Page 3 of 14 FirstFirst 123456713 ... LastLast
Results 31 to 45 of 202

Thread: The daughter paradox

  1. #31

    Re: The daughter paradox

    Quote Originally Posted by skbuncks View Post
    Emma then Sue
    Sue then Emma
    Steve then Emma
    Emma then Steve

    ...which gives the probability of having two girls as 1/2.
    No it doesn't - the first two of your four possibilities have a probability of 1/6 each, whereas the others have probability of 1/3 each. We want the probability of two girls, so it is: 1/6 + 1/6 in 1/6 + 1/6 + 1/3 + 1/3 which is 1 in 3.

    The name is a complete red herring - the probability is determined by gender alone, so any additional labeling cannot affect the probability.

  2. #32
    Senior Member
    Join Date
    Apr 2009
    Location
    Chesterfield, UK
    Posts
    110

    Re: The daughter paradox

    Quote Originally Posted by brianp View Post
    No it doesn't - the first two of your four possibilities have a probability of 1/6 each, whereas the others have probability of 1/3 each. We want the probability of two girls, so it is: 1/6 + 1/6 in 1/6 + 1/6 + 1/3 + 1/3 which is 1 in 3.

    The name is a complete red herring - the probability is determined by gender alone, so any additional labeling cannot affect the probability.
    That's what I was trying to get at with the coins (probably not very well). You have put it more succinctly.
    I had it all...but I threw it away. Then I found it again, so that's alright.

    Blog thingy

    Me on Twitter

  3. #33
    Appreciative guest
    Join Date
    Apr 2007
    Location
    London
    Posts
    2,544
    Blog Entries
    2

    Re: The daughter paradox

    Well blow me down but specifying the name really does seem to make the difference stated in the OP.

    I've worked out the probability tree for all permutations assigning an arbitrary value n for the probability that the child is named Emma.

    That's a little tricky because you have to account for that fact that the probability of both daughters being called Emma is zero. If you don't do that then it doesn't work the same.

    Anyway full working shown in the attached PDF
    daughter paradox.pdf

  4. #34
    Appreciative guest
    Join Date
    Apr 2007
    Location
    London
    Posts
    2,544
    Blog Entries
    2

    Re: The daughter paradox

    Quote Originally Posted by brianp View Post
    The name is a complete red herring - the probability is determined by gender alone, so any additional labeling cannot affect the probability.
    That's what I thought at first. And to be sure if you make the name independent such as with two class mates who might well both be called Emma then what you're saying is true.

    The additional bit of information your given is that both daughters won't both be called Emma.

    If it's then a given that one daughter is called Emma that means that the other child is either a boy of any name or a girl not called Emma.

    Think about it. If the other sibling is a boy then the first child has more chance of being called Emma than if the other sibling were a girl (who might herself be an Emma)

  5. #35
    eliminate the impossible
    Join Date
    Sep 2009
    Posts
    777

    Re: The daughter paradox

    Wow Matt, where do you find the time ...

  6. #36

    Re: The daughter paradox

    Quote Originally Posted by Matt View Post
    Think about it. If the other sibling is a boy then the first child has more chance of being called Emma than if the other sibling were a girl (who might herself be an Emma)
    But the difference is miniscule. There are hundreds of potential girls names and if we rule out one of those names for the second girl, there are still hundreds. The probability of two girls given that one is Emma and the other not Emma, is still (to a close approximation) 1 in 3.

  7. #37
    Hero member
    Join Date
    Jan 2008
    Location
    London
    Posts
    2,417

    Re: The daughter paradox

    Quote Originally Posted by Matt View Post
    That's what I thought at first. And to be sure if you make the name independent such as with two class mates who might well both be called Emma then what you're saying is true.

    The additional bit of information your given is that both daughters won't both be called Emma.

    If it's then a given that one daughter is called Emma that means that the other child is either a boy of any name or a girl not called Emma.

    Think about it. If the other sibling is a boy then the first child has more chance of being called Emma than if the other sibling were a girl (who might herself be an Emma)
    Can't fault the workings, but consider this - X has 2 children, one is a daughter, what is the probability of two daughters? We know she has one daughter, we shall label that daughter D1, now what is your probability? Can it really change simply because of this manouver?

    I think the problem is applying mathematical reasoning and forgetting genetics. If we know two male offspring are excluded, then you have 3 possible combinations of genetic outcomes BE, EB, or GG(one of which happens to be called Emma).

  8. #38
    Senior Member
    Join Date
    Apr 2009
    Location
    Chesterfield, UK
    Posts
    110

    Re: The daughter paradox

    Quote Originally Posted by Pebble View Post
    I think the problem is applying mathematical reasoning and forgetting genetics. If we know two male offspring are excluded, then you have 3 possible combinations of genetic outcomes BE, EB, or GG(one of which happens to be called Emma).
    I think you're right. The problem is asking for the genetic outcome (probability of 2 girls when we know there is at least one). The name of the girl(s) is irrelevant to the genetic possibilities which remain as BG, GB, GG.
    I had it all...but I threw it away. Then I found it again, so that's alright.

    Blog thingy

    Me on Twitter

  9. #39
    Appreciative guest
    Join Date
    Apr 2007
    Location
    London
    Posts
    2,544
    Blog Entries
    2

    Re: The daughter paradox

    Quote Originally Posted by brianp View Post
    But the difference is miniscule. There are hundreds of potential girls names and if we rule out one of those names for the second girl, there are still hundreds. The probability of two girls given that one is Emma and the other not Emma, is still (to a close approximation) 1 in 3.
    But if that's the case where's the flaw in my calculations?

    I can accept that my simplified explanation may be wrong but I hang my hat on the full statistical anaysis I provided.

  10. #40
    Hero member
    Join Date
    Jan 2008
    Location
    London
    Posts
    2,417

    Re: The daughter paradox

    Quote Originally Posted by Matt View Post
    But if that's the case where's the flaw in my calculations?

    I can accept that my simplified explanation may be wrong but I hang my hat on the full statistical anaysis I provided.

    There is nothing wrong with the statistics, the analysis has just has been applied to the wrong question. It is not the naming of a child that determines the possibility of having a given number of same sex offspring.

  11. #41
    Appreciative guest
    Join Date
    Apr 2007
    Location
    London
    Posts
    2,544
    Blog Entries
    2

    Re: The daughter paradox

    I think that though the n is small for the general population, it hangs on the fact that whatever the size of n, the chance of having a daughter called Emma is doubled if you have two daughters rather than one.

    This is better than the Monty Hall problem.

  12. #42
    Appreciative guest
    Join Date
    Apr 2007
    Location
    London
    Posts
    2,544
    Blog Entries
    2

    Re: The daughter paradox

    Quote Originally Posted by Pebble View Post
    There is nothing wrong with the statistics, the analysis has just has been applied to the wrong question. It is not the naming of a child that determines the possibility of having a given number of same sex offspring.
    Don't you agree that the chance of having a daughter called Emma is doubled if you have twice as many daughters?

    If my answer is correct but a correct answer to the wrong question that I have to ask what is the question that I've correctly answered and how does that differ from the question in the OP
    Last edited by Matt; 17th July 2010 at 01:29 PM.

  13. #43
    Hero member
    Join Date
    Jan 2008
    Location
    London
    Posts
    2,417

    Re: The daughter paradox

    Quote Originally Posted by Matt View Post
    Don't you agree that the chance of having a daughter called Emma is doubled if you have twice as many daughters?
    Yes, but every daughter has a name, so the same logic applies to every daughter in the world.

    Quote Originally Posted by Matt View Post
    If my answer is correct but a correct answer to the wrong question that I have to ask what is the question that I've correctly answered and how does that differ from the question in the OP
    The problem is the assumption that Emma is a unique individual, in reality a daughter is a unique individual, this one simply happens to be labeled Emma. It follows that the probabilites remain dependent on the liklihood of having a given distribution of XY or XX chromasomes among your children, not what happens to those individuals after birth in terms of identifiers.

  14. #44
    Pontificator-in-Chief Admin's Avatar
    Join Date
    May 2006
    Posts
    5,352
    Blog Entries
    2

    Re: The daughter paradox

    I started trying this out using playing cards - 2 queens and 2 kings and using the queen of spades as 'Emma'. But it got rather tedious.

    So I wrote a small program to simulate the choices:

    Here: http://www.ukskeptics.com/births.php

    You can choose the number of pairs of children (up to 30,000)

    The percentage of girls whose name is Emma. If you choose 100% it means that any girl/boy combination will result in the girl being named Emma and in a girl-girl pair one of them will definitely be called Emma.

    If you choose 50%, 20% etc. then the chances of a girl being named Emma reduce accordingly.

    If you tick 'display output' it outputs a table of all the combinations made.

    It turns out that whatever combination of choices you make, 'Emma' will be paired with another girl 33% of the time and with a boy 67% of the time.

    Unless I've made an error.

    Here's the page code for any programmers to inspect (yes it's probably poorly coded as I did it ad hoc off the top of my head - so no criticism required ):

    Code:
    <!doctype html public "-//W3C//DTD HTML 4.0 Transitional//EN">
    <html>
    <head>
    <title> New Document </title>
    <meta name="Generator" content="EditPlus">
    <meta name="Author" content="">
    <meta name="Keywords" content="">
    <meta name="Description" content="">
    </head>
    <body>
    <?php if($_POST['no_of_pairs']>30000){$_POST['no_of_pairs']=30000;}//prevents server throwing out an error ?>
    <form name="form1" method="post" action="">
      <table width="95%" border="1" align="center" cellpadding="2" cellspacing="2" bordercolor="#000000" bgcolor="#aabbaa">
        <tr align="center">
          <td width="33%">Number of pairs of children 
            <input name="no_of_pairs" type="text" id="no_of_pairs" value="<?php if(isset($_POST[no_of_pairs])){echo($_POST[no_of_pairs]);} else {echo("10");} ?>" size="5" maxlength="5"></td>
          <td width="34%">Percentage changed to Emma 
            <select name="percent_of_emma" id="percent_of_emma">
              <option value="1" <?php if($_POST[percent_of_emma]==1){echo("selected");} ?>>100</option>
              <option value="2" <?php if($_POST[percent_of_emma]==2 || $_POST[percent_of_emma]==""){echo("selected");} ?>>50</option>
              <option value="3" <?php if($_POST[percent_of_emma]==3){echo("selected");} ?>>33</option>
              <option value="4" <?php if($_POST[percent_of_emma]==4){echo("selected");} ?>>25</option>
              <option value="5" <?php if($_POST[percent_of_emma]==5){echo("selected");} ?>>20</option>
              <option value="6" <?php if($_POST[percent_of_emma]==6){echo("selected");} ?>>17</option>
              <option value="10" <?php if($_POST[percent_of_emma]==10){echo("selected");} ?>>10</option>
              <option value="20" <?php if($_POST[percent_of_emma]==20){echo("selected");} ?>>5</option>
              <option value="50" <?php if($_POST[percent_of_emma]==50){echo("selected");} ?>>2</option>
              <option value="100" <?php if($_POST[percent_of_emma]==100){echo("selected");} ?>>1</option>
              <option value="200" <?php if($_POST[percent_of_emma]==200){echo("selected");} ?>>0.5</option>
              <option value="1000" <?php if($_POST[percent_of_emma]==1000){echo("selected");} ?>>0.1</option>
            </select></td>
          <td width="33%">Display output 
            <input name="display_output" type="checkbox" id="display_output" value="checked" <?php if($_POST[display_output]=="checked"){echo("checked");} ?>></td>
        </tr>
      </table>
      <p align="center"> 
        <input type="submit" name="Submit" value="Submit">
      </p>
    </form>
    <p>
      <?php
    if ($_POST[Submit])
    {
    $count=$_POST[no_of_pairs];//number of pairs of births
    if($count==0){$count=1;}
    for($i=0; $i<$count; $i++)
        {
            $first_born=rand(1,2);
            $second_born=rand(1,2);
            if($first_born==1){$first_child="Boy";}
            if($first_born==2){$first_child="Girl";}
            if($second_born==1){$second_child="Boy";}
            if($second_born==2){$second_child="Girl";}
    //randomly assigns first/scond born as boy or girl
            if($first_child == "Girl" || $second_child == "Girl")
                //if at least one is a girl, give the random option to name one Emma
            {
                $change=rand(1,$_POST[percent_of_emma]);//second figure can be changed to reduce the occurences percentage
                if ($first_child=="Boy" && $change==1){$second_child="Emma"; $change=0;}
                if ($second_child=="Boy" && $change==1){$first_child="Emma"; $change=0;}
                if ($first_child=="Girl" && $second_child=="Girl" && $change==1)
                {
                    $choice=rand(1,2);
                    if($choice==1){$first_child="Emma";}
                    if($choice==2){$second_child="Emma";}
                }
            }
        $numbers[$i][0]=$first_born;
        $numbers[$i][1]=$second_born;
        $output[$i][0]=$first_child;
        $output[$i][1]=$second_child;    
    //numbers array keeps track of male/female. Output keeps track of boy or girl/emma
        }
    for($i=0; $i<$count; $i++)
        {
            if ($output[$i][0]=="Emma" || $output[$i][1]=="Emma")
            $end=1;//routine to count how may boys/girls Emma matches with
            {
                if($output[$i][0]=="Emma" && $output[$i][1]=="Boy" && $end==1){$emma_boy+=1;$end=0;}
                if($output[$i][0]=="Emma" && $output[$i][1]=="Girl" && $end==1){$emma_girl+=1;$end=0;}
                if($output[$i][0]=="Boy" && $output[$i][1]=="Emma" && $end==1){$emma_boy+=1;$end=0;}
                if($output[$i][0]=="Girl" && $output[$i][1]=="Emma" && $end==1){$emma_girl+=1;}
            }
        }
    $pop=$count*2;//the population is double the number of pairs of children!
    for($i=0; $i<$count; $i++)
        {
            if ($numbers[$i][0]==1){$no_boys+=1;}
            if ($numbers[$i][0]==2){$no_girls+=1;}
            if ($numbers[$i][1]==1){$no_boys+=1;}
            if ($numbers[$i][1]==2){$no_girls+=1;}
            if ($output[$i][0]=="Emma"){$no_emma+=1;}
            if ($output[$i][1]=="Emma"){$no_emma+=1;}
            if ($numbers[$i][0]==1 && $numbers[$i][1]==1){$boy_boy+=1;}    
            if ($numbers[$i][0]==1 && $numbers[$i][1]==2){$boy_girl+=1;}
            if ($numbers[$i][0]==2 && $numbers[$i][1]==1){$girl_boy+=1;}    
            if ($numbers[$i][0]==2 && $numbers[$i][1]==2){$girl_girl+=1;}        
        }
    }
    ?>
    </p>
    <table width="75%" border="1" align="center" cellpadding="4" cellspacing="4">
      <tr> 
        <td width="50%"><strong>Population</strong></td>
        <td width="25%" align="right"><?php echo($pop); ?>&nbsp;</td>
        <td width="25%"><div align="center"><strong>Percentage</strong></div></td>
      </tr>
      <tr> 
        <td>Number of boys</td>
        <td align="right"><?php echo($no_boys); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($pop))$var=round($no_boys/$pop*100,2); echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>Number of girls</td>
        <td align="right"><?php echo($no_girls); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($pop))$var=round($no_girls/$pop*100,2); echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>No of Emmas</td>
        <td align="right"><?php echo($no_emma); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($no_girls))$var=round($no_emma/$no_girls*100,2); echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>&nbsp;</td>
        <td align="right">&nbsp;</td>
        <td align="right">&nbsp;</td>
      </tr>
      <tr> 
        <td>Boy-Boy</td>
        <td align="right"><?php echo($boy_boy); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($count)){$var=round($boy_boy/$count*100,2);echo($var);} ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>Boy-Girl</td>
        <td align="right"><?php echo($boy_girl); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($count))$var=round($boy_girl/$count*100,2);echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>Girl-Boy</td>
        <td align="right"><?php echo($girl_boy); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($count))$var=round($girl_boy/$count*100,2);echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>Girl-Girl</td>
        <td align="right"><?php echo($girl_girl); ?>&nbsp;</td>
        <td align="right">
          <?php if (isset($count))$var=round($girl_girl/$count*100,2);echo($var); ?>&nbsp;
        </td>
      </tr>
      <tr> 
        <td>&nbsp;</td>
        <td align="right">&nbsp;</td>
        <td align="right">&nbsp;</td>
      </tr>
      <tr>
        <td>Emma matched with Boy</td>
        <td align="right"><?php echo($emma_boy); ?>&nbsp;</td>
        <td align="right"><?php if (isset($no_emma))$var=round($emma_boy/$no_emma*100,2); echo($var); ?>&nbsp;</td>
      </tr>
      <tr>
        <td>Emma matched with Girl</td>
        <td align="right"><?php echo($emma_girl); ?>&nbsp;</td>
        <td align="right"><?php if (isset($no_emma))$var=round($emma_girl/$no_emma*100,2); echo($var); ?>&nbsp;</td>
      </tr>
    </table>
    <?php
    if ($_POST[display_output]=="checked")
    {
    echo("<br><br><br><div align=\"center\"><b>Data output table</b><br></div>");
    echo("<table align=\"center\" border=\"1\">\n<tr>");
    $cols=12;//number of table columns
    for($i=1; $i<=$count; $i++)
        {
            print("<td>".$output[$i-1][0]."-".$output[$i-1][1]."</td>");
            if ($i<$count && $i%$cols==0){echo("</tr>\n<tr>");}
        }
    echo("</tr></table>");
    }
    ?>
    </body>
    </html>
    So, it does seem as if knowing the name of one girl doesn't make the reduce the increase the probability of her having another sister from 33% to 50% after all.
    .

  15. #45
    Hero member
    Join Date
    Jan 2008
    Location
    London
    Posts
    2,417

    Re: The daughter paradox

    I agree,

    In essence inheritance for each individual child is an independent event. So for any two offsrping related or unrelated the ratio of possible outcomes are 1 M+M, 2 M+F and 1 F+F. The order is immaterial.

    Now because we are dealing with a single family instead of describing the frequencies as above we describe the outcomes as MM, MF, FM, FF. We have made a fatal error - introducing the concept of order of events suggesting linkage that does not exist (MF = FM in terms of sex ratio).

    Having snuck in permutations instead of combinations by this back door we then compound the error by attribuiting the same property to daughters (D1 & D2). Thus we end up considering that and given daughter (named Emma) must be D1 or D2, if D1 then she has two chances of having a brother and one chance of having a sister, if D2 then she can only have a sister. Since she is either D1 or D2, then we incorrectly assume meeting them in the order D2 then D1 is different to D1 then D2 hence 2/4 chances of having a sister.

    The name is irrelevant, as long as the order is unknown.

Similar Threads

  1. Paradox of Nothingness short version
    By Alumno deVerum in forum Religion/Atheism/Mysticism/Philosophy
    Replies: 164
    Last Post: 10th March 2008, 11:40 PM
  2. Paradox Of Nothingness Long Version
    By Alumno deVerum in forum Religion/Atheism/Mysticism/Philosophy
    Replies: 7
    Last Post: 12th November 2007, 07:58 AM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •