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Ratios and Tendency posted:Paradoxes are just a consequence of fuzzy language, they really aren't very interesting.  Whatup my dude. I agree! But it can be hard to articulate just how the language is being slippery, which can be pretty interesting.prick with tenure posted:Say I tell my logic class, which meets once each weekday, that they will have a quiz next week and that it will be a surprise - that is, on the day I give the quiz no student, no matter how rational, will come to class rationally justified in feeling sure that the quiz is scheduled for that day. A student responds that my promise is impossible to fulfill, and reasons as follows. She can know now that the test can't be next Friday, since if next Friday rolls around and I haven't given the quiz yet, she will walk in knowing that the quiz will take place that day. A Friday surprise quiz is thus an impossibility. With a Friday surprise quiz eliminated, she can go on to eliminate a Thursday surprise quiz, since if Thursday rolls around and the quiz hasn't happened yet, the class will know the quiz must happen that day, as it can't happen on Friday for the reasons above. She claims she can eliminate surprise quizzes on Wednesday, Tuesday and Monday following the same train of thought. I nod, and when the next week comes I give a quiz on Wednesday, evidently to everyone's surprise. What's wrong with the student's reasoning? Her argument, cleaned up a bit, is something like the following: Proposition 1 1) Assume the quiz is not given on Monday, Tuesday, Wednesday, or Thursday. 2) if 1), then it is certain that a quiz will be given on Friday, and therefore a surprise quiz cannot happen on Friday. Proposition 2 1) Assume the quiz is not given on Monday, Tuesday, or Wednesday, and the quiz cannot be given on Friday because of Prop 1. 2) If 1), then it is certain that the quiz will be given on Thursday, and therefore a surprise quiz cannot happen on Thursday. Proposition 3 1) Assume the quiz is not given on Monday or Tuesday, and the quiz cannot be given on Thursday or Friday because of Prop 2. 2) If 1), then it is certain that the quiz will be given on Wednesday, and therefore a surprise quiz cannot happen on Wednesday. Proposition 4 1) Assume the quiz is not given on Monday and the quiz cannot be given on Wednesday, Thursday or Friday because of Prop 3. 2) If 1), then it is certain that the quiz will be given on Tuesday, and therefore a surprise quiz cannot happen on Tuesday. Proposition 5 1) Assume the quiz cannot be given on Tuesday, Wednesday, Thursday or Friday because of Prop 4. 2) If 1), the quiz will be given on Monday, and therefore a surprise quiz cannot happen on Monday. Conclusion: If Props 1-5 obtain, then on no day during the week can there be a surprise quiz.  The problem with all that is that propositions 2-5 involve causal chains that reach into the future and generate contradictions. For Prop 5, for instance: The P --> Q that the quiz will be given on Monday relies on a chain of reasoning that starts with the assumption (Prop 1, point 1) that the quiz has not been given on Monday, Tuesday, Wednesday or Thursday. It's the same causation problem for Props 4, 3, and 2. So while she can reason correctly that the surprise quiz can't happen on Friday, she can't parley that into a surprise quiz being impossible for the other days of the week. Of course, the real problem is that she's an insufferable freshman who just discovered that logic is a thing when she skimmed the syllabus last class*. *Haha no, she didn't look at the syllabus.
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Sep 1, 2016 15:50
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Ytlaya posted:This is the argument I was trying to make (that it's only impossible on Thursday given that it definitely can't occur Friday), but it seems like the causal chain isn't really an issue because the fact that it can't be on Friday is definitely known regardless of any other conditions (that is, it's always true). So if it reaches Thursday you definitely know it can't be held Friday, meaning that you'd know it was Thursday (thereby invalidating Thursday as a possibility since you'd know it would be held once that day arrived). Also I don't think the Wikipedia page lists it as a solution, so I imagine there's some problem with that reasoning even if it's not what I just mentioned. You're falling victim to the same logical blind spot as the hypothetical student. How do you know the quiz can't happen on Friday? (a) Because if it hasn't happened on Monday, Tuesday, Wednesday, or Thursday, it has to happen on Friday so it wouldn't be a surprise quiz if it did. How do you know the quiz can't happen on Thursday? (b) Because if it hasn't happened on Monday, Tuesday, or Wednesday, and it can't happen on Friday because (a), it has to happen on Thursday. Therefore if it happened on Thursday it wouldn't be a surprise quiz But (a) stipulates that the quiz hasn't happened on Thursday and (b) requires that it does. (a) and (b) contain contradictory premises so (b) can't follow from (a) and the resolution to the paradox is that the student was running with an invalid argument. I don't think it's any deeper than that, though I'm willing to be convinced and THAT might be an interesting discussion. =) edit: Cleaned up some language to make the argument more clear. wateroverfire fucked around with this message at 18:51 on Sep 1, 2016 |
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Sep 1, 2016 18:45
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Phyzzle posted:A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being T. When his wife tells him "I can see one of them, and it's H.", TT is eliminated and the possibilities are only TH or HT.
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Sep 5, 2016 17:09
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Phyzzle posted:Or HH. Oh, yup. Oh, I see what happened.. A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being H. A man flips two coins, and they both roll under a desk. His wife looks under the desk and says, "It's dark, but I can see one of them, and it's H". The other one now has a 67% chance of being T. For the man. Since he doesn't know which coin came up H, there are three equal possibilities for him: HH, TH, and HT. But since his wife is looking directly at one specific coin rather than hearing a report of "at least one H", the other coin still has a 50% chance of being H. But only for her. Something seems amiss here . . . wateroverfire fucked around with this message at 17:35 on Sep 5, 2016 |
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Sep 5, 2016 17:30
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Phyzzle posted:A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being T. You have the following possible combinations: TT TH HT HH In the first case, the man sees one coin land H so the probability the other is T is 50%. So far so good. In the second case, the wife reports at least one coin came up H, elminiating TT and leaving TH HT HH Which you interpreted as being a 67% chance of T (prompting the WTF, this doesn't seem right moment). But 67% is the probability that at least one of two unknown flips comes up T, and in this case we know the result of one flip was H (whether the husband saw it or not, assuming he can trust his wife). Conditional on that information, only one flip is in doubt and the chance of T is 50%. Or you could think of it this way: One of TH or HT is impossible if we know the result of one of the throws was H, because TH requires that the second coin comes up heads while HT requires that the first one does. If you eliminate one or the other, the possibilities are TH, HH or HT, HH. 50% chance the other coin came up T either way.
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Sep 5, 2016 18:11
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King of Bleh posted:If the coins are marked in some way, then the wife is looking at "one specific coin" and the probabilities are as you said, due to the two parties having different information. If they aren't, then your statement of "specific coin" is false, and the probabilities are symmetric for both parties. I think the probabilities are the same for both parties regardless of marked coins.
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Sep 5, 2016 18:50
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King of Bleh posted:If one is red and one is blue, the wife has two bits of information and can rule out 2/4 possibilities. (both aren't tails, red isn't tails because she's looking at it). I think that second part is not correct. The possibilities for 2 independant coin flips before the event are: pre:Coin 1 Coin 2 H H H T T H T T He knows that TT is impossible. But he also knows that one of TH or HT is impossible given that one coin is, concretely, H. So if it happens to be Coin 1 that's H, the possibilities left are HH, HT. If it happens to be Coin 2, the possibilities are HH, TH. In either case the probability that the other coin comes up T is 50%, so it doesn't matter which coin is observed. Alternately: The probability that at least 1 coin comes up T from two independant flips is 67%, but once one coin is revealed to be heads there is only 1 uncertain flip left, so the probability is 50%. wateroverfire fucked around with this message at 20:16 on Sep 5, 2016 |
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Sep 5, 2016 20:13
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VitalSigns posted:No the probability of getting at least one heads is 75%. Yeah, I am a dumb. VitalSigns posted:If you look at one coin at random then the probability the other is heads is 50% And this is right. VitalSigns posted:If someone looks at both coins and deliberately shows you a tails, the probability the other is heads is 67% But I'm not following you here. Each flip is an independent event, right? If someone shows you that one flip was T, the probability that the other flip is H is still 50%. That's what it means for events to be independent of one another.
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Sep 6, 2016 16:09
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botany posted:HH IDK, Think of it this way. I show you a T, and we arbitrarily call that coin Coin 1. The viable options are: TT TH But not HT, because we are staring at Coin 1 and it is T rather than H. The probability that the other coin is H is 50% based on the viable options. So then we say no, that T is arbitrarily Coin 2. Then the viable options are: TT HT But not TH, because we are staring at Coin 2 and we know it is T. So the probability that the other coin is H is 50% based on the viable options. So whether Coin 1 or Coin 2 is revealed, the probability of the other coin being H is 50%. The outcome of the other flip is independent of the flip we observed.
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Sep 6, 2016 17:05
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botany posted:alright honestly this has been explained over and over in the thread, but the example here is so simple that you can actually just do it yourself. Yeah but my dude, that does not address the probability in question.
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Sep 6, 2016 17:18
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VitalSigns posted:I flip two coins. Hear me out, man, and follow the logic below to the end. I know this looks intuitive but it's wrong. TH and HT are outcomes that become mutually exclusive when information is added about the outcome of one of the coins - even if we don't know which coin we have information about. So say you tell me "at least 1 coin is T" and show me a T, so we end up with the table of results in your quote above. 1) If Coin 1 is T, the only possible outcomes are TH and TT. This is correct and should also be intuitive if you think about it, but if you disagree then show how and let's talk it through. 2) If Coin 2 is T, the only possible outcomes are HT and TT by the same reasoning. Therefore, the probability that we are in a world in which the other coin is H is the following: P(Coin you showed is Coin 1) * P(TH conditional on Coin1 being T) + P(Coin you showed is Coin 2)*P(HT conditional on Coin 2 being T) We have only two coins so P(Coin you showed is Coin 1) = 50% P(Coin you showed is Coin 2) = 50% Following 1), P(TH conditional on Coin1 being T) = 50% Following 2), P(HT conditional on Coin 2 being T) = 50% Plug everything in and 50%*50% +50%*50% = 25% + 25% = 50% probability the other coin is an H. Proving what we know from having defined the coin flips as independent events. The above is NOT the Monty Hall problem in concept. In Monty Hall, the values of the "flips" are contingent - there is only 1 Car and the other two "flips" must be Goat. So revealing a Goat gives you information about the potential values of the other "flips" where revealing a T in the coin flip problem gives you nothing. edit: Notice, if you work through the formulas, it doesn't matter how I come by the information that one coin is T. If you pick a coin at random and it happens to be T, the results tables are the same and the probability calculation is the same 50%. If you deliberately show me a T, the results tables are the same and the probability calculation is the same 50%. If you deliberately show me Coin 1 that happens to be a T, the results tables are the same and the P(Coin you showed me is Coin 2) is 0% while the P(Coin you showed me is Coin 1) is 100%. The probability calculation works out to the same 50%. If you deliberately show me Coin 2 that happens to be a T, the results tables are the same and the P(Coin you showed me is Coin 1) is 0% while the P(Coin you showed me is Coin 2) is 100%. The probability calculation works out to the same 50%. wateroverfire fucked around with this message at 14:56 on Sep 7, 2016 |
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Sep 7, 2016 14:34
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Phyzzle posted:It's this step here. P(Coin you showed is Coin 2) is not 50%. What you're talking about is adding two pieces of information to what I know that are not present in the original problem. 1) That VS shows me the first coin that pops T. 2) That I know when VS skips coin 1 (or, alternately, that I know which coin VS is showing me). Conditional on those two pieces of info, I know that when VS shows me Coin 2 the probability that Coin 1 is H is 100% since the only possibility is HT. If VS shows me Coin 1, the probability that Coin 2 is H is 50% since the two possibilities are TH and TT. But that is a totally different problem then the one we've been analyzing. =P
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Sep 7, 2016 15:11
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Buried alive posted:This is literally the exact same reasoning as this: That simulation is not brute forcing the thing in question, so the problem is that you and VS are both bad at stats in this instance. Follow the mathematical argument and find the problems in it, if any, rather than trying ot intuit your way through problems that even trained statisticians gently caress up on because they're counterintuitive.
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Sep 7, 2016 15:15
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Phyzzle posted:No, piece (1) is not being added: whatever coin he grabs first becomes #1 (if they are not marked), so you knew that. Ok, sure, if you like. So he grabs a coin that popped T and shows it to me, and we call that Coin 1. P(Coin he shows me is Coin 1) = 100%. P(Coin he shows me is coin 2) is 0% so we drop that term. The possible outcomes are TH, TT as before. P(TH conditional on Coin 1 being T) = 50% (obvious, right? Two possible outcomes, of which that is one) P(Coin he shows me is Coin 1) * P(TH conditional on Coin 1 being T) = 100% * 50% = 50%
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Sep 7, 2016 15:32
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twodot posted:There really shouldn't be any discussion on "if you take a set of random events and remove all events with a particular quality, does that gently caress with the distribution of that set?" yes of course it does. I do think people are doing a bad job of stating in English whether they are talking about a pre-poisoned set or not. It doesn't actually matter what the distribution of the set is, if the question we're debating is "what is the probability of a coin flipping heads conditional on knowing the result of some other coin flip where the two are independant events". People are doing a really bad job being responsive to the question they think they're being responsive to.
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Sep 7, 2016 15:37
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Phyzzle posted:Yes, that's the case where grabs a coin that popped T and shows it to you. Which is not the only possible case. What other case would you lik to consider?
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Sep 7, 2016 16:30
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rudatron posted:Your choice between coin 1 or 2 is only 50 50 in the TT case - in the other two cases, you have no agency over which coin you choose, so your logic is bunk from the start. Dude where do I even start with this... I'm just going to go with 1 * (1/2)/(1/3) = 3/2 and not 2/3, and it should be obvious why that result cannot be a probability for anything whatsoever, for right now.
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Sep 7, 2016 16:41
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Phyzzle posted:All of them. Every possible way for one coin to be Heads must be considered to find the probability of one coin being Heads. Like he grabs one coin, and it's Heads, so he then grabs another. Do the math and show me what you're talking about. But again...just to not get too far afield... The very elemental probability question we've somehow managed to spend like 3 pages talking about is simply "Given two independant coin flips, what is the probability that one coin is H given that we know the other is T".
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Sep 7, 2016 16:46
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twodot posted:There's a 25% chance they refuse to show you the coin in the first place you've been ignoring. Consider this: I flip a bunch of coins and eliminate all HH and TT pairs from my set (instead of merely all HH pairs), given I show you a H, the chance of the other coin being a T is 100%. But literally none of that is responsive to the question, which is a very straight-forward exercise in probability, "Given two independant coin flips, what is the probability that one coin is H if we know the other is T".
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Sep 7, 2016 16:49
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VitalSigns posted:Maybe stating the reverse of the problem would help clarify this possible misunderstanding. Yeah, I see where you're going.
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Sep 8, 2016 18:31
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