It’s the Friday Puzzle!

95

I was walking along the street and saw a woman playing with two children of different ages.  One of the children was a girl.  What are the chances that both children were girls?

As ever, please do NOT post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

95 comments on “It’s the Friday Puzzle!

  1. Dave says:

    Its a 50-50 decision from me

    • Brayden says:

      What a retard….

      “As ever, please do NOT post your answers, but do say if you think you have solved the puzzle and how long it took.”

    • Roland says:

      Brayden: Dave didn’t give an answer (IMHO), he just doubts between two answers, which are equally possible to him ;-) (at least that is how I read his response)

    • Braydon says:

      The trick here is not to assume that the woman and children are related. On that basis its simply an issue of gender birth rates. No doubt the pedants out there will quote the male/female ratio to 23 decimal places…….

    • deepfield says:

      …when 7 decimal places are quite sufficient.

    • Kristian says:

      @Braydon make sure you check back on monday for the solution :)

    • @Brayden Did you REALLY just drop the R-Bomb?!? I thought the clientele of Richard’s blog were above that.

  2. SilverMarc says:

    Got it quickly (being awake at 2:30am leaves plenty of space for thinking!) but, did I get it right, or do I just -think- I got it?

  3. Roland says:

    Wasn’t there a similar question 1 or 2 years ago?
    And I’m almost certain that Monday the discussions will come again when the answer is given :-) (as it is not obvious to everyone)

    • Adam says:

      He’s had similar questions before, and yes they always lead to the same discussions once he gives his answer.

  4. Anonymous says:

    75%

  5. Todio says:

    Immediately came up with an answer then smacked my forehead realizing my basic mistake and came up with the correct one. Could be because it’s 2:30 am here?

  6. Far too easy. Got the answer instantaneously. Might not be the right one, of course, as it relies on the precise wording of the question.

  7. The Masked Twit says:

    It depends.
    Was the girl born on a Tuesday?

    • ivan says:

      Indeed. She was born at 1113 am exactly and has green eyes, a slight squint and a web between two toes on her left foot. And if you think this has changed the probabilities, you need to consider the difference between extracted information and voluntarily given information in changing prior probabilities. There is an excellent article, in the context of playing bridge (where it matters) by Richard Pavlicek here:
      http://www.rpbridge.net/7z75.htm

    • ivan says:

      Indeed, if you don’t know what day she was born on, you could always ask.
      (Note for others, the joke here is that, in relation to a slightly different problem from the one presented here, the information “she was born on a Tuesday” apparently changes the answer, according to the standard analysis. But in reality for information really to change the probabilities you need to understand the context in which the information was provided.)

  8. Cherrynford says:

    Solved! 20 secondes. Have a nice week end everybody!

  9. mike rablins says:

    took me 1 minute including a double check but I have heard something similar before so avoided the tea.

  10. Justin says:

    I’ve seen this one before, so the solution almost instantly, followed by the slowly growing dread of the long arguements that will eventuate on Monday.

  11. John says:

    Much easier than the Tuesday Boy problem. 5 seconds.

  12. Bob O'H says:

    Uh oh. There are a lot of variants on this question, and they always end up I’m arguments that revolve around the precise wording. So this one looks straightforward, but let’s see what happens on Monday.

  13. Solved :) . Very easy. Took me 20 sec to read the question exactly.. after that it was a breeze.

  14. Stupid Illiterite Sociopath says:

    Were the children called “Frequentism” and “Subjectivism”?

  15. Thinker says:

    Could you tell what sex both were and are telling us one was a girl or could you only tell the sex of one of them ?

  16. Tom Cantwell says:

    Got it after a minute of reading and re-reading.

  17. Mad Kev says:

    These questions are getting worse.

  18. Drew says:

    If someone said “I’ve been to Egypt twice. One time I went to the pyramids.” then what are the chances they went to the pyramids the other time? But if someone said that what were they most likely to have actually meant?

    My wife (a statistician) disagrees with my answer by 50%.

    (This reminds me of the “35cents in two coins but one of the coins isn’t a quarter” riddle.)

  19. Paul Durrant says:

    Because I’ve come across similar probability puzzles in the past, I have an answer that I think is right, in under a minute.

  20. Braydon.. says:

    The trick here is not to assume that the woman and children are related. On that basis its simply an issue of gender birth rates. No doubt the pedants out there will quote the male/female ratio to 233 decimal places…….

    • Anonymous says:

      You also have to consider the difference in mortality rates between girls and boys. Of course we have to average it on all of childhood because we don’t know the ages of the children.

    • Nick says:

      Plus there’s social conditioning that comes with age. Small boys and girls play together. As they get older they often drift into boys-only and girls-only groups. Then at teenage, start to notice each other again. Would that affect the stats I wonder?

    • Why would you assume they were related?

  21. Dharmaruci says:

    the big clew is in the sentence “One of the children was a girl”

    imagine a similar situation where i got to the Doctor and say “help me, one of my legs is broken.” what should we deduce about my other leg?

    • Ben Saunders says:

      I assume we’re to understand this as ‘at least one of the children is a girl’ rather than as ‘exactly one of the children is a girl’. In your doctor case then the answer is nothing, since you haven’t said anything about whether or not your other leg is broken.

    • Anon says:

      The other leg has been amputated?

  22. It depends of you’re doing your probability theory right or not. Bayes Bayes Bayes.

  23. Tom says:

    Quite quickly, but it took several minutes to check my calculations.

  24. Tony says:

    There is one correct answer and several incorrect answers. I have two possible answers. What is the chance that one of these is incorrect…?

  25. Ian Reeve says:

    Are we ignoring social considerations? Such as, is a grown woman more likely to be playing with a two girls than a boy and a girl?

  26. Yat says:

    Sooooo poorly worded ! Assumptions need to be made about why the guy says “one of the children was a girls” anyway : Was he looking for a girl, or did he just state the sex of the first child he saw ? In my opinion, the second option seems more plausible, because nothing is stated in the problem about him being specifically looking for a girl. So I’m assuming he just picked a random child and stated its sex.

    This leads to the answer which will be revealed as “wrong” on monday. But I’m ok with it, the question is wrong.

    • J says:

      ‘Assumptions need to be made about why the guy says “one of the children was a girls” anyway’

      No they don’t. You don’t have this information; therefore, you cannot use it. It’s true that having an answer to that question would impact the probability, but that doesn’t mean that you get to assume an answer.

  27. Alex says:

    This is an old martin Gardner riddle. Very nice!

  28. Michael Sternberg says:

    There’s the surface answer from plain statistics, made under reasonable assumptions, which I got after about a minute. I had to get the “obvious” answer out of my head first.

    Then there are the *alternative* answers, some are fun to contemplate, the others are rather more nitpicking.

  29. mittfh says:

    I’ve got one answer, which assumes the children aren’t necessarily related to the woman. If it’s to be assumed they are related, that gives a different answer.

    There’s also the possibility that at least one of the children is an androygnous / effeminate / crossdressing boy, or an androgynous / butch / tomboyish girl, which would have the potential to really screw up the probabilities :D

  30. Craggus says:

    Once again, a classic puzzle that has been worded badly…

  31. Photon Boy says:

    About a second to reject the obvious answer in favour of a less obvious answer, then another minute to ditch that, in favour of an even less intuitive answer.

    Don’t know why some posters are discussing whether the woman is related to the children, it makes absolutely no difference. Nor do we need to make any assumptions about why the guy says “One of the children was a girl.” as one poster has indicated.

    • Yat says:

      Yes we do, because there can be at least two reasons why he said that:
      -he was checking specifically if there was at least a girl (why ? I don’t know… he likes little girls ?)
      -he just randomly stated the sex of one child
      Both reasons lead two different answers. You just need to count the combinations.

  32. Photon Boy says:

    There can be 153 different reasons why he said that, but it does not change the probability of both children being girls.

    • Yat says:

      If you actually counted the combinations, it would be obvious to you that it actually does.
      If he specifically wants to spot a girl, then there are 4 combinations, 3 of which match the problem, 2 of which have a boy.
      If he randomly choses, then there are 8 combinations, 4 of which match the problem, 2 of which have a boy.
      Just do the math. Or make a simulation, if you like. The reason why he said “One of the children was a girl” actually is the key to the problem.

    • Photon Boy says:

      The problem does not say that the man randomly chooses boy/girl so why infer it? All the specific information we have is that one of the children is a girl. We work out the probability from there. Anything else is an overcomplication of the problem, making unnecessary assumptions.

    • Yat says:

      The beauty of subjectivity.

      The unnecessary assumption, for me, is to assume that Richard was, before meeting the woman, actively looking for a girl. I don’t know why he would do that, it seems very arbitrary, so it seems like a heavy assumption to me.
      On the other hand, assuming that Richard just picks one child and states its gender seems much more natural to me.

      The point is, whatever you think, both are assumptions, and they give different results, so the problem is ambiguous.

    • Photon Boy says:

      For me the problem is UNambiguous, because I am not making any unwarranted assumptions.

    • Yat says:

      Yes you are, I just explained why, you are just blatantly ignoring my arguments…

    • Photon Boy says:

      You are making assumptions Yat. I am doing the maths based on what information is given, and no more.
      I suspect you will still be discussing this on Monday. Have a good weekend.

    • Yat says:

      I am not making assumption, I am just stating that when you solve the problem you are inherently making an assumption (When I give an answer I am too, because I have to). You think you are only using the information you have, but you actually are adding your personal assumptions. The fact that you don’t see it does not change it as a fact. You assume that Richard was specifically looking for a girl. I don’t see why it would necessarily be the case, this is just an assumption.

      Deal with it.

    • Photon Boy. says:

      I am not assuming that Richard was “specifically looking for a girl” whatever you choose that to mean. I am using the information in the question, “one of the children was a girl”. Nothing more.

      Simples.

      Enjoy your day.

    • Yat says:

      So you assume that if there were a boy and a girl, he would necessarily have told that there was a girl, and would never have said that there was a boy.

      this is a big assumption, what’s sad is that you don’t even see it.

    • Photon Boy. says:

      No Yat, I’m not assuming this either.

      You seem to have a strange debating style. Accuse someone of assuming something that they haven’t assumed – then try to put them down, or tell them that they don’t understand the problem.

      I think you will have to verbally spar with someone else because, frankly, I have better things to do.

      Enjoy the rest of the debate. No doubt I will read some more of your posts on Monday.

      Cheerybye

    • Yat says:

      Just because you don’t understand the implications of the choice you are doing while you solve the problem doesn’t mean there aren’t any.

      Concerning the assumption I am talking about in my last comment, you need to make it to get to your result. If Richard was able to say “One of the children was a boy”, then in the boy+girl scenario it would have been a 50% chance, and it the end we have the 8 combinations I was talking about earlier, which leads the answer to be inevitably that there is a 50% chance that both children are girls (which is not the answer which will be given on Monday, hence the assumption that Richard could only talk about girls)

      It is sad that you prefer to blame everything on the way I debate rather than actually trying to understand what I am saying, and keep on ignoring everything.

      Nothing will be different on Monday, everybody knows what Richard’s answer will be, and I already stated that in my first post.

    • Photon Boy. says:

      Yat, youve done it again.
      “you need to make it to get to your result”.
      I don’t believe I have told you what result I have got (that would be a spoiler). You are making yet another assumption. LOL

      See Ya.

      PS I do understand what you are saying. I just don’t agree with you.

    • Yat says:

      I think that I could allow myself to give more details about my reasoning by giving the alternate solution. As it is NOT the solution that Richard is waiting for, it shouldn’t be considered as a spoiler.

      So, let’s NOT make any assumption. More specifically, let’s not assume that Richard, for some unknown reason, is only able to talk about girls. So if we don’t make this assumption, that means that if the woman was with a boy and a girl, he could as well have said “One of the children was a boy”. Without additional information, nothing prevents this case from happening, this is a 50% chance. So let’s see what all the combinations are, I will regroup similar combinations for clarity.

      1) The woman has two girls, Richard says “One of the children was a girl”. probability 1/4
      2) The woman has two boys, Richard says “One of the children was a boy”. probability 1/4
      3) The woman has a boy and a girl (p=1/2), Richard says “One of the children was a girl”. probability 1/4
      4) The woman has a boy and a girl (p=1/2), Richard says “One of the children was a boy”. probability 1/4

      Only cases 1 and 3 match the problem statement. The probability that both children are girls is (1/4)/(1/4+1/4). That is 1/2.

      Now, what I am saying is that, to get to the expected answer (that I won’t give here because that would be a spoiler), you need to assume that whenever Richard sees a woman with a girl and a boy, he would only be able to say “One of the children was a girl”. Why would it be the case ? This is the assumption I am talking about. Get it now ?

    • Yat says:

      No, this was not an assumption : you wrote “There can be 153 different reasons why he said that, but it does not change the probability of both children being girls.”, which is just wrong. The protocol is the key.

    • Hugh Janus says:

      Yawn

    • Anon says:

      I’m assuming loads of stuff and it feels great

  33. Juan AR says:

    I only see one possible answer. I can’s see the doubt. If you throw a coin…

  34. Anonymous says:

    the key is the difference between the following: “one was a girl. what are the chances that both are girls?” and “one was a girl. what are the chances the other was a girl?”

  35. Wigan FC says:

    Er …. Kinda amounts to the same thing.

  36. Furie says:

    Can we get information on how well the person who saw the children explains themselves? If they have language problems then this may be unsolvable.

  37. Steve Jones says:

    Lot of room for argument on this one as (perhaps deliberately) Richard has not explicitly stated all the required information. It may seem to be a trivial problem, but there are important issues.

    For instance, was did he choose only to pose the given question after pre-selecting a woman with two children of a chosen combination, or was it a random choice of a woman with two children? Did he then choose to report the sex of the first child at random or, did he have a protocol where he’d report there was a girl if either was.

    There’s is lots of literature on this type of question. It is far from trivial, and the question posed is ambiguous.

    • Yat says:

      I really like the way this is worded :
      “Did he then choose to report the sex of the first child at random or, did he have a protocol where he’d report there was a girl if either was.”

      Now concerning the hypothesis of pre-selecting a woman with two children of a chosen combination, I think this would be a huge assumption…

  38. Philipp says:

    Knew this beforehand, since I’m a mathematician, it is a nice puzzle.

  39. Anders says:

    “Solved” it fairly quickly, Solved in quote marks because I don’t think it really counts as solving it when I’ve seen the puzzle before and really only remembered the solution.

    Obligatory spanner in the works: “girl” originally just meant “child” and did not imply a gender, so chances are 100% that both children were children

  40. Laith says:

    this one seems obvious, but I bet something in the exact wording is going to trip me up.

  41. Hugh Janus says:

    Straightforward explanation here:

    The Boy or Girl paradox surrounds a well-known set of questions in probability theory which are also known as The Two Child Problem,[1] Mr. Smith’s Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled The Two Children Problem, he phrased the paradox as follows:
    Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
    Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
    Gardner initially gave the answers 1/2 and 1/3, respectively; but later acknowledged that the second question was ambiguous.[1] Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[3] and Nickerson.[4]
    Other variants of this question, with varying degrees of ambiguity, have been recently popularized by Ask Marilyn in Parade Magazine,[5] John Tierney of The New York Times,[6] and Leonard Mlodinow in Drunkard’s Walk.[7] One scientific study[2] showed that when identical information was conveyed, but with different partially ambiguous wordings that emphasized different points, that the percentage of MBA students who answered 1/2 changed from 85% to 39%.
    The paradox has frequently stimulated a great deal of controversy.[4] Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view. The paradox stems from whether the problem setup is similar for the two questions.[2][7] The intuitive answer is 1/2.[2] This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sex of the second child (i.e., boy and girl),[2][8] and that the probability of these outcomes is absolute, not conditional.[9]
    Contents [hide]
    1 Common assumptions
    2 First question
    3 Second question
    4 Analysis of the ambiguity
    5 Bayesian analysis
    6 Variants of the question
    7 Psychological investigation
    8 See also
    9 References
    10 External links
    [edit]Common assumptions

    The two possible answers share a number of assumptions. First, it is assumed that the space of all possible events can be easily enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}.[10] This notation indicates that there are four possible combinations of children, labeling boys B and girls G, and using the first letter to represent the older child. Second, it is assumed that these outcomes are equally probable.[10] This implies the following model, a Bernoulli process with :
    Each child is either male or female.
    Each child has the same chance of being male as of being female.
    The sex of each child is independent of the sex of the other.
    In reality, this is an incomplete model,[10] since it ignores (amongst other factors) the fact that the ratio of boys to girls is not exactly 50:50, the possibility of identical twins (who are always the same sex), and the possibility of an intersex child. However, this problem is about probability and not biology. The mathematical outcome would be the same if it were phrased in terms of a coin toss.
    [edit]First question

    Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
    Under the forementioned assumptions, in this problem, a random family is selected. In this sample space, there are four equally probable events:
    Older child Younger child
    Girl Girl
    Girl Boy
    Boy Girl
    Boy Boy
    Only two of these possible events meet the criteria specified in the question (e.g., GG, GB). Since both of the two possibilities in the new sample space {GG, GB} are equally likely, and only one of the two, GG, includes two girls, the probability that the younger child is also a girl is 1/2.
    [edit]Second question

    Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
    This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2. Specifically, Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways:
    From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.[citation needed]
    From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.[3][4]
    Grinstead and Snell argue that the question is ambiguous in much the same way Gardner did.[11]
    For example, if you see the children in the garden, you may see a boy. The other child may be hidden behind a tree. In this case, the statement is equivalent to the second (the child that you can see is a boy). The first statement does not match as one case is one boy, one girl. Then the girl may be visible. (The first statement says that it can be either.)
    While it is certainly true that every possible Mr. Smith has at least one boy – i.e., the condition is necessary – it is not clear that every Mr. Smith with at least one boy is intended. That is, the problem statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified as having a boy this way.
    Commenting on Gardner’s version of the problem, Bar-Hillel and Falk [3] note that “Mr. Smith, unlike the reader, is presumably aware of the sex of both of his children when making this statement”, i.e. that ‘I have two children and at least one of them is a boy.’ If it is further assumed that Mr Smith would report this fact if it were true then the correct answer is 1/3 as Gardner intended.
    [edit]Analysis of the ambiguity

    If it is assumed that this information was obtained by looking at both children to see if there is at least one boy, the condition is both necessary and sufficient. Three of the four equally probable events for a two-child family in the sample space above meet the condition:
    Older child Younger child
    Girl Girl
    Girl Boy
    Boy Girl
    Boy Boy
    Thus, if it is assumed that both children were considered while looking for a boy, the answer to question 2 is 1/3. However, if the family was first selected and then a random, true statement was made about the gender of one child (whether or not both were considered), the correct way to calculate the conditional probability is not to count the cases that match. Instead, one must add the probabilities that the condition will be satisfied in each case[11]:
    Older child Younger child P(this case) P(“at least one boy” given this case) P(both this case, and “at least one boy”)
    Girl Girl 1/4 0 0
    Girl Boy 1/4 1/2 1/8
    Boy Girl 1/4 1/2 1/8
    Boy Boy 1/4 1 1/4
    The answer is found by adding the numbers in the last column wherever you would have counted that case: (1/4)/(0+1/8+1/8+1/4)=1/2. Note that this is not necessarily the same as reporting the gender of a specific child, although doing so will produce the same result by a different calculation. For instance, if the younger child is picked, the calculation is (1/4)/(0+1/4+0+1/4)=1/2. In general, 1/2 is a better answer any time a Mr. Smith with a boy and a girl could have been identified as having at least one girl.
    [edit]Bayesian analysis

    Following classical probability arguments, we consider a large Urn containing two children. We assume equal probability that either is a boy or a girl. The three discernible cases are thus: 1. both are girls (GG) – with probability P(GG) = 0.25, 2. both are boys (BB) – with probability of P(BB) = 0.25, and 3. one of each (G.B) – with probability of P(G.B) = 0.50. These are the prior probabilities.
    Now we add the additional assumption that “at least one is a girl” = G. Using Bayes Theorem, we find
    P(GG|G) = P(G|GG) * P(GG) / P(G) = 1 * 1/4 / 3/4 = 1/3.
    where P(A|B) means “probability of A given B”. P(G|GG) = probability of at least one girl given both are girls = 1. P(GG) = probability of both girls = 1/4 from the prior distribution. P(G) = probability of at least one being a girl, which includes cases GG and G.B = 1/4 + 1/2 = 3/4.
    Note that, although the natural assumption seems to be a probability of 1/2, so the derived value of 1/3 seems low, the actual “normal” value for P(GG) is 1/4, so the 1/3 is actually a bit higher.
    The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a bit sticky. Just how do we know that “at least” one is a girl? One description of the problem states that we look into a window, see only one child and it is a girl. Sounds like the same assumption…but…this one is equivalent to “sampling” the distribution (i.e. removing one child from the urn, ascertaining that it is a girl, then replacing). Let’s call the statement “the sample is a girl” proposition “g”. Now we have:
    P(GG|g) = (P(g|GG) * P(GG) / P(g) = 1 * 1/4 / 1/2 = 1/2.
    The difference here is the P(g), which is just the probability of drawing a girl from all possible cases (i.e. without the “at least”), which is clearly 0.5.
    The Bayesian analysis generalizes easily to the case in which we relax the 50/50 population assumption. If we have no information about the populations then we assume a “flat prior”, i.e. P(BB) = P(GG) = P(G.B) = 1/3. In this case the “at least” assumption produces the result P(GG|G) = 1/2, and the sampling assumption produces P(GG|g) = 2/3, a result also derivable from the Rule of Succession.
    [edit]Variants of the question

    Following the popularization of the paradox by Gardner it has been presented and discussed in various forms. The first variant presented by Bar-Hillel & Falk [3] is worded as follows:
    Mr. Smith is the father of two. We meet him walking along the street with a young boy whom he proudly introduces as his son. What is the probability that Mr. Smith’s other child is also a boy?
    Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is 1/2 and, when making the most natural assumptions, this is correct. However, someone may argue that “…before Mr. Smith identifies the boy as his son, we know only that he is either the father of two boys, BB, or of two girls, GG, or of one of each in either birth order, i.e., BG or GB. Assuming again independence and equiprobability, we begin with a probability of 1/4 that Smith is the father of two boys. Discovering that he has at least one boy rules out the event GG. Since the remaining three events were equiprobable, we obtain a probability of 1/3 for BB.”[3]
    Bar-Hillel & Falk say that the natural assumption is that Mr Smith selected the child companion at random but, if so, the three combinations of BB, BG and GB are no longer equiprobable. For this to be the case each combination would need to be equally likely to produce a boy companion but it can be seen that in the BB combination a boy companion is guaranteed whereas in the other two combinations this is not the case. When the correct calculations are made, if the walking companion was chosen at random then the probability that the other child is also a boy is 1/2. Bar-Hillel & Falk suggest an alternative scenario. They imagine a culture in which boys are invariably chosen over girls as walking companions. With this assumption the combinations of BB, BG and GB are equally likely to be represented by a boy walking companion and then the probability that the other child is also a boy is 1/3.
    In 1991, Marilyn vos Savant responded to a reader who asked her to answer a variant of the Boy or Girl paradox that included beagles.[5] In 1996, she published the question again in a different form. The 1991 and 1996 questions, respectively were phrased:
    A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who’s giving them a bath. “Is at least one a male?” she asks him. “Yes!” she informs you with a smile. What is the probability that the other one is a male?
    Say that a woman and a man (who are unrelated) each has two children. We know that at least one of the woman’s children is a boy and that the man’s oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys?
    With regard to the second formulation Vos Savant gave the classic answer that the chances that the woman has two boys are about 1/3 whereas the chances that the man has two boys are about 1/2. In response to reader response that questioned her analysis vos Savant conducted a survey of readers with exactly two children, at least one of which is a boy. Of 17,946 responses, 35.9% reported two boys.[10]
    Vos Savant’s articles were discussed by Carlton and Stansfield[10] in a 2005 article in The American Statistician. The authors do not discuss the possible ambiguity in the question and conclude that her answer is correct from a mathematical perspective, given the assumptions that the likelihood of a child being a boy or girl is equal, and that the sex of the second child is independent of the first. With regard to her survey they say it “at least validates vos Savant’s correct assertion that the “chances” posed in the original question, though similar-sounding, are different, and that the first probability is certainly nearer to 1 in 3 than to 1 in 2.”
    Carlton and Stansfield go on to discuss the common assumptions in the Boy or Girl paradox. They demonstrate that in reality male children are actually more likely than female children, and that the sex of the second child is not independent of the sex of the first. The authors conclude that, although the assumptions of the question run counter to observations, the paradox still has pedagogical value, since it “illustrates one of the more intriguing applications of conditional probability.”[10] Of course, the actual probability values do not matter; the purpose of the paradox is to demonstrate seemingly contradictory logic, not actual birth rates.
    [edit]Psychological investigation

    From the position of statistical analysis the relevant question is often ambiguous and as such there is no “correct” answer. However, this does not exhaust the boy or girl paradox for it is not necessarily the ambiguity that explains how the intuitive probability is derived. A survey such as vos Savant’s suggests that the majority of people adopt an understanding of Gardner’s problem that if they were consistent would lead them to the 1/3 probability answer but overwhelmingly people intuitively arrive at the 1/2 probability answer. Ambiguity notwithstanding, this makes the problem of interest to psychological researchers who seek to understand how humans estimate probability.
    Fox & Levav (2004) used the problem (called the Mr. Smith problem, credited to Gardner, but not worded exactly the same as Gardner’s version) to test theories of how people estimate conditional probabilities.[2] In this study, the paradox was posed to participants in two ways:
    “Mr. Smith says: ‘I have two children and at least one of them is a boy.’ Given this information, what is the probability that the other child is a boy?”
    “Mr. Smith says: ‘I have two children and it is not the case that they are both girls.’ Given this information, what is the probability that both children are boys?”
    The authors argue that the first formulation gives the reader the mistaken impression that there are two possible outcomes for the “other child”,[2] whereas the second formulation gives the reader the impression that there are four possible outcomes, of which one has been rejected (resulting in 1/3 being the probability of both children being boys, as there are 3 remaining possible outcomes, only one of which is that both of the children are boys). The study found that 85% of participants answered 1/2 for the first formulation, while only 39% responded that way to the second formulation. The authors argued that the reason people respond differently to this question (along with other similar problems, such as the Monty Hall Problem and the Bertrand’s box paradox) is because of the use of naive heuristics that fail to properly define the number of possible outcomes.[2]
    [edit]See also

    Necktie paradox
    St. Petersburg paradox
    Sleeping Beauty problem
    Monty Hall problem
    Two envelopes problem
    [edit]References

    ^ a b Martin Gardner (1954). The Second Scientific American Book of Mathematical Puzzles and Diversions. Simon & Schuster. ISBN 978-0-226-28253-4..
    ^ a b c d e f g h Craig R. Fox & Jonathan Levav (2004). “Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Conditional Probability”. Journal of Experimental Psychology 133 (4): 626–642. doi:10.1037/0096-3445.133.4.626. PMID 15584810.
    ^ a b c d e Maya Bar-Hillel and Ruma Falk (1982). “Some teasers concerning conditional probabilities”. Cognition 11 (2): 109–122. doi:10.1016/0010-0277(82)90021-X. PMID 7198956.
    ^ a b c Raymond S. Nickerson (May 2004). Cognition and Chance: The Psychology of Probabilistic Reasoning. Psychology Press. ISBN 0-8058-4899-1.
    ^ a b Ask Marilyn. Parade Magazine. October 13, 1991; January 5, 1992; May 26, 1996; December 1, 1996; March 30, 1997; July 27, 1997; October 19, 1997.
    ^ Tierney, John (2008-04-10). “The psychology of getting suckered”. The New York Times. Retrieved 24 February 2009.
    ^ a b Leonard Mlodinow (2008). Pantheon. ISBN 0-375-42404-0.
    ^ Nikunj C. Oza (1993). “On The Confusion in Some Popular Probability Problems”. CiteSeerX: 10.1.1.44.2448.
    ^ P.J. Laird et al. (1999). “Naive Probability: A Mental Model Theory of Extensional Reasoning”. Psychological Review.
    ^ a b c d e f Matthew A. CARLTON and William D. STANSFIELD (2005). “Making Babies by the Flip of a Coin?”. The American Statistician.
    ^ a b Charles M. Grinstead and J. Laurie Snell. “Grinstead and Snell’s Introducti

  42. Hamilcar Barca says:

    I’m in the “so poorly worded that its ambiguity leaves no definite answer” camp. It depends upon whether “One of the children was a girl.” Implies that the other is not, or whether it’s merely an observation on the status of one child.

    Since the second interpretation leads to a statistical morass of birthrates and child mortality, I’ll get all Occam’s-y and choose the first.

  43. JohnLoony says:

    Is it Ghostbusters 2?

  44. Nelis says:

    Ambiguous – because context is everything!

  45. Ken Haley says:

    The question is ambiguous, and here’s a simple explanation of how (without revealing the answer): (1) If the person knows that one child is a girl and knows nothing about the other one, you get one answer. But (2) if the person saw both children and is now declaring that one of them is a girl, you get a different answer.

    • Kristian says:

      I don’t understand your explanation. I get to the same answer with your (1) and (2).
      If you’re talking about specificity – that in one case both child A and child B can be the girl we know about (non-specific) and in the other case only child A is the girl we know about (specific) – then I think it’s pretty clear that the puzzle seeks a solution for a non-specific case. The girl is not identified by any means (looks, furthest to the left, name, etc).

  46. Kristian says:

    Took me a minute, drawing it out all pretty. A few parts of the puzzle I find very similar to the Monty Hall problem, I wonder if it’s actually the same puzzle in a disguise.

  47. jonny says:

    Wow people are really over-thinking this. Isn’t it just a gambler’s fallacy question…?

    • Kristian says:

      I think there’s a little more to it than that, but I could be wrong. Check back on monday! :)

  48. There was an initial reaction which screamed out to me. Then I realized it was wrong. (Saying it louder does not make you right- even for the voices in my head). The problem from the wikipedia article is not the same as this problem. I have my answer and I will wait until Monday to see if I am right or not.

  49. Niva says:

    Quite interesting to see how people can take apart a relatively straight-forward puzzle and turn it into something else. This is a well-known puzzle and the solution is only surprising to someone who’s not familiar with probabilities or someone who jumps to the “obvious” answer.

  50. Deja says:

    Think I got it. 1 minute. Probably Wrong
    tho lol

  51. Occamsrazor says:

    Is her name Florida?

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