sartre
Footballguy
I have to get home so I can't address this specifically, but my position hasn't changed. I don't understand your insistence to NOT try this with a real deck of cards, it could save many of us so much time and energy.
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Logic problem (1 Viewer)
- Thread starter biggamer3
- Start date
bostonfred
Footballguy
I'm not sure I have enough energy left for this thread to try to come up with a new example you can disagree with. I guess you'll have to continue disagreeing with the one I said.
kutta
Footballguy
We would break even.The bold part is the problem. If you only look at one card, you cannot be certain that "there is at least on black card." Do you see that in your exact case above, sometimes you will peek at a red card in the RB or BR pairs. In that case you would have to say we have one red card. However, if you looked at both cards, in all the RB and BR pairs, you could then say that you "have at least one black card" in ALL those cases.
The issue is, by looking at one card, you will not "get" ALL the cases of BR and RB. By looking at both cards you will.
The 2/3 answer assumes that you count the black cards (or red cards, whichever we are looking for) in all the BR and RB cases. The 1/2 answer assumes that half the time you will pick black in the RB and BR cases, and half the time you will pick red.
FatMax
Member
He addressed this. If he picks a red card first, then there would be a 2/3 chance that the second card is black. I still haven't decided if I agree with that or not.
kutta
Footballguy
No there wouldn't!Tell me why this is wrong:
BB
RB
BR
RR
Pick a stack. Pick a card. What are the odds that the card chosen is opposite from the other one.
In half the cases you will pick BB or RR, you will then flip a card, and and the other card will be the same.
In half the cases you will pick RB or BR, you will then flip a card, and the other cards will be different.
That's the problem. That's it!
You can add words around and say you flipped the first card, the second card, that a moderator flipped the card, it just doesn't matter. It is what it is.
bostonfred
Footballguy
The only contentious thing about sweet J's hypotheticals is when he sometimes suggests that when one of a pair of unknown but equally likely trials is revealed, the odds of the other become 2/3. This is the only part that anyone's debating anymore, and even then, it's only so that sweet J can see the distinction between "revealing that one card is red" and "knowing that at least one of the pair is red". If there were no distinction, then I could sit at a blackjack table*, and bet $100 on the first hand. If I win, I bet $10 on the second hand, because, according to sweet J, I'd now be 2/3 likely to have the opposite result on the second hand. If, on the other hand, I lost, I'd increase my bet to $1000, because I'd now be 2/3 likely to win the next hand. So if sweet J were correct, then my system would work as follows: After playing 60 hands (or 30 such pairs of hands) you'd expect me to have won the first bet 30 times, for a gain of $3000, after which I'd have lost the second bet 2/3 of the time, losing $200, and winning it 1/3 of the time, winning $100. For a net gain of $2900. But far more profitable, I'd have lost the first hand 30 times, for a total loss of $3000. But I'd have quickly recouped my bet when I won the following bet for $1000 2/3 of the time, for a gain of $20,000, while losing 1/3 of the time, for a loss of $10,000, for a total $7000. So for every 60 hands, I'd win $9900. Of course, logic dictates that that wouldn't happen. sweet J would correctly point out that we should set the formula up like this:Win $100, lose $10 15 times for a net gain of $1,350Win $100, win $10 15 times for a net gain of $1,650Lose $100, win $1000 15 times for a net gain of $13,500Lose $100, lose $1000 15 times for a net loss of $16,500Which would net out to exactly zero. So where did we go wrong here? The problem is that simply knowing the result of the first hand is not enough for me to say, aha, the odds of the second hand are now 2/3 likely to be the opposite. The second hand is still 50/50 to win or lose, regardless of the outcome of the first hand. I suppose that if you were in cahoots with the dealer, and the dealer could somehow tip you off that you were absolutely, positively going to win at least one of the next two hands, then you would be foolish not to put as much money on each of them as you could. But if the dealer reveals the first of two trials, and says, lose, that does not mean that the second is now 2/3 likely to win. * For the purposes of this example, we're assuming the house has no edge and that a blackjack hand is 50/50, which it isn't due to blackjacks, split hands, double downs and pushes. You could use some other game like roulette or baccarat if you'd prefer; the point is the same.
bostonfred
Footballguy
Another distraction for this thread:
The house deals out two cards. They look at the two cards. If both cards are black, they flip them over, and throw them out. Otherwise, they have to play. They then offer you odds to play a game. Which of the following games should you be willing to play?
A) The dealer chooses a card to flip over. You win $19 if it's red, and lose $10 if it's black
B) The dealer flips over the card on the left. You win $19 if it's black, and lose $10 if it's red.
C) The dealer only flips over the card on the left if it is red. Otherwise, they reshuffle. You then bet $10 that the card on the right is red, losing $10 if it's black.
Also, what are the "true odds" on each of those games?
The house deals out two cards. They look at the two cards. If both cards are black, they flip them over, and throw them out. Otherwise, they have to play. They then offer you odds to play a game. Which of the following games should you be willing to play?
A) The dealer chooses a card to flip over. You win $19 if it's red, and lose $10 if it's black
B) The dealer flips over the card on the left. You win $19 if it's black, and lose $10 if it's red.
C) The dealer only flips over the card on the left if it is red. Otherwise, they reshuffle. You then bet $10 that the card on the right is red, losing $10 if it's black.
Also, what are the "true odds" on each of those games?
Death Bytes
Footballguy
24 pages?!?!?
Biggest. Fishing. Trip. EVAH!
Biggest. Fishing. Trip. EVAH!
Sweet J
Footballguy
I see my problem. I am assuming equal distribution of BB, RB, and BR when he tells me that "at least one of the cards is black." While it is true that there is an equal distribution of the four different piles, fatguy is saying "you can only assume equal distribution if he always lets you know that there is a black card in the pile when you chose the BR and RB piles. Otherwise, you will discard the choice half of the time." So, in our original problem that is stated "you have two cards/kids, and at least one is black/boy, what is the other," the answer is 2/3 chance the other is the opposite. But, my hypo as clarified is stated "you have two cards and I choose one and look at it, what is the other?" Which appears to be 5050. I can assume I'll look at the BB pile 100 times, but I can't assume that in the 100 times I look at the RB pile, they will always be Black. On the contrary, only 50.
So, yes, the wording as I clarified the hypo appears to be wrong. fatguy, I owe you $10, and you only have to show me a boobie shot of your wife instead of full frontal.
Sweet J
Footballguy
This is the "classic" two thirds answer. You are keeping all sets that the RR, RB, and BR. They have a 1/3 probability of each, so you are 2/3 more likely to get RB or BR than RR. This is where I made my mistake. I was really, really, wrong.
kutta
Footballguy
Sweet J
Footballguy
Sweet J
Footballguy
I owe my boss a lot of money for lost time. Ugh.Let's go get a beer.
kutta
Footballguy
This was the most fun I've had wasting time in quite awhile.I owe my boss a lot of money for lost time. Ugh.Let's go get a beer.
fatguyinalittlecoat
Footballguy
You live somewhere in the DC area, right? I'll take my $10 in the form of beer. Time and place TBA. And if you play your cards right, maybe I can accommodate your desire for nudity.
rick6668
Footballguy
Salvation lies within.
bostonfred
Footballguy
I can be in DC by midnight
sartre
Footballguy
bf, are you now in agreement that the problem as stated in the OP has the answer 50% ? You understand the arguments so well, but I thought you were still hung up on that last piece about how we treat the father's decision to pick a child to talk about as random.In your example of the football game, of course you are introducing a bias that is not present in the OP. I was asking you if my understanding of your example is correct, because if it is, then it has no bearing on the OP.
sartre
Footballguy
You owe my boss too.I owe my boss a lot of money for lost time. Ugh.Let's go get a beer.
bostonfred
Footballguy
Lac
No, I still think the OP should be answered as 2/3, or perhaps as somewhere between 2/3 and 50/50, because of the ambiguity. You cannot explicitly say that this is 50/50 without knowing that the methodology used to determine which child to talk about is truly random, but you also cannot call it 2/3 if we used a random methodology.Scientific American's answer was that the wording of a problem almost identical to this was too ambiguous for statisticians to come to a single conclusion, and I agree with this. I understand why someone might feel that either 50/50 or 2/3 is the correct answer, though, and I don't think you're wrong if you feel strongly about it.
kutta
Footballguy
I think the way the OP is worded we can assume the son just came up in random conversation, which would make it 50/50. The man was not asked if he had a boy, in which case it would for sure be 2/3.For example, getting back to the card example, if I look at both of them and say "one of them is black," when I could have just as easily said, "one of them is red," then I think we still have a 50/50 situation. If I look at both of them and someone asks me if at least one of them is black I must say yes in all cases but the RR case.Same with the OP. The man talked about his boy. He could have talked about his girl if he was in the BG or GB group. But he chose to talk about his boy. So I say the OP is 50%.Said one more way, if someone says, "tell me the sex of one of your children" and I say "boy," it is still 50/50 the other one is a girl. If someone says, "do you have a boy" and I say "yes," it is 2/3 that the other one is a girl.I'm a bit tired so this might be gobbly #####, so I'll have to check back in the AM to see who agrees and who doesn't.I might agree with fred that an argument could be made for 2/3 in the OP, and I don't know if it is worth the effort to debate it much more.But, hey, I'm up for it if the rest of you are...
sartre
Footballguy
Scientific American is actually clear about this, and do not suggest that statisticians can't "come to a single conclusion". They say the classic wording (of all sets with at least one of...) yields the correct answer 2/3. They also say if the sets are not restricted and the information is offered randomly then the answer is clearly 1/2. What they do suggest is that even the best of mathemeticians can get tripped up on the wording.There's no ambiguity in the OP. The puzzle is not flawed or invalid. It is just different that the one most of us are familiar with.bostonfred said:
You are holding youself back from taking that last step. Your assertion is that maybe you met this guy at a football game, and know that he has one son before we even get to the first word of the puzzle. That's the intellectual equivalent of this:
I flip two coins. The first one I see land, is heads. What is the probability that the second is tails?
Well... these coins could be wieghted. The first one did come up heads, after all. Therefor we can only assume the probability is somewhere between 0 and 50. The question is too ambiguous and we don't know whether the coins are fair or not.
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sartre
Footballguy
If you really want to get hung up on this, you should change your answer to "somewhere between 0 and 66%". If the methodology is that "all fathers with at least one son must always talk about a son first" then you get your 66%If the methodology is that "all fathers will pick a child at random" then you get 50%If the methodology is that "all fathers with at least one girl must always talk about a girl first" then you get 0%Of course there is no hint, no shred of information in the OP that there is any 'methodology' rule at all. Absent that, we assume normal distribution from a universal sample space.bostonfred said:
sartre
Footballguy
I tried to explain this back on post 901.If we start at the back end of the question, Sweet J asks, "what are the odds on the other?" We can't identify "the other" since we have made no distinction of a primary subject. Saying "at least one is a boy" is assigning a property to the pair of children, and distinguishes neither.
Other properties you could assign to the pair:
Both are tall
Neither are Cajun
Inserting these into the hypo, you can see the inconsistency:
Joe has two kids, both are tall. what are the odds on the other?
Joe has two kids, neither are cajun. what are the odds on the other?
Put another way, "at least one" is the logical equivalent of "one or more". (In the case of a pair, "one or both" is also logically equivalent.) Making the substitution:
Joe has two kids, one or both is a boy. what are the odds on the other?
People might think I am "mincing words", but being clear on whether we are discussing the set, or the individuals, is crucial to solving a problem like this. This difference is really what separates the classical puzzle from the one in the OP:A man has two children. At least one is a boy. What are the odds he has a daughter? The answer is 2/3. The entire time, we are discussing his {set} of children, but never address them specifically
A man has two children. One of them is a boy. What are the odds the other one is a girl? We start off talking about the {set}, but as soon as we get to the red part, we are now speaking in terms of individual events.
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Sweet J
Footballguy
I'm on mt bby (in a boring meeting) so forgive the shorthand:
The reason the OP is problematic is the same reason I lost my "card" bet with fatguy: if the "moderator" "looks" at both cards (or both children), you will get 2/3 because you *always* include the boy/red in *all* situations of BG GB (or RB BR). That is the only way to get, assuming 100 views, 25/75 percentage. You must,*every* time there is at least one boy, say "there is at least one boy.". That way, you will get a girl 50/75 times.
This is why we needed the moderater to look at both cards. And why I lost. If you pick one of the two cards randomly, you won't include red half the time you have RB BR, and you will end up 50/50.
If you understand that (and I finally do). Then you understand that dad must mention that he has a son EVERY time he has a son (that would be equivalent to a moderater "looking at botg cards). If the dad choses one of the kids to talkb about randomly (or by hair color, or whatever), than the answer will be 50/50, because that is equivalent to the moderator looking at just one of the cards randomly. And we proved that such was 50/50.
So, if we say "a man has two kids. *He randomly picks one of his kids to talk about.* He talks about his son. What are the odds of the other being a girl?". Is 50/50.
But if we say: "A man has two kids. *Every time he has at least one son, he will talk about him.* he talks about his son. What are the odds of a daughter?". The answer is 2/3.
To avoid having to "interpret" this problem, it is berter to say: "a man has two kids. At least one is a boy. What are the odds that th other is a girl." This way we CAN assume that the "moderater" has looked at both kids. (Or making the argument to the contrary is more strained).
Did I sum that up correctly? Anyone disagree?
The reason the OP is problematic is the same reason I lost my "card" bet with fatguy: if the "moderator" "looks" at both cards (or both children), you will get 2/3 because you *always* include the boy/red in *all* situations of BG GB (or RB BR). That is the only way to get, assuming 100 views, 25/75 percentage. You must,*every* time there is at least one boy, say "there is at least one boy.". That way, you will get a girl 50/75 times.
This is why we needed the moderater to look at both cards. And why I lost. If you pick one of the two cards randomly, you won't include red half the time you have RB BR, and you will end up 50/50.
If you understand that (and I finally do). Then you understand that dad must mention that he has a son EVERY time he has a son (that would be equivalent to a moderater "looking at botg cards). If the dad choses one of the kids to talkb about randomly (or by hair color, or whatever), than the answer will be 50/50, because that is equivalent to the moderator looking at just one of the cards randomly. And we proved that such was 50/50.
So, if we say "a man has two kids. *He randomly picks one of his kids to talk about.* He talks about his son. What are the odds of the other being a girl?". Is 50/50.
But if we say: "A man has two kids. *Every time he has at least one son, he will talk about him.* he talks about his son. What are the odds of a daughter?". The answer is 2/3.
To avoid having to "interpret" this problem, it is berter to say: "a man has two kids. At least one is a boy. What are the odds that th other is a girl." This way we CAN assume that the "moderater" has looked at both kids. (Or making the argument to the contrary is more strained).
Did I sum that up correctly? Anyone disagree?
kutta
Footballguy
I think that's pretty good. So it sounds like you think the solution to the OP should be 50/50?
sartre
Footballguy
sartre
Footballguy
You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
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FatMax
Member
The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
sartre
Footballguy
Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
Sweet J
Footballguy
I think the OP leaves out crucial info. So the answer depends on your interpretation of the question. (I know from the OPKs other responses, he clearly intended the question to be a simple "there is at laest one boy" type question, taking out the issue of whether the dad mentions one of his two kids randomly).
So, to answer your question, kutta, I would say if we assume dad "talks about one of his children randomly," and mentions his son. Then the answer is 50/50. On the same lines, if we assume that the phrase "he talks about his son" is a construct of the question meant to imply simply that the father has at least one boy, then the answer is 2/3.
If I was a teacher who gave this question on a test, and got a simple response of either 50/50 or 2/3, I would be forced to give credit for both. But I would give extra credit to someone who identified the problem that the reader. Is forced to make an assumption.
I can undersrand why you would answer 50/50. But it still requires an assumption, just like 2/3 requires an assumption. I do NOT think the answer is "somewhere between 50/50 and 2/3" because it has to be one or the other. I'd say "the answer is either one or other, depending on a necessary assumption."
So, to answer your question, kutta, I would say if we assume dad "talks about one of his children randomly," and mentions his son. Then the answer is 50/50. On the same lines, if we assume that the phrase "he talks about his son" is a construct of the question meant to imply simply that the father has at least one boy, then the answer is 2/3.
If I was a teacher who gave this question on a test, and got a simple response of either 50/50 or 2/3, I would be forced to give credit for both. But I would give extra credit to someone who identified the problem that the reader. Is forced to make an assumption.
I can undersrand why you would answer 50/50. But it still requires an assumption, just like 2/3 requires an assumption. I do NOT think the answer is "somewhere between 50/50 and 2/3" because it has to be one or the other. I'd say "the answer is either one or other, depending on a necessary assumption."
Sweet J
Footballguy
Age is important only because if you DO know that the child we are talking about is "the older one" then the answer can never be 2/3, even if worded correctly.
So, if he said: "a man has two children. At least one is a boy. It is the older child." Then the answer can't be 2/3.
I think he mentioned the fact that you didn't know the age as a clumbsy way od saying "at least one is a boy."
So, if he said: "a man has two children. At least one is a boy. It is the older child." Then the answer can't be 2/3.
I think he mentioned the fact that you didn't know the age as a clumbsy way od saying "at least one is a boy."
FatMax
Member
I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
FatMax
Member
That isn't the only clumsy thing. :smoo:
kutta
Footballguy
I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
FatMax
Member
I don't see a difference between the two. Both are 2/3.I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
kutta
Footballguy
It is getting down to the real nitty gritty, and I'm still struggling with it myself. But since we beat this horse for this long we might as well go a bit further.I would contend that as long as I have an option to tell you about either my boy or my girl (or my red card or my black card), then we still have a 50/50 situation. So if I look at two cards and have the CHOICE to tell you about either one, and I say one of them is black, the the odds are 50/50 that the other one is red. This assumes that I was just as likely to tell you about a girl as I was a boy (or my red card as opposed to my black card).If however, I do NOT have a choice to reveal whichever one I want, and I am instead asked if I have a black card (or a son) and I must answer correctly, then the odds of the other being a girl (or a red card) is 2/3.So if someone says that a guy has two kids and at least one of them is a boy, he peeked at both kids (or cards) and decided to tell me about the boy. I would say this results in a 50/50 situation.I don't see a difference between the two. Both are 2/3.I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
FatMax
Member
I don't disagree with you on principle, but I think you have it ###-backwards. Satre's question would be 2/3. Yours (might be) 50%, based on what you say. There is no choice in satre's classical version.It is getting down to the real nitty gritty, and I'm still struggling with it myself. But since we beat this horse for this long we might as well go a bit further.I would contend that as long as I have an option to tell you about either my boy or my girl (or my red card or my black card), then we still have a 50/50 situation. So if I look at two cards and have the CHOICE to tell you about either one, and I say one of them is black, the the odds are 50/50 that the other one is red. This assumes that I was just as likely to tell you about a girl as I was a boy (or my red card as opposed to my black card).If however, I do NOT have a choice to reveal whichever one I want, and I am instead asked if I have a black card (or a son) and I must answer correctly, then the odds of the other being a girl (or a red card) is 2/3.So if someone says that a guy has two kids and at least one of them is a boy, he peeked at both kids (or cards) and decided to tell me about the boy. I would say this results in a 50/50 situation.I don't see a difference between the two. Both are 2/3.I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
kutta
Footballguy
How about this.A guy looks at both cards and is going to tell you about them. He is holding a BR. He sees a red Corvette drive by and says, "one of the cards is red." The choice for him to reveal which card he is holding is random. He is not asked if he has a red card, he randomly chooses to reveal it. I would contend this is still a 50/50 situation.If someone says, "do you have a red card?" and he says "yes," then it is 2/3 that the other one is black.
kutta
Footballguy
So in Satre's version, let me ask it this way.A guy looks at your family and sees you have two kids. He then walks up and tells me, "at least one kid is a boy." I would say the odds are still 50/50 the other one is a girl.Because if one was a girl, the man could have just as easily revealed to me that "at least one kids is a girl." I think when you have the choice to reveal either the boy or the girl, you are still in a 50/50 situation.Just as in this situation. A guy looks at two cards and knows both of them. He then lays them both face down and picks one and tells me, "at least one is 'the suit I picked.'" What are the odds the other card is the opposite. It will be 50/50.Just knowing both cards doesn't matter. HOW you choose to reveal one of the cards is what is important.I don't disagree with you on principle, but I think you have it ###-backwards. Satre's question would be 2/3. Yours (might be) 50%, based on what you say. There is no choice in satre's classical version.It is getting down to the real nitty gritty, and I'm still struggling with it myself. But since we beat this horse for this long we might as well go a bit further.I would contend that as long as I have an option to tell you about either my boy or my girl (or my red card or my black card), then we still have a 50/50 situation. So if I look at two cards and have the CHOICE to tell you about either one, and I say one of them is black, the the odds are 50/50 that the other one is red. This assumes that I was just as likely to tell you about a girl as I was a boy (or my red card as opposed to my black card).If however, I do NOT have a choice to reveal whichever one I want, and I am instead asked if I have a black card (or a son) and I must answer correctly, then the odds of the other being a girl (or a red card) is 2/3.So if someone says that a guy has two kids and at least one of them is a boy, he peeked at both kids (or cards) and decided to tell me about the boy. I would say this results in a 50/50 situation.I don't see a difference between the two. Both are 2/3.I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
FatMax
Member
I don't disagree. It goes back to the BB problem that eventually settled this thread. But what you typed in not a version of Satre's classical."A man has two children. At least one is a boy. What are the odds that he has a daughter?"Nobody offers anything here. It is just known information.So in Satre's version, let me ask it this way.A guy looks at your family and sees you have two kids. He then walks up and tells me, "at least one kid is a boy." I would say the odds are still 50/50 the other one is a girl.Because if one was a girl, the man could have just as easily revealed to me that "at least one kids is a girl." I think when you have the choice to reveal either the boy or the girl, you are still in a 50/50 situation.Just as in this situation. A guy looks at two cards and knows both of them. He then lays them both face down and picks one and tells me, "at least one is 'the suit I picked.'" What are the odds the other card is the opposite. It will be 50/50.Just knowing both cards doesn't matter. HOW you choose to reveal one of the cards is what is important.I don't disagree with you on principle, but I think you have it ###-backwards. Satre's question would be 2/3. Yours (might be) 50%, based on what you say. There is no choice in satre's classical version.It is getting down to the real nitty gritty, and I'm still struggling with it myself. But since we beat this horse for this long we might as well go a bit further.I would contend that as long as I have an option to tell you about either my boy or my girl (or my red card or my black card), then we still have a 50/50 situation. So if I look at two cards and have the CHOICE to tell you about either one, and I say one of them is black, the the odds are 50/50 that the other one is red. This assumes that I was just as likely to tell you about a girl as I was a boy (or my red card as opposed to my black card).If however, I do NOT have a choice to reveal whichever one I want, and I am instead asked if I have a black card (or a son) and I must answer correctly, then the odds of the other being a girl (or a red card) is 2/3.So if someone says that a guy has two kids and at least one of them is a boy, he peeked at both kids (or cards) and decided to tell me about the boy. I would say this results in a 50/50 situation.I don't see a difference between the two. Both are 2/3.I would disagree that the way satre wrote the problem results in the classic 2/3 answer.If I look at two of my cards and tell you, "at least one of them is black" that really means nothing other than I decided to tell you that one of the cards was black. I could have said one of the cards was red. I think a better wording of the classic problem is:A man has two kids. If you ask him if at least one is a boy and he says "yes," what are the odds he also has a daughter.I agree. What I'm saying is that the classical problem that you wrote would have a different answer if you knew the boy or girl was older. That's all.The way you wrote the problem is better.Wait... are we back on "age is a magical property"?You don't need the older/younger information, it just helps folks understand that individuals are being addressed, as opposed to the set. The bare bones classic problem can be simply stated:A man has two children. At least one is a boy. What are the odds that he has a daughter?The older/younger parts are not red herrings in the classic problem. It is vitally important information. I think in this case, it was an unintentional red herring because of the ambiguity of the OP.You know I was thinking, that perhaps the OP came to the FFA with this because someone posed the question to him, and told him his 2/3 answer was incorrect. That's why he included the "not a trick question" (it is) and "you don't know which is older or younger" (red herring) bits.Pure speculation on my part, of course.We should all go back and read the sub-title of this thread one more time.
sartre
Footballguy
You would?
sartre
Footballguy
sartre
Footballguy
Again, you have to be careful with the phrasing."a man has two children. At least one is a boy. It is the older child." is the same as"a man has two children. One or both are boys. It is the older child."
