2004 Well, I don't think that you can make a reasonable analysis assuming every game is a 50/50 proposition. We know that isn't the case - you'd be hard pressed to find someone make you an even money bet on next week's San Fran/Buffalo game. In this case, you have to factor in the lines on the games. Here's my projections: Buff at SF - Buffalo by 13 Pitt at Buff - Pitt 3 (if 1 seed isn't wrapped up) Buff 1-3 (if Pitt wraps up one seed) Denver at Tennessee - Denver -3 Indy at Denver - Denver -3 (if Indy can't improve their position, which is very likely...if this game were in week 16, Indy would be a three point favorite - this is really a killer for the Bills) Houston at Jacksonville - Jags -5.5 Jacksonville at Oakland - Jacksonville -6 (if they can win to get in - a scenario that will inflate the line....normally that's a Jacks -3) Baltimore at Pittsburgh - Pittsburgh -7 Miami at Baltimore - Baltimore -12 There's also the Jets scenario.... Pats at Jets - New England -3 Jets at St. Louis - Jets -8 Since Denver and Jacksonville are favored in both their games, there's a higher than 25% probability (nominal coin flipping odds) that either one will win both. I'd put the Jags somewhere around 50% and the Broncos at 35%. I'd put Baltimore at around 15%. The Bills are highly dependent on the value of the game to Pittsburgh. I think the Bills have a 20% shot or so at getting into this thing. Link to comment Share on other sites More sharing options...
MDH Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] You have to love someone insulting other's education when the answer they come to isn't even close to being the correct answer. Try taking a Stats class before coming back to sit on your high horse. Link to comment Share on other sites More sharing options...
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