11 ms·

When intuition and math probably look wrong

25 points by 3hoss 16y ago

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
dkarl 16y ago
Another riddle where the math is easy, but translating the riddle into math is hard. Which is to say, it's not really a math riddle at all. In fact, the riddle is complicated by the fact that it's posed as a riddle -- if you had to answer the question "in real life," you would probably know how you came about the information, so the problem would be straightforward.
dododo 16y ago
isn't this the same kind of "intuitive" reasoning that fails at simpson's paradox? http://en.wikipedia.org/wiki/Simpsons_paradox http://en.wikipedia.org/wiki/Simpsons_paradox conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) < p(boy = 2 | boy >= 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)
- bad_user 16y ago
  You know, the problem with that conditional probability is that the sex of the second child is in no way conditioned by the sex of the first child, so ... p(boy = 2 | boy >= 1) = p(any child = boy) And this was the original problem that led them to the 33% probability.
  - dododo 16y ago
    in the sample space i was intending, that would not be the case. i was imagining boy as a random variable that counts instances in an order tuple of genders (the underlying sample space). you're right, without this explicit construction, it's problematic.
Daniel_Newby 16y ago
The answer given is not even wrong. The statement "I have two children, one of whom is a son born on a Tuesday" is semantically ambiguous. It can mean (1) "I have two children, and the quantity of them who are males born on a Tuesday is exactly one", (2) "I have two children, at least one of whom is a male born on a Tuesday", or even (3) "I have two children, and the maximum quantity of males born on the same Tuesday is one".
- shasta 16y ago
  I agree with you that it's not even wrong, but I disagree with your reasoning. I think (2) is the natural intended meaning of the question. The problem I have is attempting to assign a probability to something that isn't reasonably known to be based on randomness. Did the questioner arrive at this statement by picking an arbitrary child and then declaring his gender and day of birth? Or did he pick his favorite child and declare his gender and day of birth? Would he have used this same question if they had both been born on Tuesday, or would have have picked a different distinguishing feature?
RiderOfGiraffes 16y ago
I know Gary, the person who presented this, and I was there when it happened. He fully intended this to ignite the argument it has. Firstly, as presented it is clearly ambiguous. It is intended to be ambiguous, but in such a way that people who are familiar with the original version will get suckered into believing that it's well formed. Secondly, if presented precisely, the answer usually given is either 13/27 or 1/2, depending on which version. Finally, this is like the Monty Hall problem all over again. There are people arguing vehemently and without listening at all, demonstrating clearly that they are excellent at missing the point. In case you're wondering, here's one statement and answer. Suppose on knock on people's doors and ask - Do you have exactly two children? If they answer no, I move on. If they answer yes I then ask - Is at least one of them a boy born on a Tuesday? If they say no, I move on. If they look surprised and say "Yes," what is the probability that they have two boys? Answer: 13/27. Yes, it really is. If you replace the second question with "Is at least one of them a boy with red hair, left-handed, plays piano, was born on Tuesday, and has a cracked left upper incisor" then if the answer is "Yes" then the probability of both children being boys is almost exactly 50%. If, instead, you replace the second question with "Is at least one a boy" then the probability of two boys is 1/3. Finally, suppose you see a parent that you know has two children in the park with a boy. Now the probability of two boys is 50%, because, assuming uniform probabilities, having two boys makes it more likely you see them with a boy. tl;dr: It's hard, and depends precisely on the assumptions you make.
nazgulnarsil 16y ago
I had a math teacher that got mad when I used the word "intuitively". Probably because most people don't understand that when your intuition doesn't match the mathematical result you have to hit your intuition with a hammer until it does.
seanMeverett 16y ago
This is why I'm glad I was a mathematics major :)

25 points by 3hoss 16y ago

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

dkarl 16y ago

Another riddle where the math is easy, but translating the riddle into math is hard. Which is to say, it's not really a math riddle at all. In fact, the riddle is complicated by the fact that it's posed as a riddle -- if you had to answer the question "in real life," you would probably know how you came about the information, so the problem would be straightforward.

dododo 16y ago

isn't this the same kind of "intuitive" reasoning that fails at simpson's paradox? http://en.wikipedia.org/wiki/Simpsons_paradox http://en.wikipedia.org/wiki/Simpsons_paradox conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) < p(boy = 2 | boy >= 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)

bad_user 16y ago
You know, the problem with that conditional probability is that the sex of the second child is in no way conditioned by the sex of the first child, so ... p(boy = 2 | boy >= 1) = p(any child = boy) And this was the original problem that led them to the 33% probability.
- dododo 16y ago
  in the sample space i was intending, that would not be the case. i was imagining boy as a random variable that counts instances in an order tuple of genders (the underlying sample space). you're right, without this explicit construction, it's problematic.

Daniel_Newby 16y ago

The answer given is not even wrong. The statement "I have two children, one of whom is a son born on a Tuesday" is semantically ambiguous. It can mean (1) "I have two children, and the quantity of them who are males born on a Tuesday is exactly one", (2) "I have two children, at least one of whom is a male born on a Tuesday", or even (3) "I have two children, and the maximum quantity of males born on the same Tuesday is one".

shasta 16y ago
I agree with you that it's not even wrong, but I disagree with your reasoning. I think (2) is the natural intended meaning of the question. The problem I have is attempting to assign a probability to something that isn't reasonably known to be based on randomness. Did the questioner arrive at this statement by picking an arbitrary child and then declaring his gender and day of birth? Or did he pick his favorite child and declare his gender and day of birth? Would he have used this same question if they had both been born on Tuesday, or would have have picked a different distinguishing feature?

RiderOfGiraffes 16y ago

I know Gary, the person who presented this, and I was there when it happened. He fully intended this to ignite the argument it has. Firstly, as presented it is clearly ambiguous. It is intended to be ambiguous, but in such a way that people who are familiar with the original version will get suckered into believing that it's well formed. Secondly, if presented precisely, the answer usually given is either 13/27 or 1/2, depending on which version. Finally, this is like the Monty Hall problem all over again. There are people arguing vehemently and without listening at all, demonstrating clearly that they are excellent at missing the point. In case you're wondering, here's one statement and answer. Suppose on knock on people's doors and ask - Do you have exactly two children? If they answer no, I move on. If they answer yes I then ask - Is at least one of them a boy born on a Tuesday? If they say no, I move on. If they look surprised and say "Yes," what is the probability that they have two boys? Answer: 13/27. Yes, it really is. If you replace the second question with "Is at least one of them a boy with red hair, left-handed, plays piano, was born on Tuesday, and has a cracked left upper incisor" then if the answer is "Yes" then the probability of both children being boys is almost exactly 50%. If, instead, you replace the second question with "Is at least one a boy" then the probability of two boys is 1/3. Finally, suppose you see a parent that you know has two children in the park with a boy. Now the probability of two boys is 50%, because, assuming uniform probabilities, having two boys makes it more likely you see them with a boy. tl;dr: It's hard, and depends precisely on the assumptions you make.

nazgulnarsil 16y ago

I had a math teacher that got mad when I used the word "intuitively". Probably because most people don't understand that when your intuition doesn't match the mathematical result you have to hit your intuition with a hammer until it does.

seanMeverett 16y ago

This is why I'm glad I was a mathematics major :)