What is the probability that a random two-child family has at least one girl?

In summary, the problem is asking for the probability of a family having a girl when there is at least one boy, which is equivalent to the probability of two randomly chosen children being of different genders, excluding the case of two girls. The answer is 2/3, as there are three possibilities and only one of them results in two children of different genders. Additionally, the discussion about the boy/girl ratio in a population is dependent on factors such as the number of children parents want and potential methods for choosing the sex of a child, but assuming a 50% chance for each gender, the expected number of boys in a family before a girl is born is 1.
  • #1
Howers
447
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?"
 
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  • #2
Howers said:
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
Though this might be counter-intuitive (which is why it is a famous problem) I don't see the paradox.
 
  • #4
CompuChip said:
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
humanino said:
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
davee123 said:
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
davee123 said:
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
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
 
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  • #9
Howers said:
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
 
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  • #10
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.
 
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  • #11
jimmysnyder said:
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
CompuChip said:
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
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
Howers said:
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
jimmysnyder said:
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
jimmysnyder said:
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
Howers said:
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.
 
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  • #18
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
Defennder said:
but never more girls than boys.

False - exactly 50% families have more girls then boys.
 
  • #20
Borek said:
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.
 
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  • #21
Yeah I overlooked those with one female child. A mistake.
 
  • #22
seeing as we are looking at families with atleast one boy... the probability is only dependant on the second child. 50%
 
  • #23
shamrock5585 said:
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
ok... i see your reasoning but I am 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 doesn't 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
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
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
NeoDevin said:
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... there's a .25 chance of both... then a .25 chance boy boy and .25 girl girl
 
  • #28
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
i think I am just confusing myself here because i can see both yours and green zach's answers as being correct
 
  • #30
shamrock5585 said:
i think I am 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:
Green Zach said:
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
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
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
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
Im pretty sure that is what he was implying
 
  • #35
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 don't 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 sense lol
 
<h2>1. What is the probability of having at least one girl in a two-child family?</h2><p>The probability of having at least one girl in a two-child family is 3/4 or 75%. This is because there are four possible outcomes for a two-child family: two boys, two girls, a boy and a girl, and a girl and a boy. Out of these four outcomes, only one has no girls, so the probability of having at least one girl is 3/4.</p><h2>2. Does the gender of the first child affect the probability of having at least one girl in a two-child family?</h2><p>No, the gender of the first child does not affect the probability. Each child has an equal chance of being a boy or a girl, so the probability of having at least one girl is the same regardless of which child is born first.</p><h2>3. How does the probability change if the family has more than two children?</h2><p>The probability of having at least one girl increases as the number of children in the family increases. For a three-child family, the probability is 7/8 or 87.5%. For a four-child family, the probability is 15/16 or 93.75%. This trend continues as the number of children increases.</p><h2>4. Is the probability affected by the parents' gender?</h2><p>No, the probability is not affected by the parents' gender. The probability of having at least one girl in a two-child family is solely based on the genders of the two children, not the genders of the parents.</p><h2>5. Can the probability be calculated for families with more than two children?</h2><p>Yes, the probability can be calculated for families with any number of children. It follows a similar pattern as the two-child family, where the probability increases as the number of children increases. However, the calculations become more complex as the number of children increases, so it may be easier to use a probability calculator or chart for larger families.</p>

1. What is the probability of having at least one girl in a two-child family?

The probability of having at least one girl in a two-child family is 3/4 or 75%. This is because there are four possible outcomes for a two-child family: two boys, two girls, a boy and a girl, and a girl and a boy. Out of these four outcomes, only one has no girls, so the probability of having at least one girl is 3/4.

2. Does the gender of the first child affect the probability of having at least one girl in a two-child family?

No, the gender of the first child does not affect the probability. Each child has an equal chance of being a boy or a girl, so the probability of having at least one girl is the same regardless of which child is born first.

3. How does the probability change if the family has more than two children?

The probability of having at least one girl increases as the number of children in the family increases. For a three-child family, the probability is 7/8 or 87.5%. For a four-child family, the probability is 15/16 or 93.75%. This trend continues as the number of children increases.

4. Is the probability affected by the parents' gender?

No, the probability is not affected by the parents' gender. The probability of having at least one girl in a two-child family is solely based on the genders of the two children, not the genders of the parents.

5. Can the probability be calculated for families with more than two children?

Yes, the probability can be calculated for families with any number of children. It follows a similar pattern as the two-child family, where the probability increases as the number of children increases. However, the calculations become more complex as the number of children increases, so it may be easier to use a probability calculator or chart for larger families.

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