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pair "boy-boy" would have to be entertained twice by the person that does not know whether the boy is first or second. Alternatively, you would have to consider birth order irrelevant in the cases of "girl-boy" and "boy-girl", reducing this alternative to a single observation, "a boy, being either first or second born, has a sister". By either method, the intuitive answer becomes the "right" answer-- 1/2.
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of girl then boy thus reducing the number of possibilities to boy then boy or boy then girl, that is 1 of 2 options and conversely if the girl was born second we can remove the possibility of girl then boy and once again we are left with only 2 options boy boy or boy girl and the probability must be 1/2.
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This problem falls victim to what's known as the
Gambler's Fallacy. There is no link between the one birth and the other, each event has its own probablity of occuring. Just because the first child is born a boy it's no more likely that the next child born will be a girl. You can ignore completely
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It seems to me the odds are still misspecified as 1/3. When you encounter a random pair of "boy-boy" with birth order not specified, then it would seem this "random occurence" should be double weighted. Not knowing whether the boy encountered is the first or the second born would suggest the random
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The second mistake in this problem is that when introducing the 1/3 solution they introduce birth order as a restriction. Let's assume that the second child is a girl and the first born, then it must be the case that the boy was born first and if he was born first than you can eliminate the choice
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Actually, this possibility is not eliminated, because all that was stated was that there is one son, not whether that son is first or second. It would be different if we had 2 children and randomly revealed the gender of one of them. However, since it was revealed that there was a son, regardless
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In this case when knowing that the family has at least a boy, it's clear that we take out the two girls case. Now it simply depends if we know or not if the boy was first born. If we know, then it's a 50/50 probability of the other being a boy. If we don't know, then it's clearly 1/3 since it may
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It's quite simple, it depends on the point of reference. If we calculate in the most basic way what's the probability of a child's gender being boy or girl at birth there's 50/50 (not taking into account special cases). THEN, if we calculate what is the probability of a family with two kids having
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The entry is incorrect. The problem stems from confusing the probability that a 2-child family with at least one son also has a daughter (2/3), and the probability that a son in a two-child family has a sister (1/2). These are not at all the same questions and have different answers.
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I have one more problem too. Actually I have thought of that 2 births scenario, but I dropped that idea when I realized that I have no data about the probability of having a single birth or 2 births. Should that count in the question which may make the answer incorrect?
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the gender of the first child when considering the probability of the gender of the second child. It is the same basic fallacy that a team that hasn't won in a long time is "due" or that they must lose because they have been losing for so long.
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Now we might have 2 different cases: the one above with 3 possibilities or the one stated in the article, which actually has 4 possibilities: two boys, two girls, first boy then girl, first girl then boy.
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The conclusion is that the question is intentionally poorly formulated just to make you answer with 50/50 and then be amazed by a simple logic, founded on lack of details in the question.
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We are given that one of the children is a boy. Thus, only one of the four possibilities -- two daughters -- is eliminated. Three possibilities with equal probabilities (1/3) remain.
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The tag on the page says it needs a cleanup, but I think it's worse than that. The whole article seems rather arbitrary and random itself. How is this topic different from
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to test this. Exclude all entries of 2-2 (daughter-daughter) and total up how many are both sons (1-1) vs. how many have one son and one girl regardless of order (1-2
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The possibility of first a girl, then a boy is also eliminated. So it's still a chance of 1/2. Or maybe that was the point, the article isn't very clear. --
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or any of a list of different articles that already exist? Brain teaser is just anohter way of saying something already discussed elsewhere.
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If we encounter someone with two children, given that at least one of them is a son, what is the probability that the other is also a son?
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This article has been a random mess for over a decade. It is unnecessary and should probably just be redirected to
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of birth order or whether or not there is a daughter, the odds the second child is a son are reduced to 1/3.
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Definitely this is not a brain teaser and should not appear in this article. At least not in this format.
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two boys, is clearly 1/3 (again no special cases). (possibilities: two boys, two girls, boy and girl)
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Indeed the example is poorly formulated and it's done like that for the sake of a trick question.
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fall in one of the following cases: two boys, first boy then girl, first girl then boy.
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Maybe I'm just being stupid, but the example seems wrong:
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