Calculating the Probability of Two Boys for Math-Obsessed Friend

[JeffJo]

The question further seems to imply that only the limited information given in the question itself should be used when solving the problem.

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 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.

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?

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.

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.

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.

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:

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?

He gave the answers as 1/2 and 1/3. But he retracted the second one six months later:

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.

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.

 

[Jimmy Snyder]

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.

This answer suffers from the same error as the 13/27 answer. That is, you have generalized Tuesday, but you have not generalized the number of children. What if the question in the OP is a particularization of the general question "I have n children. One is a {boy/girl} born on {S/M/T/W/T/F/S}, what is the probability I have n {boys/girls}."

 

[Dadface]

If two people were to ask you the same question and each one used a different day of the week, then you would know for sure which universe you were dealing with. If they asked you using the same sex and day of the week, then you could, with some confidence know which universe, but not be completely sure. However, if only a single person asks you, then you have no way of knowing which universe.

What is meant by a universe as used here?As I see it at present the original question refers to a universe which is just a family with two children and the other information(boy born on Tuesday) is not needed to get the real answer.

 

[JeffJo]

This answer suffers from the same error as the 13/27 answer. That is, you have generalized Tuesday, but you have not generalized the number of children. What if the question in the OP is a particularization of the general question "I have n children. One is a {boy/girl} born on {S/M/T/W/T/F/S}, what is the probability I have n {boys/girls}."

Then you need to know the probabilities P(n children) for all n>0 to answer. But the cases where you have n or m children, where n<>m, can not overlap and so can be treated independently. Which is what I did, and you are calling an "error." It is not.

The cases where the man says "I have two including a b3" and "I have two including a b4" can overlap, so treating them independently is an error.

+++++

There are many red herrings that can be chased in these problems. Other sized families. Why woudl a parent make this odd statement? More boys are born than girls, so the answer can't be based on P(boy)=1/2. There is a correlation between genders in siblings due to identical twins, so a far more complicated approach should be used. More children are born on weekdays because cesearian sections are scheduled, seldom of weekends. Some parents are more proud of one particular child of the two, so unknown biases exist.

What I find happens when people who want to defend the 1/3 or 13/27 answers can't find a hole in my arguments, is that they start chasing such red herrings. Please don't do that. There is a simple, incontrovertible demonstration of what the answer needs to be:

P(two boys) = sum(i=1 to 7, P(two boys|parent says "one is a boy born on DAY(i))*P(parent says "one is a boy born on DAY(i))

where sum(i=1 to 7,j=1 to 2, P(parent says "one is a SEX(j) born on DAY(i)) = 1
​

Yes, I did build "parent of two" into that; all the parents mentioned have two children. See above for justificaiton. I also built in the fact that a parent of two says "one is a SEX(j) born on DAY(i)", so that the second equation holds. That is also reasonable, since we aren't interested in any parents who don't, and you can adjust the probabilities for any situation you think applies within that condition. And I left out the "j=1 to 2" part in the first equation, since the conditional probability is always 0. Technically it should be in there, but it takes less space if I leave it out, and contributes nothing.

Do we all agree that P(two boys)=1/4? This is the unconditional probability (except the conditions that they have two children, and make the odd statement). In order for P(two boys|parent of two says "one is a BOY born on TUESDAY) to equal 13/27, you need to accept one of three absurdities. Either

1. P(two boys) is not 1/4
2. P(two boys|parent of two says "one is a BOY born on DAY(i)) is different for different values of i
3. P(parent of two says "one is a SEX(j) born on DAY(i)) is different for different values of i and j.

If #2 and #3 are false - meaning the respective probabilities are all equal - then P(two boys|parent of two says "one is a BOY born on DAY(i))=13/27 for all i and j, and P(parent of two says "one is a BOY born on DAY(i))=1/14 for all i and J. That makes P(two boys)=13/54. But if any of those probabilities are different, the problem has to establish why in what is known as the a priori state. That's the state before the known condition "parent of two says 'one is a BOY born on TUESDAY'" is applied, which is the state these probabilities must evaluated in. So you can't use the given values of i and j to deduce anythign about the probabilities.

 

[BobG]

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.

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:

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.

I can agree with this point in theory, but Gardner's statement is slightly detached from reality.

I think it's safe to say that questions asked in brain teasers are asked in a "Clue" type environment in which you're looking to eliminate possibilities. In other words, if the person answered "I have a girl", you would have narrowed things down to a different probability than if the person answered "I have a boy" and would be on a different path. Only the path you're currently on is relevant.

Just like in the game where you have a person pick a number from 1 to 100 and guess his number in no more than seven tries. There may be a nearly 50/50 chance of getting higher or lower (allowing for 1 guess being correct), but each answer eliminates nearly 50% of the possibilities.

Actually, maybe the brain teaser is more detached from reality, since there's just no way you're going to get perfectly random answers in the real world. Trying to figure out how to test this to make my point kind of tilts me at least a little bit towards what Gardner said, since asking humans just isn't as easy as using cards (the truth of the Monty Hall problem is easily provable by simulating the exercise with 3 cards). You'd have to do something weird, such as having the parent flip a coin to decide which child's sex they'd tell me about to make the brain teaser probabilities work out. Otherwise you'd have to resort to saying that asking lousy subjects caused the results of your real world test to be unrealistic.

But brain teasers exist in a universe where native tribes can live happily with rampant infidelity in their tribe for generations until a missionary tells them at least one person is unfaithful and then the tribe has to respond with a mass murder of all the men. I think the original question can be safely answered assuming a perfectly logical, yet perfectly random universe.

 

[JeffJo]

I can agree with this point in theory, but Gardner's statement is slightly detached from reality.

No, it is the one that is grounded in reality. It is the interpretation that "We have knowledge of only one of two genders in this family, and it is male" is 100% erquivalent to "this family has at least one boy" that is detached. Since, by definition, we know that our knowledge is incomplete, we have to consider that a familiy might have a boy, but we don't know it.

Again, look at my arguments using the law of total probability. I'll even express it a different way:

I have two children, and one is the gender I have written down and sealed in this envelope. What is the probability I have one boy and one girl (I'm reversing it for ease of comparison)? If I reveal the gender I wrote, and it is "Boy," what is the probability? If I reveal it, and is is "Girl," what is the probability?​

I think everybody agrees the first probability I asked for is 1/2. The second is a conditional probability, which can be expressed P(I have one of each|I wrote down "boy"). The third is P(I have one of each|I wrote down "girl").

I think everbody will also agree that the last two probabilities have to be the same. Let's call that value Q.

One rule of probability, called Bayes' Rule, is that P(I have one of each|I wrote down "boy")*P(I wrote down "boy")=P(I have one of each and I wrote down "boy")=Q*P(I wrote down "boy"). Similarly P(I have one of each|I wrote down "girl")*P(I wrote down "girl")=P(I have one of each an I wrote down "girl")=Q*P(I wrote down "boy").

Adding these two together gives P(I have one of each and I wrote down "boy")+P(I have one of each and I wrote down "girl") = Q*[P(I wrote down "boy")+P(I wrote down "girl")]. But another rule, called the law of Total Probability, says that P(I have one of each and I wrote down "boy")+P(I have one of each and I wrote down "girl")=P(I have one of each). The same rules says P(I wrote down "boy")+P(I wrote down "girl")=1. So P(I have one of each)=Q. But we know what P(I have one of each) is: 1/2. So Q=1/2.

That's a very long-winded way of saying that the second and third probabilities have to be the same, and that since there are only two possibilites for what I wrote down, the answer is the same regardless of what I wrote down. So teh answer has to be the same for all three.

I've repeated this argument about five different ways now. Nobody has tried to address any of them, because thay are all based on indistutable logic. So why can you not believe it is the right answer?

I think it's safe to say that questions asked in brain teasers are a
sked in a "Clue" type environment

Why? And please consider that you might think this because it gets the answer you want to be right; rather than to deduce this first, and use that to derive the answer. Also note that this issue is only important to probabiltiy probles (because you need to be concerned with what could have happened, but didn't) and even then, only when the possibilities overlap. So you can't compare it to most brain teasers you've seen.

In other words, if the person answered "I have a girl", you would have narrowed things down to a different probability than if the person answered "I have a boy" and would be on a different path. Only the path you're currently on is relevant.

I agree that if a person answered such a question, the answer is 1/3. I just see no reason, except to force the answer to be 1/3, to think they answered "Do you have a boy?" It is backwards to assume that the answer defines the unknown question that was asked. It is why Gardner said the question was ambiguous.

While we don't have to make brain teasers realistic, we do need to assume they are consistent. There are lots of questions we could assume were asked, but if we assume it was "Do you have a boy," then we also need to assume some people in the same universe are asked "do you have a girl?" So anybody who answered "no" is eliminated. But while none of the mixed families will answer "no," half will answer "yes" to a different question, and we still get 1/2 as the correct answer.

 

[Jimmy Snyder]

What is meant by a universe as used here?As I see it at present the original question refers to a universe which is just a family with two children and the other information(boy born on Tuesday) is not needed to get the real answer.

There are two universes. One universe is a room A filled with fathers who have two children, one of which is a boy born on Tuesday. The other universe is a room B filled with father who have two children, but not necessarily a boy born on Tuesday. Note that some of the fathers in room B have a boy born on Tuesday, but not all of them do. Someone from room A or room B greets you. You don't know which room they came from. They ask you the question in the OP. You still don't know which room they came from.

 

[Dadface]

There are two universes. One universe is a room A filled with fathers who have two children, one of which is a boy born on Tuesday. The other universe is a room B filled with father who have two children, but not necessarily a boy born on Tuesday. Note that some of the fathers in room B have a boy born on Tuesday, but not all of them do. Someone from room A or room B greets you. You don't know which room they came from. They ask you the question in the OP. You still don't know which room they came from.

Ah I get it.Thank you Jimmy.I don't think it makes a difference to the main point I am trying to put across.

 

[JeffJo]

Imagine a rather unique casino game. The players have two cards, one marked "Doubles" and one "Not Doubles" on the face-down side, but identical on the other. After the croupier rolls two standard dice behind a screen, he announces "One of the dice landed on a ..." and names a number. Each player can wager $1 on Doubles of this number, or Not Doubles, by pushing out a dollar and the appropriate face-down card. Then the dice are revealed, and all players' cards. If bets were placed on both options, the winners split the losers' money. If not, everybody takes back their bets.

Obviously, the croupier needs a strategy for deciding what number to announce when two different numbers come up. This strategy affects the probabilities after each number he announces. For example, if he always announces the highest number, the probability of doubles is 1/11 if he announces a six, but 1/1 if he announces a one.

Interestingly, if everybody else knows what that strategy is and bets accordingly (i.e., gambling on the longer odds a proportionate number of times), but you don’t, you still will break even by betting on doubles once every six games. And I can show you how to make an Excel spreadsheet that simulates it. This means the average probability of doubles is 1/6, even if you don't know the strategy. It may differ on individual rolls; but you can't know how so your best - and correct - estimate is the average.

This strategy represents what you all have been calling a "universe." This example shows that you don’t need to know what the universe is, to answer the question "what is the probability that two '1 in N' chances match, if I tell you what one of them is, based on the knowledge you have." It is 1/N, not 1/(2*N-1). The only point to the "universe" is that if you know what it is, you can use that to refine your answer. But if you don't, you can't assuem there is a non-standard on in a brain teaser.

If a man tells you "I have two children, and one of them is a boy," but you don’t know why he would tell you about one gender over another (i.e., what "universe" you are in), the probability of two boys is 1/2. Not 1/3. It is only if you know, for a fact, that he must tell you about a boy if he has one, that your probability is 1/3.

 

[Dadface]

I meant to write another post in this thread about the "Boys Puzzle" but I have been stopped in my tracks(temporarily possibly) after reading more about the "Boy Girl Paradox" which was first published by Martin Gardener in 1959.The original question was phrased as follows.

Mr Smith has two children.At least one of them is a boy.What is the probability that both children are boys?

Apparently the paradox arises from the fact that the answer could be 1/2 or 1/3 depending on how it is found out that one child is a boy.Look at the information in the question:

1.Mr Smith has a son
2.Mr Smith has a second child

We know that,regardless of how the information was obtained,the son has one sibling.We further know that there is a fifty percent chance that the sibling is a boy and a fifty percent chance that it is a girl.The answer is 1/2.Where's the paradox?

(considering age differences makes no difference because we know,without being told,that the son is either the oldest or the youngest child)

 

[JeffJo]

We know that, regardless of how the information was obtained, the son has one sibling. We further know that there is a fifty percent chance that the sibling is a boy and a fifty percent chance that it is a girl. The answer is 1/2. Where's the paradox?

If the family has two boys, which one is the "the son" that I colored in red? If you can't identify him, you have to consider the chances of the two children together.

Essentially, you are treating the problem statement as "I know about one particular child of Mr. Smith's. That child is a boy." When that is the problem statement, the answer is indeed 1/2 because the un-pictured child has a 1/2 chance to be a boy, as you reasoned. But that is assuming information not included in the problem. You are assuming a specific child is identified.

Others treat it as "I know about both of Mr. Smith's children. The pair includes at least one boy." When that is the problem statement, the answer is 1/3 because 3/4 of all the possible families include a boy, and 1/4 include two, so the answer is (1/4)/(3/4)=1/3. But this also assumes information not included in the problem. It assumes you actually know both ngenders: you need to, in order to always find a boy when one is there. It is implicit that you can know about one boy, but not both. It also assumes the intent to look for a boy, something you can't deduce from the statement alone.

Most people who take the second approach accuse those who answer "1/2" of taking the first. That's why it is usually compared to the Mr. Jones question. While many do exactly that - as did you - that isn't the only way to get 1/2. I prefer to think of it as "I know a gender that exists in Mr. Smith's family." This assumes nothing, since all that is really implicit in the statement is that you know a gender. This is more complicated to solve, because you have to allow for the possibility that you might know "girl." In fact, the whole reason conditional probability is unintitive is that you have to allow for things you know didn't happen. But the short of it is that in the 1/2 that have one boy and one girl, it is just as likely to know about the girl as the boy.

An example of "knowing a gender" might be if a new family moves into the three-bedrooom house next door. You might deduce they have a boy by seeing a boy's bicycle in the driveway, but you can't associate it with a specific child.

+++++

But since you are interested in history, I have traced what may (it's speculation, but it seems plausible) be the history of the Tuesday Boy Problem, and how top-level experts can't always agree.

In his 1988 book "Innumeracy," Professor John Allen Paulos of Temple University included a variant of the Two Child Problem as an example, paraphrased: "If Myrtle is a girl from a family of two, what is the probability she has a brother?" His answer was 2/3 (note that he asks for the probability if a mixed family, not two girls).

J. L. Snell (Dartmouth) and R. Vanderbei (Princeton) pointed out in a 1995 article titled "Three Bewitching paradoxes" that, by giving the girl an uncommon name, he inadvertantly changed the problem. The probability Myrtle has a sister should be 2/(4-p), where p is the probability a girl is named Myrtle (this is the same formula that would be used for the Tuesday Boy question, if it asked fro a mixed family). In his 2008 book "The Drunkard's Walk," Leonard Mlodinow (Stanford, and a frequent collaborator with no less than Stephen Hawkings) used the same problem, about a girl named Florida.

Snell and Vanderbei ignored the illogic of having two girls named Myrtle in the same family, and included that case in their count. Mlodinow argued that the factor for it depends on p^2, which is negligibly small compared to the other kinds of families. Which is wrong: compared to the fraction of the famlies he is counting, it depends on p^1.

Giulio D'Agostini (Universit`a “La Sapienza” and INFN, Rome, Italy) attempted to correct for that factor, but did it wrong by disallowing only girls named Myrtle/Florida. He allowed two girls of any other name. But at least he got the right answer, 1/2, which is quite trivial to prove by a different method!

Define the following events: M2 is the event where a family of 2 children includes a girl named Myrtle. MO is the event where she is the older sibling, and in MY she is the younger. Finally, MB is the event where she has a brother. Everybody will agree that the probability Myrtle has a brother, given that she is either the older or younger sibling, is 1/2. Which is the first statwement in this progression:

1. P(MB|MO)=1/2
2. P(MB|MY)=1/2
3. P(MO|M2)=Q (Most will say it is actually 1/2. I use a variable because I don't need to know it, and the error I'll demonstrate makes it a little more than 1/2.)
4. P(MY|M2)=1-Q (That is, MO and MY represent all possibilites in M2, and do not overlap)
5. P(MB|MO)*P(MO|M2) = P(MB and MO|M2) = Q/2
6. P(MB|MY)*P(MY|M2) = P(MB and MY|M2) = (1-Q)/2
7. P(MB and MO|M2) + P(MB and MY|M2) = P(MB|M2) = Q/2 + (1-Q)/2 = 1/2.
8. QED.

The error that Snell, Vanderbei, and Mlodinow make is that by allowing two girls named Myrtle/Florida in the same family, MO and MY overlap. Both P(MO|M2) and P(MY|M2) are equal to Q=2/(4-P), which is greater than 1/2. But they still use equation #7, which is invalid if MO and MY overlap. As a result, they get P(MB|M2)=2/(4-P), or the probability Myrtle has a sister is (2-P)/(4-P).

And what is event worse, is that my derivation above is wrong. Specificalky, equation #2 is wrong. #1 is right because the gender of a second child is independent of both the gender and name of the first child, but the name (not gender) of a second child will depend on the name of the first child, if the gender is the same. I won't go into details unless asked, but it turns out that P(Myrtle has a brother) is approximately equal to 1/2+[P(a girl receives the "average" name)-P(a girl gets named Myrtle)]/8. In other words, for uncommon names, the probabiltiy is greater than 1/2!

 

[Dadface]

JeffJo,thanks for your replies.Consider the following:

1.Mr Smith has a son.
2.Mr Smith has two children

The two children must be siblings and therefore the Smith children must be either brother with brother or brother with sister.It follows that if there is a son the probability of the second child being a son is 1/2.

This is an attempt to structure my reasoning such that I don't need to identify any of the children.

 

[JeffJo]

JeffJo,thanks for your replies.Consider the following:

1.Mr Smith has a son.
2.Mr Smith has two children

The two children must be siblings and therefore the Smith children must be either brother with brother or brother with sister. It follows that if there is a son the probability of the second child being a son is 1/2.

This is an attempt to structure my reasoning such that I don't need to identify any of the children.

Correction: it must be brother with brother, older brother with younger sister, or younger brother with older sister. Each of those cases is equally likely to exist. The fact that you cannot specify the relative ages for the first group changes how you account for them. This is the point mathematicians are trying to demonstrate with this problem.

But they overlook that "girl" cannot apply to that case, but can apply to the others. That also changes how you account for them. The answer is 1/2, but for reasons different than you propose.

 

[JeffJo]

For some reason, I can't edit posts today. I misremembered where Leonard Mlodinow teaches. He's at Caltech, not Stanford (there is another professor who is part of the story - because he is one of the few who acknowledges the ambiguity in these problems - at Stanford).

 

[Dadface]

Correction: it must be brother with brother, older brother with younger sister, or younger brother with older sister. Each of those cases is equally likely to exist. The fact that you cannot specify the relative ages for the first group changes how you account for them. This is the point mathematicians are trying to demonstrate with this problem.

But they overlook that "girl" cannot apply to that case, but can apply to the others. That also changes how you account for them. The answer is 1/2, but for reasons different than you propose.

Thanks for your feed back JeffJo.When I wrote post 62 I forgot that age differences need to be considered.What a dope I am.I wasn't familiar with these type of problems before but now I'm hooked.

 

[Dadface]

Can someone confirm,or otherwise that with problems of this type we must use the information contained in the problem statement only,even though this might be ambiguous and open to different interpretations,and that we are not allowed to bring our own extra knowledge to the task?.Consider the boy girl paradox.We don't need to be told that the two siblings have different ages,this is knowledge we bring to the task.We don't need to be told that if there are two boys they can be identified because they have numerous other differences in addition to having different ages but this is knowledge it seems we are not allowed to bring to the task.The restrictions imposed on what we can and cannot use seem arbitary and artificial.Consider this...we are required to suspend all knowledge that if there are two boys they will have different age related bodily conditions but retain the knowledge that they(siblings) have different ages.

 

[JeffJo]

Can someone confirm, or otherwise that with problems of this type we must use the information contained in the problem statement only, ...

Of course. But like Inigo Montoya said, I don’t think that means what you think it means.

We know the family has two distinct children. That means that they can be differentiated from one another by somebody, but not necessarily by us. And we do know this, whether or not it is stated in the problem.

Age is a very convenient tool to use to express the difference in terms of an order. But it isn't the only one we could use. I prefer to alphabetize the first name of each child's best friend, or order them clockwise (relative to mother) as they sit around the dinner table. As long as it is unique (which I suppose friends' names might not be, but let's assume they are) and independent of gender, all that matters is that the order exists.

We don’t even have to know how the order applies to any family; we use it only to calculate the proportions of the various groups of families we need to keep track of. Here's some simpler examples:

• A six-sided die has two red sides, and four white sides.; The red sides are opposite each other. Other than that, no side can be distinguished from another. What is the probability that a red side will come up on a roll? Is it 1/2, because there are two colors? Or 4/6=2/3, because there are four red sides out of six, even though we can't distinguish them?
• Two normal dice are completely indistinguishable. What is the probability their sum is 7 when you roll them? Is it 1/11, because there are 11 different totals that could come up? Or 3/21=1/7, because there three unordered pairs of numbers that add up to 7, out of 21 possible unordered pairs? Or 6/36=1/6, because there are 36 possible ordered pairs, and 6 of them total 7?

The last answer is right for each, of course. The different groupings exist, and we know they exist, even though it might be ambiguous what group a particular roll belongs to. The problem with how this Paradox is normally presented, is that it makes age sound like an we need ultimately to know the age, and we don't.

+++++
But you can't add information to a problem if it isn’t there, or clearly implied. That is why the Two Child Problem is generally ambiguous, but the best answer is 1/2.

When a man walks up to you and says "Mr. Smith has two children, and one is a boy," do you know, for a fact, how he decided to tell you that? In particular, do you have any reason to believe one of these possible reasons (and there are others) is preferable to another? (Note that you have to assume he never intended to tell you all the information that applies to Mr. Smith, since there would be no probability problem then. That makes all possibilities a little unrealistic, so don’t be surprised by it.)

1. He knows about only one child of Mr. Smith's.
2. He picked a gender at random from what is either one, or two, that exist in the family.
3. He is predisposed to mention boys, and so the tells you about a boy if he can, and a girl if he can't.
4. He is predisposed to mention girls, and so the tells you about a girl if he can, and a boy if he can't.

He could say "one is a boy" with any of these, so you can't pick one that must be true. So most wordings of the problem are ambiguous. But most of the reasons that you can put in the list require adding some information. Does the problem say he knows about both of Mr. Smith's children (as #2, #3, and #4 require)? Some do, but the Mr. Smith one doesn't. Does it say he is predisposed toward one gender or the other? Very few do. And the 1/3 answer requires that the answer to both questions be "yes." Without both, the answer is 1/2.