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Probability
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9/23/2010 12:44:40 AM Posted: 7 years ago 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 3/44 Then you're drawing 2 cards from 43 which is 2/43 Then you're drawing 1 card from 42 which is 1/42 You'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 1/21.5 Repeating 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.... 1/98,771,297.6 So 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? 
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9/23/2010 4:10:09 AM Posted: 7 years ago Come on DDO.
Wherez all teh genius? 
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9/23/2010 4:18:38 AM Posted: 7 years ago I 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 otherwiselaw abiding people break specific laws en masse, it's usually a fault that lies with the law.  Unknown 
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9/23/2010 5:27:32 AM Posted: 7 years ago At 9/23/2010 12:44:40 AM, tvellalott wrote: 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. 
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9/23/2010 5:29:54 AM Posted: 7 years ago 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. 
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9/23/2010 5:33:16 AM Posted: 7 years ago At 9/23/2010 5:29:54 AM, tvellalott wrote: 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. 
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9/23/2010 5:34:35 AM Posted: 7 years ago Unless the initial draw counts as one of the 6 trials, in which case adjust accordingly.

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9/23/2010 6:17:44 AM Posted: 7 years ago At 9/23/2010 5:29:54 AM, tvellalott wrote: okay say that card you are going for is the 2H the chance that you will draw the 2H when drawing 3 cards from a deck of 44 cards is this.. 1st draw 1/44 2nd draw 1/43 3rd draw 1/42 add those together and you have a 1/129 chance of drawing the 2H doing that six times would be.. 1/774 chance of the 2H being one of the cards you draw in each of the 6 trials unless I did it wrong... i don't quite remember exactly it MIGHT be.. 1/79464 and 1/476784 
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9/23/2010 6:40:45 AM Posted: 7 years ago Ehhh may have done that horribly wrong, didn't take into account the 3 cards.
3/44 3/43 3/42 then the next draws would depend on the cards 3/44 3/43 3/42 3/44 3/43 2/42 3/44 2/43 2/42 3/44 2/43 1/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 Then for the NEXT selection EACH of those ^^^ then have chances 3/44 3/43 3/42 3/44 3/43 3/42 3/44 3/43 2/42 3/44 2/43 2/42 3/44 2/43 1/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 3/44 3/43 2/42 3/44 3/43 2/42 3/44 2/43 2/42 3/44 2/43 1/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 3/44 2/43 2/42 3/44 2/43 2/42 3/44 2/43 1/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 3/44 2/43 1/42 3/44 2/43 1/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 2/44 2/43 2/42 2/44 2/43 2/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 2/44 2/43 1/42 2/44 2/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 2/44 1/43 1/42 2/44 1/43 1/42 1/44 1/43 1/42 1/44 1/43 1/42 1/44 1/43 1/42 Then 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.. 
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9/23/2010 6:49:05 AM Posted: 7 years ago Gah, 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 cards The odds of getting one card out of the original three is 3/44 BUT, 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 3/44 This is why I'm so confused. 
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9/23/2010 6:50:55 AM Posted: 7 years ago At 9/23/2010 6:40:45 AM, xxdarkxx wrote: O_O !!! 
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9/23/2010 6:57:51 AM Posted: 7 years ago At 9/23/2010 6:50:55 AM, tvellalott wrote:At 9/23/2010 6:40:45 AM, xxdarkxx wrote: exactly lol, imagine actually DOING the math for each of those posibilites through 6 cycles... basically their are alot of answers 
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9/23/2010 8:02:56 AM Posted: 7 years ago Eh so you don't pick the card  nvm me then. ^^

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9/23/2010 11:55:35 AM Posted: 7 years ago At 9/23/2010 5:29:54 AM, tvellalott wrote: 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. 
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9/23/2010 1:37:01 PM Posted: 7 years ago At 9/23/2010 12:44:40 AM, tvellalott wrote: 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 AND 40/43 Non Matching Card AND 3/42 MATCHING OR 41/44 No match AND 3/43 Matching AND 40/42 No Match OR 3/44 Match AND 41/43 No Match 40/42 No Match And then take the above ^ 6 So: ( (3/44 * 41/43 * 40/42 ) + (41/44 * 3/43 * 40/42) + (41/44 * 40*43 * 3/42 ) ) ^ 6 IF, 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 * 39/42 So, the probability of you drawing AT LEAST one card is: ( 1 (41/44 * 40/43 * 39/42) ) ^ 6 
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9/23/2010 1:56:06 PM Posted: 7 years ago You guys have way too much time on your hands. I think the correct answer is "Who the f@ck cares."

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9/23/2010 2:35:47 PM Posted: 7 years ago At 9/23/2010 11:55:35 AM, Kleptin wrote:That's exactly what I was going to say when he didn't replace it.At 9/23/2010 5:29:54 AM, tvellalott wrote: +Over 9000 to Kleptin. I miss the old members. 
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9/24/2010 4:16:06 AM Posted: 7 years ago 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... SO 41/44 x 40/43 x 39/42 which is... 2,665/3,311 so... 1  2,665/3,311 which is... 646/3311 So in the end your answer is... 646/3311^6 or 72676684723410496/1317513532079237931361 which is... 1/18128.420924721165848894299464091 WIN!!! 
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9/24/2010 8:31:54 AM Posted: 7 years ago At 9/24/2010 4:16:06 AM, tvellalott wrote: 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 coauthored "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. 
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9/24/2010 4:23:39 PM Posted: 7 years ago At 9/24/2010 8:31:54 AM, RoyLatham wrote:At 9/24/2010 4:16:06 AM, tvellalott wrote: 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. That's really fascinating. 
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9/26/2010 2:28:12 AM Posted: 7 years ago No one in this thread has done this problem correctly.
tvell, your solution does not work because it only checks to make sure that at least one of the cards repeats in every trial, but it does NOT make sure that that there is a card that was present in all six trials. For instance, if A, B and C were your cards: Trial 1: contains C, Q, X Trial 2: contains B, P, Z Trial 3: contains A, C, G Trial 4: contains C, L, N Trial 5: contains A, M, O Trial 6: contains B, R, T You see, this outcome and many others like it are contained in your probability. However, this outcome does NOT fit the question. Sure, in each round, at least one of the cards is there, but there is not one card that comes up in all six trials. To do that, you need to use some combinatorics [there's another way, but it's pretty much the same thing in concept form rather than mathematical form]. First, let me clarify something. I took the question to mean that the original pick of three cards was the first draw, and then there are five more after it. I will therefore be using the exponent of 5 rather than 6. But I'll include answers for both. X: That at least one of the cards from the first trial repeats in all five subsequent trials. We want to find P(X). A: That one specific card from the first draw is present in all five subsequent draws. B: That a different specific card from the first draw is present in all five subsequent draws. C: That the other specific card from the first draw is present in all five subsequent draws. You can think of them as first, second, third if you want, but I wanted to avoid that language because order does not matter. We can now put P(X) in terms of other things. P(X) = P(A) + P(B) + P(C)  P(A and B)  P(A and C)  P(B and C) + P(A and B and C) If this confuses you, think of a venn diagram. When you add P(A) and P(B), you've added that space between them twice, so you have to subtract it out. After adding all three circles and subtracting all three meetings between circles, you have to readd in the center piece. Here's a visual aid. http://merit.oryo.us... Because there's nothing special about A vs B vs C, P(A) = P(B) = P(C) P(A and B) = P(A and C) = P(B and C) So now to simplify: P(X) = 3P(A)  3P(A and B) + P(A and B and C) Now just to get the formulas for the stuff that remains. P(A) = [(43 C 2) / (44 C 3)]^5 P(A and B) = [42 C 1 / (44 C 3)]^5 P(A and B and C) = [41 C 0 / (44 C 3)]^5 43 C 2 / 44 C 3 = 3/44 42 C 1 / 44 C 3 = (3*2)/(44*43) = 3/946 41 C 0 / 44 C 3 = (3*2*1)/(44*43*42) = 1/13244 If you need an explanation, look up some stuff on combinatorics. I guess if you ask a specific question about why I put what I put, I'll give you an answer when I see your question. P(X) = 3(3/44)^5  3(3/946)^5 + (1/13244)^5 In the end, using five trials after the original one gets you an overall answer of ~4.4x10^6, or more precisely, 900595797560267 / 203735103722310426112. Using six trials after the original, as you seem to want to, yields ~3.01x10^7, or 406619086992323389 / 1349133856849139641713664. 
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9/27/2010 4:26:10 PM Posted: 7 years ago At 9/26/2010 2:28:12 AM, omelet wrote: That does sound right. The only bit I don't get is the combinatorics stuff. I'll have to read up. Thanks for adding. :) I like mathematics. I wish I'd paid more attention in school. 
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9/27/2010 6:05:35 PM Posted: 7 years ago At 9/27/2010 4:26:10 PM, tvellalott wrote:I'll try to give at least a brief explanation.At 9/26/2010 2:28:12 AM, omelet wrote:The only bit I don't get is the combinatorics stuff. I'll have to read up. Thanks for adding. :) 44 C 3 is, by definition, the number of combinations of 3 cards you can possibly pick from a deck of 44. When you want to figure out how many of those combinations have one specific card, then you just put that one card into its spot and see how many combinations you can make from there. At this point, there are 2 empty spots you're filling, and 43 cards to pick from. So essentially, the number of combinations that contain the "A" card is going to be 43 C 2. Since all of the combinations are equally likely, the probability that the A card will show up is the number of combinations in which the A card shows up divided by the total number of combinations. Which is where the 43 C 2 / 44 C 3 comes from. Since that's the probability for the A card coming up in a single trial, you have to put it to the 5th power if you want the probability that it happens in every trial (you'd use the 6th power if you took the question to mean 6 trials after the first draw). I followed a similar format with the other probabilities. 42 C 1 is the number of combinations that exist with two cards already in their spots, because then you're just picking one card out of 42 for the last spot. I had at first simply written simply as 42, which is correct, but I like consistency, so I stuck with combination notation. 41 C 0 is similarly the number of combinations where three specific cards are picked. There are zero spaces left to fill, and 41 cards left to pick from. Really, I could have just written this as 1, because there's only one combination with all three cards [it's the combination of those three cards], but again I thought I'd show some more consistency. In retrospect, this might have just added confusion, hence me trying to explain rather than expecting you to figure it out yourself. 
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9/27/2010 6:14:28 PM Posted: 7 years ago How did OP pick the number 44 for the number of cards?
in the blink of an eye you finally see the light 
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9/29/2010 9:31:00 PM Posted: 7 years ago At 9/27/2010 6:14:28 PM, wjmelements wrote: A friend of mine posed the question on Facebook. She is a teacher and was interested. Lots of people said "(3/44 x 2/43 x 1/42)^6" but I knew that was wrong. I've been searching for the correct answer. omelet seems to have nailed it. 