Calculating the Probability of Two Boys for Math-Obsessed Friend

[peteb]

Your new mathematics-obsessed friend says to you, "I have two children.
One is a boy born on Tuesday. What is the probability I have two boys?"

The first thing you think to ask him is, "What the heck does Tuesday have to do with it?"
"Everything!", he replies...

So what is the probability?

 

[davee123]

Tuesday does indeed have nothing to do with it, assuming he's truly anal, or without further information. Hence, the probability, assuming his statements are true, is 1/3.

However, using intuition, I might infer from the inclusion of "Tuesday", that if it's important, that he uses it as a way of distinguishing his children. If so, further inferring that the day of the week of a birth is generally a less useful statistic than gender, I might conclude that he has twin boys, one born on Tuesday, and the other born on either Monday or Wednesday. Hence, the day of the week becomes more relevant as a distinguishing factor, whereas if the children were years apart or were of different genders, one probably wouldn't use the day of the week as a method of distinguishing them. Hence, the probability of him having 2 boys would increase to some higher number, depending on how sure of myself I am in inferring my friend's likely precedence on distinguishing his children.

DaveE

 

[peteb]

The day is important, and the answer is not 1/3. Keep trying...

Pete B

 

[Jonathan Scott]

The day is important, and the answer is not 1/3. Keep trying...

Pete B

My attempt:

Firstly, from the point of view of the mathematician, the answer is that the probability is either 0 or 1, so that's one possible answer.

If "Tuesday" was just a piece of information that was provided randomly about one child, it would provide no information about the other child, so "Tuesday" could not have any significance and the probability of both being boys would simply be 1/2.

The only way that "Tuesday" can have significance is if there is an unstated assumption that it provides some information about the other child, presumably that the other child is NOT "a boy born on a Tuesday" (so the other child is either a girl or a boy born on some other day of the week). As this is unstated, I consider it cheating, but in that case I think the probability of the other one also being a boy, and hence two boys, would be 6/13.

This is not the same as knowing that there are two children and obtaining the answer to "is at least one of them a boy", in which case the chances of two boys would of course be 1/3.

 

[davee123]

This is not the same as knowing that there are two children and obtaining the answer to "is at least one of them a boy", in which case the chances of two boys would of course be 1/3.

I don't follow here-- Why not? The probability given 2 children (assuming 50% chance of having either gender in an individual case, and no hermaphrodites or otherwise) is 1/4 of 2 boys, 1/4 of 2 girls, and 1/2 of having one of each gender. Knowing that 1 is a boy eliminates the possibility of having 2 girls, meaning that there's a 1/3 chance of having two boys.

As stated, if you wantonly assume that one of them being a "boy born on a Tuesday" means the other child is NOT a "boy born on Tuesday", then you could get some other outcome. But that's an assumption. The other one could just as well be another boy born on Tuesday, a girl born on Tuesday, a boy born on Friday, or a girl born on Monday.

DaveE

 

[Ygggdrasil]

Two cases: 1) the elder child is a boy born on Tuesday, 2) the younger child is a boy born on Tuesday.

case 1) There are 7 possible scenarios for the younger child to be a girl (girl born on any day of the week) and 7 possible scenarios for the younger child to be a boy.

case 2) There are 7 possible scenarios for the older child to be a girl, but only six possible scenarios for the older child to be a boy (the case where both children are boys born on Tuesday was already covered above).

Therefore, the probability of having two sons is 13/27 or just under 50%. See the link below for a useful graphical explanation.

http://sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong

 

[D H]

"What the heck does Tuesday have to do with it?"
"Everything!", he replies...

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

 

[Jonathan Scott]

As usual, it depends on the implicit question, which defines the space of other possible answers, as described in the sciencenews article. If you asked someone "is your first child a boy or a girl?" and "on what day of the week was the child born?", then neither answer would tell you anything at all about the second child.

 

[peteb]

Ygggdrasil has the correct answer and solution. It is worth noting that the item of interest here is not unique in that the boy was born on Tuesday. Any other specified day would produce the same probability. Also consider that other conditionals, such as a statement that the boy was "born in February", would also produce a probability that differs from the unconditional problem statement. Thus the intuitive component is found in recognizing that it is the presence of any conditional of this type that alters the result, not just the one specific conditional mentioned in this example.


Pete B

 

[D H]

Correct. This oldie but goodie admits a lot of variations, each of which is an interesting case study regarding how/why we are so bad at computing probabilities -- and such easy marks for con men.

 

[Proton Soup]

lessee... the possibilities originally were 14 for child1 (boy or girl times 7 days of week) and 14 for child2. and 14x14 gives a space of 196 combinations. 1 in 14 have child1 as boy-tuesday. 1 in 14 have child2 as boy-tuesday. exactly 1 has both child1 and child2 as boy-tuesday. so... it should be (1/14 +1/14)*196 - 1 = 28-1 = 27 possibilities. so, a chance of 1/27 that both are boys with at least 1 born on tuesday.

edit:^yeah, i did that wrong i think. 27/196 maybe? i need more coffee. and got something else i need to work on, too. i'd like to spend some more time thinking about that solution in the article, but I'm still feeling like it's answering a question that's not being asked at this point.

but... the way the question is worded, knowing that one child was born on tuesday does not reduce the space you're looking at. it is extraneous information. you could also say that one has blue eyes, webbed feet, and a freckle on his nose. it does not matter. all we care about is the sex. and regarding sex, we know that one is a boy (ergo, not both girls). that leaves three possibilities, of which only one is BB. so 1/3.

 

[skeptic2]

Suppose instead of the original question, the friend asks, "I have two children.
One is a boy born on Tuesday. What is the probability the other is a boy?"

Or perhaps "I have one child who is a boy born on Tuesday and my wife is pregnant with another child. What are the chances that child will be a boy?"

 

[Proton Soup]

Suppose instead of the original question, the friend asks, "I have two children.
One is a boy born on Tuesday. What is the probability the other is a boy?"

Or perhaps "I have one child who is a boy born on Tuesday and my wife is pregnant with another child. What are the chances that child will be a boy?"

the first doesn't specify which child is a boy. so the question is really, what are the odds that both are boys, given that I've excluded the possibility of girl-girl. 1/3.

the second specifies the first is a boy. and that gives you no information about the future event. each new birth has the same odds: 1/2. this is also why schemes to pick lottery numbers do not work. each new lottery is an independent event.

 

[davee123]

That's... quite odd, but just to prove it to myself, I wrote a program to simulate it, and indeed, it's true. I'm not quite sure I've quite wrapped my head around why specifying the date of birth ought to affect anything, but indeed, it does. As apparently would specifying any other equally arbitrary statistic (time of day, day of the year, etc).

DaveE

 

[mystermenace]

Ygggdrasil has the correct answer and solution. It is worth noting that the item of interest here is not unique in that the boy was born on Tuesday. Any other specified day would produce the same probability. Also consider that other conditionals, such as a statement that the boy was "born in February", would also produce a probability that differs from the unconditional problem statement. Thus the intuitive component is found in recognizing that it is the presence of any conditional of this type that alters the result, not just the one specific conditional mentioned in this example.


Pete B

Suppose: "I have two children. One is a boy born on some day. What is the probability I have two boys?" Does this change the answer?

 

[davee123]

Suppose: "I have two children. One is a boy born on some day. What is the probability I have two boys?" Does this change the answer?

I think that's the part that trips people up:

"I have two children".

Probability of two boys is currently 1/4.

"One of my children is a boy".

Probability of two boys is now 1/3.

"The boy in question was born on a particular day of the week. I haven't told you which day that is, and it goes without saying that of course he was born on a particular day. But in just a second, I'm going to tell you which day that was, but I haven't yet."

Probability of two boys is still 1/3.

"The boy was born on a Tuesday".

Probability of two boys is now 13/27.

Effectively, the way I'm imagining it, by specifying that one of the boys was born on a Tuesday, you've just now eliminated ALL of the possibilities of having NO boys born on Tuesdays, which is a hefty portion of options-- hence bringing the probability a lot closer to 1/2.

DaveE

 

[hamster143]

Suppose you meet 196 mathematicians, each one of them picks one of his kids at random and asks you the question.

98 of them will say "one of them is a girl".

84 of them will say "one of them is a boy born on Monday / Wednesday / ..."

Of the remaining 14, 7 will have two boys and 7 will have a boy and a girl.

The answer is 1/2, as to be expected.

 

[JeffJo]

Your new mathematics-obsessed friend says to you, "I have two children.
One is a boy born on Tuesday. What is the probability I have two boys?"

It's 1/2. And no, I am not missing a subtle point about probability. You are.

Tuesday does indeed have nothing to do with it, assuming he's truly anal, or without further information. Hence, the probability, assuming his statements are true, is 1/3.

And the answer to the problem you reduced it to, by omitting "Tuesday" from peteb's question, is also 1/2.

Everything indeed. There is a big difference between

1. "I have two children. One is a boy."
2. "I have two children. The older of the two is a boy."
3. "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.

You are right in your assessment at the end, but you need to add more to your list.

1. A: "I have two children." B:"Is one of them is a boy?" A: "Yes."
2. A: "I have two children." B:"Is one of them is a boy born on a Tuesday?" A: "Yes."

The answer, when these two are the condition, is 1/3 and 13/27, respectively. The answer when any of yours is the condition is 1/2. The reason they are different, is that my condition #1 is different than your condition #1, and my #2 is different than your #3.

There could be some people who would respond "Yes" to my #1, but who, for whatever unspecified reason they are telling you what they told you in your list, would not say "One is a boy." They would say "One is a girl." Since you provided no reason for why they would choose one statement over the other, we can only assume both are equally likely when both are possible. So, while it is true that there are twice as many parents of two who have a boy and a girl, as those who have two boys; the same number in each group would tell you "One is a boy." The same arguments apply to the Tuesday version.

And the really ironic thing, is that the people who get the wrong answer for the Monty Hall Problem you mentioned are taking the same approach to it, as you do to this one. As Fox and Levav write, they are "subjectively partitioning the sample space into n interchangeable events, editing out events that can be eliminated on the basis of conditioning information, counting remaining events, then reporting probabilities as a ratio of the number of focal to total events."

So, for your first condition, you partition the sample space into the four interchangeable events BB, BG, GB, and GG; edit out the GG event which is the only one that is impossible if you have a boy; and report the probability as the ratio of focal event left (BB) to total events left (BB, BG, GB). Just like with Monty Hall they partition the sample space into the three interchangeable events where the prize is behind D1, D2, or D3; edit out the D3 (or whatever) event which is the only one that is impossible after you see what door the host opens; and report the probability as the ratio of focal event left (D1) to total events left (D1, D2).

The technically-correct answer is not the ratio of the number of cases, but the ratio of the sum of the probabilities that each of the appropriate cases would produce the observed result. In the Two Child Problem, if P is the probability a parent of a boy and a girl would say "One is a boy," the answer is 1/(1+2P). In the Monty Hall Problem, if P is the probability Monty Hall would open Door #3 when he could also open Door #2, the answer is 1/(1+P). And in the Tuesday Boy problem, the answer is (1+12P)/(1+26P) for a similar P. If P=1 in any of these, you get the "wrong" answer that Fox and Levav talk about. If P=1/2, you get the right answer.

The difference in my list of statements, is that P=1. In yours, it is really ambiguous what P is, but we can't assume anything except 1/2.

Thus the intuitive component is found in recognizing that it is the presence of any conditional of this type that alters the result, not just the one specific conditional mentioned in this example.

No. It is requiring the presence of the conditional information that alters the probability. If a parent is selected at random from a group of parents who have two children and at least one boy, the answer is 1/3. If a parent is selected at random from a group of parents who have two children, and one gender is observed, the answer is 1/2 regardless of what that gender is.

"The boy in question was born on a particular day of the week. I haven't told you which day that is, and it goes without saying that of course he was born on a particular day. But in just a second, I'm going to tell you which day that was, but I haven't yet."

Probability of two boys is still 1/3.

"The boy was born on a Tuesday".

Probability of two boys is now 13/27.

If this were true, then the answer would be 13/27 regardless of what day you mention. And if it is 13/27 regardless of what day you mention, it is also 13/27 for the first question. This is called the Law of Total Probability. If {X1, X2, ..., XN} are independent events that represent all possibilities, then for any event Y, P(Y)=P(Y|X1)*P(X1)+P(Y|X2)*P(X1)+...+P(Y|XN)*P(XN). If all the conditional probabilities are equal to the same P, then P(Y)=P because P(X1)+P(X2)+...+P(XN)=1.

 

[Dadface]

What is special about information which is passed on but already known?

Your friend asks the following question

I have two children.
One is a boy.
What is the probability I have two boys?

After coming up with an answer you might then realize that the boy referred to must have been born on a certain day of the week.Your friend didn't mention that because it is a fact that is obvious.Armed with this new information,which was inherent in the question all along,how,if at all,does your answer now change?

 

[JaredJames]

EDIT: I just read this post by davee and it makes sense now.

Effectively, the way I'm imagining it, by specifying that one of the boys was born on a Tuesday, you've just now eliminated ALL of the possibilities of having NO boys born on Tuesdays, which is a hefty portion of options-- hence bringing the probability a lot closer to 1/2.

 

[peteb]

Here is the source I had for this puzzle, including the solution:

http://tinyurl.com/468zcyn


Pete B

 

[Dadface]

(for the sake of clarity let's assume that the moment of birth can be pinned down to an instant eg the instant that the baby starts to take his first breath)
Following on from my post above you now realize that the birth must have been within a certain hour of a certain day and within a certain minute of that hour and so on.How do the probabilities change as the amount of implicit information increases that is with the time interval approaching zero.

 

[Jimmy Snyder]

Here is a brute force proof that the probability is 13/27. First of all, I use the following notation: b1 means a boy born on Sunday, b2 means a boy born on Monday, ... b7 means a boy born on Saturday. g1 means a girl born on Sunday, etc. To any child we can attach anyone of 14 labels b1, ..., b7, g1, ... g7 and they occur with equal probability. If there are two children, there are 14 x 14 = 196 equally probable outcomes which I have listed below:

b1b1 b1b2 b1b3 b1b4 b1b5 b1b6 b1b7 b1g1 b1g2 b1g3 b1g4 b1g5 b1g6 b1g7
b2b1 b2b2 b2b3 b2b4 b2b5 b2b6 b2b7 b2g1 b2g2 b2g3 b2g4 b2g5 b2g6 b2g7
b3b1 b3b2 b3b3 b3b4 b3b5 b3b6 b3b7 b3g1 b3g2 b3g3 b3g4 b3g5 b3g6 b3g7
b4b1 b4b2 b4b3 b4b4 b4b5 b4b6 b4b7 b4g1 b4g2 b4g3 b4g4 b4g5 b4g6 b4g7
b5b1 b5b2 b5b3 b5b4 b5b5 b5b6 b5b7 b5g1 b5g2 b5g3 b5g4 b5g5 b5g6 b5g7
b6b1 b6b2 b6b3 b6b4 b6b5 b6b6 b6b7 b6g1 b6g2 b6g3 b6g4 b6g5 b6g6 b6g7
b7b1 b7b2 b7b3 b7b4 b7b5 b7b6 b7b7 b7g1 b7g2 b7g3 b7g4 b7g5 b7g6 b7g7
g1b1 g1b2 g1b3 g1b4 g1b5 g1b6 g1b7 g1g1 g1g2 g1g3 g1g4 g1g5 g1g6 g1g7
g2b1 g2b2 g2b3 g2b4 g2b5 g2b6 g2b7 g2g1 g2g2 g2g3 g2g4 g2g5 g2g6 g2g7
g3b1 g3b2 g3b3 g3b4 g3b5 g3b6 g3b7 g3g1 g3g2 g3g3 g3g4 g3g5 g3g6 g3g7
g4b1 g4b2 g4b3 g4b4 g4b5 g4b6 g4b7 g4g1 g4g2 g4g3 g4g4 g4g5 g4g6 g4g7
g5b1 g5b2 g5b3 g5b4 g5b5 g5b6 g5b7 g5g1 g5g2 g5g3 g5g4 g5g5 g5g6 g5g7
g6b1 g6b2 g6b3 g6b4 g6b5 g6b6 g6b7 g6g1 g6g2 g6g3 g6g4 g6g5 g6g6 g6g7
g7b1 g7b2 g7b3 g7b4 g7b5 g7b6 g7b7 g7g1 g7g2 g7g3 g7g4 g7g5 g7g6 g7g7

Note that of these 196, only 27 have a b3, that is, a boy born on Tuesday. I have colorized them. Of these 27 colorized pairs, only 13 have a second boy. I have colored them blue. Thus with the given inputs, the probability is clearly 13/27.

 

[JaredJames]

Perfect explanation Jimmy.

I was fairly ok with it before but that's hit the nail on head.

Nice one!

 

[hamster143]

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.

 

[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 . 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 believe 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 question 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

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.