4 ms·

Ok, I was about to rage about yet another article going on about the Two Children problem and getting it wrong by leaving out a whole host of children (i.e. chi

by bena 16y ago

Ok, I was about to rage about yet another article going on about the Two Children problem and getting it wrong by leaving out a whole host of children (i.e. children in families of more or less than two children). Then it surprised me by not only acknowledging it, but acknowledging that the the 50% answer is correct when we are selecting from an arbitrary family (as the original problem is usually presented). It then goes on to acknowledge that the 33% answer would be correct if we specifically choose a two child family that fits the parameters of the problem beforehand. It's all about why the information was selected.

bad_user 16y ago
> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order. So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl So if you're told one of them is a boy ... boy, girl boy, boy And that's it ... a 50% probability that both children are boys.
- RiderOfGiraffes 16y ago
  You can't mean that. There are four possibilities: Older child boy, younger child boy Older child boy, younger child girl Older child girl, younger child boy Older child girl, younger child girl Of those only one is precluded by saying (at least) one child is a boy.
  - bad_user 16y ago
    No there aren't ... notions of order like older/younger don't enter the equation (as the problem was stated). Oh well, I guess this is what the article is talking about :)
    - RiderOfGiraffes 16y ago
      Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it. Suppose I roll two dice until at least one of them shows a 6. What's the odds of both being 6's? I've said nothing about the red die versus the blue die, but the underlying truth requires that the situations are kept separate. It's only - as far as we know - in quantum mechanics where you deliberately lose the distinction. I've done these as real world experiments as I explore them with kids, and I have a lot of direct experience. If you disagree then I'd be delighted to gamble with you.
      - bad_user 16y ago
        > Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it. You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head? Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin. That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.
        RiderOfGiraffes 16y ago
        Here's the problem as Gary originally stated it, and as the article quotes it: > I have two children, one of whom is a son born > on a Tuesday. What is the probability that I > have two boys? You say: the problem as stated in the article is that you know one coin is going to be a head, No. The point of the article is that you don't know how or why you are given this information. Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3. I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous. It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.
- RiderOfGiraffes 16y ago
  Actually, I'm going to stop now: http://xkcd.com/386/ http://xkcd.com/386/ It's pretty clear I won't convince you. Let me leave you with these questions: If I toss two coins until at least one shows a Head, what's the probability that both are Heads? If I roll two dice until at least one shows a 6, what's the probability that both are 6's? If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both are red? Are you sure?
  - bad_user 16y ago
    It's been fun :-) > If I toss two coins until at least one shows a Head, what's the probability that both are Heads? I answered in my last reply ... P(head_a AND head_b | head_b) = 1/3 The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that). P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2 QED :)
    - RiderOfGiraffes 16y ago
      I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking about something completely different, and I really don't understand what you're saying. And that's the last I'll say.
      - bad_user 16y ago
        Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until this happens normally this would be 3/4 Making the problem ... P( blue AND red | blue OR red) = P( blue OR red | blue AND red ) * P( blue AND red ) / P(blue OR red) (applied bayes) = 1 * P(blue) * P(red) / 1 But ... P(blue OR red) = P(blue) + P(red) - P(blue) * P(red) = 1 => P(blue) * P(red) = 1 - P(blue) - P(red) = 1 - 1/4 - 1/4 = 1/2 Show me the error. [EDITED] ... modified the stuff as I've totally fucked up the previous version :) ... as I said, I'm rusty

by bena 16y ago

bad_user 16y ago

> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order. So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl So if you're told one of them is a boy ... boy, girl boy, boy And that's it ... a 50% probability that both children are boys.

RiderOfGiraffes 16y ago
You can't mean that. There are four possibilities: Older child boy, younger child boy Older child boy, younger child girl Older child girl, younger child boy Older child girl, younger child girl Of those only one is precluded by saying (at least) one child is a boy.
- bad_user 16y ago
  No there aren't ... notions of order like older/younger don't enter the equation (as the problem was stated). Oh well, I guess this is what the article is talking about :)
  - RiderOfGiraffes 16y ago
    Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it. Suppose I roll two dice until at least one of them shows a 6. What's the odds of both being 6's? I've said nothing about the red die versus the blue die, but the underlying truth requires that the situations are kept separate. It's only - as far as we know - in quantum mechanics where you deliberately lose the distinction. I've done these as real world experiments as I explore them with kids, and I have a lot of direct experience. If you disagree then I'd be delighted to gamble with you.
    - bad_user 16y ago
      > Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it. You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head? Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin. That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.
      - RiderOfGiraffes 16y ago
        Here's the problem as Gary originally stated it, and as the article quotes it: > I have two children, one of whom is a son born > on a Tuesday. What is the probability that I > have two boys? You say: the problem as stated in the article is that you know one coin is going to be a head, No. The point of the article is that you don't know how or why you are given this information. Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3. I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous. It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.
RiderOfGiraffes 16y ago
Actually, I'm going to stop now: http://xkcd.com/386/ http://xkcd.com/386/ It's pretty clear I won't convince you. Let me leave you with these questions: If I toss two coins until at least one shows a Head, what's the probability that both are Heads? If I roll two dice until at least one shows a 6, what's the probability that both are 6's? If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both are red? Are you sure?
- bad_user 16y ago
  It's been fun :-) > If I toss two coins until at least one shows a Head, what's the probability that both are Heads? I answered in my last reply ... P(head_a AND head_b | head_b) = 1/3 The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that). P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2 QED :)
  - RiderOfGiraffes 16y ago
    I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking about something completely different, and I really don't understand what you're saying. And that's the last I'll say.
    - bad_user 16y ago
      Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until this happens normally this would be 3/4 Making the problem ... P( blue AND red | blue OR red) = P( blue OR red | blue AND red ) * P( blue AND red ) / P(blue OR red) (applied bayes) = 1 * P(blue) * P(red) / 1 But ... P(blue OR red) = P(blue) + P(red) - P(blue) * P(red) = 1 => P(blue) * P(red) = 1 - P(blue) - P(red) = 1 - 1/4 - 1/4 = 1/2 Show me the error. [EDITED] ... modified the stuff as I've totally fucked up the previous version :) ... as I said, I'm rusty