This is a question I originally posted on Math Stackexchange. I've just seen a very good discussion of Monty Hall problem, and someone mentioned the "siblings" paradox. I've had some success explaining the Monty Hall, but ran into some tough resistance in the siblings paradox. A summary of the arguments I faced is below.

So here is the problem:

There is a well-known paradox (mentioned, among other sources, by Martin Gardner).

A) You know that I have two children, and the oldest is a daughter. What is the probability that both are daughters? (let's ignore the possibility of identical twins, which may muddle the question; and, for simplicity, assume that probability that a child is a daughter is exactly 1/2). This is very easy to answer.

B) Now, suppose you know that at least one child is a daughter (but you don't know if it's the oldest one or not). What is the probability that both are daughters?

The common way to answer is: you got extra information that the number of daughters, let's call it D, is at least one. Thus, $P(D=2 | D \ge 1)= P(D=2,D \ge 1)/P(D \ge 1)=(1/4)/(3/4)=1/3$.

Another way to answer (B) is to consider all possible outcomes.

The paradox is that the answers in (A) and (B) are different!

However, a student told me that this is a wrong way to look at it. How do you know that at least one child is a daughter? Perhaps you overheard me saying "my daughter ... (just had a birthday or something)". But in this case you were more likely to overhear in the first place, if I had had two daughters, and not one. This messes up the probabilities calculated above.

To put it another way: let's generate the list of random 2-child families. Let's randomly "overhear" the gender of one random child in each family on the list. [This is what I foolishly suggested the student do!] Ignore sons and concentrate on daughters. Calculate which proportion of the families we picked had two daughters. Then your answer is 1/2 again.

Can you describe a realistic situation in which you can come up with the information "at least one girl", and obtain the "right" answer of 1/3?

  • 1
    $\begingroup$ What about a parent in a toy store buying two toys: one which is girl-only and one which is gender-neutral? $\endgroup$
    – dtldarek
    Commented Jun 18, 2014 at 1:35
  • 4
    $\begingroup$ @dtldarek She could have a boy with an atypical gender identity :) $\endgroup$
    – Ryan Reich
    Commented Jun 18, 2014 at 15:36
  • $\begingroup$ In a shop selling birthday cards, the mother tells me that she wanted to buy two cards for both her children, whose birthday is just three days apart (by coincidence), but she only found one that she liked, and it says "Happy birthday, daughter". Now she'll have to go to another shop... $\endgroup$
    – gnasher729
    Commented Jul 2, 2014 at 16:39
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    $\begingroup$ @gnasher729: it really depends on why it was hard to find one she liked for the other child. Maybe she was looking for a very specific kind of card for the other child, or maybe that child is very picky. If the reason for the difficulty is something specific to that child, the probability will no longer be 1/3. $\endgroup$ Commented Jul 11, 2014 at 13:26
  • $\begingroup$ You find a partially damaged family tree in your attic. You see that your grandmother had a sibling, but that part of the picture is ripped off so you cannot see any information about the sibling. $\endgroup$ Commented Feb 29, 2016 at 0:39

4 Answers 4


I think there are a few independent questions here that can be considered teaching mathematics (I will not broach the subject of how to solve the problem). First, there is:

How can you state this problem unambiguously without using artificial language?

Indeed, as I see it, the reason this problem sounds paradoxical is that the phrase "one of them is a daughter" is used slightly differently in English than in mathematics:

  • In English, to say that "you know that one of his children is a daughter" can mean either that you know the formal statement $\exists i (\text{$x_i$ is a daughter})$ (where the children are $x_1$ and $x_2$) or that you know its instantiation $\text{$x_i$ is a daughter}$ for one value of $i$. In fact, the latter is preferred, since this sentence seems to default to talking about "one of his children" as a place in the family filled by a person, rather than the person themselves.

  • In math, the phase defaults to the former statement (quantified), though because of the linguistic ambiguity, if the latter is meant it is often written in the modified form "you know that a particular one of his children is a daughter".

Unfortunately, there is no common modified form that indicates that the quantified statement is the one intended, which leads to lawyerly language in attempting to specify it, and this makes the problem sound unnatural.

The second problem is really annoying, in my experience.

How can you present an everyday model for the precise statement of this problem, averting any possible objection along the lines of "who would do that" or "that would never happen"?

This is always the danger of trying to meet someone halfway from mathematical abstraction to familiar reality: if they care, it's never far enough, and if they don't care, then there was no point in abandoning abstraction in the first place.

Here is a realistic scenario that seems to work: you see a man in his early thirties waiting outside a schoolyard at the end of the school day, obviously to pick up his child. You chat, and he comments that his wife had to take "our other kid" to the doctor today. Soon, a girl comes out and meets him, and they depart. What's the probability that their "other kid" is a girl?

I think this covers all bases. The point of the age and setting being specified is to create an image in the mind of the listener that nothing can be deduced about the relative likelihood of meeting a particular child in this place. A man of that age probably has two kids in a similar school year, so you can't tell by looking whether you're seeing the older or the younger one (kids always get sick at that age!). The implication is that both would have been there but for illness. The language is neutral, and it doesn't seem to me that there is any greater chance of a child of a particular gender or (in that range) age getting sick.

This relies heavily on interpretation, but then, as discussed above, that's the whole point and problem.

Here is a scenario that actually models the precise statement. It is quite difficult to naturally talk indirectly about an unspecified child!

You meet a pregnant woman that you don't know and, as you talk, you politely inquire about her family. She tells you it's her third child, and jokes that although the house will be a lot louder, at least it won't be as bad as with three boys. You guess this means that the newborn will be a girl, but she says that she didn't want to learn the gender in advance, so doesn't know. What's the probability that her current two children are both girls?

Looking past the fluff, I believe this story says exactly and no more than that she has two children, and one of them is a girl.

  • $\begingroup$ Ryan, thanks for your discussion. However, doesn't your situation imply that the probability is now 1/2? Since we have seen a specific child and that's the crux of the paradox. $\endgroup$
    – PA6OTA
    Commented Jun 17, 2014 at 17:23
  • $\begingroup$ No. If you like, imagine that instead this story was told to you by me, who saw the child; thus, it was only reported to you that there is one female child. The crux is that there are four possible families: BB, BG, GB, and GG, of which the first is out; each has probability 1/4, so the last three have conditional probability 1/3 each. Since you don't recognize the child, you can't distinguish BG from GB, which is precisely where seeing a particular known individual in the family would make the answer 1/2. $\endgroup$
    – Ryan Reich
    Commented Jun 17, 2014 at 17:31
  • $\begingroup$ I have to side with PA6OTA on this one. See my answer. $\endgroup$ Commented Jun 18, 2014 at 12:10
  • $\begingroup$ @Will Yes, I see that you both are right. Getting this to work is proving to be quite a challenge. $\endgroup$
    – Ryan Reich
    Commented Jun 18, 2014 at 14:42
  • $\begingroup$ I have reverted the edit that removed the struck-out text because both the comments and the other answer refer to it. $\endgroup$
    – Ryan Reich
    Commented Jun 18, 2014 at 16:50

I see three issues:

  1. the limitations of the English language,
  2. information leakage,
  3. sampling.

Issue 1 was addressed by Ryan Reich. In brief, there is no natural English phrase corresponding to the mathematician's "I have two children, at least one of whom is female." Real people just don't talk that way.

Issue 2: in real life, when people talk about their children, they reveal information about them. Any story that allows one to conclude that a specific individual child is female requires one to assign a probability of $1/2$ to the other child being female. This includes Ryan Reich's example, so I have to side with PA6OTA on that one. The child in the schoolyard is a specific individual, as is the child visiting the doctor. Each independently has a $50\%$ chance of being female. Learning that that the one in the schoolyard is female does not influence the probability that the one at the doctor is.

Any story one might make up is likely to be isomorphic to Gardner's older child / younger child problem. You sit next to a stranger on the bus and start talking about your families. She tells you she has two children but doesn't specify female or male. Later in the conversation she mentions that there was a birthday party for one of the children yesterday, saying "she got lots of presents." Now you know that one of the children is female. But you know more than that: you know that the child whose birthday it was yesterday is female. (Maybe both had birthdays yesterday? Real people would have mentioned that, but even if not, see below.) Even if your traveling companion's reference to "she" was disconnected from any story and gave you no specific information whatsoever about her child, real people will be thinking of a particular child when they use the pronoun "she". This allows you to conclude that a particular child—the one your companion was thinking of when she used the word "she"—is female. The probability that the other child—the one she wasn't thinking of—is female is therefore $1/2.$

Returning to the birthday example, it is possible, using mathematician-speak, to partially, but not completely specify an individual. This is the "Tuesday" problem: a mathematician says to you "I have two children, at least one of whom is a girl born on Tuesday." In some ways this is like the problem with answer $1/3:$ we know that at least one of the children is a girl, but not which specific child. We do know something else however, namely that a girl was born on Tuesday. Having been born on Tuesday is not enough, in contrast to my earlier examples, to specify her as a unique individual. After all, this is a mathematician speaking, and the other child might also have been born on Tuesday. On the other hand, Tuesday is only one of seven days of the week, so the probability that the other child was born on Tuesday is small. So it is likely, but not certain, that this information uniquely specifies the child. If you do the calculation, you find that this partial specification is enough to drive the probability that the other child is a girl from $1/3$ almost, but not quite, to $1/2.$

This is what I referred to as the information leakage issue: in any real example, there is likely to be $100\%$ information leakage: enough information will leak to specify the female child as a unique individual. But as the Tuesday example shows, even partial leakage is enough to change the probability away from $1/3.$

Issue 3 is the one identified by your student. Any real-life story is likely to involve implicitly sampling from a population, and you need to have a decent model of the sampling procedure to do any calculations. I actually had never thought about this issue in quite this form. Your student is extremely perceptive and obviously thought very hard about the problem. This is the kind of student we like to have!

My conclusion: get away from stories involving individuals, and make the sampling procedure clear. Here's my attempt: you have an internship with the census bureau and your division is doing an analysis of two-child families. One of your colleagues has a hypothesis about two-boy families, and takes all of the files for those families to her office in order to perform an analysis. You pick a random file from those remaining in the filing cabinet. What is the probability that you pick a two-girl family?

  • $\begingroup$ I really like the discussion of the Tuesday problem, which sounds familiar now that you mention it. Probability is truly a very subtle subject. $\endgroup$
    – Ryan Reich
    Commented Jun 18, 2014 at 15:05
  • $\begingroup$ Thank you Will. It seems that all these paradoxes exist in a gap between the everyday language and statistical language. When you do expose the sampling, you reduce this gap, and the answer kind of becomes obvious. $\endgroup$
    – PA6OTA
    Commented Jun 18, 2014 at 15:58

As with the Monty Hall problem, it is important to know exactly what information you have. Let's say I offer a parent a present of 100 dollars if they have two girls, and 10 dollars if they have at least one girl. If that parent answers "I have at least one daughter", the probability for having two daughters is practically zero (because a parent with two daughters would have said "two girls" to get the 100 dollars, and not the ten dollars).

Now let's say I have prepared cards saying "I have two boys", "I have a boy and a girl", "I have two girls", "I have at least one boy", "I have at least one girl". I ask the parent to pick cards at random until one with a true statement comes up. The card says "I have at least one girl". With two girls, the chance of giving this answer is 50%. With a boy and a girl, the chance of giving this answer is 33.33%. A bit of maths shows the chance of having two girls is (1/4) / (1/4 + 1/6) = 60%.

  • $\begingroup$ Is your first paragraph simply illustrating the $100 deal will draw out the more detailed "I have two"? $\endgroup$ Commented Jun 19, 2014 at 15:50
  • $\begingroup$ @Joe That was my interpretation too. $\endgroup$
    – Ryan Reich
    Commented Jun 19, 2014 at 18:35
  • $\begingroup$ @joe and ryan: Yes, exactly. But I think there are other situations where an English speaking parent with two girls would be very unlikely to say "I have (at least) one daughter" if indeed they have two. This might different in another language. $\endgroup$
    – gnasher729
    Commented Jul 2, 2014 at 16:44

Given two genders (daughters and sons), and two birth slots, there are four permutations of gender and birth order. They are, 1) DD, 2) DS, 3) SD, 4) SS. They all have equal chances of occurring, after "rounding."

If you know that at least one of the children is a daughter, the three allowable permutations are, 1) DD, 2) DS, and 3) SD. The fourth permutation, SS, has been eliminated from consideration. "Both daughters" is the first permutation, one out of three allowable ones, and the chances of that are one third.

If you know that the oldest child is a daughter, the allowable permutations are "only," 1) DD and 2) DS. (The third permutation has been eliminated, because this is a more restrictive condition than the previous one.)

The chances of DD, or both daughters, is one out of two in this restricted space.

  • 1
    $\begingroup$ The question does not ask for a solution, but a realistic scenario that creates the specified question. $\endgroup$
    – Tommi
    Commented Feb 25, 2016 at 8:17
  • $\begingroup$ @TommiBrander: There are only four realistic "scenarios." If you go through them one by one and make the logical eliminations, you get the solution. My answer starts with "scenarios," ends with solution, and give you both. $\endgroup$
    – Tom Au
    Commented Feb 25, 2016 at 14:07

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