3 ms·

Am I being really stupid here? I keep getting 13/27, both by Baye's Theory P(BTB?∩B?BT | BT??∩??BT) = P(BT??∩??BT | BTB?∩B?BT) · P(BTB?∩B?BT) / P(BT??∩??BT) =

• hanoz 7y ago
  You are correct! Most people tend to say 1/2, or if they have some knowledge of probability but dismiss the Tuesday aspect then 1/3. It does rely on the problem being interpreted as equivalent to the dialogue I have posted below, but even when so framed, many people will refuse to believe it.
• username90 7y ago
  You are not correct, but you are not entirely wrong. This is a badly made retake of the original two problems where you had one where you just mention that you have a boy, and the other where you mention that the older is a boy. The older being a boy removes 2 cases while 1 being a boy just removes 1 case so they are different. But in this case mentioning the weekday doesn't matter at all. To make it clearer take this script "I have 2 kids, one is a boy, that boy's birthday is {{ boy's birthday }}". In that case it is 1/3. Take this script on the other hand "I have 2 kids {{ if 1 is a boy born on tuesday print ', one which is a boy born on Tuesday' }}". That is what you and he meant it. But it is wrong since it is not the only way to read it, and arguably it is the less correct way to read it. This HN thread is full of people who don't know maths who think they know it and are trying to teach others...
  • hanoz 7y ago
    You're on a hiding to nowhere attempting to frame this in terms what the subject does or doesn't decide to mention. You say: > This is a badly made retake of the original two problems where you had one where you just mention that you have a boy, and the other where you mention that the older is a boy. The older being a boy removes 2 cases while 1 being a boy just removes 1 case so they are different. But in this case mentioning the weekday doesn't matter at all. This doesn't make sense because if you concede that being told at least one child is a boy yields a two boy probability of 1/3, and you also concede that then being told the elder child is a boy changes that probability to 1/2, there's no reason not to concede that being told instead that at least one child is a boy born on a Tuesday gives 13/27 as Sean1708 has demonstrated. Otherwise you're applying your "decides to mention" augment differently in each case. Anyway, it's mathematically more interesting to dispense with all the intent and instead consider the matter as an enquiry, which you may well argue is not equivalent to the original pronouncement (for reasons of intent), but nevertheless is an interesting counter intuitive observation in its own right: (a) We have a man we know has two children. The probability he has two boys is then 1/4. (b) If we ask him if at least one of his children is a boy and he says yes, the probability he has two boys is now 1/3. (c) If instead we ask him if at least one of his children is a boy born on a Tuesday and he says yes, then the probability he has two boys is 13/27. (d) If instead we ask him if at least one of his children is a boy born on Christmas Day and he say yes, the probability moves even closer to 1/2. The exact value is left as an exercise to the reader. i.e. the probability tends towards 1/2 as the additional information narrows the boy being referred to down to a specific child. So: (e) "is at least one child a boy with this fingerprint? - yes", or: (f) "is the elder child a boy - yes" will both get you to p = 1/2.
    • username90 7y ago
      > Otherwise you're applying your "decides to mention" augment differently in each case. You don't see the difference between saying "the older one is a boy" and "I have a boy born on a Tuesday"? The first one selects first by the property of Age and then states their gender. The second selects by the property of gender and then states which weekday they were born.