tazinib1 Posted June 16, 2012 Share Posted June 16, 2012 "Peanut" scolds teacher for homework assignment Pretty funny and Tillman handled it pretty well IMO Quote Link to comment Share on other sites More sharing options...
White lightning Posted June 16, 2012 Share Posted June 16, 2012 Peanut is truly one of the good guys. I enjoy his style of play and his sense of humor. He will be one of those former Bears with a radio/tv gig once his playing days are over. Quote Link to comment Share on other sites More sharing options...
nuke'em ttg Posted June 16, 2012 Share Posted June 16, 2012 Quote Link to comment Share on other sites More sharing options...
Big John Posted June 16, 2012 Share Posted June 16, 2012 So an Illinois teacher realizes the truth. Quote Link to comment Share on other sites More sharing options...
Bronco Billy Posted June 16, 2012 Share Posted June 16, 2012 [JackNicholson] YOU CAN'T HANDLE THE TRUTH! [/JackNicholson] Quote Link to comment Share on other sites More sharing options...
ABearWithFurniture Posted June 16, 2012 Share Posted June 16, 2012 Bear up! Quote Link to comment Share on other sites More sharing options...
stevegrab Posted June 18, 2012 Share Posted June 18, 2012 I would have been more impressed if he actually correctly solved the math problem but am ok with his "100%" answer. Quote Link to comment Share on other sites More sharing options...
The Irish Doggy Posted June 18, 2012 Share Posted June 18, 2012 IIRC, 0.16% and 59% Quote Link to comment Share on other sites More sharing options...
stevegrab Posted June 18, 2012 Share Posted June 18, 2012 IIRC, 0.16% and 59% The first one (winning all 4 makes sense), 20% chance (0.2) of winning each game, .2 x .2 x .2 x .2 = .0016 or .16% But winning any one of the 4, with a 20% chance to win any game, wouldn't those add up, giving them an 80% chance (4 chances at 20% each) of winning any one game. Just curious how you're getting to 59%. Its been ages since I dealt with statistical probabilities so I may be way off. (I do know enough to know that one outcome does not affect the others. If you toss a coin 10 times and it comes up heads each time, the chance of tails on the next toss is still 50%. There are no odds in your favor that it starts coming up tails.) Quote Link to comment Share on other sites More sharing options...
Papajohn Posted June 18, 2012 Share Posted June 18, 2012 But winning any one of the 4, with a 20% chance to win any game, wouldn't those add up, giving them an 80% chance (4 chances at 20% each) of winning any one game. No, because then six tries would equal 120%. It is a fairly easy problem to solve with an advanced calculator or computer program. Quote Link to comment Share on other sites More sharing options...
Big John Posted June 18, 2012 Share Posted June 18, 2012 (edited) The first one (winning all 4 makes sense), 20% chance (0.2) of winning each game, .2 x .2 x .2 x .2 = .0016 or .16% But winning any one of the 4, with a 20% chance to win any game, wouldn't those add up, giving them an 80% chance (4 chances at 20% each) of winning any one game. Just curious how you're getting to 59%. Its been ages since I dealt with statistical probabilities so I may be way off. (I do know enough to know that one outcome does not affect the others. If you toss a coin 10 times and it comes up heads each time, the chance of tails on the next toss is still 50%. There are no odds in your favor that it starts coming up tails.) Doggy is correct. The probably of 0 wins is 41%, so the probably of at least 1 win is 100%-41%, or 59% (exactally 1 win is also 41%, 2 wins is 15.36%, 3 wins is 2.56% and 4 wins is 0.16%) P(k out of N) = ( N! / k!(N-k)! ) (pk)(qN-k) where: N = the number of opportunities for event x to occur; k = the number of times that event x occurs or is stipulated to occur; p = the probability that event x will occur on any particular occasion; and q = the probability that event x will not occur on any particular occasion. Edited June 18, 2012 by Big John Quote Link to comment Share on other sites More sharing options...
Chief Dick Posted June 18, 2012 Share Posted June 18, 2012 Doggy is correct. The probably of 0 wins is 41%, so the probably of at least 1 win is 100%-41%, or 59% (exactally 1 win is also 41%, 2 wins is 15.36%, 3 wins is 2.56% and 4 wins is 0.16%) P(k out of N) = ( N! / k!(N-k)! ) (pk)(qN-k) where: N = the number of opportunities for event x to occur; k = the number of times that event x occurs or is stipulated to occur; p = the probability that event x will occur on any particular occasion; and q = the probability that event x will not occur on any particular occasion. Beat me to it. Quote Link to comment Share on other sites More sharing options...
Scorcher Posted June 18, 2012 Share Posted June 18, 2012 Doggy is correct. The probably of 0 wins is 41%, so the probably of at least 1 win is 100%-41%, or 59% (exactally 1 win is also 41%, 2 wins is 15.36%, 3 wins is 2.56% and 4 wins is 0.16%) P(k out of N) = ( N! / k!(N-k)! ) (pk)(qN-k) where: N = the number of opportunities for event x to occur; k = the number of times that event x occurs or is stipulated to occur; p = the probability that event x will occur on any particular occasion; and q = the probability that event x will not occur on any particular occasion. I wonder if the girl that got Peanut's autograph knows this? Quote Link to comment Share on other sites More sharing options...
Papajohn Posted June 18, 2012 Share Posted June 18, 2012 I wonder if the girl that got Peanut's autograph knows this? Maybe the teacher thought the answer to the second question was 80%. Wouldn't surprise me. Quote Link to comment Share on other sites More sharing options...
MikesVikes Posted June 19, 2012 Share Posted June 19, 2012 (edited) Maybe the teacher thought the answer to the second question was 80%. Wouldn't surprise me. Yea, a math teacher probably would have to ask any random Huddler for help on this one. Edited June 19, 2012 by MikesVikes Quote Link to comment Share on other sites More sharing options...
Papajohn Posted June 19, 2012 Share Posted June 19, 2012 (edited) Yea, a math teacher probably would have to ask any random Huddler for help on this one. The average 6th grade math teacher probably would. They might be able to google it, but I doubt they will. That would require an effort! Edited June 19, 2012 by Papajohn Quote Link to comment Share on other sites More sharing options...
The Irish Doggy Posted June 19, 2012 Share Posted June 19, 2012 But winning any one of the 4, with a 20% chance to win any game, wouldn't those add up, giving them an 80% chance (4 chances at 20% each) of winning any one game. Just curious how you're getting to 59%. Its been ages since I dealt with statistical probabilities so I may be way off. (I do know enough to know that one outcome does not affect the others. If you toss a coin 10 times and it comes up heads each time, the chance of tails on the next toss is still 50%. There are no odds in your favor that it starts coming up tails.) Doggy is correct. The probably of 0 wins is 41%, so the probably of at least 1 win is 100%-41%, or 59% Yeah, the trick to the question is realizing the probability of at least one win is equal to 1 - probability of no wins or 1 - the probability of all four GB wins (same thing). I doubt a middle school teaches the more advanced formula that BJ referenced. That some good there. So, you do 1 - .84. You can also look at flipping a coon twice as an example. What is the probability of at least one heads? The four possible outcomes are: H H H T T H T T Three of the four results have at least one heads (75%). Same as 1 - probability of both tails (.5 x .5 = 25%). Quote Link to comment Share on other sites More sharing options...
stevegrab Posted June 19, 2012 Share Posted June 19, 2012 Doggy is correct. The probably of 0 wins is 41%, so the probably of at least 1 win is 100%-41%, or 59% (exactally 1 win is also 41%, 2 wins is 15.36%, 3 wins is 2.56% and 4 wins is 0.16%) P(k out of N) = ( N! / k!(N-k)! ) (pk)(qN-k) where: N = the number of opportunities for event x to occur; k = the number of times that event x occurs or is stipulated to occur; p = the probability that event x will occur on any particular occasion; and q = the probability that event x will not occur on any particular occasion. I used to be pretty good at math, then I got into computer programming, and seem to have forgotten quite a bit since then. Quote Link to comment Share on other sites More sharing options...
chester Posted June 19, 2012 Share Posted June 19, 2012 The problem, John, is that you're smarter than any potential employers. When you find the one that can keep up with you and appreciates your talents, you're set! Quote Link to comment Share on other sites More sharing options...
CaP'N GRuNGe Posted June 20, 2012 Share Posted June 20, 2012 I never did care for sadistics. Quote Link to comment Share on other sites More sharing options...
Ditkaless Wonders Posted June 20, 2012 Share Posted June 20, 2012 Yeah, the trick to the question is realizing the probability of at least one win is equal to 1 - probability of no wins or 1 - the probability of all four GB wins (same thing). I doubt a middle school teaches the more advanced formula that BJ referenced. That some good there. So, you do 1 - .84. You can also look at flipping a coon twice as an example. What is the probability of at least one heads? The four possible outcomes are: H H H T T H T T Three of the four results have at least one heads (75%). Same as 1 - probability of both tails (.5 x .5 = 25%). Flipping coons is dangerous. Quote Link to comment Share on other sites More sharing options...
Bronco Billy Posted June 21, 2012 Share Posted June 21, 2012 Flipping coons is dangerous. Quote Link to comment Share on other sites More sharing options...
Bronco Billy Posted June 21, 2012 Share Posted June 21, 2012 You can also look at flipping a coon twice as an example. What is the probability of at least one heads? The four possible outcomes are: H H H T T H T T Three of the four results have at least one heads (75%). Same as 1 - probability of both tails (.5 x .5 = 25%). So...the Packers are the heads and the Bears are the tails? Got it! Quote Link to comment Share on other sites More sharing options...
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