CosmicBills Posted December 20, 2004 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!
Guest Guest Posted December 20, 2004 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?
daquixers Posted December 20, 2004 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 ??
JimBob2232 Posted December 20, 2004 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
CosmicBills Posted December 20, 2004 Author 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.
CosmicBills Posted December 20, 2004 Author 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!
SDS Posted December 20, 2004 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%.
JimBob2232 Posted December 20, 2004 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...
CosmicBills Posted December 20, 2004 Author Posted December 20, 2004 And its sure alot better than I would have thought we had back in october... 168189[/snapback] Amen! GO BILLS!
SDS Posted December 20, 2004 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....
CosmicBills Posted December 20, 2004 Author 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.
SDS Posted December 20, 2004 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.
CosmicBills Posted December 20, 2004 Author 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.
JimBob2232 Posted December 20, 2004 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....
SDS Posted December 20, 2004 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...
Guest TSquareBoston Posted December 20, 2004 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.
ATBNG Posted December 20, 2004 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.
MDH Posted December 20, 2004 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.
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