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Steve Martin Inner circle 1119 Posts 
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Your question is answered in my "coin game 2".
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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Your coin game #2 says that I'll tell you the first flip was heads. But in my scenario, I'm not telling you that you're seeing the first coin. I'm just saying that I'll show you one of the coins. Isn't that more like coin game #3? If I show you, for instance, a head, what you've learned is that at least one of the coins is heads; if I show you a tail, you've learned that at least one coin is tails.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Steve Martin Inner circle 1119 Posts
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The significant point in "coin game 2" is that I tell you the result for a *specific* coin.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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Is there a difference, in terms of information, between showing the person either one of the coins and telling the person, "At least one of the coins is a _____"?
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Steve Martin Inner circle 1119 Posts
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Absolutely. Yes.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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I think that's the part we disagree with. If I flip two coins and show you whichever coin I want, I think that's the same thing. If I show you a head, you're learing that at least one of the coins is heads. You're not learning what the other coin is, but you can eliminate the T-T pair. Similarly, if I show you a tail, you're learning that at least one of the coins is tails. In the case where it's tails, now you know that the choices are down to H-T or H-H.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Steve Martin Inner circle 1119 Posts
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This is really the crux of the puzzle.
If you show me a head then, yes, I am learning that at least one of the coins is heads - but I am ALSO learning that it is THAT coin! If you only tell me that one of the coins is a head, I am none the wiser as to which of the coins it is. That is why it makes a difference. P.S. When you said "In the case where it's tails, now you know that the choices are down to H-T or H-H", I think you meant to say "In the case where it's HEADS, now you know that the choices are down to H-T or H-H".
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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Quote:
On 2006-03-08 22:18, Steve Martin wrote: The P.S. is right, of course. With respect to looking at the coin, now we're at the real crux of our disagreement, which is, I imagine, about as far as we can go with it. I think that seeing a coin doesn't add to the information you have; you're just getting visual confirmation of the information you have, anyway. I apologize for the tone of my earlier posts; asking for clarification of your ideas, as these last posts have done, would have been both more polite and more productive. Very interesting thread.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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TomasB Inner circle Sweden 1144 Posts
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Quote:
On 2006-03-08 19:45, LobowolfXXX wrote: It's not about knowing how _many_ are mixed but which ones are mixed. Also you seem to divide with the total number of games played. It might sound like a paradox but conditioned he has named F you are right 2/3 of the time if you guess a mixed pair and conditioned those times he names M you are also right 2/3 of the time, but if you want the total probability to be right if you allow him to name either sex it's 1/2. But the condition that he names M is the scope of the posed question. You could play it like this: Listen to what he names at the first pair, then say "pass" every time he mentiones the other gender. You will win 2/3 of the time you play. I'm speaking of the classic problem with the two dogs. You'll stumble on it in almost any beginner's probability course. This and the pen game (but with slips of papers with digits on them) was in my 8:th grade math book's probability section to describe conditional probability. /Tomas |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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In the original problem, we are given nothing about how we came about the information; it just says "You know the sex of one dog; it is male." Assuming random distribution of individual dogs, we know that 1/2 of the "dog pairs" are MF.
If one takes the position that the answer to the question "You know that one dog is male; what are the chances that the other dog is male?" is 1/3, then one must also take the position that the answer to the question "You know that one dog is female; what are the chances that the other dog is female?" is also 1/3. The above paragraph indicates that the odds of a mixed pair are 2/3 when we're told that one dog is male, and also that the odds of a mixed pair are 2/3 when we're told that one dog is female. Now, say we visited every 2-dog household in the world, with the stipulation that when we get there, we'll be told the sex of one of the dogs. 100% of the cases will be covered by the following two situations: 1) We're told that one of the dogs is male 2) We're told that one of the dogs is female Accepting the 2/3 position, then, we would have to conclude that the odds of any given pair being mixed are 2/3, because the odds are 2/3 100% of the time (the sum of times that we're told one is male, and the times we're told one is female). But we know this is not true. The odds of any pair being mixed are not 2/3; they're 1/2. The only way the odds are 2/3 - 1/3 is if the person telling us the sex commits to which sex he'll tell us about, up front. That is to say: If the scenario is: "If there's a male dog in the next pair we find, I'll tell you," then you get to the house and the person says, "One dog is male." The odds that the other dog is female are, indeed, 2/3 - 1/3. BUT if the scenario is: "When we next get to a pair of dogs, I'll tell you what the sex of one of them is," then you get to the house and the person says, "One dog is male," then the odds that the other dog is female are 1/2 - 1/2. The difference in this case is that the FF scenario should not be disregarded. Let's say I'm flipping pairs of coins behind a screen for all eternity. If I commit to tell you every time one is heads, and only when one is heads, then 2/3 of the time, there will be a mixed pair. BUT if I commit to tell you about one of the coins, at my discretion after looking at both flips, there will be a mixed pair only 1/2 the time. We know going in that pairs of coins that are flipped will be mixed 1/2 the time. Surely, my looking at both coins and naming one of them cannot change this. I will gladly give, say, 3-2 money odds on an experiment of this nature to anyone who thinks the odds are really 2-1. The original problem posted in this thread said: "You know the sex of one of the dogs; it is male." This phrasing does not indicate which of the two situations exists. Whether knowing the sex of one of the dogs is constrained (You only learn about one of the dogs if it's male) or unconstrained (The person who told you the sex of one of the dogs was going to tell you, regardless) is critical to this problem.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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TomasB Inner circle Sweden 1144 Posts
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Quote:
On 2006-03-09 01:01, LobowolfXXX wrote: You are absolutely correct in this. So it's obviously the first scenario that is what you should focus on in this problem. In other words: Out of ALL the possible scenarios where he is at all ABLE to say that one dog is Male, the other one will be Female 2/3 of the time. /Tomas |
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magicjohn2278 Special user Isle of Man UK 544 Posts
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As you will have seen from my previous posts, I’ve been having a few problems understanding this puzzle. Even though I now understand it, thanks to all the people who have been kind enough to patiently explain it to me step by step, I’ve still been puzzled for a while WHY this problem works. BUT FINALLY, I’VE GOT IT!
It’s all to do with not being able to distinguish between the two male dogs – If we can, then the problem doesn’t always work. Correct me if I am wrong, but as I understand it, the argument goes, if you have a male and a female dog, they are different, so they could be FM or MF. (FM<>MF) First dog could be male, second dog female, or first dog female, second dog male, because male and female are obviously different. (F<>M and M<>F) But if you have two male dogs you can only have MM, you can’t argue that you could have two different states (First dog Male, second dog Male or First dog Male, second dog Male) because they are the same thing. Am I right so far? So logically, MM is the same as MM, because M and M are the same (M=M). If M and M were different, this wouldn’t be true, so they aren’t. They are the same. Obviously, using elementary maths and a bit of substitution, (x = 3, is the same as saying “x is 3”), then if M=M (and we know it is), then M is M. Actually we knew this before when we agreed that M and M were the same. Which proves that the man in the problem, who told us that he has two identical male dogs, must be mistaken, if both his dogs are the same then they must both be the same dog. Which is impossible, so in fact it is obvious from the information given in the question, that he lied, as he can only own ONE dog. (You can’t buy the same dog twice. (-Actually, thinking about it, you can, but not if you already own it! (-Actually, thinking about it, if it was lost or stolen, you can do this too, but you would still only have one dog. (-Actually, thinking about it, this isn’t true either, you could already own another one……Hmmm…… – Anyway, that’s another thread.)))) Assuming that I haven’t missed anything here, I think this conclusively shows that you were all wrong, as the probability of his other dog being male is obviously zero. He hasn’t got another dog. QED. So many thanks to all of you, for helping me to clear that up. Now with my new-found knowledge and skills, I’m off to consider a couple of similar problems, that I have never fully understood, even after I had been told the answers. 1) What is the difference between a duck? 2) Why is a mouse when it spins? 3) Have I been looking at this thread for too long? …………..I think I may need help… Anyone? |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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Like many problems of this nature, the framing of the question is the important thing. Five billion posts ago, I said something to the effect of: "When you know that at least one dog is male, you've not only eliminated the FF pairs, you've eliminated some of the MF pairs." Steve then reasonably asked which pairs I meant by that, and I didn't give him a good answer, which is unfortunate, because I think it's in the answer to this question that any apparent disagreement may lie. We were discussing the situation in which I undertake to name the sex of one of the dogs each time (either male or female). What happens is that 1/2 the time I have an MF pair, I'll say, "At least one of the dogs is male," and 1/2 the time I have an MF pair, I'll says, "At least one of the dogs is female." So even though the MF pair is twice as likely as either of the other pairs, by having to choose which dog to tell you about, I'm cutting into those probabilities by 50%.
Let's say we have two people, X & Y, and one of them always assums, after being told the sex of one of the dogs, that the other will be the opposite sex, based on the 2/3-1/3 rationale. Call him X. Y, on the other hand, just goes for the same sex every time. Z blindfolds both of them, and tells both of them the sex of one dog every time, from random pairs. Call the older dog A, and the younger dog B. In the long run, I think we all agree there are 4 outcomes that will occur in roughly equal probabilities. A will be male or female, combined with B's being male or female. How do X and Y do in our game, then? Scenario 1: Both dogs are male. Z says, "At least one dog is male." X guesses that the other dog will be female; Y guesses that the other dog will be male. At this point, Y is 1 for 1, and X is 0 for 1. Scenario 2: Both dogs are female. Z says, "At least one dog is female." X guesses that the other dog will be male; Y guesses that the other dog will be female. X is 0 for 2; Y is 2 for 2. Scenario 3: Dog A is male; Dog B is female. Z says whatever he wants. X guesses that the opposite is true for the other dog; Y guesses that the same is true for the other dog. X is up to 1 for 3; Y is down to 2 for 3. Scenario 4: Dog A is female; Dog B is male. Z says whatever he wants. X guesses that the opposite is true for the other dog; Y guesses that the same is true for the other dog. X is up to 2 for 4; Y is down to 2 for 4. If the nature of the game is, "You will learn the sex of one of the dogs every time," both strategies are equally effective.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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Quote:
On 2006-03-08 22:18, Steve Martin wrote: Another day, Steve!!  How about these two scenarios - for each case, assume the coins are a penny and a nickel.
1) I tell you that if the nickel is heads, I'll show it to you. 2) I tell you that if at least one coin is heads, I'll show it to you. 3) I tell you that I'll show you one of the coins, regardless of what it is. In each case, I flip both coins behind a screen, then show you a nickel. It is heads. Bear in mind, per your last post, that you are now learning WHICH coin is showing heads. Are the odds that the other coin is a penny the same in each scenario? If not, which is different?
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Steve Martin Inner circle 1119 Posts
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Lobowolf,
Your thinking on this is excellent. Your post of March 9, 1.01am, in particular, is extremely thought-provoking and I believe you are correct in what you say. You have altered the way I look at the puzzle. Thanks. Steve --- "Like many problems of this nature, the framing of the question is the important thing. Five billion posts ago, I said something to the effect of: "When you know that at least one dog is male, you've not only eliminated the FF pairs, you've eliminated some of the MF pairs." Steve then reasonably asked which pairs I meant by that, and I didn't give him a good answer, which is unfortunate, because I think it's in the answer to this question that any apparent disagreement may lie. We were discussing the situation in which I undertake to name the sex of one of the dogs each time (either male or female). What happens is that 1/2 the time I have an MF pair, I'll say, "At least one of the dogs is male," and 1/2 the time I have an MF pair, I'll says, "At least one of the dogs is female." So even though the MF pair is twice as likely as either of the other pairs, by having to choose which dog to tell you about, I'm cutting into those probabilities by 50%." Yes - I agree with that. You have explained it very well. I now see that the phrasing of the question is significant.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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magicjohn2278 Special user Isle of Man UK 544 Posts
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Quote:
On 2006-03-09 08:47, LobowolfXXX wrote: Yes, I quite agree. When we have MM, MF, FM and FF, and we are told that one dog is male, then it MUST be eitHer the first or the second - say it is the first... ... The guy who says he has a FM pair is lying, we know he has one male dog, and it is the first, then he can't own a female at all. His only dog must be male. Likewise, the guy with the FF pair is mistaken, as both his dogs are the same dog (see my previous post!) so he only owns a female. So we have reduced the possibllities to: M MF F F ..interestingly, giving a zero probability that "the other dog is male" as I explained in my last post! Thanks for backing me up! |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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John...my brain hurts!!
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Steve Martin Inner circle 1119 Posts
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Quote:
On 2006-03-09 08:53, LobowolfXXX wrote: I am being cheeky here, but to answer your question: in each case, the odds that the other coin is a penny is 100%
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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[quote]On 2006-03-09 02:03, TomasB wrote:
Quote:
You are absolutely correct in this. So it's obviously the first scenario that is what you should focus on in this problem. /Tomas I actually think that it's the OTHER scenario that's more interesting and should be focused on; the MATH is simple, but the information implications are more complex and interesting. Of all the dog-pairs in the world that include at least one male dog, how many are MF? Clearly, 2/3. Of all the dog-pairs in the world that include at least one female dog, how many are MF? Again, 2/3. BUT it's an extremely attractive fallacy to go from those two propositions to the idea that if I know the sexes of two dogs, and tell you one of them, you'll be any better placed (than you were before I told you anything) to tell me the sex of other one. That's actually where it starts to get interesting. The key to THAT is this: Under those parameters, you're not learning ANYTHING that helps you. I could just decide, and not tell you, that I'll name the sex of the first dog I see. Then when I ask you the sex of the "other" dog, the real question is, "What's the sex of whichever dog I don't see first?" Not 2/3!
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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LobowolfXXX Inner circle La Famiglia 1196 Posts
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[quote]On 2006-03-09 09:12, Steve Martin wrote:
Quote:
HAHAHA. Oops.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Posted: Mar 9, 2006 03:00 am 
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