Probability Brain Teaser II

A few days ago I posted a probability brain teaser which was getting me in a spin.

A family has 2 children. When you knock on the door one of the children, a boy, opens the door. What is the probability that the other child is a boy?

There were many interesting responses and as I expected, I had a mixture of two answers: a third and a half.

That which gives an answer of a third

When I first told this problem over dinner, I was told the answer was a third. The explanation is as follows:

You know that a boy opened the door. That means there is at least one boy. So out of the four possible permutations: boy-boy, boy-girl, girl-boy, girl-girl only the first three are possible. You could essentially rephrase the question as “There are two children in the family and it is not the case that both are girls”.

In the three of the permutations which are now possible (boy-boy, boy-girl, girl-boy), only one gives two boys. So the likelyhood of there being two boys is 1/3rd. Hence, the probability of the other child being a boy is a third.

That which gives an answer of a half

My immediate response was that the gender of the two children are independent and that the gender of one has no bearing on the other.

I also pursued a different technique which also lead to the answer of a half. There are two children in the family: giving the possible combinations of boy-boy, boy-girl, girl-boy and girl-girl. The probability of each is 25% or 1/4.

There are three possible ways that could lead to a boy opening the door. If there are two boys in the family then a boy will most definitely open the door. If there is one boy and one girl in the family, the chance that the boy will open the door in each case is only a half.

I’ve illustrated this information in a probability tree diagram where the probabilities shown are cumulative.

Tree Diagram

The total probability of a boy-boy combination with a boy opening the door is 25%. The total probability of there being one boy and one girl, and a boy opening the door is also 25%.

The probability of a 1 boy, 1 girl combination is hence the same as the probability of 2 boys.

After thinking both of these arguments through, I can’t actually see the logical fallacy which exists in either. They can’t both be right however. It would be great to hear your thoughts once again!

19 thoughts on “Probability Brain Teaser II

  1. There seems to be some problem with the second analysis. The probability graph shows the case when we don’t know anything else than the probabilities of the events. In other words, its missing the information that one of the child’s is a boy.

    What we need is a conditional probability. We need to find the probability of the other child being a boy, given one of them is a boy. In the graph terms, since we already know that one of the child is a boy, the probability for the edge ‘BG -> Boy Opens’ becomes 0(as it already has happened). The only case possible for BG pair is for a ‘Girl Opens’ event to happen. Similarly, ‘GB -> Boy Opens’ also will have probability 0. And now the probabilities for ‘Boy Opens’ would be equal to the previous observation.

  2. It’s not a third, it’s a half. You’re looking at a very simple problem in a complicated fashion. The gender of the two children are independant, there is a 50% chance of the second being male or female.
    It’s called the gamber’s fallacy.

  3. “In the three of the permutations which are now possible (boy-boy, boy-girl, girl-boy)”

    But the problem is that you cannot take it as permutations. The order of the siblings doesn’t matter – so the two last combinations – boy-girl and girl-boy are the same, which gives us only two combinations boy-boy and boy-girl.

    Since the beginning there were only three combinations – boy-boy, girl-girl, and girl-boy, since the order does not matter. Eliminating girl-girl by opening the door gives us only two combinations. Ergo, 50%.

  4. Good point. But boy-girl and girl-boy are different. Let’s pretend we had two coins: for the first coin we could have either heads or tails. Similarly for the second. There are four combinations with equal probabilities: (head-head, head-tail, tail-head, tail-tail). The chances of getting one head and one tail are twice that of getting two heads.

    There are two ways we can look at this problem and I hope the parallels with the above problem won’t be two tricky to follow.

    Situation 1: We throw two coins up in the air at the same time. One of them is a head: what is the probability that the other is also a head? 1/3. Because you could have (head-head, head-tail and tail-head).

    Situation 2: We throw one coin up in the air and you get a head. When you later throw a second coin would have a 50/50 chance of being head or tail because the two throws are independent.

    Perhaps what is tricky in this scenario is to ask yourself which of these situations is most and best applicable to this situation. On immediate thoughts the second situation seems more applicable but after thinking it through, I can certainly see the logic in the first one as well.

  5. The first one is IMHO wrong, because the question changes when you get the result of the first coin. The question is no longer – “What is the probability of getting the desired result from the combinations of the set {Head, Tail}?” but it changes to “What is the probability of getting a result of Head from the set {Head, Tail} knowing that the previous independent result was Head?”

    If the first question were still the case, the first coin throw would have eliminated one result, and the probability would be indeed 1/3. But since the question changes, we no longer expect the result to be a pair of results, but just a single result – head or tail. And that is independent from the first throw.

    To give a graphic example:

    We throw two coins in the air. We expect one of four results:

    Head Tail
    Tail Head
    Head Head
    Tail Tail

    Each of these has 1/4 probability. The first coin lands and it’s head. Now the wrong thing to do is to eliminate Tail Tail and say, “well, there are only three equally probable possibilities left now.”

    We have changed our perception of the problem, so we have to change the result set. This is actually similar to the Monty Hall problem, only we have only one element in the result set now – head or tail.

    So, if order is important, we remove all the results from the first set that are incorrect, and those are:

    Tail Head
    Tail Tail

    Ergo, we are only left with.

    Head Head
    Head Tail

    I’m sorry if this is long and not really straightforward, but I always have problems getting my points across and I didn’t really think this through when I started typing.

  6. It’s 50%. In the question, position is not a factor so you’re dealing with combinations not permutations. Girl-boy and boy-girl are the same thing.

    Even if position was a factor, the answer would be 50%. Lets presume the question was:

    A family has 2 children. When you knock on the door the first child you see is a boy, and what is the possibility that the next child you see is a boy?

    In this case, girl-boy would not be possible (since we’re presuming that position 1 has to be the boy that opened the door). So if it were a permutation, the possible permutations would be:


    and hence 50%. In your first conclusion, you are mixing the importance of the position.

  7. 50%.

    Your tree is a decent analysis, I think.

    You start out with four equally likely possibilities, which are hard to see, but here’s an important point: there is no sex-based determiner as to which child will open the door. So let’s suppose that we decide that the older child is the one who will answer the door. This doesn’t change the probabilities (but if you have doubts, you can repeat the “experiment” having the younger child answer the door, and add the results together).

    So here are the four equally likely combinations/outcomes:

    1. Two boys. older boy answers door.
    2. Boy, then girl. boy answers door.
    3. Girl, then boy. girl answers door.
    4. Two girls. older girl answers door.

    So when you see that the child who opens the door is a boy, you have eliminated two possible combinations: the one where a girl answered the door but there is also a boy, and the one with two girls. And you’re left with the first two, with equally likely possibilities for the other child.

    Now this should not be confused with another probability brain teaser. Suppose you ask the following questions to a man: 1. How many children do you have; and 2. Is at least one of them a son? If the answers to those are “2” and “yes”, then the probability that he has two sons is 1/3. To bring this back to the original problem, this might be a bit like supposing that you were in a country where everyone had two children and the custom was for the oldest son – if there is one – to be the one to answer the door. Here, there is a sex-based determiner on the information you receive, so you can only eliminate possibility 4 above.

    @Ghassan: your first analysis is flawed (yet, somehow, you came up with the correct answer!) as I’ve hopefully demonstrated above, you have to count permutations, because it is the permutations, not the combinations, which give you the starting probabilities. But then, when you get your information (boy answers door), then, probabilistically speaking, two of the possibilities go away.

  8. buddies, try these links and references, probably the explanation there will clear things out.

    questions 2 at:

    chap4, section 4.1, exercise 9. the solution is there in the answer book, available form same page.

  9. Some very insightful comments!

    Bob: A very good analysis but I’m still not quite sure whether you can make the statement or assumption of oldest or youngest child opening the door. By saying that the oldest (or youngest) child opens the door, you effectively predetermine it. I think this changes the problem and the sample space (see Ciju’s link to MathForum).

  10. Its a simple application of Bayes’ theorem!

    Prob(A|B) = Prob(B and A)/Prob(B) (off the top of my head).

    Should work out as a half if i did it right in my head. Not much logic needed, but it does require the assumption of prob(boy) = 1/2.

  11. The answer is 1/2.

    The problem Cow has stated is not the same as those on the links provided by ciju.

    To use the coin analogy. If we flip a coin twice, f1 and f2, there are two conditional probabilities we can consider:

    (1) P( f2=H | f1=H ) = 1/2, and
    (2) P( f1=f2=H | at least one of f1 and f2 is H) = 1/3.

    on ciju’s links (2) is considered. Here we are considering (1). The answer is 1/2.


  12. In the final analyses, the question is ambiguous. We mustn’t apply any formula to a problem that still defies definitive terms. Let me state the problem two ways to illustrate the semantic difficulty of this problem.

    Problem 1:

    In order to find a family, you randomly select a father of only two children, at least one of which is a boy. Therefore, it is equally probable that any of the three families below is the family indexed to the father.


    Consequently, there is a 2/3 probability that the family indexed to the randomly selected father has a girl. That is the way in which you have interpreted the problem.

    Problem 2:

    In order to find a family, you randomly select a *boy* from a set of families with only two children, at least one of which is a boy.

    The boy can live within one of three type families:


    NOTE: Since 50% of all boys from families with only two children have a brother, there is a 50% chance that the randomly selected boy has a brother. With that in mind we can more easily find the confusion.

    Wherein the confusion lies: Although 2/3 of all 2 children households with at least one boy have a girl, only ½ of randomly selected boys from such households have a sister and ½ don’t. Therefore, when the family in question is indexed to a randomly selected boy, the probability of a two child family with at least one son having a girl is only 50%. I must believe you will agree with that.

    With that in mind, let’s look at how the problem is usually stated: “If a family has two children and at least one of them is a boy, what is the probability that the family also contains a girl?

    Given the problem as stated, we need to know how we came across this family in question before we begin to solve the problem. Is the family in question that has two children, with at least one of whom is a boy, indexed to a randomly selected boy who has a 50% chance of having a brother? Or is the family indexed to a randomly selected father?



  13. Ron is likely wrong.

    Is the question ambiguous? Not really.

    “A family has 2 children. When you knock on the door one of the children, a boy, opens the door. What is the probability that the other child is a boy?”

    To me, and I’d like to see someone diagram that sentence otherwise you have selected a family the moment you knock on their door. At the point the boy opens the door you know you are looking at one of three possible cases BB, GB or BG. Since in only one of those options the boy has a brother.

    The only assumptions you are making are that the sentence implies the following information (denoted in square brackets):

    “A [randomly selected] family has 2 children. When you knock on the door [of their house] one of the children, a boy, opens the door. What is the probability that the other child is a boy?”

    There are lots of other verbal hints in the question to establish it’s context. For example if it’s really possible to be talking about a “randomly selected boy from a family of two” here then describing that a boy answering the door is linguistically and logically weird. What’s the alternative? A girl answers the door…at which point we have no question. We already know that a boy is in the house (since we are selecting by boys here) and we know the family has two children and we now know one is a girl.

  14. Jonathan,

    I’m not sure what you were addressing but you didn’t interact with my post. In the problem you presented, the probability is 50%.

    The boy can be *one* of four boys.

    1. He can be the oldest boy in a family of two boys.

    *B* B

    2. He can be the youngest boy in a family of two boys.

    B *B*

    3. He can be the oldest in a family of one boy and one girl.

    *B* G

    4. Or he can be the youngest in a family of one girl and one boy.

    G *B*

    If the child is older than his sibling, then family scenarios 1 and 3 are equally possible, yielding a 50% probability. If the child is younger than his sibling, then family scenarios 2 and 4 are possible, yielding a 50% probability.


  15. Ron,

    In your last comment, you’ve added an extra conditions: the boy is either the oldest or the youngest. In this situation, the boy is either the oldest or the youngest so I believe by adding this extra condition you are change the problem.

  16. Go back and read the thread. I showed the problem two ways and Jonathan did not address my points. I then addressed Jonathan’s post.


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