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Inspired by this question about the Monty Hall problem, here's a deeper look at another well-known counter-intuitive probability problem which is often stated in a way that leaves subtle ambiguities:

Version 1

You meet a woman, who tells you: "I have exactly two children. One of them is a girl."

Version 2

You meet a woman, who tells you: "I have exactly two children. The eldest is a girl."

Version 3

You meet a woman, who tells you: "I have exactly two children. One of them is a girl." You ask- "Could you please tell me specifically a child of yours who is a girl?", and she answers "The eldest is a girl."

The question is, in all three of these cases, what is the probability that both of the woman's children are girls. Assume:

  • She only tells the truth
  • She always answers any question you ask to the best of her ability
  • "One of them is a girl" is to be interpreted literally. It doesn't mean "exactly one of them is a girl".

As well as just giving a numerical answer for each version, also explain any apparent contradictions between the answers, and any hidden ambiguities in the question.

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Ben Aaronson
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    I hope we are assuming that I have two children means I have exactly two children. – ghosts_in_the_code Apr 27 '15 at 15:05
  • @ghosts_in_the_code Hah, nice spot. I'll fix that! – Ben Aaronson Apr 27 '15 at 16:07
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    This puzzle is clearly makebelieve. No mother of two always tells the truth. – Ian MacDonald Apr 27 '15 at 17:23
  • This seems very related to the Tuesday Boy problem. – QuestionC Apr 27 '15 at 18:17
  • @QuestionC Possibly! There are quite a few formulations of this, as I say the intention of this question is to dive into a subtlty usually glossed over. – Ben Aaronson Apr 27 '15 at 18:21
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    One important detail missing from this puzzle (and most similar puzzles) is what the person would have said in the counterfactual situation. For example, in version 2, are the woman's choices "The eldest is a girl" and "The eldest is a boy", or could she have said "The youngest is a girl"? – user2357112 Apr 27 '15 at 21:33
  • @user2357112 That's an important observation! – Ben Aaronson Apr 27 '15 at 21:53
  • I think this question is ambiguous because it doesn't specify the motivation for each woman to make her particular statement. You could remove the ambiguity by wording it in this way: You meet three women, each of which has exactly 2 children. To each you ask questions which each woman honestly answers. 1) Is at least one of your children a girl? Yes. 2) Is your eldest child a girl? Yes. 3) Is at least one of your children a girl? Yes. Tell me specifically one that is a girl, eldest or youngest? Eldest. This question/answer clarifies why each woman says what she says. Btw 1/3, 1/2, 1/3. – JS1 Apr 28 '15 at 06:29
  • I still think it's poorly phrased. If order would be important the question should be phrased: "What gender is the younger child" and extra information that they are not same age. In first case we don't know if the girl is the younger or elder, so we have 3 possibilities. If the question only asks about the gender, it will always be 1/2 probability. – Zikato Apr 28 '15 at 08:08
  • @Zikato Try the following experiment: Flip a pair of coins, keep it if at least one of the pair is heads, otherwise discard it. Repeat a few times. You'll find that on average, the result where both coins show heads occurs 1/3 times, not 1/2. This is a relatively well-known counterintuitive result and you can probably find plenty of explanations on google. – Ben Aaronson Apr 28 '15 at 09:45
  • @JS1 You should make that an answer, rather than a comment! – Ben Aaronson Apr 28 '15 at 09:46
  • Probably needs some caveats about monozygotic twins. – Taemyr Apr 28 '15 at 11:06
  • Question: are these three different women or do they represent the same woman? – tfitzger May 08 '15 at 03:24

5 Answers5

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There are four possibilities for the genders of her children. These are MM, MF, FM, and FF. Since order matters (eldest/youngest), MF and FM are two different cases. Therefore, the answers to the question are as follows.

Version 1:

Knowing that one child is a girl, we can eliminate MM as an option. We are left with MF, FM, and FF, out of which the probability of both being female is 1/3.

Version 2:

Knowing specifically that the eldest child is a girl, we can eliminate MM and MF as options. This leaves us with FM and FF, out of which the probability of both being female is 1/2.

Version 3:

This version starts the same as Version 1, eliminating MM as a possibility. When she answers that her eldest child is a girl, she eliminates MF from contention. This again leaves us with FM and FF, out of which the probability of both being female is 1/2. (Note: I'm still thinking about this scenario.)

Bailey M
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    Here's a potential twist for Version 3 from someone who teats Monty Hall Problems like magic eyes (I can get it for a while but then I lose it): Do you have to factor in the odds that the child she chooses to specifically tell you about is the eldest child? She could have FM in which case she had to tell you about the eldest (FME) and she could have FF in which case she could have told you about either (FFE or FFY). – Engineer Toast Apr 27 '15 at 15:43
  • My mind has been boiling about that prospect since I read the question. I have to imagine that's part of the catch, but I just can't quite get around to the numbers in my head. – Bailey M Apr 27 '15 at 15:49
  • surely the equivalent would be if Q3 was 'can you tell me the sex of your eldest child'? and she answered girl – Ewan Apr 27 '15 at 15:59
  • The difference is that (I assume) there's a true random chance of which female daughter she picks. In the situation that only her eldest child is a daughter, it's a 100% chance she chooses her. If both children are daughters, then there's a 50% chance of her choosing either. I can't bridge the gap between that and the overall probability though. – Bailey M Apr 27 '15 at 16:01
  • This is the closest so far (I actually expected more people to know the traditional answers to the first two situations, and not fall into the usual trap. The comments do seem to have driven out the confusion caused by the third answer contrasted with the first and second, so hopefully somebody will resolve it! That's really the essence of this question – Ben Aaronson Apr 27 '15 at 16:08
  • How did you determine that "order matters" from the question "what is the probability that both of the woman's children are girls"? – Mark N Apr 27 '15 at 17:24
  • Agreed, where does it say that order matters? She might just give you useless extra information. – Zikato Apr 28 '15 at 06:37
  • @Zikato Order is important as soon as we talk about eldest/youngest daughter. Bailey's reasoning is flawless in version 1 and 2. – Narmer Apr 28 '15 at 07:57
  • The question is, and I quote: "What is the probability that both of the woman's children are girls". We know one of the children is girl in all three versions. So the question might as well be "what is a probability of one child being a girl". The question IS NOT "what is the probability of younger child being a girl". – Zikato Apr 28 '15 at 08:03
  • The question could alternatively be framed as "what is the probability that the eldest child and youngest child are both girls?" The fallacy is assuming that having a girl first and a boy second is the same scenario as having a boy first and a girl second, which they can't be. Try to picture the scenario with heads/tails on a coin and the distinction of first flip/second flip and you may be able to visualize it better. – Bailey M Apr 28 '15 at 12:52
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Too ambiguous

I think this question is ambiguous because it doesn't specify the motivation for each woman to make her particular statement. For example, why does woman number one volunteer that one of the children is a girl? If this were the situation:

Questioner: Think of one of your children and tell me the gender.
Woman: One of the children is a girl.

Then the answer would be 1/2 because it would be equivalent to the "eldest girl" response.

But if this were the situation:

Questioner: Is at least one of your children a girl?
Woman: One of the children is a girl.

Then the answer would be 1/3.

Rewording the questions

You could remove the ambiguity by changing the scenario to be the following:

You meet three women, each of which has exactly 2 children. You ask questions which each woman honestly answers. For each woman, what are the chances that both of her children are girls?

Woman 1:

You: Is at least one of your children a girl?
Woman 1: Yes, at least one of my children is a girl.

Woman 2:

You: Is your eldest child a girl?
Woman 2: Yes, my eldest child is a girl.

Woman 3:

You: Is at least one of your children a girl?
Woman 3: Yes, at least one of my children is a girl.
You: Tell me specifically one that is a girl, eldest or youngest? If you have two girls, flip a fair coin to decide which to tell me.
Woman 3: Eldest.

Now with the situation clarified, the answers should be:

1/3, 1/2, 1/3. For Woman 3, the second question doesn't actually add any information because she definitely has a daughter, and so her saying "eldest" or "youngest" doesn't change anything. It does collapse the possibilities to MF and FF, but in the MF case, she was forced to say "eldest" with 100% probability whereas for the FF case, she had a 50/50 chance to say either youngest or eldest. So even though the FM MF and FF cases were each at 1/3 before the second question, the MF case is now twice as likely as the FF case due to the 100% vs 50% response.

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JS1
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    This hits the nail on the head. The difference between the answers to the 2nd and 3rd version highlight that it doesn't make sense to think of the answer solely in terms of the statement the woman makes. The information she communicates necessarily has to include how she decides what to tell you. In fact, in most "neutral" assumptions of how she decides what to tell you for version one (e.g. she picks a child at random and tells you the gender), the probability is 1/2 – Ben Aaronson Apr 28 '15 at 11:12
  • I am not sure I agree with your reasoning in case three. You assume that in the FF case she would choose uniformely at randoml, but give no basis for that assumption. My guess is that with your phrasing mothers with two girls would answer Eldest in far more than 50% of the cases, since that was the first option given. – Taemyr Apr 28 '15 at 11:18
  • @Taemyr Good point, I'll update my answer to disambiguate that case. – JS1 Apr 28 '15 at 11:59
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Assuming the chances of her having a boy or girl is equally likely, there is a

50% chance of her having two female children.

This problem is similar to stating that John has a coin that he flipped heads. If John flipped another coin, what is the chance it was also head.

The only change in your statements is the order of the flips (or children) which doesn't matter since the determination of the outcomes are independent.

Mark N
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Borrowing heavily from the comments in Bailey M's answer:

Version 3

There is a 1/3 probability that she has 2 girls.

Explanation:

We know that the eldest is a girl, meaning only FM and FF are answers, with equal probability. If the answer is FF, then she randomly chooses between telling us the eldest is a girl or the youngest is a girl. In other words, given that the eldest is a girl, we have the following possibilities: (1) The youngest is a boy and she tells us the eldest is a girl (50%) (2) The youngest is a girl and she tells us the eldest is a girl (25%) (3) The youngest is a girl and she tells us the youngest is a girl (25%) Given that she has told us that the eldest is a girl, we remove the 3rd possibility, leaving us with a 50/75 probability that the youngest is a boy and a 25/75 probability that the youngest is a girl. Reduce those, and we get a 2/3 chance that she has 1 girl, and a 1/3 chance that she has 2 girls.

This seems a little fuzzy to me, but it's the only difference I can see between versions 2 and 3. Sorry for the bad formatting, I couldn't figure out how to do a list inside spoilers.

VictorHenry
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Version 1

You don't have any information about the second child, so it can be male or female with the same probability $\frac1 2$. Note that saying that "one is female" doesn't mean "only one is female"!

Version 2

Again, you have no information about the second child, the probability of being female is $\frac1 2$

Version 3

You know that one child is female. After the second statement, we have the following cases:
- Eldest female-Youngest female
- Eldest female-Youngest male
None of these is more probable that the other, so we have again a $\frac12$ chance of both being females.

leoll2
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  • To the serial downvoter: could you please motivate your vote? Where do you think I'm wrong? – leoll2 Apr 27 '15 at 19:23
  • I don't think I'm a serial downvoter, but I downvoted this one because your reasoning is wrong. Take a look at, e.g. Baily's answer for the reasoning on the first version in particular. – Ben Aaronson Apr 27 '15 at 21:54
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    Answer I can agree on. Baily's answer assumes that order is important even though it's never mentioned in the original question. – Zikato Apr 28 '15 at 06:41
  • @Zikato "Eldest is female" distinguishes between boy-girl and girl-boy. Hence order matters. – Taemyr Apr 28 '15 at 11:12
  • Order absolutely doesn't matter, it's a tricky statement! Consider "the one with red hair is female" instead of "the eldest is female", nothing changes! In fact, what do you know about the second child, except that he/she is the youngest? Nothing! Does age influence the gender? – leoll2 Apr 28 '15 at 11:17
  • @BenAaronson FM is the same of MF! What's the difference between FM and MF? How can you tell them? Once you know that a child is female, what do you know about the other's gender? – leoll2 Apr 28 '15 at 11:21
  • @leoll2 in that case the ordering is by hair color rather than by age. - Asuming you know that one is red haired and one is not. Age does not influence the gender, unless you know that the oldest is female. – Taemyr Apr 28 '15 at 11:21
  • @leoll2 In the unordered case you have to take note that FM/MF is twice as likely as FF. If you doubt this look at the distibution of number of heads when you flip pairs of coins. – Taemyr Apr 28 '15 at 11:24
  • @Taemyr You're saying that age influences gender when you know that the oldest is female. Would the problem be different if it was "the youngest is female"? – leoll2 Apr 28 '15 at 11:25
  • @leoll2 You can simulate this with coins quite easily. Do you agree that if you flip two coins, "one of each" is twice as likely as "both tails"? (Note that nowhere have I mentioned order) – Ben Aaronson Apr 28 '15 at 11:26
  • @leoll2 It's a bad formulation on my part. Age never influences gender. Age influences what you know about gender if you know that the eldest is female. The situation is symetrical if it where the younger. – Taemyr Apr 28 '15 at 11:27
  • If I have a female child, what's the probability that my next child will be female? It's still 1/2! – leoll2 Apr 28 '15 at 11:28
  • MF is twice as likely than FF when you don't have any information about none of the two children, but you have here! – leoll2 Apr 28 '15 at 11:29
  • @leoll2 Okay, so you start with four possibilities: one of each (prob 1/2), two girls (prob 1/4), two boys (prob 1/4). Now I tell you that at least one is a girl. The only one of those possibilities which is changed is two boys, which gets removed. That leaves "one of each" twice as likely as "two girls". You can simulate this with coins- just put four pairs on the table in the starting proportion. Then add the information that at least one is tails, removing any pairs this invalidates and leaving the rest. – Ben Aaronson Apr 28 '15 at 11:30
  • Perhaps one of these links may explain it better than me: http://news.bbc.co.uk/1/hi/programmes/more_or_less/8735812.stm http://math.stackexchange.com/questions/15055/in-a-family-with-two-children-what-are-the-chances-if-one-of-the-children-is-a http://mathforum.org/dr.math/faq/faq.boy.girl.html – Ben Aaronson Apr 28 '15 at 11:33
  • @BenAaronson There is the trick! When you know that "one is female", the probabilities 1/4 and 1/2 aren't valid anymore! Consider this: having two children with same gender is 1/2, having children of different genders is 1/2. Once you know that a child has a gender, what's the probability that the second has same gender? 1/2 of course! – leoll2 Apr 28 '15 at 11:35
  • @leoll2 Sorry, I don't really know how else to explain it beyond suggesting you simulate it yourself with coins. If place four pairs of coins on the table corresponding to your starting probability, then modify them to take into account the new information that at least one is tails, you'll find you don't remove either of the "one of each" pairs. If instead you said "the leftmost coin is tails" or "the shiniest one is tails" or whatever, you'd (on average) remove one of the "one of each" pairs. – Ben Aaronson Apr 28 '15 at 12:08
  • @leoll2 Start with a set of 100 families with 2 children. Choose this set such that 25 have 2 girls, 25 have 2 boys and 50 have one boy and one girl. Do you agree that the distribution of these families matches those in the general population? Now remove all the families that have no girls. Select a family at random from the remaining 75. What is the probability that the chosen family has two girls? – Taemyr Apr 29 '15 at 07:48