65 ms·

I don't agree; Jeff was not giving a quote. Instead it is just the relevant information abstracted from whatever the person said. By choosing the quote you did

I don't agree; Jeff was not giving a quote. Instead it is just the relevant information abstracted from whatever the person said. By choosing the quote you did, you have added more information to the problem (at least when reading it with conversational English). I think this would be a better quote of what the person might have said:"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" Pulling out the information corresponding to gender and family size would give only the information given in Jeff's post. When applying math to the real world, you have to pull out the important information and deal with just that information. But here you are doing the opposite--trying to find a real world situation that applies to the math problem. In my opinion, your example does not quite apply. (I don't know how the probabilities change when you account for hermaphrodites, but if it changes significantly enough so that approximately 66% is a bad answer, I would find that very interesting!)

seano 18y ago
"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" - given just the information in your quote, the odds are 50% of a boy and a girl.
- dinosaur 18y ago
  Can you give me your reasoning? I think the only conclusions you can get from that quote is that the person has two children, and at least one of those is a girl. Do you disagree with that? If I am correct about that, then it matches the conditions discussed in the article and the answer would be 2/3 for a boy and a girl.
  - seano 18y ago
    The difference is that you have specified that the child picked up from the police station is a girl, thus only the sex of the other child is unknown. This other child is either a boy or a girl, presumably with a 50/50 chance either way, resulting in a 50% chance of one being a boy and one being a girl. Concretely, using a capital letter to denote the sex of the child picked up from the station, there are only two possible permutations Gg and Gb, each of equal probability, and 50% of those are girl and boy (Gb). On the other hand if we only know that at least one is a girl we have the permuations, gg, gb, bg - resulting in the 66% chance.
    - DougBTX 18y ago
      How much do you need to know about the person to know which child is which? I don't quite understand it, but apparently anything which can be used to distinguish the children will do. Possibilities with two children: Gg, Bg, Gb, Bb If one of them has a distinguishing mark, they have an apostrophe: (in jail, has red hair, or born first) G'g, B'g, G'b, B'b, Gg', Bg', Gb', Bb' Then note that the marked one is a girl: G'g, G'b, Gg', Bg' So, there is a 50% change that the children are a boy and a girl. Only if there is no way to distinguish them, do you get the 66% behaviour, where the set is: Gg, Bg, Gb
      - seano 18y ago
        No, that is not it. With G'g and Gg' you are repeating the same permutation in your set above! It makes no difference if they have distinguishing marks or not. It matters if you are told that a particular child is a girl (50%) or if you are only told that at least one child is a girl (66%).
        DougBTX 18y ago
        How can you be told information about a particular child if you have no way to distinguish them? I partially understand your point about G'g vs Gg' now: if having a prime is the only way to distinguish the children, then G and g must be indistinguishable, so G = g.
        seano 18y ago
        Think about it this way: I take two coins out of my pocket and hold them inside my hand so that neither of us has seen them. I show you the coin in my left hand and you see that it is tails, what are the odds of the coin in the other hand being heads? 50%. This is the chance of a head/tail combo in this case. I now put the coins back into my pocket, shuffle them about, and again take them out inside my hands. This time I look inside both my hands, not letting you see, and tell you (truthfully) that at least one is tails. Given that information, you can deduce three mutually exclusive possibilities each of equal probability - both are tails, only the coin in my right hand is tails or only the coin in my left hand is tails. Hence we have the odds in this situation of 2/3 for a head/tail combo. It is easy to see that the first situation is akin to knowing that a particular child is female, whilst the second is akin to knowing that at least one of the children is female. Also, in either case it does not matter if the coins are distinguishable - one could be a euro and the other a pound.

seano 18y ago

"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" - given just the information in your quote, the odds are 50% of a boy and a girl.

dinosaur 18y ago
Can you give me your reasoning? I think the only conclusions you can get from that quote is that the person has two children, and at least one of those is a girl. Do you disagree with that? If I am correct about that, then it matches the conditions discussed in the article and the answer would be 2/3 for a boy and a girl.
- seano 18y ago
  The difference is that you have specified that the child picked up from the police station is a girl, thus only the sex of the other child is unknown. This other child is either a boy or a girl, presumably with a 50/50 chance either way, resulting in a 50% chance of one being a boy and one being a girl. Concretely, using a capital letter to denote the sex of the child picked up from the station, there are only two possible permutations Gg and Gb, each of equal probability, and 50% of those are girl and boy (Gb). On the other hand if we only know that at least one is a girl we have the permuations, gg, gb, bg - resulting in the 66% chance.
  - DougBTX 18y ago
    How much do you need to know about the person to know which child is which? I don't quite understand it, but apparently anything which can be used to distinguish the children will do. Possibilities with two children: Gg, Bg, Gb, Bb If one of them has a distinguishing mark, they have an apostrophe: (in jail, has red hair, or born first) G'g, B'g, G'b, B'b, Gg', Bg', Gb', Bb' Then note that the marked one is a girl: G'g, G'b, Gg', Bg' So, there is a 50% change that the children are a boy and a girl. Only if there is no way to distinguish them, do you get the 66% behaviour, where the set is: Gg, Bg, Gb
    - seano 18y ago
      No, that is not it. With G'g and Gg' you are repeating the same permutation in your set above! It makes no difference if they have distinguishing marks or not. It matters if you are told that a particular child is a girl (50%) or if you are only told that at least one child is a girl (66%).
      - DougBTX 18y ago
        How can you be told information about a particular child if you have no way to distinguish them? I partially understand your point about G'g vs Gg' now: if having a prime is the only way to distinguish the children, then G and g must be indistinguishable, so G = g.
        seano 18y ago
        Think about it this way: I take two coins out of my pocket and hold them inside my hand so that neither of us has seen them. I show you the coin in my left hand and you see that it is tails, what are the odds of the coin in the other hand being heads? 50%. This is the chance of a head/tail combo in this case. I now put the coins back into my pocket, shuffle them about, and again take them out inside my hands. This time I look inside both my hands, not letting you see, and tell you (truthfully) that at least one is tails. Given that information, you can deduce three mutually exclusive possibilities each of equal probability - both are tails, only the coin in my right hand is tails or only the coin in my left hand is tails. Hence we have the odds in this situation of 2/3 for a head/tail combo. It is easy to see that the first situation is akin to knowing that a particular child is female, whilst the second is akin to knowing that at least one of the children is female. Also, in either case it does not matter if the coins are distinguishable - one could be a euro and the other a pound.