Boy Girl Paradox

  • Thread starter Howers
  • Start date
  • #1
Howers
446
3
Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"
 

Answers and Replies

  • #2
davee123
668
4
Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

2/3

Options:
Older is a boy, younger is a boy - 1/4
Older is a girl, younger is a girl - 1/4
Older is a boy, younger is a girl - 1/4
Older is a girl, younger is a girl - 1/4

So, removing the 4th case from possibility, the odds of having two different gendered children, without knowing which is older is left as 2 of the 3 remaining possibilities. Hence, since each of the possibilities is equally probable, it's 2/3.


DaveE
 
  • #3
CompuChip
Science Advisor
Homework Helper
4,306
49
Though this might be counter-intuitive (which is why it is a famous problem) I don't see the paradox.
 
  • #4
humanino
2,461
8
Though this might be counter-intuitive (which is why it is a famous problem) I don't see the paradox.
I think there is another more-or-less equivalent problem formulated in a more interesting way.

A king of a certain distant country has decided that he wants more men in the population for military purposes. He thus decides of a new law enforcing that a couple is allowed to have only one girl. What is the boy/girl ratio resulting in the population ? (assuming we wait for several generations, after the last person born before the law has deceased, to get a stable asymptotic result for instance).

A family could have say 7 boys, the eighth kid being a girl preventing any further child in the family.
 
  • #5
davee123
668
4
A king of a certain distant country has decided that he wants more men in the population for military purposes. He thus decides of a new law enforcing that a couple is allowed to have only one girl. What is the boy/girl ratio resulting in the population ?

Interesting. I think you're looking for 50%.. But that's making some assumptions.

Technically, I think we need more information regarding how many kids the parents typically want. That is, parents may not WANT (or be able to have!) 6 children. So even though they might theoretically get 5 boys and then 1 girl, they may stop after (say) the 3rd son, and lower the ratio because they didn't keep procreating until they got to their natural limit.

DaveE
 
  • #6
humanino
2,461
8
But that's making some assumptions.
Good points. So let us assume that they make as many children as necessary for them to reach their first girl. Technically, this is crazy because there is a probability that they have one billion boys and then one girl. But this is a small probability :smile:
 
  • #7
Jimmy Snyder
1,031
19
Technically, I think we need more information regarding how many kids the parents typically want. That is, parents may not WANT (or be able to have!) 6 children. So even though they might theoretically get 5 boys and then 1 girl, they may stop after (say) the 3rd son, and lower the ratio because they didn't keep procreating until they got to their natural limit.
If parents voluntarily stop having children, it would have no effect on the ratio.
If parents who give birth on Tuesdays are not allowed to have more children, it would have no effect on the ratio.
If you pick parents at random and tell them that them must stop having children, it would have no effect on the ratio.
All of these things just reduce the number of people who are allowed to have more children. None of them has any effect on the ratio.
 
  • #8
Borek
Mentor
29,003
3,668
Code:
#include <stdlib.h>
#include <stdio.h>

typedef int BOOL;

int main(int argc, char* argv[])
{
  int iTotal = 0, iBoys = 0;

  while (1)
  {
    BOOL bBoy;
    do
    {
      bBoy = rand() % 2;
      if (bBoy) iBoys++;
      iTotal++;
    } while (bBoy);
    printf("%lf\n",double(iBoys)/iTotal);
  }

  return 0;
}

0.5 it is :wink:

Borek
 
Last edited by a moderator:
  • #9
chemisttree
Science Advisor
Homework Helper
Gold Member
3,688
691
Here is a very famous problem:

A random two-child family with at least one boy is chosen. What is the probability that it has a girl? An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"

That would depend on whether the parents have read Dr. Shettles book, "How to Choose the Sex of Your Baby." Success rates vary but it sounds like it is at least a testable theory.

http://www.fertilityfriend.com/Faqs/Gender-Selection-The-Shettles-Method.html

and here.
https://www.physicsforums.com/showthread.php?t=58126&highlight=Shettles
 
Last edited:
  • #10
CompuChip
Science Advisor
Homework Helper
4,306
49
Now let me post my argument (I don't know if it's correct):
The chance of a family having exactly n boys when the chance for a boy or a girl is 1/2 every time (as in, binomially distributed): [tex]\left( p_\mathrm{boy} \right)^n \cdot p_\mathrm{girl}[/tex].
So expectation value of the number of boys in the family, assuming every family will have 0, 1, 2, ... boys until they eventually get a girl:
[tex]E = \sum_{n = 0}^\infty n \left( \frac{1}{2} \right)^{n + 1} = 1[/tex]
So we expect every family to have 1 boy, before they get a girl. That means that 1/2 of all families, and therefore of the population, will be male.
 
Last edited:
  • #11
humanino
2,461
8
If parents voluntarily stop having children, it would have no effect on the ratio.
If parents who give birth on Tuesdays are not allowed to have more children, it would have no effect on the ratio.
If you pick parents at random and tell them that them must stop having children, it would have no effect on the ratio.
All of these things just reduce the number of people who are allowed to have more children. None of them has any effect on the ratio.
The easiness with which you answer problems sometimes suggests me that they must really appear trivial to you. In that case, I must say I am not fully satisfied with your answer, which seems to miss the point, although the final conclusion is correct.

There are several ways to answer more or less rigorously as always. A pedestrian way would be the following. The possible family configurations are obviously
(1) G
(2) BG
(3) BBG
(4) BBBG
...
(N) BBBBB...G (N boys)

It is not obviously trivial from this that the ratio 1:2 is conserved. It seems, and that is the initial goal of the king by issuing the law, that there will be more boys. It is of course very easy to count them and find out that it does not work.

The probability for line (N) to occur is (1/2)^N, so the probability weighted total number of individual in this table is
[tex]T=\sum_{n=1}^{\infty}\frac{n}{2^{n}}[/tex]
while the weighted total number of boys is
[tex]B=\sum_{n=2}^{\infty}\frac{n-1}{2^{n}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}[/tex]
from which we find the desired result, that the ratio will indeed be 1:2.
 
  • #12
Howers
446
3
Though this might be counter-intuitive (which is why it is a famous problem) I don't see the paradox.

Because the answer is cleary 1/2, but probability dictates 2/3. Which is it?
 
  • #13
Jimmy Snyder
1,031
19
Look at it this way. In the first year, every couple has a child. Half the kids are boys, half are girls. Now don't let the parents of girls have any more children. Of those parents that cannot have more children half (0) have boys and half (0) have girls. Of those parents that are allowed to have children half have boys and half have girls. Keep on doing this. Every year, the same number of boys are being born as girls. The only thing that was accomplished is that fewer and fewer people are allowed to have children. How this increases the number of boys is a mystery known only to the king.
 
  • #14
Jimmy Snyder
1,031
19
Because the answer is cleary 1/2, but probability dictates 2/3. Which is it?
I think this answer belongs in a different thread. In this problem, as in the other one, probability gives the correct answer. Math, it works *itches.
 
  • #15
Defennder
Homework Helper
2,591
5
Look at it this way. In the first year, every couple has a child. Half the kids are boys, half are girls. Now don't let the parents of girls have any more children. Of those parents that cannot have more children half (0) have boys and half (0) have girls. Of those parents that are allowed to have children half have boys and half have girls. Keep on doing this. Every year, the same number of boys are being born as girls. The only thing that was accomplished is that fewer and fewer people are allowed to have children. How this increases the number of boys is a mystery known only to the king.
That's a great way to look at it.
 
  • #16
Howers
446
3
Look at it this way. In the first year, every couple has a child. Half the kids are boys, half are girls. Now don't let the parents of girls have any more children. Of those parents that cannot have more children half (0) have boys and half (0) have girls. Of those parents that are allowed to have children half have boys and half have girls. Keep on doing this. Every year, the same number of boys are being born as girls. The only thing that was accomplished is that fewer and fewer people are allowed to have children. How this increases the number of boys is a mystery known only to the king.

This is not the same. Here you are preventing GG while allowing BB. If I called you and said of my two kids one is a girl, are you telling me that you'd need to flip 3 coins to decide what the other one was?

Math works. But are we applying it correctly?
 
  • #17
Jimmy Snyder
1,031
19
This is not the same. Here you are preventing GG while allowing BB. If I called you and said of my two kids one is a girl, are you telling me that you'd need to flip 3 coins to decide what the other one was?

Math works. But are we applying it correctly?
Sorry, I was answering a different question. I agree, that among families with two children, not both girls, and one of them a boy, 2/3 have a girl. Indeed, the 'one of them a boy' part is meant to confuse you. If they are not both girls, then one of them is a boy. The question I was answering was whether you could get more boys by allowing only certain people to have babies.
 
Last edited:
  • #18
Defennder
Homework Helper
2,591
5
You know, I was thinking about the puzzle of the king's decree that no family can have more than 1 girl and found it quite puzzling that when you look at each family you'll notice that they either have the same number of boys and girls, or more boys than girls, but never more girls than boys. This strongly suggests that if we add up all the families together and if within each discrete family unit, there are never more girls than boys but may well have , shouldn't we expect the ratio to be tilted in favour of males? But of course this isn't the case, as we can see from the above.
 
  • #19
Borek
Mentor
29,003
3,668
but never more girls than boys.

False - exactly 50% families have more girls then boys.
 
  • #20
vanesch
Staff Emeritus
Science Advisor
Gold Member
5,092
18
False - exactly 50% families have more girls then boys.

Yup, that's the trick! Half of the families started out with a girl, had to stop, and so only had a girl and no boy. Maybe their second would have been a boy, but they weren't allowed to have it.

The king should have allowed all families to procreate indefinitely, but order to kill all new-born girls that weren't the first in the family. Then the production rate would still be 50/50, but as only girls are killed, and not boys, this tilts over the balance.

The essential reason is that each birth is a statistically independent draw from all other outcomes, and each has a probability 50% to be a boy, and 50% to be a girl. So no matter how you organize these drawings (per family, when they can have more or not), their average will always be 50/50.
 
Last edited:
  • #21
Defennder
Homework Helper
2,591
5
Yeah I overlooked those with one female child. A mistake.
 
  • #22
shamrock5585
199
0
seeing as we are looking at families with atleast one boy... the probability is only dependant on the second child. 50%
 
  • #23
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
seeing as we are looking at families with atleast one boy... the probability is only dependant on the second child. 50%

This is in answer to the original problem, I presume. The answer to that problem is 2/3. See the answer in the second post. It has hidden text, which you can expose by dragging your mouse over what appears to be white space. You are mistakenly applying the statistical principle of indifference to a situation where it does not apply.
 
  • #24
shamrock5585
199
0
ok... i see your reasoning but im looking at it from a different point of view... check my reasoning... you have a 50% shot at getting a boy or a girl... so your chances of getting a boy boy family is .25 (.5 * .5) your chances of having a boy girl family are .25 (.5 * .5) so your chances of having a boy boy family and your chances of having a boy girl family are the same so it doesnt matter if you have an older boy younger girl or older girl younger boy. you will have equal number of boy boy and boy girl families. so now you only have 2 choices (50%)
 
  • #25
NeoDevin
306
2
Your probabilities don't add up right. There is a .25 chance of boy boy, and a .25 chance of girl girl, and a .5 chance of boy girl. There is twice the chance of getting boy girl, than boy boy. Eliminating the chance for girl girl does not change this ratio.
 
  • #26
Green Zach
86
0
two children. 100% chance of one boy. 50% chance of another boy, 50% chance of a girl. if one boy is a constant than that would just be like saying that it is raining outside all the time and their is a 50 50 chance i will carry an umbrella. its always raining so i will always have a 50% chance of carrying an umbrella much like their will be a 50% chance that there will be a girl. even case to case this should work therefore there is a 50% chance of a girl.
 
  • #27
shamrock5585
199
0
Your probabilities don't add up right. There is a .25 chance of boy boy, and a .25 chance of girl girl, and a .5 chance of boy girl. There is twice the chance of getting boy girl, than boy boy. Eliminating the chance for girl girl does not change this ratio.

the probabilities do add up if you count your chances of a boy girl family or a girl boy family... theres a .25 chance of both... then a .25 chance boy boy and .25 girl girl
 
  • #28
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
shamrock, you are making a classical statistical error: You are assuming two events are equiprobable when in fact they are not. This is a very common mistake, and the "paradox" here hinges on people making this classical error. There is no paradox; the odds are indeed 2/3 that the other child is a girl.
 
  • #29
shamrock5585
199
0
i think im just confusing myself here because i can see both yours and green zach's answers as being correct
 
  • #30
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
i think im just confusing myself here because i can see both yours and green zach's answers as being correct

Green Zach fell into the trap of assuming equiprobable events:
two children. 100% chance of one boy. 50% chance of another boy, 50% chance of a girl.
The assumption of equiprobability is just that, an assumption. Whenever one applies this assumption one should also do some sanity check regarding the validity of making this assumption. The assumption makes sense in the case of a coin toss or the roll of a single die. The assumption does not make sense and is not valid when one looks at the problem of the sum of the rolls of two dice. There are 11 outcomes, the numbers 2 to 12. However, the 11 events are not equiprobable. The probability of rolling a seven is six times that of rolling snake eyes. Similarly, the assumption simply is not valid in the problem at hand.
 
  • #31
regor60
101
0
Assumption of equal probability: the sun will rise tomorrow or it won't, therefore there is a 50% chance the sun won't rise...
 
  • #32
CompuChip
Science Advisor
Homework Helper
4,306
49
regor, that argument will work with anything.
If I throw a die, I will get 6 or I won't. Therefore, the probability of getting 6 is 50%.
I don't think I'd agree :)
 
  • #33
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
Exactly. Regor gave an example where the assumption of equiprobability is obviously invalid. The same applies to a the throw of a single die.
 
  • #34
shamrock5585
199
0
Im pretty sure that is what he was implying
 
  • #35
Green Zach
86
0
ok i think i understand where i went wrong so tell me if my new idea makes sence. their is a 25% chance of getting girl girl, 25% chance of getting boy boy (it would be like getting lucky twice that's why its 0.25 but were all smart here) and 50% chance of getting boy girl (the most probible outcome... just like if you flip a coin 20 times, its most likeley that you will get around 10 on one side and 10 on the other.).

so now we have boy boy and girl girl at 0.25 and boy girl at 0.5. becusae the families must have at least one boy, girl girl is eliminated. now the numbers dont add up to 1.0 becsaue one of the variables was eliminated. Now the probibilities are out of three. 1/3 = 0.25 or 25%.

therefore becasue we are left with 0.25 boy boy and 0.5 girl boy, that translates into a 1/3 chance of getting boy boy and 2/3 chance of getting boy girl.

please tell me if this makes any sence lol
 

Suggested for: Boy Girl Paradox

  • Last Post
Replies
1
Views
776
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
6K
  • Last Post
2
Replies
45
Views
4K
  • Last Post
3
Replies
97
Views
13K
  • Last Post
Replies
4
Views
2K
Top