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#1




The “Boy or Girl paradox”
I wasn't sure which forum to post this question in, but i figured it's more appropriate for IMHO. The "Boy or Girl paradox" is a probability 'paradox' that seems to rely on ambiguity, but, more generally, value judgment regarding how the probability of an outcome should be assessed. The problem is normally stated as follows:
We may remove ambiguity by making it clear that the person is simply being told about the situation (without seeing any of the children) and asked to determine each probability based on the information given. The standard answers given to the two questions are 1/2 and 1/3 respectively. The first one is easy and it appears to simply serve as a way to mislead the person into evaluating the second one incorrectly. It is the second question that causes controversy, but it is usually agreed, given the way it is worded, that 1/3 (and not 1/2) is the correct answer. (The explanation is given in the wiki link above.) The problem, however, is that the second question is easy for one to interpret as essentially saying: “Mr. Smith has two children. One of them is a boy. What is the probability that the other one is a boy?” (If one strips out redundant words). This alternative phraseology seems to indicate that one is being asked to give the gender probability of a very specific child, thus making the answer appear to be 1/2. So even though the two phrases are reductively identical, they lead to very different outcomes. And yet the answer could still be 1/2 (instead of 1/3) depending on the manner in which one finds out (or is informed) that one of the children is a boy. For example, if the person is told: “I saw Mr. Smith with one of his two children yesterday. The child was a boy. What’s the probability that both his children are boys?” Then, in this case, one also has to take into account the probability of Mr. Smith having been seen with a male child, so long as there was no gender bias on his part in choosing with child to be out with. Thus, the correct answer would be 1/2 (not 1/3). So the ‘paradox’ seems to be based on linguistic ambiguity. Yet, I notice that the paradox gets even more interesting when you remove underlying assumptions that ignore the possibility of the children having been adopted or them being twins. Thus, it becomes a true valuejudgment paradox that has nothing to do with language. Although in the original version of the problem, the first (initial) question, by using the phrase “the older child is a girl”, appears to preclude any consideration of the possibility of twins when trying to answer the second (followup) question, yet, other variants of the problem don’t actually do that (like the BarHillel & Falk variant). Also, none of the versions – including the original – preclude the possibility of adoption. Biological progeny is merely assumed. The twin possibility is dismissed in the wiki article by saying: Quote:
To make it clear that the standard answer given to these questions is based (casually) on the actual facts of reproduction, consider this hypothetical example: Suppose you lived in a town where people had children only by adopting them from an orphanage filled with kids from another faraway place (where the people weren’t infertile). What now happens to the probability? Basically, there are five ways that one could end up adopting two children, one of whom is a boy: 1. Adopt a boy first. Adopt a girl at a later time. 2. Adopt a girl first. Adopt a boy at a later time. 3. Adopt a boy first and another boy at a later time. 4. Adopt a boy and girl at the same time. 5. Adopt two boys at the same time. From these options, the BB probability would clearly be 2/5. But wait! Does the actual sequence in which the parent adopted children actually matter? For example, should BG and GB be treated as separate outcomes? If we assume that the parent had the children biologically, the intrinsic uncertainty of a particular gender being conceived becomes important. Therefore it can be treated like a coinflip problem. But in the adoption case, the potential parents are fully in control and can do anything they want. For example, they could walk out with a boy and then, two seconds later, decide to go back and get a girl to be his sister. There isn’t any ‘intrinsic uncertainty’ except in the final outcome of their actions: whether they end up with two boys or with a mixed pair. In other words, the actual probability for BB should be 1/2. This problem, taking into consideration both scenarios, can be modeled with the goldcoin silvercoin analogy mentioned by the article. For example:
Is the probability different in both cases? In other words, does it make a difference that he takes them out one at a time as opposed to taking them out at the same time? And if so, under what conditions? Last edited by Gorilla; 05182015 at 06:00 AM. 
#2




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2. Yes. Probability absolutely depends on one's knowledge (or ignorance). But in puzzles where no further knowledge is given, none should be assumed. 3. No. The assumption that one of K exclusive possibilities has probability 1/K has no basis. Otherwise: "The sun will either rise tomorrow or it won't. So it's a 50:50 chance." 
#3




The probability doesn't depend on how the children are born, or anything about time order. It depends on how you choose the family to look at. Are you selecting it a) by family or b) by child
Suppose you have a list of all families in your street who have two children. Lets say there's 40 families, to make things nice and easy. Someone crosses off from your list all the twogirl families and invites you to pick a surname at random. What's the probability that you just chose a twoboy family? 1/3, assuming that your families are equally divided between GG, GB, BG and BB. But now, suppose you choose the family by putting all the boys' names in a hat, drawing one out and saying 'ok, what's the probability that this family (that the boy is in) is a BB family?' Now the answer is 1/2. Why? Because the family has two different ways to be chosen  by choosing the name of Boy 1, or the name of Boy 2. The GB and GB families only have one way to be chosen. Or, to put it another way, you had 40 boy names in your hat, and 20 of them belonged to boys in the 10 BB families. Your "I saw a parent out with their boy, what's the probability it's a two boy family?" example is basically an example of my second way of choosing. A parent of two boys out in public with one child, is twice as likely to be out with a boy specifically, as a parent with a boy and a girl is. 
#4




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GG GB BG BB With (pass me a bottle) Mr. Jones, you're knocking out two possibilities, so you have this: GG GB But with the Smiths, you're knocking out just one: GB BG BB In both cases, the remaining possibilities are still equally likely. 


#5




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3. That’s the point I was making. Quote:
Otherwise, how would you answer the goldsilver coin analogy? Quote:

#6




The adoption scenario is a furphy. The important thing about the original puzzle is not that there are four different combinations, but that there are four different combinations that we know empirically have roughly equal weight.
If you really played out your adoption scenario AND had the constraint that "adopt a two boys at the same time" happened exactly as often as "adopt a girl first then a boy" and all the other combinations, then yes, your probabilities would be totally different. Because you'd be choosing out of a population which had quite different familial proportions to our world. There's nothing particular to "answer" about the gold coin situation. Both probabilities are the same. Time of choosing doesn't matter  method of choosing can. In that case the methods are both the same (random dip out of bag) Last edited by Aspidistra; 05182015 at 09:54 AM. 
#7




You're really making a mess of the daughter/son problem. It's intended to be a question about two independent events, each of which is 50% one outcome, 50% another. Since human children are not born equally male/female (there are more males), and are not independent (sex of offspring seems at least somewhat based on the mother, despite what your bio teacher taught  prevailing theory is that the woman's hormones favor or disfavor male sperm), you can see that the original question was not supposed to be a technical one, but just an easy illustration of what you are to take as two independent events, each of which has an equally likely outcome.
Your bulleted questions about silver/gold are both 1/2  in both you said "one of them is gold," not "the one you drop on the table is gold" or anything that eliminates one of the 4 possible outcomes. 
#8




Now I am confusing myself. Let's say I have new neighbors and I know only that they have two kids roughly the same age. The next day I see that one of the kids has left a boy's bike in my yard, so I know that at least one of the kids is a boy. I resolve to find out more so I watch the neighbor's house until I see a child come out. It is a boy. What are the odds it is his bike in my yard?
3 out of 4...right? 
#9




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Probability is hard. 


#10




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#11




As septimus said earlier, just because you can break it down into mutually exclusive outcomes doesn't mean you can automatically assume that all of them are equally likely. You could stipulate, as part of the conditions of the problem, that they are, but that would be a constraint.

#12




I never understand this problem either..
If the time of adoption doesn't matter (and I don't see why it would), why does birth order matter? 
#13




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Assuming equally likely outcomes, you get this: OLDER CHILD BOY, YOUNGER CHILD BOY OLDER CHILD BOY, YOUNGER CHILD GIRL OLDER CHILD GIRL, YOUNGER CHILD BOY OLDER CHILD GIRL, YOUNGER CHILD GIRL If you say 'the older child is a girl,' you eliminate two possibilities, leaving the bottom two. The younger child is equally likely boy or girl. If you say, 'at least one is a boy,' you only eliminate the very last outcome. The other 3 are equally likely, and two of those are a boy/girl scenario. Last edited by bup; 05182015 at 05:12 PM. Reason: stupid formatting f'ed up my nice diagram 
#14




By the way, I did a poll on the question (except at least one daughter), and the results are very nearly 2/3, 1/3.



#15




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Why is birth order a relevant way of organising the problem? 
#16




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However, these statements are NOT equivalent in the precise language used in logic and statistics. They're only equivalent in the imprecise everyday language at the local pub. It's like the logical qualifier "If and only if" provides a level of precision that we'd never see in everyday language, but which is necessary when discussion logic or math. "If" and "If and only if" (and for that matter "only if" by itself) actually do have subtly different meanings. (See wikipedia if you're not sure of that.) But, again, if you go down to your local pub the guys are going to say "if" and they'll say that whether they mean "if" "only if" or "if and only if" because saying "if and only if" at a pub will get you punched in the nose. 
#17




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It would be an equivalent problem if we were taking about flipping two coins. Suppose one of the coins is a penny and the other is a nickel. Then we have four possibilities: penny comes up heads, nickel comes up heads penny comes up heads, nickel comes up tails penny comes up tails, nickel comes up heads penny comes up tails, nickel comes up tails If you say "The penny came up heads" you eliminate the bottom two outcomes. If you say "At least one of the coins came up heads" you eliminate only the last one. 
#18




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It's no different than if you flipped two coins. It doesn't matter what order you flipped them, but it's convenient to refer to the first coin flip as A and the second one as B. You could choose a different way of organizing the children, as long as it's not based on what you are trying to discover. You could use tallest first or lightest skin first or youngest first or the first one you see when you go into their house (assuming you can't tell if they were male or female.) All that matters is that you preserve some sort of order, because there were two distinct probability events. To have no order would require you to combine them, treating them as if they are dependent on one another. But each probability is an independent event. So you get 1/2 * 1/2 = 1/4. Last edited by BigT; 05182015 at 05:57 PM. 
#19




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Probability is FUN 


#20




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1/4 of all families have 2 boys 1/4 of all families have 2 girls 1/2 of all families have one of each Now, like you say the reason that the population is structured like this is because the families were generated by random independent pickings. But for the purpose of this problem, we don't actually have to care. We could say "that's what the proportions look like, because logic" or "that's what the populations look like because I went out and surveyed 1000 families"  either way, as long as you accept the validity of the proportions above, you can do the rest of the problem. 
#21




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It's not like you actually did go out and do a survey, because, as stated upthread, the chances are not actually 50/50. You derived it from the assumptions. 
#22




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The biological assumption (ignoring unusual cases like twins) is actually a very good stance to take purely for mathematical purposes. This is because it is exactly identical to a cointoss probability problem involving two coins. The boygirl paradox is just a more interesting way of presenting it. It is merely coincidental that, in the real world, the assumptions are pretty safe to make because they correspond with the large majority of actual cases. Quote:
“A man with two kids was seen outside with his son. What’s the probability that both his children are boys?” and “A man with two kids was seen outside with his son. What’s the probability that the very specific child at home is a boy?” Give exactly the same results. In other words, once you have already seen one kid, you might as well just focus on the other kid alone and ask “what’s the probability that that kid is a boy?” If you expand the problem to a scenario involving three children, you will still get the same identical outcomes: “A man has three children. One day, he was seen with his son. What’s the probability that all three of his children are boys?” The probability of him having being seen with a boy instead of a girl is 1 if from a BBB family. For BBG, GBB or BGB families, the probabilities are 2/3 each. For BGG, GGB or GBG families, the probabilities are 1/3 each. Therefore the probability of him being from a BBB family is 1/(1+2+1) = 1/4. On the other hand, if you ignore the first child and simply ask “what is the probability that the two specific children at home are boys?” then the options are BB, BG, GG and GB. Thus the probability is still 1/4. Therefore the normal intuitive answer (for the twochild case) of 1/2 is correct even though one may not be aware of all the considerations that go into it. 
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