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#1
05-18-2015, 05:56 AM
 Gorilla Guest Join Date: Jun 2014 Posts: 41

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:
• 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?

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 value-judgment 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 (follow-up) question, yet, other variants of the problem don’t actually do that (like the Bar-Hillel & 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:
 ...However, this problem is about probability and not biology. The mathematical outcome would be the same if it were phrased in terms of a gold coin and a silver coin.
And yet, by treating BG and GB as two separate outcomes, the problem is being treated precisely as being about human reproduction. Of course, the twin possibility could be dismissed on the basis of the fact that twins are relatively rare. However, once you say “this problem is not about biology” you effectively put the twin possibility on an equal footing with the rest of the outcomes - since you can’t consider the biological frequency of each option. Besides, twins are actually fairly common, at least in some parts of the world. Thus, it can be treated as a serious outcome – unlike, say, intersex children. (Incidentally, the coin analogy mentioned in the quote would produce an answer of 1/2 and not 1/3, given that the question is not detailed enough to provide a reason to treat gold-silver and silver-gold outcomes as being any different.) Note that the issue surrounding the probability is about conception and not about the process of birth-giving. Otherwise, since twins are always born one at a time, it would amount to the same process as when two non-twin siblings are born. In other words, the real issue is about the probability of two male children having being conceived.

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 coin-flip 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 gold-coin silver-coin analogy mentioned by the article. For example:
• Someone opens a bag of gold and silver coins, each of the same quantity, takes out two coins one at a time and holds them in his hand. One of them is gold. What is the probability that both are gold?
• Someone opens a bag of gold and silver coins, each of the same quantity, takes out two coins at the same time and holds them in his hand. One of them is gold. What is the probability that both are gold?

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; 05-18-2015 at 06:00 AM.
#2
05-18-2015, 06:34 AM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 15,629
Quote:
 Originally Posted by Gorilla [1] it is usually agreed, given the way it is worded, that 1/3 ... 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. [2] 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 ... [3] Basically, there are five ways that one could end up adopting two children, one of whom is a boy: ... From these options, the BB probability would clearly be 2/5.
1. No. The correct answer remains 1/3 even with the alternate phraseology.

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
05-18-2015, 06:37 AM
 Aspidistra Member Join Date: Feb 2001 Location: Melbourne, Australia Posts: 4,527
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 two-girl families and invites you to pick a surname at random. What's the probability that you just chose a two-boy 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
05-18-2015, 06:49 AM
 RTFirefly Member Join Date: Apr 1999 Location: Maryland Posts: 34,794
Quote:
 Originally Posted by Gorilla The problem is normally stated as follows: 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?
If all we know is that the Smiths and the Joneses each have two children, there are four equally probable possibilities for each family, with the sex of the older child first:

GG
GB
BG
BB

With (pass me a bottle) Mr. Jones, you're knocking out two possibilities, so you have this:

GG
GB
BG
BB

But with the Smiths, you're knocking out just one:

GG
GB
BG
BB

In both cases, the remaining possibilities are still equally likely.
#5
05-18-2015, 09:36 AM
 Gorilla Guest Join Date: Jun 2014 Posts: 41
Quote:
 Originally Posted by septimus 1. No. The correct answer remains 1/3 even with the alternate phraseology. 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."
1. Yes. But that doesn’t stop it from misleading someone into thinking it to be 1/2. It is the process of automatically breaking down the first phraseology, in one’s mind, into the second one that might cause that error. Note that the original phrase is generally regarded to be ambiguous even though it has one correct answer.
3. That’s the point I was making.

Quote:
 Originally Posted by Aspidistra 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 two-girl families and invites you to pick a surname at random. What's the probability that you just chose a two-boy 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.
A similar thing is written in the wiki article as an explanation for the ambiguity. Like I said above, the probability depends on how you find out that one child is a boy, but it also depends on the assumptions being made about how the family had the children. If all the children were adopted, then the choices would only be BG and BB families (in equal proportions). The adoption scenario doesn’t produce separate “BG” “GB” categories. That is only true for the biological case. And if you assume that the children could be identical twins, then you have a separate category called (BB) which is different from BB.

Otherwise, how would you answer the gold-silver coin analogy?

Quote:
 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.
Yes, it’s basically the same thing. I should have made it clear that the answer is 1/2 for a very different reason, i.e. not simply because the chances of having a boy as opposed to a girl is 1/2.
#6
05-18-2015, 09:51 AM
 Aspidistra Member Join Date: Feb 2001 Location: Melbourne, Australia Posts: 4,527
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; 05-18-2015 at 09:54 AM.
#7
05-18-2015, 10:26 AM
 bup Guest Join Date: Sep 1999 Location: glenview,il,usa Posts: 11,905
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
05-18-2015, 10:59 AM
 Biotop Charter Member Join Date: Mar 2000 Location: Faber, VA Posts: 6,837
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
05-18-2015, 11:33 AM
 Biotop Charter Member Join Date: Mar 2000 Location: Faber, VA Posts: 6,837
Quote:
 Originally Posted by Biotop 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?
Correction. 2 out of 3 actually seems right.

Probability is hard.
#10
05-18-2015, 12:52 PM
 Gorilla Guest Join Date: Jun 2014 Posts: 41
Quote:
 Originally Posted by Aspidistra 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.
Of course it is important that they all have equal weight. But the point is that there are four different combinations in the first place because of certain assumptions being made, including the one that ignores the "furphy" option.

Quote:
 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 doesn't have to be any 'constraint' in order to treat them as being equally likely. It is the final outcome (two children) that matters. Placing constraints would of course change the probability of each outcome. But without any constraints (or assumptions), they would be the same.

Quote:
 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)
This is about method of choosing, not about time. The answers would be different if the person picks at least one of the coins without looking into the bag (i.e: 'blind'). In this case, it is identical to the original boy-girl paradox. But if he deliberately chooses which two coins to select, then the answers to both questions are the same. That's why it is identical to the adoption example. Now you can see why there is no need for "constraints" for any of the adoption scenarios. As you yourself just said, "time of choosing doesn't matter."
#11
05-18-2015, 01:07 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,534
Quote:
 Originally Posted by Gorilla There doesn't have to be any 'constraint' in order to treat them as being equally likely.
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
05-18-2015, 04:50 PM
 Weedy Guest Join Date: May 2009 Location: Sydney, Australia Posts: 1,521
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
05-18-2015, 05:09 PM
 bup Guest Join Date: Sep 1999 Location: glenview,il,usa Posts: 11,905
Quote:
 Originally Posted by Weedy If the time of adoption doesn't matter (and I don't see why it would), why does birth order matter?
It doesn't really. It's just that, the way you ask the question, you can eliminate one of the 'quadrants.'

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; 05-18-2015 at 05:12 PM. Reason: stupid formatting f'ed up my nice diagram
#14
05-18-2015, 05:16 PM
 bup Guest Join Date: Sep 1999 Location: glenview,il,usa Posts: 11,905
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
05-18-2015, 05:40 PM
 Weedy Guest Join Date: May 2009 Location: Sydney, Australia Posts: 1,521
Quote:
 Originally Posted by bup It doesn't really. It's just that, the way you ask the question, you can eliminate one of the 'quadrants.' 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.
But why are there 4 quadrants in the first place? Why not 3 possibilities (gg,gb,bb) and you eliminate 1?

Why is birth order a relevant way of organising the problem?
#16
05-18-2015, 05:49 PM
 dracoi Guest Join Date: Dec 2008 Posts: 8,867
Quote:
 Originally Posted by Weedy 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?
Maybe your confusion is just with informal uses of the terms. If you go down to your local pub and a guy says "One of my kids is a boy" and the other guy says "At least one of my kids is a boy" then there's a good chance they actually mean the same thing.

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
05-18-2015, 05:54 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,534
Quote:
 Originally Posted by Weedy But why are there 4 quadrants in the first place? Why not 3 possibilities (gg,gb,bb) and you eliminate 1? Why is birth order a relevant way of organising the problem?
Birth order is just one handy way of distinguishing between the two kids. For two kids, there are two possible ways of having one be a boy and the other be a girl, as compared to only one way of having them both be boys.

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 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
05-18-2015, 05:55 PM
 BigT Guest Join Date: Aug 2008 Location: "Hicksville", Ark. Posts: 33,457
Quote:
 Originally Posted by Weedy But why are there 4 quadrants in the first place? Why not 3 possibilities (gg,gb,bb) and you eliminate 1? Why is birth order a relevant way of organising the problem?
It's not birth order. It's just that there are two events, and sorting them by the order in which they originally happened is convenient. You have the first born, which is one 50/50 event. Then you have another child, which is another 50/50 event.

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; 05-18-2015 at 05:57 PM.
#19
05-18-2015, 07:34 PM
 Aspidistra Member Join Date: Feb 2001 Location: Melbourne, Australia Posts: 4,527
Quote:
Originally Posted by Biotop
Quote:
 Originally Posted by Biotop Originally Posted by Biotop View Post 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?
Correction. 2 out of 3 actually seems right.

Probability is hard.
2 out of 3 is right, but possibly not for the reason you think it is. Because the probability is also 2 out of 3 that he's got a brother.

Probability is FUN
#20
05-18-2015, 07:44 PM
 Aspidistra Member Join Date: Feb 2001 Location: Melbourne, Australia Posts: 4,527
Quote:
 Originally Posted by BigT It's not birth order. It's just that there are two events, and sorting them by the order in which they originally happened is convenient. ... 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.
Actually, to solve this particular problem, you can dispense with an ordering altogether if you want. The only core fact you need to know is how much of your population falls into each group.

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
05-19-2015, 01:18 AM
 BigT Guest Join Date: Aug 2008 Location: "Hicksville", Ark. Posts: 33,457
Quote:
 Originally Posted by Aspidistra Actually, to solve this particular problem, you can dispense with an ordering altogether if you want. The only core fact you need to know is how much of your population falls into each group. 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.
The question I was being asked was specifically the "why" part that you admit is because of the two distinct pickings.

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
05-19-2015, 06:32 AM
 Gorilla Guest Join Date: Jun 2014 Posts: 41
Quote:
 Originally Posted by Thudlow Boink 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.
The issue is not how likely each individual method of selection is (of two children). What matters is the biases of each parent(s) in terms of what combination of children they want (boy-boy, boy-girl or girl-girl). It doesn't matter how they arrive at their chosen combinations. That is what makes the adoption scenario fundamentally different from the biological one.

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 coin-toss probability problem involving two coins. The boy-girl 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:
 Originally Posted by Gorilla Yes, it’s basically the same thing. I should have made it clear that the answer is 1/2 for a very different reason, i.e. not simply because the chances of having a boy as opposed to a girl is 1/2.
I’d like to make a slight correction. While it is true that the calculations are very different, they nevertheless necessarily amount to the same thing. (Though, this is only true with the normal assumptions which makes the problem identical to a coin-toss situation). In other words, the two questions:

“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 two-child 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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