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> However, the answer is not, in fact, 1/3. Why is this? This seems like a canard to me. Here is my defense of 1/3 as a correct answer: http://gist.github.com

> However, the answer is not, in fact, 1/3. Why is this? This seems like a canard to me. Here is my defense of 1/3 as a correct answer: http://gist.github.com/578386 http://gist.github.com/578386 > Is Bayes’ theorem wrong? > No, the answer comes from an unfortunate namespace collision in the word “given”. The man “gave” us the information that he has at least one male child. By this we mean that he asserted the statement “I have at least one male child.” Now our issue is when we confuse this with being “given” that the man has at least one male child, in the sense that we should restrict to the set of universes in which the man has at least one male child. This is a very different statement than the previous one. For instance, it rules out universes where the man has two girls, but is lying to us. No, we are assuming that the givens are facts that are true. > Even if we decide to ignore the possibility that the man is lying, we should note that most universes where the man has at least one son don’t even involve him informing us of this fact, and so it may be the case that proportionally more universes where the man has two boys involve him telling us “I have at least one male child”, relative to the proportion of such universes where the man has one boy and one girl. In this case the probability that he has two boys would end up being greater than 1/3. No, we don't have to consider universes where the man has at least one male child but does not inform of us of this fact. We have a set of givens that are assumed to be true, and based on those givens and the rules of logic, we can make justifiable statements of probabilities.

aphyr 16y ago
It's a canard, but still informative. I do wish he hadn't claimed that 1/3rd is wrong by Bayesian statistics because the frequentist approach, with the same interpretation of the problem, yields exactly the same results. It's still a valuable example of how to represent unreliable measurement processes in your model, and the importance of doing so.
thwarted 16y ago
1/3 isn't correct, and I find the OP's explanation to be overly complex. The set of possibilities for two genders of two children is GG, GB, and BB. In your possibilities, GB and BG are exactly the same set (order doesn't matter in a set, only membership), so you don't have 4 possibilities, you have 3 total. Since the guy asserted that one of them is a boy, you can rule out the GG possibility. This leaves only GB and BB as possible results, both of which have a 1/2 chance of being the correct one. The guy never makes a claim that the first child or the second child is the boy (but, this doesn't change the possibility that he has two boys, it just changes which one you remove from the possibilities based on the provided information). I'm not sure that it's that people don't understand statistics (although I'm not in a position to confirm or deny that), it's that people don't understand set theory. At least if you're going to use this "genders of two children" as example.
- Tichy 16y ago
  You picked that 0.5 out of thin air, though. You forgot about BG.
  - thwarted 16y ago
    I didn't "forget" about it, I contend that case is already handled by GB; "one child is a boy and one child is a girl" is the same as "one child is a girl and one child is a boy" and can be expressed as either GB or BG, making GB and BG equivalent.
    - Tichy 16y ago
      No, choosing to not care for order changes the way probabilities have to be calculated. I suppose there was enough discussion about that on this thread already. I guess it is helpful to think about two coin flips instead of boys and girls. Not sure how it is being taught in the US. In Germany we use urn models and learn formulas for drawing balls from urns in different variations. Like returning the balls or not returning the balls, or caring for order or not. All these cases have different formulas attached...
    - jodrellblank 16y ago
      GB and BG take up more area in probability space. If you have a thousand paths in front of you, one leading to a fortune, one to a potion and the rest to a pit of death, you can't say the 998 paths are together equal to the other two just because they all end at the same place. It's much more likely you'll hit the pit of death. It's much more likely to have boy/girl children even though BG and GB look the same at the end. You get as much probability current down each of the wires BG and GB you do down the GG and BB wires, making twice as much in total for BG/GB combined. If there are four buildings, red, green, gray and gray, and you pick one at random you are more likely to end up in a gray building. You dont get to say there are three colours because I can class the two gray buildings as the same the chances are 1/3rd for each colour.
      - thwarted 16y ago
kmod 16y ago
You unfortunately fall right into the trap that he's warning about: just because you can enumerate the possible outcomes doesn't mean that they are all equally likely (cf: the Monty Hall problem). The conclusion that BB/BG/GB are all equally likely follows from the assumption that the man would definitely state that he has a male child if and only if one of those conditions were true. But what if instead we add the fact that the man would only say that he has a son if he has no daughters? This isn't contradictory with anything else in the problem, but now the answer is clearly "P(BB) = 1", which makes it hard to state that certainly "P(BB) = 1/3" when there is a consistent interpretation of the problem that gives a different answer. You should reread the way he set up the problem again: he makes the subtle distinction between things that are given to us by the omniscient problem writer and things that are given to us by characters inside the problem, for whom we have to apply Bayes Law an additional time.
- Tichy 16y ago
  I think you are confusing the likelihood of something happening with the estimate based on a given set of facts. The man doesn't have quantum children that are 1/3 boy or girl. So trivially P(the gender of the man's children is the gender of the man's children) = 1. They don't change gender depending on your guess. The only thing that changes is the trueness of your guess of the other child's gender (it changes when you change the guess) The way to get clear hear is to simulate meeting the man 1000 times, and counting how many times you would have guessed right if you said he has two boys. This yields 1/3 because you don't even guess when he has 2 girls.
Dn_Ab 16y ago
I thought on this a bit. And I do believe you are correct. Because as you note, he did in fact condition the space such that 1/3 is the correct answer and is stretching. He goes on to say "Now this means that if we want to claim that the probability that the man has two boys is , what we are really claiming is that he is equally likely to inform us that he has at least one boy, in all situations where it is true, independent of the actual gender distribution of his children. I would argue that this is quite unlikely, as if he has a boy and a girl, then he could equally well have told us that he has at least one girl, whereas he couldn’t tell us that if he has only boys". I see this point but I think this example does not lend itself to explaining the notion of defining a good prior because it is such a simple scenario - it biases responses to focus on the absurdity and unnecessarily fretful complexification. And it opens the door to all sorts of subjective objections so that nothing can actually be said. For example I can say that given this guy is talking in riddles he likes being vague and will always give me just enough true information to keep things sane but interesting. Or given he is being wishy washy then he is a deviant trying to mess up my day, very likely he has 3+ kids. A more muddled scenario would have been better for similar reasons to why humans are the hardest to animate 'correctly' - no preconceptions to distract from the presentation. Otherwise the maths is straight forward, we have been given a space to condition on: he has two kids and one of them is a boy. P( (B,B)) = 1/3 since (G,G) is eliminated. Giving a probability of 1/2 makes sense only when we are not given the key piece of information of at least one is a boy. The information we get tells us how we can construct our probability (sub)space. Consider for example I met this strange man for the first time and see that he has a son and for some reason I know he has two kids. Then the probability of two sons is 1/2. This can be seen as the fact that knowing the gender of the kid I saw says nothing of the other kid's. If we whip out some probability spaces this can be seen by thinking in terms of {(M,F), (F,M), (F,F), (M,M)} X {1,2}. Where (M,F,1) is I met the older kid who is his son first. Or (M,M,2) is I met the younger son first. Some time ago I wrote a simple finite probability space toy in F# to help in building probability intuitions by using simple spaces. Ill use it here because I think code can be clear. let Z = cartesg ["MF";"MF"; "12"] let S = definePSpaceSimp Z let pboy = conditional (fun (x:string) -> x.[0] = 'M') S let p2b =(conditional (fun (x:string) -> x.Contains("MM")) pboy).Space |> pboy.Pr > val S : ProbabilitySpace<string> = {Space = set ["FF1"; "FF2"; "FM1"; "FM2"; "MF1"; "MF2"; "MM1"; "MM2"]; Pf = <fun:definePSpaceSimp@43>; Pr = <fun:definePSpace@41>;} val pboy : ProbabilitySpace<string> = {Space = set ["MF1"; "MF2"; "MM1"; "MM2"]; Pf = <fun:filtrSpace@30>; Pr = <fun:definePSpace@41>;} val p2b : PositiveReals = 0.5 Thus we see the probability of the compound event: 2 boys given seen 1 boy is 0.5. This is because based on the information I got, the compound event I can define is very different from being given information - 2 kids, at least one boy: {(M,F), (M,M), (F,M)} >EDIT: I find it interesting that a mathematical argument is downvoted. I would appreciate any flaws to be pointed out along with a downvote so I might correct my reasoning. Currently, I do not think I made any mistakes.

aphyr 16y ago

It's a canard, but still informative. I do wish he hadn't claimed that 1/3rd is wrong by Bayesian statistics because the frequentist approach, with the same interpretation of the problem, yields exactly the same results. It's still a valuable example of how to represent unreliable measurement processes in your model, and the importance of doing so.

thwarted 16y ago

1/3 isn't correct, and I find the OP's explanation to be overly complex. The set of possibilities for two genders of two children is GG, GB, and BB. In your possibilities, GB and BG are exactly the same set (order doesn't matter in a set, only membership), so you don't have 4 possibilities, you have 3 total. Since the guy asserted that one of them is a boy, you can rule out the GG possibility. This leaves only GB and BB as possible results, both of which have a 1/2 chance of being the correct one. The guy never makes a claim that the first child or the second child is the boy (but, this doesn't change the possibility that he has two boys, it just changes which one you remove from the possibilities based on the provided information). I'm not sure that it's that people don't understand statistics (although I'm not in a position to confirm or deny that), it's that people don't understand set theory. At least if you're going to use this "genders of two children" as example.

Tichy 16y ago
You picked that 0.5 out of thin air, though. You forgot about BG.
- thwarted 16y ago
  I didn't "forget" about it, I contend that case is already handled by GB; "one child is a boy and one child is a girl" is the same as "one child is a girl and one child is a boy" and can be expressed as either GB or BG, making GB and BG equivalent.
  - Tichy 16y ago
    No, choosing to not care for order changes the way probabilities have to be calculated. I suppose there was enough discussion about that on this thread already. I guess it is helpful to think about two coin flips instead of boys and girls. Not sure how it is being taught in the US. In Germany we use urn models and learn formulas for drawing balls from urns in different variations. Like returning the balls or not returning the balls, or caring for order or not. All these cases have different formulas attached...
  - jodrellblank 16y ago
    GB and BG take up more area in probability space. If you have a thousand paths in front of you, one leading to a fortune, one to a potion and the rest to a pit of death, you can't say the 998 paths are together equal to the other two just because they all end at the same place. It's much more likely you'll hit the pit of death. It's much more likely to have boy/girl children even though BG and GB look the same at the end. You get as much probability current down each of the wires BG and GB you do down the GG and BB wires, making twice as much in total for BG/GB combined. If there are four buildings, red, green, gray and gray, and you pick one at random you are more likely to end up in a gray building. You dont get to say there are three colours because I can class the two gray buildings as the same the chances are 1/3rd for each colour.
    - thwarted 16y ago

kmod 16y ago

You unfortunately fall right into the trap that he's warning about: just because you can enumerate the possible outcomes doesn't mean that they are all equally likely (cf: the Monty Hall problem). The conclusion that BB/BG/GB are all equally likely follows from the assumption that the man would definitely state that he has a male child if and only if one of those conditions were true. But what if instead we add the fact that the man would only say that he has a son if he has no daughters? This isn't contradictory with anything else in the problem, but now the answer is clearly "P(BB) = 1", which makes it hard to state that certainly "P(BB) = 1/3" when there is a consistent interpretation of the problem that gives a different answer. You should reread the way he set up the problem again: he makes the subtle distinction between things that are given to us by the omniscient problem writer and things that are given to us by characters inside the problem, for whom we have to apply Bayes Law an additional time.

Tichy 16y ago
I think you are confusing the likelihood of something happening with the estimate based on a given set of facts. The man doesn't have quantum children that are 1/3 boy or girl. So trivially P(the gender of the man's children is the gender of the man's children) = 1. They don't change gender depending on your guess. The only thing that changes is the trueness of your guess of the other child's gender (it changes when you change the guess) The way to get clear hear is to simulate meeting the man 1000 times, and counting how many times you would have guessed right if you said he has two boys. This yields 1/3 because you don't even guess when he has 2 girls.

Dn_Ab 16y ago

I thought on this a bit. And I do believe you are correct. Because as you note, he did in fact condition the space such that 1/3 is the correct answer and is stretching. He goes on to say "Now this means that if we want to claim that the probability that the man has two boys is , what we are really claiming is that he is equally likely to inform us that he has at least one boy, in all situations where it is true, independent of the actual gender distribution of his children. I would argue that this is quite unlikely, as if he has a boy and a girl, then he could equally well have told us that he has at least one girl, whereas he couldn’t tell us that if he has only boys". I see this point but I think this example does not lend itself to explaining the notion of defining a good prior because it is such a simple scenario - it biases responses to focus on the absurdity and unnecessarily fretful complexification. And it opens the door to all sorts of subjective objections so that nothing can actually be said. For example I can say that given this guy is talking in riddles he likes being vague and will always give me just enough true information to keep things sane but interesting. Or given he is being wishy washy then he is a deviant trying to mess up my day, very likely he has 3+ kids. A more muddled scenario would have been better for similar reasons to why humans are the hardest to animate 'correctly' - no preconceptions to distract from the presentation. Otherwise the maths is straight forward, we have been given a space to condition on: he has two kids and one of them is a boy. P( (B,B)) = 1/3 since (G,G) is eliminated. Giving a probability of 1/2 makes sense only when we are not given the key piece of information of at least one is a boy. The information we get tells us how we can construct our probability (sub)space. Consider for example I met this strange man for the first time and see that he has a son and for some reason I know he has two kids. Then the probability of two sons is 1/2. This can be seen as the fact that knowing the gender of the kid I saw says nothing of the other kid's. If we whip out some probability spaces this can be seen by thinking in terms of {(M,F), (F,M), (F,F), (M,M)} X {1,2}. Where (M,F,1) is I met the older kid who is his son first. Or (M,M,2) is I met the younger son first. Some time ago I wrote a simple finite probability space toy in F# to help in building probability intuitions by using simple spaces. Ill use it here because I think code can be clear. let Z = cartesg ["MF";"MF"; "12"] let S = definePSpaceSimp Z let pboy = conditional (fun (x:string) -> x.[0] = 'M') S let p2b =(conditional (fun (x:string) -> x.Contains("MM")) pboy).Space |> pboy.Pr > val S : ProbabilitySpace<string> = {Space = set ["FF1"; "FF2"; "FM1"; "FM2"; "MF1"; "MF2"; "MM1"; "MM2"]; Pf = <fun:definePSpaceSimp@43>; Pr = <fun:definePSpace@41>;} val pboy : ProbabilitySpace<string> = {Space = set ["MF1"; "MF2"; "MM1"; "MM2"]; Pf = <fun:filtrSpace@30>; Pr = <fun:definePSpace@41>;} val p2b : PositiveReals = 0.5 Thus we see the probability of the compound event: 2 boys given seen 1 boy is 0.5. This is because based on the information I got, the compound event I can define is very different from being given information - 2 kids, at least one boy: {(M,F), (M,M), (F,M)} >EDIT: I find it interesting that a mathematical argument is downvoted. I would appreciate any flaws to be pointed out along with a downvote so I might correct my reasoning. Currently, I do not think I made any mistakes.