4 ms·

No, that is not what this says. What this says is that there is ambiguity in the description of the situation. Is the question about the probability of a second

No, that is not what this says. What this says is that there is ambiguity in the description of the situation. Is the question about the probability of a second girl in the set of all families with 2 children and at least one boy, or is the question about the probability of a second girl in the set of all families with 2 children given that we know one of them is a boy. They are different sets of families. That is the point of this paradox. This has nothing to do with quantum mechanics.

pg_bot 8y ago
Say my wife tells me that the neighbor has 2 kids and one of them is a boy what is the likelihood that their other child is a girl? Given what I know about the current situation, the probability that their other child is a girl is 66.666% If I look out the window to that neighbor's yard and see a boy playing outside, the probability that their other child is a girl drops to 50%. How I know the information about on data point one affected the likelihood of an unrelated one. What seems like identical knowledge on the outside, changes outcomes based on how I accrued that knowledge. If that doesn't mimic the double-slit experiment, I don't know what does.
- perl4ever 8y ago
  It doesn't make sense to me that in order to get the probability of 2/3, you implicitly consider "a boy and a girl" to be different from "a girl and a boy". Somehow, you're saying that seeing the boy means you have ruled out the possibility of "a girl and a boy". But the ordering is in no way related to some arbitrary observation. It's just lexical.
  - pg_bot 8y ago
    Here's the math, I will break it down based on Boy/Girl and seen. For the first case here are the different combinations for two children based on sex (let's say left denotes older and right denotes younger to distinguish the difference between girl and a boy and boy and a girl.) B B B G G B G G These outcomes are each equally likely, but since we know that one has to be a boy we can eliminate the (G G) case leaving us with two chances out of three that the other child is a girl. The other case can be viewed as follows (S) denotes seen. B(S) B B B(S) B(S) G B G(S) G(S) B G B(S) G(S) G G G(S) I can eliminate all cases where I see a girl out the window, leaving me with the following equally likely cases: The neighbor has two boys and I see the older boy. The neighbor has two boys and I see the younger boy. The neighbor has an older boy whom I saw and younger girl. The neighbor has an older girl and a younger boy whom I saw. The neighbor has a girl in 50% of the remaining possible cases.
    - deleted 8y ago
      [deleted]
    - perl4ever 8y ago
      Suppose they are twins and therefore there is no older and younger. Does that change the probability from 2/3 in the first case?
      - reitanqild 8y ago
        I guess this is one of a few place to answer mu. One of the twins is still considered to be born first and therefore older :-)
        perl4ever 8y ago
        I wrote a program and successfully got the right answers experimentally for each case, so I'm not doubting the answers, but I don't buy the explanation. While you may argue that one twin is always born first, this is an idealized mathematical problem, so it can't depend on particular details of physical reality in that way. There is nothing logically preventing them from being the same age, so it can't be pivotal to solving the problem. And in fact, I experimentally got the right result without any reference to ages.
        zazen 8y ago
        So, you don't actually have to regard BG and GB as separate cases if you don't want to. People like to write the outcome space as ordered pairs BB, BG, GB, GG (ordered by age or whatever else), because this has the advantage that all four outcomes are equiprobable, so you can calculate probabilities just by counting. If you prefer to regard the outcome space as unordered pairs {B,B}, {B,G}, {G,G} this is absolutely fine: you just have to bear in mind that the prior probability of {B,G} is 1/2, whereas it's 1/4 for the other pairs, so a tiny bit more calculation might be required. P({B,G}| not {G, G}) = P({B,G})/(P{B,B} + P{B,G}) = (1/2)/((1/2) + (1/4)) = 2/3 as with the other method.
      - rcxdude 8y ago
- pmikesell 8y ago
  No, that's not what this is about. Probability means this: "in X tries, how likely is it for Y to happen". The ambiguity here is this: what is the set of X? Is the set of all families with two children at least one boy, or the set of all families with 2 children of which you have determined one is a boy. Those are different sets. It doesn't matter how you discovered the information. The apparent paradox is all grammar and set description. Think of it this way - you want to test this out and actually run this experiment? Great! What's the experiment? Are you starting with all families with 2 children, or all families which 2 children at least one boy. >> How I know the information about on data point one affected the likelihood of an unrelated one No, it doesn't.

pg_bot 8y ago

Say my wife tells me that the neighbor has 2 kids and one of them is a boy what is the likelihood that their other child is a girl? Given what I know about the current situation, the probability that their other child is a girl is 66.666% If I look out the window to that neighbor's yard and see a boy playing outside, the probability that their other child is a girl drops to 50%. How I know the information about on data point one affected the likelihood of an unrelated one. What seems like identical knowledge on the outside, changes outcomes based on how I accrued that knowledge. If that doesn't mimic the double-slit experiment, I don't know what does.

perl4ever 8y ago
It doesn't make sense to me that in order to get the probability of 2/3, you implicitly consider "a boy and a girl" to be different from "a girl and a boy". Somehow, you're saying that seeing the boy means you have ruled out the possibility of "a girl and a boy". But the ordering is in no way related to some arbitrary observation. It's just lexical.
- pg_bot 8y ago
  Here's the math, I will break it down based on Boy/Girl and seen. For the first case here are the different combinations for two children based on sex (let's say left denotes older and right denotes younger to distinguish the difference between girl and a boy and boy and a girl.) B B B G G B G G These outcomes are each equally likely, but since we know that one has to be a boy we can eliminate the (G G) case leaving us with two chances out of three that the other child is a girl. The other case can be viewed as follows (S) denotes seen. B(S) B B B(S) B(S) G B G(S) G(S) B G B(S) G(S) G G G(S) I can eliminate all cases where I see a girl out the window, leaving me with the following equally likely cases: The neighbor has two boys and I see the older boy. The neighbor has two boys and I see the younger boy. The neighbor has an older boy whom I saw and younger girl. The neighbor has an older girl and a younger boy whom I saw. The neighbor has a girl in 50% of the remaining possible cases.
  - deleted 8y ago
    [deleted]
  - perl4ever 8y ago
    Suppose they are twins and therefore there is no older and younger. Does that change the probability from 2/3 in the first case?
    - reitanqild 8y ago
      I guess this is one of a few place to answer mu. One of the twins is still considered to be born first and therefore older :-)
      - perl4ever 8y ago
        I wrote a program and successfully got the right answers experimentally for each case, so I'm not doubting the answers, but I don't buy the explanation. While you may argue that one twin is always born first, this is an idealized mathematical problem, so it can't depend on particular details of physical reality in that way. There is nothing logically preventing them from being the same age, so it can't be pivotal to solving the problem. And in fact, I experimentally got the right result without any reference to ages.
        zazen 8y ago
        So, you don't actually have to regard BG and GB as separate cases if you don't want to. People like to write the outcome space as ordered pairs BB, BG, GB, GG (ordered by age or whatever else), because this has the advantage that all four outcomes are equiprobable, so you can calculate probabilities just by counting. If you prefer to regard the outcome space as unordered pairs {B,B}, {B,G}, {G,G} this is absolutely fine: you just have to bear in mind that the prior probability of {B,G} is 1/2, whereas it's 1/4 for the other pairs, so a tiny bit more calculation might be required. P({B,G}| not {G, G}) = P({B,G})/(P{B,B} + P{B,G}) = (1/2)/((1/2) + (1/4)) = 2/3 as with the other method.
    - rcxdude 8y ago
pmikesell 8y ago
No, that's not what this is about. Probability means this: "in X tries, how likely is it for Y to happen". The ambiguity here is this: what is the set of X? Is the set of all families with two children at least one boy, or the set of all families with 2 children of which you have determined one is a boy. Those are different sets. It doesn't matter how you discovered the information. The apparent paradox is all grammar and set description. Think of it this way - you want to test this out and actually run this experiment? Great! What's the experiment? Are you starting with all families with 2 children, or all families which 2 children at least one boy. >> How I know the information about on data point one affected the likelihood of an unrelated one No, it doesn't.