1. ## 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...

2. Originally Posted by Nath
"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.

3. Originally Posted by Nath
"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

4. 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.

5. Originally Posted by Nath
"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.

6. ## 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
---

7. Originally Posted by Nath
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...... $\displaystyle P(A)=\frac{1}{4}$

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

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

Also note that the conditional probability of the 2nd child being a boy is...... $\displaystyle \displaystyle\frac{\left(\frac{2}{4}\right)}{\left (\frac{3}{4}\right)}=\frac{2}{3}$

8. 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.

9. Originally Posted by Nath
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, $\displaystyle D=B\cup C$)

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.

10. Originally Posted by Nath
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

11. Originally Posted by CaptainBlack
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.

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?

13. 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.

14. I give up trying to explain the same thing over and over. Flip a coin and see. You will recant.

15. Originally Posted by undefined
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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