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# Probability

 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/23/2010 12:44:40 AMPosted: 6 years agoI think I'm a maths fail, but bear with me.You draw three cards from a pack of 44 different cards.What the odds of getting one card the same from those original three if you repeated the process 6 times. (Ie. You first draw the JH, the KS and the 8D, what are the odds of drawing the 8D and any other two cards six times in a row)This was my theory...-----------------You are drawing 3 cards from 44 which is 3/44Then you're drawing 2 cards from 43 which is 2/43Then you're drawing 1 card from 42 which is 1/42You're repeating it 6 times and hoping to get ...only one card the same out of the three.If you were to draw three cards, put them back into the deck and draw another three cards, the odds of getting one card the same are 1/21.5Repeating it over and over and the number grows exponentially as you can see from this simple formula:(3/44 + 2/43 + 1/42)^6 which equals...1/21.5^6 which equals....1/98,771,297.6So if you're talking about getting one card from the original three in any group of three repeated six times, the odds are 1 in 98,000,000.------------However, two other guys have agreed it's simply 3/44^6, which is 729/7256313856 or 1/9953791.297668038.Have I massively overcomplicated it?"Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/23/2010 4:10:09 AMPosted: 6 years agoCome on DDO.Wherez all teh genius?"Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 4,597 Add as FriendChallenge to a DebateSend a Message 9/23/2010 4:18:38 AMPosted: 6 years agoI don't think you did it right, but that's only because I don't really understand what you're trying to do with the cards. Repeat the objective you want to complete by drawing the cards in the simplest terms possible.When large numbers of otherwise-law abiding people break specific laws en masse, it's usually a fault that lies with the law. - Unknown
 Posts: 6,457 Add as FriendChallenge to a DebateSend a Message 9/23/2010 5:27:32 AMPosted: 6 years agoAt 9/23/2010 12:44:40 AM, tvellalott wrote:This was my theory...-----------------You are drawing 3 cards from 44 which is 3/44Then you're drawing 2 cards from 43 which is 2/43Then you're drawing 1 card from 42 which is 1/42You're repeating it 6 times and hoping to get ...only one card the same out of the three.If you only want one of the three it's a 1/44 in 6 trials (assuming you pick one of those initial 3). The initial set of draw 3 simply tells you what card you want, it doesn't affect the probability of later trials. If you are unconcerned about which of the 3 cards you initially draw are, then the initial trial is 3/44 (still not sure why you think that initial draw is relevant), each trial after 1/44 (since you appear to choose 1 card of the 3). I'm confused as to why you are reducing the deck size each trial so I may not be sure about what you are trying to achieve.
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/23/2010 5:29:54 AMPosted: 6 years agoI'll post the original facebook post..."Ok you clever facebookians, here is a word problem for you-I have a deck of 44 cards. I shuffle the cards thoroughly and pull out three cards at random. What are the chances that one card keeps being pulled out in that combination of three, six times in a row?Seriously, I want to know. Someone smarter than me please work it out. My probability knowledge is not that extensive."Its a maths problem, not a word problem but I'll forgive her because she is cute."Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 6,457 Add as FriendChallenge to a DebateSend a Message 9/23/2010 5:33:16 AMPosted: 6 years agoAt 9/23/2010 5:29:54 AM, tvellalott wrote:I have a deck of 44 cards. I shuffle the cards thoroughly and pull out three cards at random. What are the chances that one card keeps being pulled out in that combination of three, six times in a row?So any of the 3? 3/44 in 6 trials. One of the 3, 1/44 in 6 trials. I'm assuming the card is inserted back into the deck each time.
 Posts: 6,457 Add as FriendChallenge to a DebateSend a Message 9/23/2010 5:34:35 AMPosted: 6 years agoUnless the initial draw counts as one of the 6 trials, in which case adjust accordingly.
 Posts: 3,090 Add as FriendChallenge to a DebateSend a Message 9/23/2010 6:17:44 AMPosted: 6 years agoAt 9/23/2010 5:29:54 AM, tvellalott wrote:I'll post the original facebook post..."Ok you clever facebookians, here is a word problem for you-I have a deck of 44 cards. I shuffle the cards thoroughly and pull out three cards at random. What are the chances that one card keeps being pulled out in that combination of three, six times in a row?Seriously, I want to know. Someone smarter than me please work it out. My probability knowledge is not that extensive."Its a maths problem, not a word problem but I'll forgive her because she is cute.okay say that card you are going for is the 2Hthe chance that you will draw the 2H when drawing 3 cards from a deck of 44 cards is this..1st draw 1/442nd draw 1/433rd draw 1/42add those together and you have a 1/129 chance of drawing the 2Hdoing that six times would be..1/774 chance of the 2H being one of the cards you draw in each of the 6 trialsunless I did it wrong... i don't quite remember exactly it MIGHT be..1/79464and1/476784
 Posts: 3,090 Add as FriendChallenge to a DebateSend a Message 9/23/2010 6:40:45 AMPosted: 6 years agoEhhh may have done that horribly wrong, didn't take into account the 3 cards.3/443/433/42then the next draws would depend on the cards3/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/42Then for the NEXT selection EACH of those ^^^ then have chances3/44 3/43 3/423/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 3/43 2/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 1/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 2/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 1/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 1/43 1/422/44 1/43 1/421/44 1/43 1/421/44 1/43 1/421/44 1/43 1/42Then for EACH of those they all have the possibilities that fall below them.In other words its WAY to much math that I don't want to do..
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/23/2010 6:49:05 AMPosted: 6 years agoGah, this is keeping me awake...So you draw three cards from 44.Those are the base cards.You put them back in the deck and shuffle them.Now, you deal three cardsThe odds of getting one card out of the original three is 3/44BUT, you might get two or three of the original three, you have to factor that in...so it's 3/44 + 2/43 + 1/42 (This is where I get confused, it doesn't seem right to multiply these.... anyway, once you reduce it you get 1/21.5)So you have a 1/21.5 chance of getting at least one of the original 3.Now, you're going to put the cards back in and draw another three.This is where I get messed up. Is just 1/21.5 x 1/21.5?Maybe this is where it becomes 1/21.5 x 3/44This is why I'm so confused."Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/23/2010 6:50:55 AMPosted: 6 years agoAt 9/23/2010 6:40:45 AM, xxdarkxx wrote:Ehhh may have done that horribly wrong, didn't take into account the 3 cards.3/443/433/42then the next draws would depend on the cards3/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/42Then for the NEXT selection EACH of those ^^^ then have chances3/44 3/43 3/423/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 3/43 2/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 1/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 2/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 1/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 1/43 1/422/44 1/43 1/421/44 1/43 1/421/44 1/43 1/421/44 1/43 1/42Then for EACH of those they all have the possibilities that fall below them.In other words its WAY to much math that I don't want to do..O_O !!!"Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 3,090 Add as FriendChallenge to a DebateSend a Message 9/23/2010 6:57:51 AMPosted: 6 years agoAt 9/23/2010 6:50:55 AM, tvellalott wrote:At 9/23/2010 6:40:45 AM, xxdarkxx wrote:Ehhh may have done that horribly wrong, didn't take into account the 3 cards.3/443/433/42then the next draws would depend on the cards3/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/42Then for the NEXT selection EACH of those ^^^ then have chances3/44 3/43 3/423/44 3/43 3/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 3/43 2/423/44 3/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 2/423/44 2/43 2/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/423/44 2/43 1/423/44 2/43 1/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 2/422/44 2/43 2/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 2/43 1/422/44 2/43 1/422/44 1/43 1/421/44 1/43 1/422/44 1/43 1/422/44 1/43 1/421/44 1/43 1/421/44 1/43 1/421/44 1/43 1/42Then for EACH of those they all have the possibilities that fall below them.In other words its WAY to much math that I don't want to do..O_O !!!exactly lol, imagine actually DOING the math for each of those posibilites through 6 cycles...basically their are alot of answers
 Posts: 6,457 Add as FriendChallenge to a DebateSend a Message 9/23/2010 8:02:56 AMPosted: 6 years agoEh so you don't pick the card - nvm me then. ^^
 Posts: 5,095 Add as FriendChallenge to a DebateSend a Message 9/23/2010 11:55:35 AMPosted: 6 years agoAt 9/23/2010 5:29:54 AM, tvellalott wrote:I'll post the original facebook post..."Ok you clever facebookians, here is a word problem for you-I have a deck of 44 cards. I shuffle the cards thoroughly and pull out three cards at random. What are the chances that one card keeps being pulled out in that combination of three, six times in a row?0, because the card is not replaced.: At 5/2/2010 2:43:54 PM, innomen wrote: It isn't about finding a theory, philosophy or doctrine and thinking it's the answer, but a practical application of one's experiences that is the answer. : At 10/28/2010 2:40:07 PM, jharry wrote: I have already been given the greatest Gift that anyone could ever hope for [Life], I would consider myself selfish if I expected anything more.
 Posts: 4,711 Add as FriendChallenge to a DebateSend a Message 9/23/2010 1:37:01 PMPosted: 6 years agoAt 9/23/2010 12:44:40 AM, tvellalott wrote:I think I'm a maths fail, but bear with me.You draw three cards from a pack of 44 different cards.What the odds of getting one card the same from those original three if you repeated the process 6 times. (Ie. You first draw the JH, the KS and the 8D, what are the odds of drawing the 8D and any other two cards six times in a row)This was my theory...-----------------You are drawing 3 cards from 44 which is 3/44Then you're drawing 2 cards from 43 which is 2/43Then you're drawing 1 card from 42 which is 1/42You're repeating it 6 times and hoping to get ...only one card the same out of the three.If you were to draw three cards, put them back into the deck and draw another three cards, the odds of getting one card the same are 1/21.5Repeating it over and over and the number grows exponentially as you can see from this simple formula:(3/44 + 2/43 + 1/42)^6 which equals...1/21.5^6 which equals....1/98,771,297.6So if you're talking about getting one card from the original three in any group of three repeated six times, the odds are 1 in 98,000,000.------------However, two other guys have agreed it's simply 3/44^6, which is 729/7256313856 or 1/9953791.297668038.Have I massively overcomplicated it?It depends, the phrasing is a bit ambiguous.If you have to pick three cards, and ONLY one must match the original three; and you must do this six times, the options of picking one of the three are:41/44 Non Matching Card AND40/43 Non Matching Card AND3/42 MATCHINGOR41/44 No match AND3/43 Matching AND40/42 No MatchOR3/44 Match AND41/43 No Match40/42 No MatchAnd then take the above ^ 6So:( (3/44 * 41/43 * 40/42 ) + (41/44 * 3/43 * 40/42) + (41/44 * 40*43 * 3/42 ) ) ^ 6IF, on the other hand, you can draw BETWEEN one, and three of the original cards; then the odds of NOT drawing ANY of your cards is:41/44 * 40/43 * 39/42So, the probability of you drawing AT LEAST one card is:( 1- (41/44 * 40/43 * 39/42) ) ^ 6
 Posts: 5,044 Add as FriendChallenge to a DebateSend a Message 9/23/2010 1:56:06 PMPosted: 6 years agoYou guys have way too much time on your hands. I think the correct answer is "Who the f@ck cares."
 Posts: 2,033 Add as FriendChallenge to a DebateSend a Message 9/23/2010 2:35:47 PMPosted: 6 years agoAt 9/23/2010 11:55:35 AM, Kleptin wrote:At 9/23/2010 5:29:54 AM, tvellalott wrote:I'll post the original facebook post..."Ok you clever facebookians, here is a word problem for you-I have a deck of 44 cards. I shuffle the cards thoroughly and pull out three cards at random. What are the chances that one card keeps being pulled out in that combination of three, six times in a row?0, because the card is not replaced.That's exactly what I was going to say when he didn't replace it.+Over 9000 to Kleptin.I miss the old members.
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/24/2010 4:16:06 AMPosted: 6 years agoI've very disappointed DDO...The easiest way of looking at this is to calculate the probability that you WON'T get any of the cards and minus it from one...SO41/44 x 40/43 x 39/42which is...2,665/3,311so...1 - 2,665/3,311which is...646/3311So in the end your answer is...646/3311^6or72676684723410496/1317513532079237931361which is...1/18128.420924721165848894299464091WIN!!!"Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...
 Posts: 4,488 Add as FriendChallenge to a DebateSend a Message 9/24/2010 8:31:54 AMPosted: 6 years agoAt 9/24/2010 4:16:06 AM, tvellalott wrote:I've very disappointed DDO...The easiest way of looking at this is to calculate the probability that you WON'T get any of the cards and minus it from one...You did it right. The really hard part of this problem was figuring out what it was you were asking.Here is another problem.One person in 10,000 has a certain disease. There is a test that determines whether or not you have the disease with 95% accuracy, which to say 5% of the results are false positives. You test positive for the disease. What is the probability that you actually have it?This type of problem is discussed in the excellent book, "The Drunkard's Walk: How Randomness Rules Our Lives" by physicist Leonard Mlodinow. The infamous Monty Hall Problem is in there as well. Mlodinow is good at explaining difficult things and recently co-authored "The Grand Design" with Stephan Hawkings.Since there are 500 false positives for each true positive, the odds of actually having the disease is 1 in 500.This type of calculation occurs in testing for AIDS, andalso in some Court cases. Mlodinow cites a case in which a couple was convicted on the grounds that there was only a one in million chance of of their combination of physical traits occurring in a couple. However, there would have been ten such couples in Los Angeles where the crime occurred, so their probability of being the guilty ones was one in ten, not near certainty.
 Posts: 10,865 Add as FriendChallenge to a DebateSend a Message 9/24/2010 4:23:39 PMPosted: 6 years agoAt 9/24/2010 8:31:54 AM, RoyLatham wrote:At 9/24/2010 4:16:06 AM, tvellalott wrote:I've very disappointed DDO...The easiest way of looking at this is to calculate the probability that you WON'T get any of the cards and minus it from one...You did it right. The really hard part of this problem was figuring out what it was you were asking.Sorry about that, it wasn't actually my question. With mathematics (in my case) it always seems like something i've asked makes perfect sense, but other people don't understand.Here is another problem.One person in 10,000 has a certain disease. There is a test that determines whether or not you have the disease with 95% accuracy, which to say 5% of the results are false positives. You test positive for the disease. What is the probability that you actually have it?This type of problem is discussed in the excellent book, "The Drunkard's Walk: How Randomness Rules Our Lives" by physicist Leonard Mlodinow. The infamous Monty Hall Problem is in there as well. Mlodinow is good at explaining difficult things and recently co-authored "The Grand Design" with Stephan Hawkings.Since there are 500 false positives for each true positive, the odds of actually having the disease is 1 in 500.This type of calculation occurs in testing for AIDS, andalso in some Court cases. Mlodinow cites a case in which a couple was convicted on the grounds that there was only a one in million chance of of their combination of physical traits occurring in a couple. However, there would have been ten such couples in Los Angeles where the crime occurred, so their probability of being the guilty ones was one in ten, not near certainty.That's really fascinating."Caitlyn Jenner is an incredibly brave and stunningly beautiful woman." Muh threads Using mafia tactics in real-life: http://www.debate.org... 6 years of DDO: http://www.debate.org...