I honestly am not trying to be disrespectful of anyone, when I say the interpretation you express here is a common misconception held by many. They have been known to defend it voraciously because they "know" it produces the "right" answer to this problem. They are so convinced, that they won't look at arguments against it, since they already "know" those arguments must be wrong. This has been done by students, teachers, full professors, and even Nobel Prize winners for a different problem.The question further seems to imply that only the limited information given in the question itself should be used when solving the problem.
The correct approach is that the possible information is not limited to what is in the problem statement. The specific instance of a random process that the problem exemplifies is conditioned on that information, and there is a big difference. To condition a probability on another event, you need to first enumerate all the possibilities for the process. Then the conditional probability for event E, given event C, is the probability of E and C happening together, divided by the probability of event C happening regardless of E. And the issue here is that C is not "A family has a boy born on a Tuesday," it is "A parent states one fact, of what is likely two different but equivalent facts that apply."
The other problem I refer to is the Game Show Problem. The information given in it, is that there is one car and two goats, each behind one of three numbered doors. One door is chosen by the contestant, and a different one is revealed to have a goat. If we limit the solution to using this information alone, since the car was equally likely to have been placed behind either of the two remaining doors, it seems it now has a 1/2 chance to be behind either.
This solution has been accepted by students, teachers, full professors, and Nobel Prize winners. Thousands, including many holding doctorates in Math, wrote angry letters to Marilyn vos Savant when she said, in Parade Magazine, that the answer to the problem is that the chosen door has a 1/3 chance, and the unchosen, unopened door has a 2/3 chance. Yet that is what is accepted as the correct answer.
The correct solution (different from the explanations of how the game works in the long run which are usually given) is that if the contestant chose a door with a goat, only one door could have been opened. But if the chosen door had the car, then there were two doors that could have been opened. So even if the information in the problem tells use which door was opened, the possibility of a different door being opened - contradicting the problem statement the same way the possibility of a parent saying "I have a girl" contradicts ours - means that case's chances are cut in half. The same solution applies to the Two Child Problem, except that there are two cases where other possibilities exist, that contradict the specific information in the problem statement, but not the process the statement describes or implies which produced that information.
It is correct than some have accepted that answer, just like Nobel Prize winners have said the chances in the Game Show Problem are even. The answer itself is not correct.From my understanding of previous posts it seems to be agreed that working within the implied limitations gives a probability of 13/27. Is that correct?
I haven't been very clear on addressing this because (honestly) I try to avoid my tendency to be long-winded. The issue is what you mean by "pinned down." In order for your 1/3 -> 1/2 progression to be correct, you have to have chosen the specific time interval you want before you chose the family you apply it to. A set of families that all meet that requirement has to be assembled, effectively, and you pick one of the qualified group. So if you decide "I'm look for boys" and arrange to get a family with a boy, the answer is 1/3. If you then decide you want a Tuesday Boy, you need to assmeble a different (smaller) group, pick a new family, and the answer is 13/27.A. Following on from the above ... As the birth time is pinned down more accurately to reducing time intervals the probability tends to 1/2.
But if, as seems to be a closer interpretation of "pinned down," you mean that you pick a family first and then determine a gender that exists in that family that could be either "boy" or "girl" but in this case happens to be "boy," the answer is 1/2. If you then, through a repeated series of questions, narrow the interval the includes the birth of that boy? The answer stays at 1/2 throughout.
In 1959, Martin Gardner asked this in Scientific Ameican:
He gave the answers as 1/2 and 1/3. But he retracted the second one six months later:Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
By Gardner's standards, the question that started this thread is unanswerable since it does nto tell us why we were told what we were told. If we assume there is enough information to answer it, we must assume that any similar statement could be given, and the answer is 1/2.Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.