2 ms·

Right, that's the last case. The eldest is a boy (or a girl, it doesn't matter), so the younger is equally likely to be either. The middle case is 2/3 likely t

by sago 12y ago

Right, that's the last case. The eldest is a boy (or a girl, it doesn't matter), so the younger is equally likely to be either. The middle case is 2/3 likely to be a girl. Think of the possibilities of the sex of two children (sex given in birth order) M then M, M-F, F-M, F-F. I tell you one is a boy, so we have MM, MF, FM, in two of the three cases, the other is a girl. In the first case the answer is 14/27 a girl, so almost 1/2, but not quite. To figure this out you can consider all 196 combinations of two children with birth-days-of-the-week, cross out all those that don't match the phrase "he has a son born on a Tuesday", and you're left with 27 possibilities: 14 of them are when the other is a girl (14 because she could be born first or second, and on any day of the week 2x7=14), and 13 when the other is a boy (6 when the other is an older boy born on a non Tuesday, 6 when it is a younger boy born on a non Tuesday, and 1 when both are both on a Tuesday). So that's not the intuition, as much as the answer. The intuition is that as you add more specific knowledge about which child you mean, that question contributes less to the probability of sex, and is more an irrelevant detail. Being more specific can be anything. Do it with birthday-of-the-year, (more specific than day of the week) and it is even closer to 1/2. Do it with something unique, like their phone number (or their birth order, or 'the taller one is a boy') and it is exactly 1/2. Be less specific, make it so they both toss a coin and it is 'one of the children tossed heads and is a boy' and it is 3/7.

javajosh 12y ago
Maybe I begin to see the light. First, there is an important difference between considering a random event on it's own and several random events together: in essence, when your friend says they have a boy, he is telling you a little something about two ordered events. That's why you count the BF and FB cases separately (which is the counter-intuitive part). The actual Tuesday problem requires a little more thought. Somehow that extra information makes you drop a case! I can't help but think that the ordering is the critical factor, and that it's not a coincidence that the "extra information" has to do with time. In other words, the dropped case happens because we drop the distinctive ordering when the two boys are born on a Tuesday.
- sago 12y ago
  You picked up on a really important issue that is often assumed. Thank you for not letting me just slip it under the radar without question. To answer your question: we treat the MF and FM differently, just for ease of calculation. What we're doing is taking a set of possible situations (different 2-child families), each with an attached probability (I'll come back to this), throwing away those that don't match some known information (i.e. those without a boy), taking those that do, grouping them into sets (the set of families with the other being a girl, and the set where the other is a boy), and calculating the probability the actual scenario was in each of those sets. Now the 'attached probability' probably sounded odd in there, because that's not what we've done, right? We've just been counting situations. True, but that's because we carefully started with situations that are all equally likely. So rather than bugger about with those attached probabilities, we could just use raw counts, knowing that it would work. So, if we start with four possible sexes of two children: MM, MF, FM, FF, those are all equally likely. I'm using the birth order as a way of breaking the two MF cases apart, so I've got two cases that are equally likely, letting me do the counting trick. I can just count the 2 out of 3 remaining situations that match the information, so 2/3. Easy. But I don't have to do it that way. Let's say we follow your intuition and start with three possible sex combinations: MM, MF, and FF. Now, those aren't equally likely, right? There are twice as many families with MF children as with MM children. So I need to use the attached probabilities. MM (1/4 of 2-child families), MF (1/2), FF (1/4) We can remove the FF as inconsistent with the information (as before), then the probability the parent has an MF family (i.e. their other child is a girl) is 1/2 over (1/2 + 1/4) = 2/3 as before. In fact, for any reasonable calculation based on data, we're almost certainly not going to be able to use the counting trick. In the case where I was using actual birth rate statistics for sex (rather than assuming as many girls as boys are born), I'd have to do it this way. The counting trick is really only useful for toy problems and teaching basic probability: you were right to call me out on it. --- The Tuesday's child doesn't drop a case. There are certainly seven cases where, in an MM family, the eldest is a Tue child, and seven where the youngest is, but one of those cases is shared: a case where both are Tue children. In terms of the intuition you can think of it this way, identifying by birth-day-of-the-week is identifying which child is which, so we're almost at 1/2, but there is one case (both Tue boys) where this information does nothing to distinguish them, so that tiny bit of ambiguity remains, and it can't quite reach the full 1/2, there's still a trace of the 1/3 result that came when the ambiguity was total.

by sago 12y ago

javajosh 12y ago

Maybe I begin to see the light. First, there is an important difference between considering a random event on it's own and several random events together: in essence, when your friend says they have a boy, he is telling you a little something about two ordered events. That's why you count the BF and FB cases separately (which is the counter-intuitive part). The actual Tuesday problem requires a little more thought. Somehow that extra information makes you drop a case! I can't help but think that the ordering is the critical factor, and that it's not a coincidence that the "extra information" has to do with time. In other words, the dropped case happens because we drop the distinctive ordering when the two boys are born on a Tuesday.

sago 12y ago
You picked up on a really important issue that is often assumed. Thank you for not letting me just slip it under the radar without question. To answer your question: we treat the MF and FM differently, just for ease of calculation. What we're doing is taking a set of possible situations (different 2-child families), each with an attached probability (I'll come back to this), throwing away those that don't match some known information (i.e. those without a boy), taking those that do, grouping them into sets (the set of families with the other being a girl, and the set where the other is a boy), and calculating the probability the actual scenario was in each of those sets. Now the 'attached probability' probably sounded odd in there, because that's not what we've done, right? We've just been counting situations. True, but that's because we carefully started with situations that are all equally likely. So rather than bugger about with those attached probabilities, we could just use raw counts, knowing that it would work. So, if we start with four possible sexes of two children: MM, MF, FM, FF, those are all equally likely. I'm using the birth order as a way of breaking the two MF cases apart, so I've got two cases that are equally likely, letting me do the counting trick. I can just count the 2 out of 3 remaining situations that match the information, so 2/3. Easy. But I don't have to do it that way. Let's say we follow your intuition and start with three possible sex combinations: MM, MF, and FF. Now, those aren't equally likely, right? There are twice as many families with MF children as with MM children. So I need to use the attached probabilities. MM (1/4 of 2-child families), MF (1/2), FF (1/4) We can remove the FF as inconsistent with the information (as before), then the probability the parent has an MF family (i.e. their other child is a girl) is 1/2 over (1/2 + 1/4) = 2/3 as before. In fact, for any reasonable calculation based on data, we're almost certainly not going to be able to use the counting trick. In the case where I was using actual birth rate statistics for sex (rather than assuming as many girls as boys are born), I'd have to do it this way. The counting trick is really only useful for toy problems and teaching basic probability: you were right to call me out on it. --- The Tuesday's child doesn't drop a case. There are certainly seven cases where, in an MM family, the eldest is a Tue child, and seven where the youngest is, but one of those cases is shared: a case where both are Tue children. In terms of the intuition you can think of it this way, identifying by birth-day-of-the-week is identifying which child is which, so we're almost at 1/2, but there is one case (both Tue boys) where this information does nothing to distinguish them, so that tiny bit of ambiguity remains, and it can't quite reach the full 1/2, there's still a trace of the 1/3 result that came when the ambiguity was total.