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1. Unfogged: Comment on John Tierney Is Courageous

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   fenster 10-Jul-2010

   "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?" ^{link »}

   If the first kid is a boy born tuesday there's 14 possibilities for the second kid. If the second kid is a boy born tuesday there's 14 possibilities for the first kid. But we've overcounted one situation, namely both kids are boys born on tuesday. So there's actually exactly 27 possibilities. Of those 27, we see that 13 of them have both kids being boys ^{link »}

   double boy families are about twice as likely to pass through the "boy born on a Tuesday" seive as single boy families. ^{link »}

   You've got a stadium full of two kid families, and you're making announcements over the PA. If you say, "Everyone who doesn't have at least one boy, leave," all the families with two girls leave, and everyone else stays. If you say "Everyone whose oldest kid isn't a boy, leave", then more families will leave -- the two girl families, and the girl then boy families. Your question is "What fraction of the remaining families are two boy families", and it should be clear that the answer is different depending on which class of families you ruled out (told to leave the stadium) with your initial conditions. ^{link »}

   There's a fifty percent chance that any individual birth event is the event of a boy's birth. We aren't, however, asking the question, "what was the probability that the birth of my other child was the birth of a boy", but the question, "what is the probability that I have two boys, given that I have a boy?". This is a question about the way children can be distributed across two births. The "given" clause it's what's throwing you off, making you think that we're asking about the birth of the other child considered in itself. Think of it this way: we don't even know which is the "other" child. If we did (if it were known that the elder is a boy and we're talking about the younger, or vice versa), then we really would be talking just about the event of the other birth. Since we're asking about both births, and placing a constraint on one (but which one is unknown!), we need to look at the ways someone could have two children such that at least one is a boy. There are three such ways: BB, BG, and GB. ^{link »}

   %boy=(2n-1)/(4n-1) where n is the number of equal probability intervals (7 for the day of the week formulation), and %boy=(2-p)/(4-p) for the general case where p=the probability of satisfying the "constraining" condition. Neatly shows the special cases of 1/3 for p=1 (no conditions, just "a boy") and 1/2 for p=0 (boy "uniquely" identified). ^{link »}

   For the Tuesday information to change the probability that both children are boys, you have to assume that you asked the guy "Do you have a child born on a Tuesday?" If you asked him "What day was your child born?" or you asked him nothing at all and he just volunteered the information, the probability doesn't change with the information about the day of birth. The real answer to the question is: "underspecified: what did you do to get that information?" ^{link »}

   What if the kid is there? You'd be able to see all sorts of lower probability things about him than what day of the week he was born, and that can't change the odds. ^{link »}

   If I do the stadium thing (filled with randomly selected families at the beginning, everyone who doesn't match the facts I know about my target family leaves, the probability of two boys is equal to the fraction of two-boy families left) the timing and source of the information doesn't matter. ^{link »}

   If we use your stadium scenario, after you have eliminated all families without two children and at least one boy, in one interpretation you eliminate all families without a bboat; in the other you randomly select a family and ask what day one of their sons is born. Unsurprisingly, regardless of their answer they were still selected from a crowd with 2/3 single boy families. ^{link »}

   Okay, here is where I first encountered this problem. Here is the quote where the problem is posed: "Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences: 'I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?'" ^{link »}

   The way I'm thinking of this, P(two boys) is equivalent to "If I pull one family from all the possible families in the world that satisfy what I know about this family, what's the probability that my random draw will get a two-boy family?" which is equivalent to "What fraction of all the possible families in the world that satisfy what I know about this family have two boys?" I think the way Tia is using 'success' here is that unless you decided ahead of time that you care about Tuesday-birth, then the pool of 'all possible families in the world that satisfy what I know about this family' still contains families with no Tuesday-born boys. But that (with caveats about watching me spin like a weathervane) doesn't hold: whether or not you cared about it beforehand, there is a fact of the world that states how many two-child families there are with a Tuesday boy, and another fact of how many of those are two-boy families, and given that you know that the family you're interested in is in the pool "two-child families with a Tuesday boy", P(two boys if you know that one child in the family is a Tuesday boy) has to be (number of two- boy families with a Tuesday boy)/(number of two- child families with a Tuesday boy). Any information you get about the family, regardless of when you get it or whether you care about it, affects the makeup of the population you're taking a random draw from, so it affects the probability. ^{link »}

   Dubious. ^{link »}

   But if the kid is there, P(2 kids) becomes 50%, because the prior of running into a boy kid is double for 2-boy families. ^{link »}

   ("bboat"="boy born on a Tuesday.") But in interpretation 2, the population of the stadium no longer represents your knowledge. What if Gary, unprompted, tells you "I have a disease which makes me unable to sire girls?" Does it not change the odds? ^{link »}

   here's another person making the same point ^{link »}

   People like Gary may always tell you if they have a boy born on Tuesday. Or they may not tell you sometimes, but instead tell you about their other kid. ^{link »}

   If you ask him if one of his sons was born on a tuesday, the more sons he has the more likely he is to say yes. This changes the probability. If he volunteers that one of the sons was born on a specific day, say Tuesday, this doesn't change the probabilities because he could always say some other day if that were true for the son. Would he be more likely to say he had a son born on Any specific day, say Tuesday, if he had two sons? I don't think so. ^{link »}

   suppose you run into a man and a young boy. He tells you "this is my son, and I have one other child". What's the probability that he has two boys? ^{link »}

   If you just meet someone who tells you "I have two children; at least one is a boy" then the answer is obviously 1/3. But once you see the boy -- this one is a boy -- then it would seem the probability of the other being a boy is 1/2. On the other hand, given the initial question that has the answer 1/3, you know there is a boy, so in some sense you haven't gained new information by seeing a particular boy. I'm confused. ^{link »}

   People will say "one of them is a boy" 3/4 of the time, but the child they're with is a boy only 1/2 the time. So P1=.25/.75 but P2=.25/.50 . ^{link »}

   Suppose he said "I've got five boys born on Tuesdays, and six children total. What is the probability that I've got six boys?" Now it seems more clear (to me, at least) that the Tuesday information is meaningful, because if the extra child is a boy, there are a lot more ways to distribute the Tuesday birthdays, but if the extra child is a girl, the Tuesday birthdays are forced. ^{link »}

   The "seeing the boy" situation really does need a bit more specification. Using the stadium starting with 2-child families. Case 1: Every family that *could* show a boy stays. Answer: 1/3 Case 2: Families stayed if some random selection process (such as birth order) yielded a boy. Answer: 1/2 ^{link »}

   Assume that right after Foshee steps down from the microphone, a second person with at least one boy gets up and says "I have two children. One is a boy born on a Monday. What is the probability I have two boys?" Then a third gets up and says "I have two children. One is a boy born on a Friday. What is the probability I have two boys?" This continues until everyone in the room with at least one boy, 30 people in all, has stepped up. Then we see how many of them actually have two boys. Are you expecting the answer to be 10, or 15? And if the former, how can the probability for Foshee or any other individual one have been roughly 1/2? Is there anything in his question to indicate you're not just seeing one instance of this potential parade of parents? You need something that does that to change the probability. The only way it would be different is if a call went out in advance for someone with a boy born specifically on a Tuesday (or a weekday, or a day starting with "T"), in which case a two-boy family would be more likely to have one and therefore to be picked to get up in front of the room. ^{link »}

   The question about any probability problem is, "what constraints does this information place on the probability space?" This is related to Tia's "what question did you ask to get this information" / "is this question being used as a filter." In LB's stadium example, "everybody who isn't part of a family with two children, at least one of which is a boy born on Tuesday" uses both bits (boy and Thursday) as filters. The second party game in 798 is a way to get that information without using it as a filter. One other way to look at the value of the information is, "am I getting more information about an entity which was selected by some other process?" (in the 798 example, the drawing of balls is the process that selects which child you get information about, further information about the child doesn't narrow the possible options) ^{link »}

   The important question, Tia's question, is whether that day was specified in order to select the person. If it was, then yes, we're in the ~1/2 situation. If it wasn't, we're in the standard 1/3 situation. And the person's naming a day doesn't indicate which situation we're in. ^{link »}

   Tia and Mr. Blandings are wrong. Information is information, whether you asked for it or not. ^{link »}

   We know of the two child families with at least one boy 1/3 will have two boys. So when we select a family from that pool and meet one randomly chosen child who happens to be a boy, one might think that we have to conclude there is only a 1/3 chance the other is as well, but this would be an error. We now are describing an experiment with three outcomes: boy-girl family presenting boy, boy-girl family presenting girl, and boy-boy family presenting boy. The probabilities of these three cases are 1/3 (2/3*1/2), 1/3 (2/3*1/2), and 1/3 (1/3*1). Given we met a boy we must be in the first or third case, and so the probability ghat the other child is a boy is 1/2 ((1/3)/(1/3+1/3)) as it should be. ^{link »}

   One could work through the Tuesday problem in a similar manner to my 840 in the case where the family chosen should tell the birth day of their son (or a son chosen at random if they have two). Counting the possible outcomes b-g family saying Monday, ...) you would find that you can just get rid of the day information. ^{link »}

   YES!! Of parents with a Tuesday boy, all of the BG parents would tell you they have a Tuesday boy. BUT: Only half of the BB parents would. The other half would tell you the day their OTHER boy was born. UNLESS you had forced them by ASKING "Do you have a Tuesday boy?" 7(Bt,G) + 7(G,Bt) + .5*6(Bx,Bt) + .5*6(Bt,Bx) + 1((Bt,Bt) = 21 cases where they TELL you "Tuesday boy." Of these, .5*6 + .5*6 + 1 = 7 are two-boy. 7/21 = 1/3!!! ^{link »}

   As I attempted to explain in 730 the "one is a boy problem" is also underspecified if the information is just volunteered as you don't know that the probability that this information will be volunteered is the same for the three cases where at least one child is a boy. For example in the cases where one child is a boy and one child is a girl the person could alternatively volunteer that "one is a girl". If they do this half the time then the probability that the other child is a boy is 1/2 as the mixed cases get half the weight of the both boys case. ^{link »}

   In hindsight: Correct, but unexplained. The key is that in the latter case, two-boy parents will only choose to mention "Tuesday" half the time. ^{link »}

   tags: jul 10, unfogged.com, annotations

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