**Replying to my comment since I can’t figure out how to edit it**

Mistake no2 is nonsense, I take that back. For anyone else like me who struggles to understand how older/younger can provide extra information, I think its easier to understand if you categorize in terms of child 1 and child 2. I.e. the 4 possible ways to construct the two children in terms of child 1 and 2 is b,b or b,g or g,b or g,g . The condition in the first question can be rephrased to, “child 1 is a boy” and the condition in the second question to “either child 1 or child 2 is a boy”. It is more clear to me this way how the first question is more selective and hence results in a larger probability. And then of course you can just change child 1 and child 2 to any arbitrary categories and the question still works. Basically, using the category of age is nice as a confusing smokescreen to make the question more interesting.

That being said, mistake no1 applies.

The solutions to questions 1 and 2 are different from what I originally wrote since I should have used P(A|D) and P(B|D) which are not both 1/2. Condition D is for question 1, the oldest child is a boy and for question 2, at least one child isa a boy.

Question 1 had its solution prior to the observation of the boy with his father being 1/2 according to condition D. Thus the answer is still 2/3 as I wrote in my first post (correct by fluke).

Question 2 has P(A) = 1/3 and P(B) = 2/3 (I’m skipping condition D and just writing P(A|D) as P(A) ). P(C) = P(A)*P(C|A) + P(B) * P(C|B) = 1/3 * 1 + 1/2 * 2/3 = 2/3
Thus, P(A|C) = P(A)*P(C|A)/ P(C) -> P(A|C) = (1/3) / (2/3) = 1/2.

So looking back, I agree with what Nathan posted!

Mistake no1:

The fact that you have observed the father walking with a boy gives you extra information. After all it could have been a girl that you saw him walking with. This calls for usage of Bayes theorem or at least the logic that is employed within Bayes theorem. I will give an explanation of how this should be properly applied in my solution at the bottom.

Mistake no2: In question 2 there are only two cases, i.e. case 1) two boys, case 2) one boy and a girl. The analysis has expanded case 2) into two parts, i.e. part 1) older boy and younger girl, part 2) older girl and younger boy. Or rather, part 1) older child 1 and younger child 2 and part 2) older child 2 and younger child 1. This is unnecessary but not wrong. The mistake is not applying the same logic to case 1). This should also be expanded to part 1) older boy 1 and younger boy 2, part 2) older boy 2 and younger boy 1. The two boys and two girls are no longer indistinguishable once you assign them the binary properties of older/younger. In other words, age is irrelevant to this question, but if it is included then it needs to be included everywhere in a consistent manner. This changes the answer to having 2/4 rows of the table being satisfied rather than 1/3. Thus, the answer according to the logic of mistake no1, is 1/2, the same as in question 1.

Correct answer with application of Bayes theorem:

There are two cases to consider, either the father has A) two boys or he has B) one boy and a girl. The probabilities of A and B with no extra information are P(A)=1/2 and P(B)=1/2 . However, we have seen that the father went for a walk with a boy. The probability of this occurring, prior to the observation, labelled P(C), can be calculated by considering the cases A and B. In case A, there are two boys, hence the probability that a boy was chosen for the walk is 1. Mathematically, this is denoted by P(C|A)=1, (the probability that C is true given that A is true equals one). In case B, there is only one boy, hence the probability that a boy was chosen for the walk is 1/2. Mathematically, this is denoted by P(C|B)=1/2 (the probability that C is true given that B is true equals 1/2). Hence, P(C) is P(A)*P(C|A) + P(B) * P(C|B) = 1/2 * 1 + 1/2 * 1/2 = 3/4. This is important because we know that this 3/4 chance ended up being the case, i.e. we observed the father walking with a boy. Thus, we are wanting to calculate P(A|C), (the probability that there are two boys, given the fact the father went for a walk with a boy). The question then is “which case contributed more to the probability P(C)?”. Case A, contributed P(A)*P(C|A) = 1/2 and case B contributed P(B) * P(C|B) = 1/4 to P(C). Therefore, case A is twice as likely as case B to be responsible for our observation that there was a father with his boy, and hence the probability of P(A|C) is 2/3.

Bayes theorem sums this up with the formula P(A|C) = P(A)*P(C|A)/ P(C) -> P(A|C) = (1/2) / (3/4) = 2/3 .

I thought the first two questions should be 2/3 and 1/2 respectively due to bayes rule. I computed Pr(two male children | met father with son with him as opposed to daughter). Pr(two male children) matches the answers on the website, and Pr(met father with son) is 3/4 and 4/6 respectively.

The mechanisms you added do indeed make the original answers correct. But the original wordings of the “at least one boy” and “Tuesday boy” problems cannot be interpreted to have these mechanisms, so they were not ambiguous. If they were, the Monty Hall Problem would have the same ambiguity.

And the point is that your incorrect interpretation has been taught so often, that people who should know better believe it, and teach it in turn.

Thank you for your comment! I think the original wording was ambiguous, and could allow for the answer to the second question to be 1/3 or 1/2, depending on interpretation. I have changed the wording to make the question more precise, which I hope makes the answer more clear, while still presenting an interesting puzzle.

The point is that the child was chosen by that means, and not because he was a boy. In another two-child family, or on another day, that method could have chosen a girl. Even if she had a brother. So you can only count half of the boy+girl families – those days where the father would choose to take a walk with his son, and not his daughter. So fully 1/2 of what you should count have two boys.

In fact, this is the same reason why the answer to the Monty Hall Problem is 2/3. Those who say 1/2 are counting all possible games where Monty Hall can open Door #3, just like you are counting all possible families where the man has a son. You should only count half of the games, or families, where a different observation could be made by the implied method.

The Tuesday Boy Problem is the same. Unless there is a reason that, say, a father of a Tuesday Boy and a Thursday Girl cannot tell you that he has a Girl, you can only count half of every cell in your table EXCEPT the one with two Tuesday Boys. The answer is (1+12/2)/(1+26/2)=(7)/(14)=1/2.

Glad you like the site, and thanks for the comment! I agree that using birth order as the distinguishing trait is somewhat arbitrary, and you could use another trait to achieve the same effect of narrowing the probability that the other child is a boy from 1/3 to 1/2. (Height or weight might be problematic since there are gender differences for those traits.)