Boys Puzzle

[Jimmy Snyder]

There's one big issue at the heart of it (just like in the Monty Hall problem), JeffJo alluded to it, and it is the one that decides whether the answer is 13/27, 1/3 or 1/2.

If your friend did not have a boy born on Tuesday, what would he have done instead?

If he would've picked one kid at random and asked the question anyway, using that kid's data, the answer is 1/2.

If he would've stayed silent, the answer is 13/27.

There's one contingincy you haven't calculated. What is the probability if we don't know what he would have done. In other words, what is the answer to the question as it was put?

 

[hamster143]

There's one contingincy you haven't calculated. What is the probability if we don't know what he would have done. In other words, what is the answer to the question as it was put?

If we don't know what he would have done, we can't put a probability on that. It's called "uncertainty". Which is different from risk/probability because it's not quantifiable. The best we can do is put probabilities on our expectations of his mental state, and use them to weigh 1/2 and 13/27 into the final number.

 

[Dadface]

It is known that the boy is born on a certain day so what difference does it make if one is told that? In fact even more details are known but unstated and the answer of 13/27 is found by using just a limited amount of the known information.Let me illustrate this by means of an example.

Let the information used be that the boy was born during the first half of the week,in other words the week is divided up into two time intervals only.Using the methods outlined here the probability works out to be 3/7=0.42
If the week is divided into seven time intervals,as in the original question,the probability works out to be 13/27=0.48
Lets not stop there and divide the week into 168 hourly time intervals.Nobody needs to be told that the birthtime(depending on how this is defined)is within the limits of a certain hour.Using hourly intervals the probability works out at 335/671=0.49.
It can be seen where this is going.If the week is divided into n time intervals then the probability is given by:
P=(2n-1)/(4n-1)
from this it is seen that as n increase P tends towards the real answer which is 1/2.
In summary,depending on the definition of birth time we know that the boy had been born on a certain day and within a certain hour of that certain day and so on.Why then use just the limited information stated in the question when there is much more information at one's disposal?

 

[JaredJames]

Why then use just the limited information stated in the question when there is much more information at one's disposal?

Yes, those factors would alter the answer, but they are only at your disposal if you have them. You don't. They aren't stated in the question so they are of no consequence to you. There isn't "much more information at one's disposal".

 

[Dadface]

Yes, those factors would alter the answer, but they are only at your disposal if you have them. You don't. They aren't stated in the question so they are of no consequence to you. There isn't "much more information at one's disposal".

But the factors are at my disposal because I know them already without being told and the factors are a matter of general knowledge and implicit in the question.Presumably this is supposed to be a real question about a real family and my answer assumes that to be so.Since when can a maths question require,without specifying, that a solver forget certain bits of information that they might otherwise bring to the task?

 

[JaredJames]

Ah, I see what you're saying. Hmm, yeah I suppose you could improve things like that.

 

[JeffJo]

Here is a brute force proof that the probability is 13/27.

Actually, it isn't. It falls into the error described by http://fox-lab.org/papers/Fox&Levav(2004).pdf [Broken]. Just read the abstract if you don't believe me.

A proper solution does not count cases, it sums the probability that each case would produce the observed result. If every case that is possible can't produce any other result, the two methods are the same. This is true for many other simple probability problems, which is why people mistakenly beleive that counting is correct. It isn't.

To do it right, assign a number to each of the 196 (originally) equally probable outcomes in your grid, representing the probability that the parent of that family would say "One is a boy born on a Tuesday."

Obvioulsy, 196-27=169 get a zero this way. We don't really want to ignore them, we add a zero for them. It amounts to the same thing, I know, but it helps conceptually to recognize that you add their contribution.

One cell, "b3b3," gets a 1 this way. Assuming the parent is going to give us some information in this form, there is only one thing that parent can say.

The rest get a number somewhere between 0 and 1, inclusive. I called it P. There are 12 P's in the blue cells you outlined, and 14 in the red cells. That makes the answer (1+12P)/(1+12P+14P), as I said before.

There's one contingincy you haven't calculated. What is the probability if we don't know what he would have done. In other words, what is the answer to the question as it was put?

(1+12P)/(1+26P). If we don't know, we can only represent the answer as a function of what it is we don't know. The problem statement does not allow us to state categorically what P is, but assuming anything other than 1/2 represents a bias on the part of the parent, and we can't assume a bias. We can assume the lack of a bias, which means P=1/2, and the answer is 1/2.

Look up the Monty Hall Problem on Wikipedia. Some mathematicians do the same thing for it. The answer depends on what bias Monty Hall has, toward opening whatever door you saw him open. The probability your door has the car is P/(1+P), and that the other door has it is 1/(1+P). Since P/(1+P)<=1/(1+P) for any P, you should always switch. But the laws in the USA require P=1/2, so the answers really are 1/3 and 2/3.

There is one other nasty side-effect if you assume P=1. When you ask the same quesiton but with "...born on a Wednesday," you need to put a different number in each cell, and the numbers in each cell can't add up to more than 1. Same for any other day. Since at most two numbers can be non-zero, they must add up to 1. So if you put a 1 in cell b3b4 for "Tuesday," you must put a 0 in it for "Wednesday." The maximum probability you can get for this second question is 11/25. Then 9/23, and 7/21, etc., for other days. But if they are all to be the same, it can only be 1/2.

All you need to do to see I am right, is to look at what I said about the Law of Total Probability. That is a relationship that cannot be broken in probability. It essentially says that the average of the answers for all possible days MUST be 1/2.

+++++

For peteb: you overlooked one part of the solution you linked to (with emphasis added):

Everything depends, [Yuval Peres of Microsoft Research] points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.

The first case corresponds to my P=1. The second, to P=1/2. What part of the statement you made do you think shoudl suggest to us that you "specifically choose him because he was a boy born on Tuesday?" Because the implication most people get, is that it is an observation.

In the simpler problem, where the dabated answers are 1/2 and 1/3, there are three approaches. The most naïve is that you are talking about a specific child, so the "other" child has a 50% chance to be a boy. The intermediate approach is that there is no "specific" child, and that is partially true. People who take this approach compare it to saying the older chidl is a boy to emphasize what choosing a specific child does. But what they are doing, instead, is choosing a specific gender, which is just as wrong. And in BG families, where there is only one boy, specifying "boy" is the same as specifying "older." The best answer is that however the statement was chosen, it was done randomly. Unless, of course, the problem statement is explicit about how one child, or one gender, was chosen.

 

[Dadface]

Everything indeed. There is a big difference between

• "I have two children. One is a boy."
• "I have two children. The older of the two is a boy."
• "I have two children. One is a boy born on Tuesday."

Most of us are pretty lousy at dealing with conditional probabilities. Many people, even many otherwise very intelligent people get the Monty Hall problem wrong, and many apparently got this one wrong as well.

A paper on why we suck at this kind of reasoning: http://fox-lab.org/papers/Fox&Levav(2004).pdf [Broken]

I have just looked up the Monty Hall problem and this makes reference to three parts,namely two goats and one car or alternatively three separate doors.The boys problem refers to,by implication,seven separate parts namely the seven days of the week.A major difference between the two problems is that the parts referred to in the M H problem are separate and indivisible whereas the parts referred to in the boys problem can be subdivided into smaller parts eg hours ,weeks and seconds.I guess that the person who wrote the boys problem did not want any solvers to carry out this subdivision and therefore reach an answer of 13/27 but the question is phrased in such a way that such subdivision is allowed.Because of this I find the question to be ambiguous and I think that it needs a careful rewriting so as to make it as the author intended.

 

[JeffJo]

I have just looked up the Monty Hall problem and this makes reference to three parts,namely two goats and one car or alternatively three separate doors.The boys problem refers to,by implication,seven separate parts namely the seven days of the week.A major difference between the two problems is that the parts referred to in the M H problem are separate and indivisible whereas the parts referred to in the boys problem can be subdivided into smaller parts eg hours ,weeks and seconds.I guess that the person who wrote the boys problem did not want any solvers to carry out this subdivision and therefore reach an answer of 13/27 but the question is phrased in such a way that such subdivision is allowed.Because of this I find the question to be ambiguous and I think that it needs a careful rewriting so as to make it as the author intended.

You are missing the point. In Gary Foshee's solution (the 13/27 one) to the problem, he is actually requiring a family to have a boy born on a Tuesday before they can be selected. Since a family with two boys is (almost) twice as likely to meet the requirement, the probability of two boys goes up from what it would be if you only required that there be a boy.

If, on the other hand, no restrictions are placed on who can be randomly selected, and then you make an observation about the family that coudl be any fact that is true, the answer is 1/2 whether or not "Tuesday" or any other subdivision is mentioned.

You are right that it should be rewritten. If the author wants the answer to be 1/3 (withjout "Tuesday") or 13/27 (with it), this requirement must be made explicit. Without that, the answer is 1/2.

 

[JaredJames]

You are missing the point. In Gary Foshee's solution (the 13/27 one) to the problem, he is actually requiring a family to have a boy born on a Tuesday before they can be selected. Since a family with two boys is (almost) twice as likely to meet the requirement, the probability of two boys goes up from what it would be if you only required that there be a boy.

But the odds of having two boys are half of having only one boy, so there's no gain in advantage.

 

[JeffJo]

But the odds of having two boys are half of having only one boy, so there's no gain in advantage.

That's already accounted for. Assuming the requirement R is "one is a boy" in the simpler problem, without "Tuesday"

• P(two boys and R) = P(R|two boys)*P(two boys) = 1*(1/4) = 1/4
• P(one boy and R) = P(R|one boy)*P(one boy) = 1*(1/2) = 1/2
• P(no boys and R) = P(R|no boys)*P(no boys) = 0*(1/4) = 0

So the answer is (1/4)/(1/4+1/2+0)=1/3.

But if you add an additional requirement that any given boy has a a small probability Q to satisfy, the two-boy family has a 2Q-Q^2 chance (for two boys that could meet it, but you don't want to double-count the Q^2 chance that both do):

• P(two boys and R) = P(R|two boys)*P(two boys) = (2Q-Q^2)*(1/4) = Q/2 - <small> = ~Q/2
• P(one boy and R) = P(R|one boy)*P(one boy) = Q*(1/2) = Q/2
• P(no boys and R) = P(R|no boys)*P(no boys) = 0*(1/4) = 0

The "~" means approximate. And now the answer is (~Q/2)/(~Q/2+Q/2+0)= ~1/2. The more accurate answer is (1-Q/2)/(2-Q/2). Which, if Q=1/7, is 13/27.

That is what happens when you require the information in the problem to be true before you select a family. If you merely observe it, represented by O,

• P(two boys and O) = P(O|two boys)*P(two boys) = 1*(1/4) = 1/4
• P(one boy and O) = P(O|one boy)*P(one boy) = (1/2)*(1/2) = 1/4
• P(no boys and O) = P(O|no boys)*P(no boys) = 0*(1/4) = 0

And the answer is (1/4)/(1/4+1/4)=1/2. But now adding in Q when you observe one boy makes no difference, sicne there is still a Q chance the boy you observe meets it:

• P(two boys and O) = P(O|two boys)*P(two boys) = Q*(1/4) = 1/4
• P(one boy and O) = P(O|one boy)*P(one boy) = (Q/2)*(1/2) = Q/4
• P(no boys and O) = P(O|no boys)*P(no boys) = 0*(1/4) = 0

And the answer is (Q/4)/(Q/4+Q/4)=1/2.

 

[davee123]

That is what happens when you require the information in the problem to be true before you select a family.

Maybe I missed something, but isn't that the point? If the information in the problem weren't true of the selected family, isn't that case necessarily not included in the probability?

DaveE

 

[JeffJo]

Maybe I missed something, but isn't that the point? If the information in the problem weren't true of the selected family, isn't that case necessarily not included in the probability?

DaveE

Necessarily, yes. The information is true of the selected family. Sufficiently, no. The information can be true even if the parent says something else. That's the problem. And the real issue is that your arguments require it to be both necessary and sufficient.

The problem statement is that a parent says "I have a boy." The issue is whether this particular parent is a representative sample of "all parents of two including a boy," or a representative sample of "all parents of two who chose to tell you 'I have a boy.'" Since there is no requirement that a parent of a boy and a girl must say "I have a boy," some might say "I have a girl" in the same situation. In fact, we should expect half of them to do so.

The issue is 100% equivilent to the controversy in the Monty Hall Problem. If "Monty Hall reveals a goat behind door X" is equivalent to "There is a goat behind door X," then the other two doors remain equally likely to have the car. Although they use many different arguments, the people who recognize that Monty Hall had a choice of two doors in one case (the contestant's door has the car), but not in the other (the contestant's door has a goat) see that counting possible cases doesn't work. You need to sum the probability that these cases would produce the observed results.

Please, rather than resorting to superficial arguments, examine what I've argued here. The Law of Total Probability says that P(two boys|one boy)=sum[D=Sunday to Saturday, P(two boys|parent says "one boy born on D")*P(parent says "one boy born on D"). Do you contest this? Because if you do, you need to revist Probability 101. It's called a "law" for a reason.

Now, do you think that P(two boys|parent says "one boy born on D") could be different for any of the seven possible values of D? I certainly hope not. You'd need to provide some reason why one day is any different than another, and there is none.

Is there some reason why you think sum[D=Sunday to Saturday, P(parent says "one boy born on D") is not 1 for our assumed situation? I agree the statement is not orthodox, but since we are given that a parent made a statement in this form, we must consider only situations where a statment in that form is made.

The end result is that P(two boys|parent says "one boy born on D") must equal P(two boys|parent says "one boy"). Under the "requirement" assumption, this is "1/3=13/27." Under the "observation" assumption, this is "1/2=1/2." Which do you prefer to defend?

Please, please, please, step back from your preconceived notions of what the answer should be. Look at what it has to be. As [PLAIN]http://fox-lab.org/papers/Fox&Levav(2004).pdf[/PLAIN] [Broken] argue, you should not "partition the sample space into n interchangeable events, edit out events that can be eliminated on the basis of conditioning information, count remaining events, then report probabilities as a ratio of the number of focal to total events." You need to sum the probabilities that each possible event could produce the observed results, instead of merely counting them. Unless you can produce evidence that the "count" must be the same as my "sum," you are not answering these issues.

 

[davee123]

The issue is 100% equivilent to the controversy in the Monty Hall Problem. If "Monty Hall reveals a goat behind door X" is equivalent to "There is a goat behind door X," then the other two doors remain equally likely to have the car.

I believe that's true in the case that Monty Hall doesn't know which door contains the car. The unstated assumption in the Monty Hall problem is that Monty knows perfectly damn well where the car is, and will intentionally show you the door that he knows contains the goat.

It seems like you might be implying something of similar distinction here, but I'm having a difficult time understanding how you might be applying that logic.

DaveE

 

[Dadface]

You are missing the point. In Gary Foshee's solution (the 13/27 one) to the problem, he is actually requiring a family to have a boy born on a Tuesday before they can be selected. Since a family with two boys is (almost) twice as likely to meet the requirement, the probability of two boys goes up from what it would be if you only required that there be a boy.

If, on the other hand, no restrictions are placed on who can be randomly selected, and then you make an observation about the family that coudl be any fact that is true, the answer is 1/2 whether or not "Tuesday" or any other subdivision is mentioned.

You are right that it should be rewritten. If the author wants the answer to be 1/3 (withjout "Tuesday") or 13/27 (with it), this requirement must be made explicit. Without that, the answer is 1/2.

I am aware of the requirements and am not missing the point(please read my posts).I'm saying that if the information is such that the time of birth can be pinned down to increasingly more precise values between limits(eg Tuesday as in the stated problem and then Tuesday morning and then Tuesday morning between 9 and 10 etc)then the probabilities as calculated by the method used here get increasingly closer to 0.5.

 

[D H]

JeffJo does have a point here. Making the question short and story-like makes the intended answer (13/27) incorrect. The probability that both children are boys is indeed 13/27 given a family randomly drawn from the set of two child families with one son born on a Tuesday. However, that is not necessarily the right universe for the question "I have two children, one is a son born on a Tuesday. What is the probability my other son is a boy?"

Suppose Monty gets tired of giving away cars to smarty-alecky mathematicians. He asks the members of the audience to raise their hand if they have exactly two children, one a boy and the other a girl. He asks one such couple to come on down and asks the couple on what day of the week the boy was born. Monty now takes you out of a sound-proofed room and tells you "This fine young couple have two children. One is a boy born on a Tuesday. What is the probability the other child is a boy?"

 

[Dadface]

What about the fact that the solver can use additional knowledge,mainly common sense and general knowledge,in order to solve the problem.If it is the intention of the question compiler that any solver should not use this general knowledge then in my opinion the boys problem is too artificial.

 

[JeffJo]

I believe that's true in the case that Monty Hall doesn't know which door contains the car. The unstated assumption in the Monty Hall problem is that Monty knows perfectly damn well where the car is, and will intentionally show you the door that he knows contains the goat.

If Monty opens a door at random, and just happens to find a goat, the probability your door has the prize is 1/2. That's not what I'm talking about.

First, let me point out that you will always know what the door numbers are when you have to decide to switch. So let's specify that you choose Door #1, and Monty opened Door #3. What these numbers are shouldn't matter, but specifying them brings the comparison closer to the Two Child Problem, where "boy" is similarly specified.

If Monty Hall is biased, and is required to open Door #3 if there is a goat behind it and the contestant did not choose it, then the probability Door #1 has is the car 1/2. There are three possibilities, one is eliminated, and the other two have equal probability. But if Monty is free to choose a door to open, he will open Door #3 100% of the time the car is behind Door #2, 0% of the time it is behind Door #3, and 50% of the time it is behind Door #1. The proper solution does not count cases, it sums these probabilities: P(D1)=(50%)/(100%+0%+50%)=1/3.

If this parent is required to say he has a boy in all cases where he has one, then the probability he has two is 1/3. There are four possibilities, one is eliminated, and the other three have equal probability. And the answer changes if additional information about a boy is presented as a requirement, since a family of two boys is almost twice as likely to meet that requirement as a family woth one boy.

But if this parent is free to choose any gender among his children, he will say "boy" 100% of the time he has two, 0% of the time he has none, and 50% of the time he has one. The proper solution does not count cases, it sums these probabilities: P(2Boys)=(100%)/(100%+0%+50%+50%)=1/2. And the answer can't change with additional information.

My point is that the presence of one value in the problem statement, from what we should consider a set of N equally-likely possibilities, cannot be taken to mean that value was required. You know what door Monty Hall opened, but you can't assume he was required to open that door if it was possible to. You were told there is a boy in the family, but you similarly can't assume the person who told you was required to say "boy" if he could. A parent of a boy and a girl should have been equally likely to say "one is a girl (born on...)," and that makes the answer 1/2 regardless of what additional information is presented.

JeffJo does have a point here. Making the question short and story-like makes the intended answer (13/27) incorrect. The probability that both children are boys is indeed 13/27 given a family randomly drawn from the set of two child families with one son born on a Tuesday. However, that is not necessarily the right universe for the question "I have two children, one is a son born on a Tuesday. What is the probability my other son is a boy?"

Remove the "necessarily" and this is correct. First off, as soon as he says "other" (colored in red), he has specified one child just as completely as he would have if he had said "My older child is a boy." The answer is 1/2. But let's assume you asked what you meant, "what is the probability I have two boys?"

Look back at Jimmy Snyder's grid in Post #23. In the universe that produces the answer 13/27, father b3b4 or b4b3 can't say "I have two children, one is a son born on a Wednesday." Assuming the father does in all other cases, the answer for the question "what is the probability I have two boys?" is probably 11/25, not 13/27. It could be something else, but it can't be 13/27. Since you can't postulate a universe where the answers are different for different days, you can't postulate the universe where the answer is 13/27. That's the point.

Suppose Monty gets tired of giving away cars to smarty-alecky mathematicians. He asks the members of the audience to raise their hand if they have exactly two children, one a boy and the other a girl. He asks one such couple to come on down and asks the couple on what day of the week the boy was born. Monty now takes you out of a sound-proofed room and tells you "This fine young couple have two children. One is a boy born on a Tuesday. What is the probability the other child is a boy?

A poorly formulated question. A probability problem has to imply a random process somehow. If you don't say what that process is, the solver has to assume there is an equal probability for what appear to be equivalent outcomes. In this question, your process did not allow the "other" child to be a boy, so the actual answer is 0. It also pre-determined that the formulation of the question would include the word "boy," and you did not convey that information, either. The person in the booth has to assume any usage of gender had to allow either possibility. Specifically:

• He can't assume you asked for parents of two, but the rest is independent of this so it doesn't matter. (And I think the fact that, when independence applies, it doesn't matter if the fact is a requirement or an observation, is what confuses people into believing it never matters.)
• He can't assume you specified the gender make-up of the family. BB, GB, BG, and GG are all equally likely.
• He can't assume you asked about the day a specific child was born. That fact could apply to any of the eight children that are possible.
• He must assume that if the equivalent facts for both children are different, that each was equally likely to be presented to him.

So the best answer based on the information you presented to him is 1/2.

 

[JeffJo]

I am aware of the requirements and am not missing the point(please read my posts).I'm saying that if the information is such that the time of birth can be pinned down to increasingly more precise values between limits(eg Tuesday as in the stated problem and then Tuesday morning and then Tuesday morning between 9 and 10 etc)then the probabilities as calculated by the method used here get increasingly closer to 0.5.

And I'm saying that it is a result of requiring a boy to be born within an interval that causes this. A family of one boy has one chance only, and the probability is Q. The proportion of families that have one boy is not included in determining this Q; the two probabilities must be multiplied together to get the probability that both happen in the same family.

But a family of two boys has nearly twice the probability of having one that meets such a requirement; 2Q-Q^2, actually. This value is also multiplied by the proportion of families with two boys to get the probability of both happening together.

The result is that the list possible families decreases as Q decreases, but the sub-list of two-boy families decreases at about half the rate of the one-boy families. So the ration of the two approaches 1/2.

But if the facts we are given are treated as an observation of a random family, there are exactly two facts that apply to any particular family. We don't have to say "boy" or "girl", but pick an applicable gender based on the two children. We don't have to say "born between 9:13PM and 9:14PM on a Thursday", but pick an applicable minute based on a child of the applicable gender.

The chances of two boys, given that we observe "a boy born between 9:13PM and 9:14PM on a Thursday" this way, is exactly 1/2. It remains exactly 1/2 no matter what size of an interval you choose, because you can always isolate the births to such an interval.

 

[Jimmy Snyder]

I get it now. Here is a simplified version of the problem.
A mathematician says "I have one child. It is a boy. What is the probability that my child is a boy."
The common sense answer is 1. However, we don't know what the mathematician would have asked if their child was a girl.
Case 1: If they would not have asked you a question, then the answer to the question they didn't ask is 14/27.
Case 2: If they would have asked "I have one child. It is a girl. What is the probability that my child is a girl." Then the answer to the question is 1/2.

Of course, I'm just joking. Look at it this way. Imagine you are in a room with 196 other people, one of each type in the grid that I made. 27 of them walk up to you and say I have two children, one is boy born on Tuesday and the other child is a {boy/girl}, then of the 27, thirteen will say that the other child is a boy, and 14 will say that the other child is a girl. There's your 13/27

If all 196 people walk up to you and say "I have two children, one is a {boy/girl} born on {S/M/T/W/T/F/S} and the other is a {boy/girl}, then 98 will say that the other child is a boy and 98 will say that the other child is a girl (it will be easier to see it if you assume that each will identify the eldest child first). There's your 1/2.

Taken in isolation, you can't tell which situation the OP is discussing. If a bunch of people ask the question, then you may get a feel for which of these two universes you live in.

 

[D H]

If Monty opens a door at random, and just happens to find a goat, the probability your door has the prize is 1/2. That's not what I'm talking about.

Good thing that, because if Monty opens a door at random my chances of winning the car is 2/3. Suppose Monty randomly opens door #3 and happens to show the car. In that case I am going to switch to door #3.

 

[JeffJo]

Good thing that, because if Monty opens a door at random my chances of winning the car is 2/3. Suppose Monty randomly opens door #3 and happens to show the car. In that case I am going to switch to door #3.

Read what I said. When (not "if") he reveals a goat, as is explicit in the problem statement, the probability is 1/2 if he chose any door you didn't choose randomly, 1/3 if he chose a goat-door you didn't choose randomly, and either 1/2 or 1 if he prefers to open a specific door if he can. If he reveals a car, it is a different problem unrelated to anything we have talked about. We don't know if you would be offered the chance to switch.

 

[JeffJo]

I get it now. ...

Taken in isolation, you can't tell which situation the OP is discussing. If a bunch of people ask the question, then you may get a feel for which of these two universes you live in.

Exactly.

Finally, imagine that you leave after the first person who walks up to you says "I have two children. One is a boy born on a Tuesday." Can you rule out either scenario as being the reason he did this? No. Can you assume it was because he was required, as in your first scenario, to tell you exactly that fact, and no other? No. Can you assume it is a random occurrence that could equally-likely be any statement of that form? Yes. The answer is 1/2. And it is also 1/2 if he only says "One is a boy."

Both problems are really ambiguous; but then, so are most probability puzzles. They don't tell you things like "Assume 50% of children are boys, 50% girls, and gender is independent in all siblings." Because those assumptions are not true in the real world. I'm not criticizing the problems for that, I'm saying that assuming equiprobability is an inherent part of these problems. There could be reasons why the probability here is not 1/2, but unless they are explicit in the problem statement, the only possible answer is 1/2.

 

[Dadface]

JeffJo,or anybody else, will you clarify something please?The original question seems to impose certain limitations:

1.The question seems to imply that consideration be given only to a particular "universe".
2.The question further seems to imply that only the limited information given in the question itself should be used when solving the problem.

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? This thread,however,has gone further with some suggesting/stating that the true probability is 1/2:

A.Following on from the above I show that if one applies a similar methodology(which is not necessarily the best methodology) as previously used but also uses extra information as drawn from general knowledge then one obtains different answers.As the birth time is pinned down more accurately to reducing time intervals the probability tends to 1/2.(actually reaching 1/2 if birth time can be defined as an instantaneous event).If I am correct the probality of 1/2 applies to the universe described in the question and possibly,by extension,to all universe.(I have yet to check my maths)

B.The same conclusion of a probability of 1/2 is reached but differently as in the event is analysed primarily by not considering the defined universe only.

Right or wrong the conclusions reached by methods A. and B. seem to breach the requirements of the original question.The conclusion reached by my method (A) goes beyond limitation 2.(see above)and the conclusion reached by method B.breaches limitations 1. and 2.

Have we moved too far away from the original question?Is the original question meaningless?Am I being daft?(the answer is yes to at least one of these questions)

 

[Jimmy Snyder]

1.The question seems to imply that consideration be given only to a particular "universe".

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.