The Two Children Problem
You meet a father and son, whose family was selected at random from among all two-child families where the eldest child is a boy. What are the chances that the man’s other child is also a boy?
The next day, you meet a different father and son, whose family was selected at random from among all two-child families with at least one boy. What are the chances that the man’s other child is also a boy?
For the purposes of both questions, assume that a child must be either a girl or a boy, that girls and boys exist in equal numbers in the population, and that the gender of a child is independent of the gender of the child’s sibling. If you think you have it figured out, read the answer and explanation below. Then scroll down for more boys and girls problems.
See Answer and ExplanationIt may seem like both questions should have the same answer, but that is not the case.
For the first question, the probability that the other child is also a boy is 1/2.
For the second question, the probability that the other child is also a boy is 1/3.
To understand why, begin with the following table, which shows the four gender-distribution possibilities for a two-child family, each of which has 25 percent probability:
| Older Child | Younger Child | |
| 1 | Boy | Boy |
| 2 | Girl | Girl |
| 3 | Boy | Girl |
| 4 | Girl | Boy |
For the first question, with the information we are given, that the older child is a boy, we have narrowed down the possibilities to #1 or #3, so there is a 1/2 probability that the other child is also a boy.
For the second question, the information we are given, that one of the children is a boy, narrows down the possibilities only to #1, #3, or #4. Among those three choices, in two cases the other child is a girl and in one case the other child is a boy, so the probability that the other child is also a boy is 1/3.
Another Boys and Girls Problem
Assume that every pregnancy has an equal chance of producing a boy or a girl, and in a certain society, every family follows the policy of having as many children as they can until they have a boy, and then they stop having children. So a family that has a boy first will have only one child, but a family might have one, two, three, or more girls before having their first boy and then having no more children. In this society, after many generations, what will the ratio of boys to girls be?
See Answer
The Four Children Problem
Now suppose you hear of a family with four children, and you do not know the gender of any of them, but you wish to guess the most probable scenario. You reason that four boys or four girls is very unlikely. Three of one gender and one of another seems unlikely too. Since the probability for each child is 1/2, you decide the most likely breakdown is two girls and two boys. Are you correct?
See Answer
Son Born on a Tuesday
Now suppose you encounter a mother and son, whose family was selected at random from among all two-child families with at least one boy born on a Tuesday. What is the probability that her other child is a boy?
Make the same assumptions as in the first problem, and also assume that births are equally likely on different days of the week.
See Answer
So while we might have been expecting the probability to be 14/28 or 1/2, we cannot count the “Tuesday boy – Tuesday boy” possibility twice. The real probability of two boys is therefore 13/27. This problem was devised by Gary Foshee.
Love your site! I disagree with your answer to the first question, though. Why is birth order relevant at all? In your chart, instead of “older child” and “younger child,” why not just “this child” and “that child?” Birth order is an arbitrary variable. You could just as easily have used height, weight, or any other distinguishing trait.
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.)
The answer to both of the Two Children problems is 1/2. It does not matter how you identify the gender of one *specific* child. It can be “eldest”, “this one with me”, “named Aaron”, or even “sits to Mother’s right at the dinner table”. See https://philpapers.org/rec/BARSTC. (And order is “important” only because it provides an identification method.)
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.
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 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.
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.
All answers are arbitrary until the extra information concerning which child the father is most likely to go on a walk with is specified. I will assume that the father is equally probable to go out with child 1 as with child 2 regardless of their age or gender. Hence, both questions are the same and have the answer of 2/3. There are two mistakes made in the analysis, no1 is common to both questions, and no2 is only for question 2.
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 .
**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!