[OT] mathematical query

apsuman

First Post
A man and a woman meet for the first time.

After some conversation the find out that they both have two children. The man shows the woman a picture of him with a son on a vacation.

Two questions:

A. What is the probability of the woman having children of two sexes (that is one boy and one girl)?

B. What is the probability of the man having children of two sexes?

g!

Edit: Yeah this is OT, sorry :).
 
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apsuman said:
A man and a woman meet for the first time.

After some conversation the find out that they both have two children. The man shows the woman a picture of him with a son on a vacation.

Two questions:

A. What is the probability of the woman having children of two sexes (that is one boy and one girl)?

B. What is the probability of the man having children of two sexes?

g!

Edit: Yeah this is OT, sorry :).

50% for both. Cause that may not be his son in the picture.
 

Re: Re: [OT] mathematical query

smetzger said:


50% for both. Cause that may not be his son in the picture.

I knew I would miss something. For purposes of this question, assume that the son pictured with the man is indeed, his.

I hope that clears it up.

g!
 

Simple:

For each of the man and the woman there are four possibilities, with equal probability of each occurring (assuming that there is truly a 50% probability that a given child is born Male or Female).

For the woman:
1.) MM
2.) MF (older is male, younger is female)
3.) FM (older is female, younger is male)
4.) FF

That's simple - there are two combinations of "one of each" and two combinations of "both the same sex" so the odds are 50-50 that she has "one of each."

So the answer to A is "50%".

For the man:
1.) MM
2.) MF (older is male, younger is female)
3.) FM (older is female, younger is male)
4.) FF

The fact that he shows a picture of him with a son eliminates possibility four, leaving us with...

1.) MM
2.) MF
3.) FM

All of these have equal probability, so there is a 67% chance that the man has children of both sexes.

It does NOT change the probability to "we know child 1 is male, whether he is older or younger, hence child 2 has a 50-50 chance of being male or female and hence there is a 50% chance of him having one of each" (a common mistake). We're assuming the picture gives us more information than it does... if he had said, "this is my oldest child and is my son" it would have eliminated possibility 3, thus lending a 50-50 chance.

THAT'S why the answer to B is 67%.

--The Sigil
 


die_kluge said:
a 50%
b 0% - men can't have children *smirk*
Men can HAVE children - they just can't BEAR children.. or birth them. ;)

I have a son. My wife bore and birthed my son.

Or are we arguing the definition of "having" as to genitive sense (he is my son because I am his father) vs. euphemistic sense (we say a woman "has" a child when she births a child)?

;)

--The Sigil
 


The Sigil said:
Simple:

It does NOT change the probability to "we know child 1 is male, whether he is older or younger, hence child 2 has a 50-50 chance of being male or female and hence there is a 50% chance of him having one of each" (a common mistake). We're assuming the picture gives us more information than it does... if he had said, "this is my oldest child and is my son" it would have eliminated possibility 3, thus lending a 50-50 chance.

THAT'S why the answer to B is 67%.

--The Sigil

The "common mistake" here is actually correct. Think about it this way, would you consider betting $2 to win $1 that the man's other child was male a fair bet? Because that's what the above reasoning implies. In these sorts of decisions how you update your probabilities based on new information is pretty important; in some decision theory problems there isn't even a good solution.

In this one though we can appeal to the brute fact that generally there's a 50/50 chance of any given child being male.

Also, consider a similar example where someone asks you what's the chance of two coin flips coming up the same. Should you revise your probabilities after the first coin flip?

edit: to fix embarassing math mistake where I said $3 to win $2
 
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For the second case break it up into two cases of equal probability.

a) It is the older child. 50% of the younger being male too.
b) It is the younger child, 50% of the older being male too.
 

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