# Probability Conundrum

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• Oct 1st 2010, 12:51 PM
Nath
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...
• Oct 1st 2010, 01:22 PM
undefined
Quote:

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.
• Oct 1st 2010, 01:38 PM
CaptainBlack
Quote:

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
• Oct 1st 2010, 03:54 PM
awkward
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.
• Oct 1st 2010, 05:06 PM
Quote:

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.
• Oct 2nd 2010, 08:31 AM
Nath
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
---
• Oct 2nd 2010, 01:16 PM
Quote:

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...... $\displaystyle \displaystyle\frac{\left(\frac{2}{4}\right)}{\left (\frac{3}{4}\right)}=\frac{2}{3}$
• Oct 3rd 2010, 05:11 AM
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.
• Oct 3rd 2010, 05:22 AM
undefined
Quote:

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.
• Oct 3rd 2010, 05:37 AM
CaptainBlack
Quote:

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
• Oct 3rd 2010, 05:57 AM
undefined
Quote:

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.
• Oct 3rd 2010, 06:16 AM
undefined

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?
• Oct 3rd 2010, 06:17 AM
Nath
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.
• Oct 3rd 2010, 06:31 AM
undefined
I give up trying to explain the same thing over and over. Flip a coin and see. You will recant.
• Oct 3rd 2010, 06:34 AM
Nath
Quote:

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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