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This is wrong. In oversimplifying your table for potential idiot readers by introducing the silent variable of older vs younger child, you have introduced orde

This is wrong. In oversimplifying your table for potential idiot readers by introducing the silent variable of older vs younger child, you have introduced order into a statistics problem wherein order is irrelevant and masked the fact that what matters is which of the children you already know to be a boy. You grant this difference for the BG pairing, thus producing two options, but not for the BB pairing. There is not ONE way of knowing this for BB, as you present in the table, but two: you know the older is the boy, or you know the younger is the boy. 2/4 = 1/2. QED. What you should have done in the first place, however (unless your aim is to produce a blog post which, apparently, can convince otherwise intelligent people that irrelevant information can magically become relevant) was simply remove the one boy from the equation and reformulate the question. What is the probability that this other child is a boy (and thus that both are boys)? 1/2. Fuck the table. Had you not used a table, this would never have happened. But it feels authoritative, right? Thank you for the psychology lesson. . . . Downvotes but not refutations, because there aren't any. I'll assume it's my tone. My opinion of the competence of the average reader on this site has plummeted reading the other comments, however, so hey, fuck you guys too. :)

klodolph 13y ago
You're oversimplifying. The problem is the interpretation of the question in precise terms. We are given some background information and then told the observation of a particular random variable. In the given article, the random variable is interpreted as "yes/no: at least one child is a boy born on a Tuesday". Rambling about "order" does not change the fact that the article is correct, given the definition of the random variable that is observed. The problem is that this interpretation is only correct if you imagine that you are always given the observation of the same random variable no matter what the genders or birthdays of the children actually are. For example, if I had two girls born on Wednesday, I would then tell you "no child is a boy born on a Tuesday". That's an unlikely scenario which is what makes it legitimately surprising. It's like the Monty Hall problem. What if the host picks the door in advance? What if the host only opens a door with a goat under the condition that you picked the door with the prize? What if the host only opens a door with a goat under the condition that you picked the wrong door? The article has problems but "wrong" is not one of them. > so hey, fuck you guys too Hacker News is a terrible place to visit if you want to take the voting system personally. Don't make this about you. It's about the post.
- gametheoretic 13y ago
  I'm not, you're overcomplicating! X is a boy. What is the chance that Y is also a boy? 1/2 or 1/3? Non-rhetorical, answer the question as posed.
  - klodolph 13y ago
    > answer the question as posed You can't give a correct answer to an incomplete question. Sit down and define the problem and if you come up with a different answer it's because you made a different assumption about the definition of the observed random variable. Let's suppose there are four universes. * Two boys (25% chance) * Two girls (25% chance) * Boy and girl (50% chance) If we assume that the observed random variable has a preimage that contains all possibilities where the statement is true, then we come to the 1/3 conclusion, which is counterintuitive but true. This is the position of the article. I have come to the same conclusion without using order, but I have used the same definition of the observed random variable. Tell me what you think the random variable is. We could assume that the random variable has a preimage which only contains the case where both children are boys. Bam, probability 100%, not 1/2 or 1/3. Or we can assome that the random variable has a preimage which only contains the case where one child is of each gender. Bam. Probability 0%. So any answer between 0% and 100% is defensible.
    - gametheoretic 13y ago
      God, I knew this was coming. Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it), let me try this instead: You misunderstand Monty Hall. I would know; I learned about it by being asked the question and getting it right, not by reading someone else's explanation (which leaves room for misinterpretation). The key to Monty Hall is none of what you stated (or rather, rhetorically asked); it's that the host must pick a door that is both a) bad and b) not yours. If your door is good, then he can leave either, but if your door is bad, then he must leave the good one. So, two possibilities: the door he left was either bad or good, BUT the rule driving his choice wasn't 50/50, so the odds of it being bad or good aren't 50/50, so neither are the odds of yours, even though there are only two left. Take a minute, think about it. Now ask yourself which "door" (read child) the father told you was a boy. There's the gotcha. Under your/the author's formulation of the question, he doesn't remove a door in the BB universe, but he does in the BG universe. There are two BG's because the table acknowledges that he can tell you which of the two is definitely a boy, and thus he has two options for doing so BUT HE DOESN'T, HE HAS ONE BUT THERE ARE TWO BG UNIVERSES, SO IT LOOKS LIKE TWO.
deleted 13y ago
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ColinWright 13y ago
Your reasoning works if the person says "The older is a boy." You claim that this is correct. Consider ... Assuming the sex of a child is 50:50, then consider all couples with exactly two children. Of those, consider only those who can truthfully say "at least one of my children is a boy." Of those couples, 1/3 will have two boys. Interpreting the initial problem in this manner, the answer of 1/3 is right. What I'd like to know is why you think the initial problem should be interpreted in any different way.
- deleted 13y ago
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  - ColinWright 13y ago
    > ... in the table, he DOES tell you that in the BG case. > Otherwise there aren't two spots; there's only one. He > introduces the older/younger thing, then fails to apply > the new information across all cases. I'm having real trouble understanding you here. The explanation lists all the possibilities, and to do so it's necessary to distinguish between the children. The most obvious way to do that is to talk about the older and younger. we know that in families with two children about half the time you have one of each sex. You only get that if you distinguish between the children in some sense so that there are four overall possibilities: BB, BG, GB, GG. If you don't distinguish between the children then there are only three possibilities: Both boys, both girls, one of each. Doing real world trials clearly shows that model to be flawed. We must distinguish between the children when enumerating cases. Forgive me if this is all obvious to you, but I honestly can't see your argument, so it's necessary to lay down much more detail to try to find where your reasoning varies from mine. So now consider the situation I laid out. Take all families with two children. There are four equally likely possibilities. Eliminate those who cannot truthfully say "At least one child is a boy." You are left with three equally likely possibilities. In only one of those do we have two boys. Thus one out of three possibilities has both children boys. The probability of both children being boys is 1/3. Can you explain where that reasoning is faulty? Also, computer simulations clearly show that under this model of what's going on, the chance of two boys is 1/3. If you explain the faulty reasoning, you'll also need to explain why the computer simulation gives the same answer.
brazzy 13y ago
You are an overconfident ignoramus. Thanks for demonstrating the Dunning-Kruger effect so impressively. The reason for having BG and GB is the fact that it's twice as likely to occur than each of GG and BB.
- gametheoretic 13y ago
  Hahahaha oh my god what a response. The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless. In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second. Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.
  - ColinWright 13y ago
    Right, now it's clear that you are answering a different question, and in that question, the answer is 1/2. Many people insist that to be the right question, because the answer matches their intuition, but it is a different question. Not recognizing the difference is what lets some poker players make money. Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two sons? (corrected in edit - thank you) Answer: 1/2. Another version: I meet a man and his son, and in conversation it emerges that he has two children. I can see he has at least one son, what are the chances he has two sons? Answer: 1/2. Another version: I meet a man and it emerges in conversation that he has two children. I observe that he is carrying a shopping bag with a child's dress in it, from which I conclude that he has at least one girl. What are the chances he has two girls? Answer: 1/3. (I had to change gender because I couldn't think of anything that would be specific to a boy.) Your dismissal of the idea that there are other interpretations than your own is disappointing. The language in which you couch your dismissals even more so. Are you like this in person?
    - 13y ago
ajanuary 13y ago
I did an experiment: I kept flipping two coins [1]. If it came up tails, tails I ignored it. If it came up heads, heads I noted it down in one column. If it came up heads, tails I noted it down in another column. After many flips the ratio was 1:3. As far as I can tell this is an accurate simulation of the problem. We have two things, coins or children, which have a 50:50 chance of being one value or another. We are told one combination of values isn't the case, so we discount them from the simulation. We then use the frequency of the target combination to approximate its probability. We haven't added any artificial ordering into the problem. I know it's not well argued logical reasoning, but it supports the articles approach. Can you point out where I tripped up in the simulation? [1] Of course I didn't actually sit there flipping coins. I wrote a program to do it. (let [sample-size 10000000] (/ (get (frequencies (take sample-size (filter #(not= % [:girl :girl]) (repeatedly #(vector (rand-nth [:boy :girl]) (rand-nth [:boy :girl])))))) [:boy :boy]) sample-size))
- icameron 13y ago
  Instead of actually programming, I flipped 2 coins this morning and left them on my desk. They are still sitting there. Either HT, TH, HH or TT I just remembered one of them is tails, so is there is a 1/2 chance that they are both tails now? Nope it's 1/3 chance they are both tails. Either HT, TH, or TT.
  - ajanuary 13y ago
    Of course that's a much simpler way to simulate the problem, but then you get people trying to argue that HT and TH should be collapsed into a single state. By using a simulation with repetition it's easier to pose the states as 'TT' and 'not TT' and use frequencies to show 'not TT' occurs more frequently than 'TT'. I find this starts to help convince people that the states shouldn't be collapsed, even if it doesn't explicitly explain why.
- gametheoretic 13y ago
  >We haven't added any artificial ordering into the problem. Yes you have! To run the experiment, you have to treat TH and HT as different but you "pretend" not to do so by adding them into the same pile. You're pretending to know that the father tells you which child is the "known boy", but by the statement of the problem, you don't know that. Why is this an error? Because the father must follow a rule - he must give you a boy, and there are two ways for him to do that in BB but you only recognize one. If you go the "table-minus" route, which both the author and your experiment do. But you can't do that without affecting the odds! By telling you they aren't both girls, he lowers the chance that either of them might be. Yes, I'm serious. Correct formulation of table: a) BG; b) GB; c) BB and he gives you the first; d) BB and he gives you the second.

klodolph 13y ago

You're oversimplifying. The problem is the interpretation of the question in precise terms. We are given some background information and then told the observation of a particular random variable. In the given article, the random variable is interpreted as "yes/no: at least one child is a boy born on a Tuesday". Rambling about "order" does not change the fact that the article is correct, given the definition of the random variable that is observed. The problem is that this interpretation is only correct if you imagine that you are always given the observation of the same random variable no matter what the genders or birthdays of the children actually are. For example, if I had two girls born on Wednesday, I would then tell you "no child is a boy born on a Tuesday". That's an unlikely scenario which is what makes it legitimately surprising. It's like the Monty Hall problem. What if the host picks the door in advance? What if the host only opens a door with a goat under the condition that you picked the door with the prize? What if the host only opens a door with a goat under the condition that you picked the wrong door? The article has problems but "wrong" is not one of them. > so hey, fuck you guys too Hacker News is a terrible place to visit if you want to take the voting system personally. Don't make this about you. It's about the post.

gametheoretic 13y ago
I'm not, you're overcomplicating! X is a boy. What is the chance that Y is also a boy? 1/2 or 1/3? Non-rhetorical, answer the question as posed.
- klodolph 13y ago
  > answer the question as posed You can't give a correct answer to an incomplete question. Sit down and define the problem and if you come up with a different answer it's because you made a different assumption about the definition of the observed random variable. Let's suppose there are four universes. * Two boys (25% chance) * Two girls (25% chance) * Boy and girl (50% chance) If we assume that the observed random variable has a preimage that contains all possibilities where the statement is true, then we come to the 1/3 conclusion, which is counterintuitive but true. This is the position of the article. I have come to the same conclusion without using order, but I have used the same definition of the observed random variable. Tell me what you think the random variable is. We could assume that the random variable has a preimage which only contains the case where both children are boys. Bam, probability 100%, not 1/2 or 1/3. Or we can assome that the random variable has a preimage which only contains the case where one child is of each gender. Bam. Probability 0%. So any answer between 0% and 100% is defensible.
  - gametheoretic 13y ago
    God, I knew this was coming. Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it), let me try this instead: You misunderstand Monty Hall. I would know; I learned about it by being asked the question and getting it right, not by reading someone else's explanation (which leaves room for misinterpretation). The key to Monty Hall is none of what you stated (or rather, rhetorically asked); it's that the host must pick a door that is both a) bad and b) not yours. If your door is good, then he can leave either, but if your door is bad, then he must leave the good one. So, two possibilities: the door he left was either bad or good, BUT the rule driving his choice wasn't 50/50, so the odds of it being bad or good aren't 50/50, so neither are the odds of yours, even though there are only two left. Take a minute, think about it. Now ask yourself which "door" (read child) the father told you was a boy. There's the gotcha. Under your/the author's formulation of the question, he doesn't remove a door in the BB universe, but he does in the BG universe. There are two BG's because the table acknowledges that he can tell you which of the two is definitely a boy, and thus he has two options for doing so BUT HE DOESN'T, HE HAS ONE BUT THERE ARE TWO BG UNIVERSES, SO IT LOOKS LIKE TWO.

deleted 13y ago

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ColinWright 13y ago

Your reasoning works if the person says "The older is a boy." You claim that this is correct. Consider ... Assuming the sex of a child is 50:50, then consider all couples with exactly two children. Of those, consider only those who can truthfully say "at least one of my children is a boy." Of those couples, 1/3 will have two boys. Interpreting the initial problem in this manner, the answer of 1/3 is right. What I'd like to know is why you think the initial problem should be interpreted in any different way.

deleted 13y ago
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- ColinWright 13y ago
  > ... in the table, he DOES tell you that in the BG case. > Otherwise there aren't two spots; there's only one. He > introduces the older/younger thing, then fails to apply > the new information across all cases. I'm having real trouble understanding you here. The explanation lists all the possibilities, and to do so it's necessary to distinguish between the children. The most obvious way to do that is to talk about the older and younger. we know that in families with two children about half the time you have one of each sex. You only get that if you distinguish between the children in some sense so that there are four overall possibilities: BB, BG, GB, GG. If you don't distinguish between the children then there are only three possibilities: Both boys, both girls, one of each. Doing real world trials clearly shows that model to be flawed. We must distinguish between the children when enumerating cases. Forgive me if this is all obvious to you, but I honestly can't see your argument, so it's necessary to lay down much more detail to try to find where your reasoning varies from mine. So now consider the situation I laid out. Take all families with two children. There are four equally likely possibilities. Eliminate those who cannot truthfully say "At least one child is a boy." You are left with three equally likely possibilities. In only one of those do we have two boys. Thus one out of three possibilities has both children boys. The probability of both children being boys is 1/3. Can you explain where that reasoning is faulty? Also, computer simulations clearly show that under this model of what's going on, the chance of two boys is 1/3. If you explain the faulty reasoning, you'll also need to explain why the computer simulation gives the same answer.

brazzy 13y ago

You are an overconfident ignoramus. Thanks for demonstrating the Dunning-Kruger effect so impressively. The reason for having BG and GB is the fact that it's twice as likely to occur than each of GG and BB.

gametheoretic 13y ago
Hahahaha oh my god what a response. The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless. In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second. Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.
- ColinWright 13y ago
  Right, now it's clear that you are answering a different question, and in that question, the answer is 1/2. Many people insist that to be the right question, because the answer matches their intuition, but it is a different question. Not recognizing the difference is what lets some poker players make money. Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two sons? (corrected in edit - thank you) Answer: 1/2. Another version: I meet a man and his son, and in conversation it emerges that he has two children. I can see he has at least one son, what are the chances he has two sons? Answer: 1/2. Another version: I meet a man and it emerges in conversation that he has two children. I observe that he is carrying a shopping bag with a child's dress in it, from which I conclude that he has at least one girl. What are the chances he has two girls? Answer: 1/3. (I had to change gender because I couldn't think of anything that would be specific to a boy.) Your dismissal of the idea that there are other interpretations than your own is disappointing. The language in which you couch your dismissals even more so. Are you like this in person?
  - 13y ago

ajanuary 13y ago

I did an experiment: I kept flipping two coins [1]. If it came up tails, tails I ignored it. If it came up heads, heads I noted it down in one column. If it came up heads, tails I noted it down in another column. After many flips the ratio was 1:3. As far as I can tell this is an accurate simulation of the problem. We have two things, coins or children, which have a 50:50 chance of being one value or another. We are told one combination of values isn't the case, so we discount them from the simulation. We then use the frequency of the target combination to approximate its probability. We haven't added any artificial ordering into the problem. I know it's not well argued logical reasoning, but it supports the articles approach. Can you point out where I tripped up in the simulation? [1] Of course I didn't actually sit there flipping coins. I wrote a program to do it. (let [sample-size 10000000] (/ (get (frequencies (take sample-size (filter #(not= % [:girl :girl]) (repeatedly #(vector (rand-nth [:boy :girl]) (rand-nth [:boy :girl])))))) [:boy :boy]) sample-size))

icameron 13y ago
Instead of actually programming, I flipped 2 coins this morning and left them on my desk. They are still sitting there. Either HT, TH, HH or TT I just remembered one of them is tails, so is there is a 1/2 chance that they are both tails now? Nope it's 1/3 chance they are both tails. Either HT, TH, or TT.
- ajanuary 13y ago
  Of course that's a much simpler way to simulate the problem, but then you get people trying to argue that HT and TH should be collapsed into a single state. By using a simulation with repetition it's easier to pose the states as 'TT' and 'not TT' and use frequencies to show 'not TT' occurs more frequently than 'TT'. I find this starts to help convince people that the states shouldn't be collapsed, even if it doesn't explicitly explain why.
gametheoretic 13y ago
>We haven't added any artificial ordering into the problem. Yes you have! To run the experiment, you have to treat TH and HT as different but you "pretend" not to do so by adding them into the same pile. You're pretending to know that the father tells you which child is the "known boy", but by the statement of the problem, you don't know that. Why is this an error? Because the father must follow a rule - he must give you a boy, and there are two ways for him to do that in BB but you only recognize one. If you go the "table-minus" route, which both the author and your experiment do. But you can't do that without affecting the odds! By telling you they aren't both girls, he lowers the chance that either of them might be. Yes, I'm serious. Correct formulation of table: a) BG; b) GB; c) BB and he gives you the first; d) BB and he gives you the second.