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Math Help - Probability Conundrum

  1. #1
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    Lightbulb Probability Conundrum

    "I have 2 children. One is female. What is the probability the other is also female?"

    This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.

    I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.

    I'm interested to get some more opinions...
    Last edited by Nath; October 1st 2010 at 01:59 PM. Reason: Trying to move to "Puzzles" thread...
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    Quote Originally Posted by Nath View Post
    "I have 2 children. One is female. What is the probability the other is also female?"

    This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.

    I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.

    I'm interested to get some more opinions...
    Perhaps it's the wording that is making you think it's 1/2.

    Assuming each child you had was equally likely to be male or female,

    MM - 1/4
    MF - 1/4
    FM - 1/4
    FF - 1/4

    So it's one of MF, FM, and FF which occur with equal probability. One out of these three leads to FF. So the answer is 1/3.

    BUT, if you label the children 1 and 2, and if you are given that child #1 is female, then the probability that child #2 is also female is 1/2.
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    Quote Originally Posted by Nath View Post
    "I have 2 children. One is female. What is the probability the other is also female?"

    This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.

    I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.

    I'm interested to get some more opinions...
    When you know that the speaker has two children you have the probability that they are BB is 1/4, GG is 1/4 and BG is 1/2.

    Now you are told that one is a girl, which means that BB is not possible, but the relative probabilities of GG and BG are unchanged, so the probability of GG is 1/3.

    Alternatively the result follows from Bayes' theorem:

    P(GG|one or more girls)=P(one or more girls|GG)P(GG)/P(one or more girls)=1.(1/4)/((1/4)+(1/2))=1/3



    CB
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    I agree with undefined that the wording is ambiguous.

    If someone told me they had two children and one is a girl, I would conclude that they mean one is a girl and one is a boy.

    But there are at least two other interpretations. (1) The child I am looking at (or thinking of) is a girl; (2) at least one child is a girl. The canonical interpretation of the problem is that (2) is intended, but it's very difficult (for me, at least) to think of a realistic situation in which you know that at least one child is a girl and that's all you know-- unless the speaker is being intentionally obscure.
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    Quote Originally Posted by Nath View Post
    "I have 2 children. One is female. What is the probability the other is also female?"

    This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.

    I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.

    I'm interested to get some more opinions...
    Here is an alternative "logical" way to consider this.


    There are 4 possibilities in regard to the children being boy or girl.
    The probability of having a boy and a girl is twice the probability of having 2 girls,
    and is also twice the probability of having 2 boys.

    (i) 1st and 2nd children are girls
    (ii) 1st child is a girl and the 2nd is a boy
    (iii) 1st child is a boy and the 2nd is a girl
    (iv) 1st and 2nd children are boys

    GG
    GB
    BG
    BB

    One is female.
    This reduces to...

    GG
    GB
    BG

    3 cases of equal probability and in only 1 of these cases is the other child also a girl.
    Hence the probability of the 2nd child also being a girl is 1/3.

    Also note that in 2 of these 3 cases, the 2nd child is a boy,
    therefore, if one of the children is a girl, the probability that the other is a boy is 2/3.
    Last edited by Archie Meade; October 2nd 2010 at 12:57 PM. Reason: original post erroneous due to non-equal probabilities
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  6. #6
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    Exclamation Solutions

    This was my solution, but alas my friend disagrees:

    I have 2 children, 1 is female. This gives four possible scenarios:

    1. The female has a younger sister
    2. The female has an older sister
    3. The female has a younger brother
    4. The female has an older brother

    Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
    ---

    My friend, with the PhD in Maths thinks the following:

    Sample space: Omega = {GG, GB, BG, BB}.
    Our sample space is composed of four events, distinct from each other, and with equal probability
    p =1/4.B = event that I have (at least) 1 girl.
    B = {GG, GB, BG}. P (B) = 3/4A = event that I have 2 girls. A = {GG}. P(A) = 1/4
    Clearly, B&A = A (i.e. the interesection between A and B is A)
    Now, by definition of conditional probailities: P(A|B)
    = P(A&B)/P(B) = P(A)/P(B) = (1/4)/(3/4) = 1/3
    ---
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    Quote Originally Posted by Nath View Post
    This was my solution, but alas my friend disagrees:

    I have 2 children, 1 is female. This gives four possible scenarios:

    1. The female has a younger sister
    2. The female has an older sister
    3. The female has a younger brother
    4. The female has an older brother

    The probability of a female having a brother is twice the probability of a female having a sister.

    Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
    ---

    The problem is that the probabilities are not 1/4, they are 1/6

    GG
    GB
    BG

    Given that the family has at least 1 girl, the probability of a child being the younger of 2 sisters is 1/6.
    The probability of the child being an older sister is 1/6.
    The probability of the child being the younger sister of a boy is 1/6 etc


    My friend, with the PhD in Maths thinks the following:

    Sample space: Omega = {GG, GB, BG, BB}.

    Our sample space is composed of four events, distinct from each other, and with equal probability

    p =1/4.

    B = event that I have (at least) 1 girl.
    B = {GG, GB, BG}. P (B) = 3/4

    A = event that I have 2 girls.
    A = {GG}. P(A) = 1/4

    Clearly, B&A = A (i.e. the intersection between A and B is A)

    Now, by definition of conditional probailities: P(A|B)= P(A&B)/P(B) = P(A)/P(B) =(1/4)/(3/4) = 1/3
    ---
    Yes, that is the way to calculate using conditional probability...

    A is the event of having 2 girls...... P(A)=\frac{1}{4}

    B is the event of having a girl................ P(B)=\frac{3}{4}

    AB is the event that the family has a girl and the family has 2 girls,
    which reduces to....the family has 2 girls and so is A

    Conditional probability is........................ P(A|B)=\displaystyle\frac{P(AB)}{P(B)}=\frac{P(A)}  {P(B)}=\frac{\left(\frac{1}{4}\right)}{\left(\frac  {3}{4}\right)}=\frac{1}{3}


    Also note that the conditional probability of the 2nd child being a boy is...... \displaystyle\frac{\left(\frac{2}{4}\right)}{\left  (\frac{3}{4}\right)}=\frac{2}{3}
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  8. #8
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    I still can't agree it's 1/3 at an intuitive level. For example, when I run a analogous scenario throwing coins, I get the same result of 1/2:

    I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

    c1 = {H1, T1}
    c2 = {H2, T2}

    Throwing both coins at the same time gives the following set of possibilities:

    {H1, H2}
    {H1, T2}
    {H2, T1}
    {T1, T2}

    So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

    {H1, T2}
    {H2, T1}
    {T1, T2}

    But if a throw of c2 gives T2, then {H2, T1} is eliminated, so there remains a 50% chance that c1 = H1 and a 50% chance that c1 = T1

    Conversely, if a throw of c1 gives T1, then {H1, T2} is eliminated, so there remains a 50% chance that c2 = H2 and 50% chance that c2 = T2.
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    Quote Originally Posted by Nath View Post
    I still can't agree it's 1/3 at an intuitive level. For example, when I run a analogous scenario throwing coins, I get the same result of 1/2:

    I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

    c1 = {H1, T1}
    c2 = {H2, T2}

    Throwing both coins at the same time gives the following set of possibilities:

    {H1, H2}
    {H1, T2}
    {H2, T1}
    {T1, T2}

    So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

    {H1, T2}
    {H2, T1}
    {T1, T2}

    But if a throw of c2 gives T2, then {H2, T1} is eliminated, so there remains a 50% chance that c1 = H1 and a 50% chance that c1 = T1

    Conversely, if a throw of c1 gives T1, then {H1, T2} is eliminated, so there remains a 50% chance that c2 = H2 and 50% chance that c2 = T2.
    Let event A = both coins are tails
    Let event B = coin 1 is tails
    Let event C = coin 2 is tails
    Let event D = coin 1 is tails or coin 2 is tails (in set notation, D=B\cup C)

    Fact: P(A|B) = 1/2
    Fact: P(A|C) = 1/2
    Fact: P(A|D) = 1/3

    This is exactly the issue I brought up in my first post.
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  10. #10
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    Quote Originally Posted by Nath View Post
    This was my solution, but alas my friend disagrees:

    I have 2 children, 1 is female. This gives four possible scenarios:

    1. The female has a younger sister
    2. The female has an older sister
    3. The female has a younger brother
    4. The female has an older brother

    Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
    ---
    They are not equally likely.

    The unambiguous was to settle this is to run a simulation:

    Code:
    >
    >function Simulate(NN)
    $
    $  nGG=0; ##numb of families with GG
    $  nG=0;  ##numb of families with at least one G
    $
    $  for idx =1 to NN ##simulate NN families
    $   ##generate a 2 child family 0=boy, 1=girl 
    $    xx=RandInt(1,2,0,1);
    $    if sum(xx)>0  ##count number with at least one G
    $      nG=nG+1;
    $    endif
    $    if sum(xx)==2  ##count numb of families with two G
    $      nGG=nGG+1;
    $    endif
    $  end ##for
    $  return nGG/nG  ##return the required fraction
    $endfunction
    >
    >Simulate(100000)
         0.332302 
    >
    The simulation also has the advantage of making the probability model explicit.

    CB
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    Quote Originally Posted by CaptainBlack View Post
    They are not equally likely.

    The unambiguous was to settle this is to run a simulation:
    Speaking of simulations.. @Nath, you might want to try flipping an actual coin to convince yourself that the result is in fact 1/3.
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    @Nath, maybe this re-framing of the problem will help you.

    Given: a couple has two children, and they are not both boys.

    (Assume that probability of having a boy is same as that of having a girl.)

    What is the probability that they are both girls?
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    With both the Fortran simulation example and the Bayes Theorem quotation, the event space is skewed because again, it assumes that BG & GB are separate events, but that GG is only 1 event, even though the 2 Gs are separate entities (i.e. G1G2 & G2G1 are separate events in the event space).

    Therefore I stick with my original proof:

    I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

    c1 = {H1, T1}
    c2 = {H2, T2}

    Throwing both coins at the same time gives the following set of possibilities:

    {H1, H2}
    {H1, T2}
    {H2, T1}
    {T1, T2}

    So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

    {H1, T2}
    {H2, T1}
    {T1, T2}

    But if a throw of c2 gives T2, then {H2, T1} is eliminated, so there remains a 50% chance that c1 = H1 and a 50% chance that c1 = T1

    Conversely, if a throw of c1 gives T1, then {H1, T2} is eliminated, so there remains a 50% chance that c2 = H2 and 50% chance that c2 = T2.
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  14. #14
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    I give up trying to explain the same thing over and over. Flip a coin and see. You will recant.
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    Quote Originally Posted by undefined View Post
    I give up trying to explain the same thing over and over. Flip a coin and see. You will recant.
    I've tried flipping the two coins from the previous example and I get the result from the previous example :-)
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