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It's more like seeing one boy actually impacts the gender of the second child, which is why the quantum experiment is so mind boggling. It doesn't seem to have

It's more like seeing one boy actually impacts the gender of the second child, which is why the quantum experiment is so mind boggling. It doesn't seem to have any resemblance with classical physics and only makes sense in quantum physics.

pg_bot 8y ago
That's exactly what the paradox states. Seeing one boy does affect the gender of the second child.
- dumbfoundded 8y ago
  It's not a paradox if you put it in less confusing language. There are 4 possibilities: GG GB BG BB GG is impossible so you have GB, BG, & BB. One of them is a B, what's the other. Well that's clearly 2 / 3 G. The other one situation is completely. It's basically just the chance of a child being a girl.
  - perl4ever 8y ago
    But are those ordered pairs or sets of two items each?
    - uryga 8y ago
      I think of them as ordered pairs, but would it change anything if they were sets? Order doesn't matter in this question.
      - dumbfoundded 8y ago
        Order doesn't matter but it helps make it more clear as why I included the same combination twice.
  - Sean1708 8y ago
    Is this basically to do with the fact that you know how likely it is to see a boy given any of the combinations? So there's a 33% chance that there are two boys and you see a boy, a 33% chance that there's a boy and girl and you see a boy, and a 33% chance that there's a boy and a girl and you see a girl. Given that you've seen the boy you've eliminated one of the outcomes so the other two outcomes are equally likely?
    - dumbfoundded 8y ago
      I'm confused by your comment but I'll try to explain more. Given one of the children is a boy. The only combinations possible are: GB, BG, & BB. Where G is Girl and B is Boy. The question is about the probability of the other child being a Girl. When we eliminate one of the B from each combination, we get: GB, BG, BB -> G, G, B We can see that 2/3 are G and 1/3 are B. Does this help?
- hexane360 8y ago
  No it doesn't. In the 2/3 case, you're given information about the whole set, not just one member. Instead of "seeing one boy" affecting the gender of the other child, it's "seeing there are more than 0 boys" affecting the gender of one of the children. Here's another example: If I flipped 100 coins, and told you there were 99 heads, what's the likelihood the other 1 was tails? The answer is 100/101, because there are 100 ways the one tail could happen, while only one way zero tails can happen. But if you're only told that the first 99 are heads, the answer is 1/2. You know the last flip has exactly the same odds as all the other ones -- the trials are independent. The difference is in one all the flips were included in the results (and the question isn't about any one specific flip), while in the other you had no information about the last flip (which the question was about). Another way of looking at it: In the first example, the question was asking about the sum of 100 trials and you were given information about 99 of those 100 trials. You were given 99/100s of the information you need to answer the question. In the second example, the question was asking about the last trial and you were given information about 99 unrelated trials beforehand. You were given none of the information you need to answer the question.

pg_bot 8y ago

That's exactly what the paradox states. Seeing one boy does affect the gender of the second child.

dumbfoundded 8y ago
It's not a paradox if you put it in less confusing language. There are 4 possibilities: GG GB BG BB GG is impossible so you have GB, BG, & BB. One of them is a B, what's the other. Well that's clearly 2 / 3 G. The other one situation is completely. It's basically just the chance of a child being a girl.
- perl4ever 8y ago
  But are those ordered pairs or sets of two items each?
  - uryga 8y ago
    I think of them as ordered pairs, but would it change anything if they were sets? Order doesn't matter in this question.
    - dumbfoundded 8y ago
      Order doesn't matter but it helps make it more clear as why I included the same combination twice.
- Sean1708 8y ago
  Is this basically to do with the fact that you know how likely it is to see a boy given any of the combinations? So there's a 33% chance that there are two boys and you see a boy, a 33% chance that there's a boy and girl and you see a boy, and a 33% chance that there's a boy and a girl and you see a girl. Given that you've seen the boy you've eliminated one of the outcomes so the other two outcomes are equally likely?
  - dumbfoundded 8y ago
    I'm confused by your comment but I'll try to explain more. Given one of the children is a boy. The only combinations possible are: GB, BG, & BB. Where G is Girl and B is Boy. The question is about the probability of the other child being a Girl. When we eliminate one of the B from each combination, we get: GB, BG, BB -> G, G, B We can see that 2/3 are G and 1/3 are B. Does this help?
hexane360 8y ago
No it doesn't. In the 2/3 case, you're given information about the whole set, not just one member. Instead of "seeing one boy" affecting the gender of the other child, it's "seeing there are more than 0 boys" affecting the gender of one of the children. Here's another example: If I flipped 100 coins, and told you there were 99 heads, what's the likelihood the other 1 was tails? The answer is 100/101, because there are 100 ways the one tail could happen, while only one way zero tails can happen. But if you're only told that the first 99 are heads, the answer is 1/2. You know the last flip has exactly the same odds as all the other ones -- the trials are independent. The difference is in one all the flips were included in the results (and the question isn't about any one specific flip), while in the other you had no information about the last flip (which the question was about). Another way of looking at it: In the first example, the question was asking about the sum of 100 trials and you were given information about 99 of those 100 trials. You were given 99/100s of the information you need to answer the question. In the second example, the question was asking about the last trial and you were given information about 99 unrelated trials beforehand. You were given none of the information you need to answer the question.