3 ms·

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 p

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
      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. I guess that's where my confusion lies. I see that the chances of picking a grey building the above is 1/2, but I assert that the set of possibilities for the children is "two girls", "different genders" and "two boys". But yours is a good explanation, I see that that "different genders" state has a double chance of occurring compared to the other two states. The Wikipedia article I cited above is interesting in that in one explanation, it bases the different results on interpretation of the givens. I see the problem stated as about a specific man's state, not the chances that a randomly chosen man/family with two children has two boys. In my case, this was a misdirection in trying to understand this.
RickHull 16y ago
> The set of possibilities for two genders of two children is GG, GB, and BB. Yes, but there are two equally probable paths to arrive at (set-theoretic) GB. Each of these paths is equally probably to the remaining paths, BB and GG. There are 4 possible paths, and (set-theoretic) GB is the result of 2 of them. Your application of set theory is inappropriate given 2 independent events.
- thwarted 16y ago
  The question isn't what is the probability of any one of the paths, the question is about the probability of the final result. A fork in the road that joins up again gives each fork equal probability of reaching the destination.
  - deleted 16y ago
    [deleted]
  - seabee 16y ago
    You can't understand the final result in isolation. Out of all men with two children, the probability that they will have two daughters is 1/4, the probability that they will have two sons is 1/4. This leaves heterogeneous offspring occurring 1/2 the time. If the man is able to make the statement "I have two kids, at least one's a boy", this puts him among the 3/4 of all men with one or two sons. The probability of a man with two sons cannot jump from 1/4 to 3/8 (half of 3/4), as you assert earlier. It's unintuitive, but it's more obvious when you negate the statement: "I have two children, but I do not have two daughters."
    - carbocation 16y ago
      Perhaps this would offer an equally appealing explanation: if the man said, "I have two children but at least my firstborn is not a girl..." then the intuitively appealing response of 1/2 becomes roughly correct.
    - thwarted 16y ago
      But we're not trying to find out if this man falls into the 3/4 of all men with one or two sons, we're trying to determine if this man's two children are both boys when we know that one of them is.
Dn_Ab 16y ago
Think of it in terms of a set of tuples and not a set. {(B,G), (G,B) , (B,B) , (G,G)}
abeppu 16y ago
1/3 would be correct if you set aside the issues about 'given' information addressed in the post. And I think the post has a valid point, that we have to be careful to appropriately understand biases and uncertainties in our data sources. But your reading of it, I think, demonstrates the real weakness of this part of the OP's discussion, namely that he's started with, and trying to expand from, a basic, 'classic' problem that people notoriously have difficulty agreeing on (e.g. http://en.wikipedia.org/wiki/Maryln_vos_Savant#.22Two_boys.22_problem http://en.wikipedia.org/wiki/Maryln_vos_Savant#.22Two_boys.2...). So as a pedagogical exercise, or as an attempt to get his point across, he has to trip over all the misunderstandings that cause people to think in the simple, classic case, that the answer is 1/2. I think he would have been better off expanding the classic coin example, only changing the story such that a naive research assistant is tasked with flipping the coin n times and bringing us the results. Then we could start by finding the probability that the coin is biased taking the research assistant at his word, and then we could reason about the probability that the research assistant either doctored the sequence when he felt they weren't 'random' enough, or the probability that he just pulled an 'HTTH' sequence out of his head without ever touching a coin.
- thwarted 16y ago
  http://en.wikipedia.org/wiki/Boy_or_Girl_paradox http://en.wikipedia.org/wiki/Boy_or_Girl_paradox has interesting content in the sections Second Question and Ambiguous Problem Statements. I'll accept that I'm wrong and that the answer is 1/3 and that I just don't understand it and that I'm supporting the point that "no one understands probability, especially me", but there has been little explanation why the birth order is even a factor. Even that Wikipedia entry labels the four possibilities with "older child" and "younger child", which seems to be extra, unneeded information when the term "at least" is used to describe the number of boys independent of order.
  - abeppu 16y ago
    Birth order per se doesn't matter; probability mass does. Birth order is an easy way to show that you're twice as likely to get a boy and a girl as you are to get two boys. But you can ignore birth order and say instead "let k be a binomially distributed random variable with n=2 and p=1/2. If you know k>=1 what is p(k=2)?". Then the answer is a straight comparison of p(k=2) to p(k=1), where we note that '2 choose 1' is 2, and '2 choose 2' is 1. Note that ordering per se isn't entering into this, but to someone without probability background, it's a bit obtuse. Describing the probability space in terms of birth order is not strictly necessary but helps a lot of people grasp the concept.

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
    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. I guess that's where my confusion lies. I see that the chances of picking a grey building the above is 1/2, but I assert that the set of possibilities for the children is "two girls", "different genders" and "two boys". But yours is a good explanation, I see that that "different genders" state has a double chance of occurring compared to the other two states. The Wikipedia article I cited above is interesting in that in one explanation, it bases the different results on interpretation of the givens. I see the problem stated as about a specific man's state, not the chances that a randomly chosen man/family with two children has two boys. In my case, this was a misdirection in trying to understand this.

RickHull 16y ago

> The set of possibilities for two genders of two children is GG, GB, and BB. Yes, but there are two equally probable paths to arrive at (set-theoretic) GB. Each of these paths is equally probably to the remaining paths, BB and GG. There are 4 possible paths, and (set-theoretic) GB is the result of 2 of them. Your application of set theory is inappropriate given 2 independent events.

thwarted 16y ago
The question isn't what is the probability of any one of the paths, the question is about the probability of the final result. A fork in the road that joins up again gives each fork equal probability of reaching the destination.
- deleted 16y ago
  [deleted]
- seabee 16y ago
  You can't understand the final result in isolation. Out of all men with two children, the probability that they will have two daughters is 1/4, the probability that they will have two sons is 1/4. This leaves heterogeneous offspring occurring 1/2 the time. If the man is able to make the statement "I have two kids, at least one's a boy", this puts him among the 3/4 of all men with one or two sons. The probability of a man with two sons cannot jump from 1/4 to 3/8 (half of 3/4), as you assert earlier. It's unintuitive, but it's more obvious when you negate the statement: "I have two children, but I do not have two daughters."
  - carbocation 16y ago
    Perhaps this would offer an equally appealing explanation: if the man said, "I have two children but at least my firstborn is not a girl..." then the intuitively appealing response of 1/2 becomes roughly correct.
  - thwarted 16y ago
    But we're not trying to find out if this man falls into the 3/4 of all men with one or two sons, we're trying to determine if this man's two children are both boys when we know that one of them is.

Dn_Ab 16y ago

Think of it in terms of a set of tuples and not a set. {(B,G), (G,B) , (B,B) , (G,G)}

abeppu 16y ago

1/3 would be correct if you set aside the issues about 'given' information addressed in the post. And I think the post has a valid point, that we have to be careful to appropriately understand biases and uncertainties in our data sources. But your reading of it, I think, demonstrates the real weakness of this part of the OP's discussion, namely that he's started with, and trying to expand from, a basic, 'classic' problem that people notoriously have difficulty agreeing on (e.g. http://en.wikipedia.org/wiki/Maryln_vos_Savant#.22Two_boys.22_problem http://en.wikipedia.org/wiki/Maryln_vos_Savant#.22Two_boys.2...). So as a pedagogical exercise, or as an attempt to get his point across, he has to trip over all the misunderstandings that cause people to think in the simple, classic case, that the answer is 1/2. I think he would have been better off expanding the classic coin example, only changing the story such that a naive research assistant is tasked with flipping the coin n times and bringing us the results. Then we could start by finding the probability that the coin is biased taking the research assistant at his word, and then we could reason about the probability that the research assistant either doctored the sequence when he felt they weren't 'random' enough, or the probability that he just pulled an 'HTTH' sequence out of his head without ever touching a coin.

thwarted 16y ago
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox http://en.wikipedia.org/wiki/Boy_or_Girl_paradox has interesting content in the sections Second Question and Ambiguous Problem Statements. I'll accept that I'm wrong and that the answer is 1/3 and that I just don't understand it and that I'm supporting the point that "no one understands probability, especially me", but there has been little explanation why the birth order is even a factor. Even that Wikipedia entry labels the four possibilities with "older child" and "younger child", which seems to be extra, unneeded information when the term "at least" is used to describe the number of boys independent of order.
- abeppu 16y ago
  Birth order per se doesn't matter; probability mass does. Birth order is an easy way to show that you're twice as likely to get a boy and a girl as you are to get two boys. But you can ignore birth order and say instead "let k be a binomially distributed random variable with n=2 and p=1/2. If you know k>=1 what is p(k=2)?". Then the answer is a straight comparison of p(k=2) to p(k=1), where we note that '2 choose 1' is 2, and '2 choose 2' is 1. Note that ordering per se isn't entering into this, but to someone without probability background, it's a bit obtuse. Describing the probability space in terms of birth order is not strictly necessary but helps a lot of people grasp the concept.