# Couple has two children

In the second problem we are told a couple has two children, one of them a girl. Drachman then asks what the odds are the other child is a boy, assuming the biological odds of having a male or female child are equal. His answer: 2 in 3. How can the gender of one child affect the gender of another? It can’t. The answer is 1 in 2. --Adam Martin and Anna Davlantes, Evanston, Illinois

The second question is much the same. The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

If we asked the parents if they had a daughter that would be correct.

If they tell us they have a daughter then we need to know what they could have told us.

1A is a female
1B is a male
2A is a female
2B is a female
3A is a male
3B is a female
4A is a male
4B is a male

So if it is equally likely they will tell us about either child then we can eliminate 4 cases and see in the other 4 cases that 50 percent of the time they have a son.

I guess this is a comment on Let’s make a Deal.

I did read that but that is a different question.

That question has 16 different cases. The cases are 4 birth types times 2 children they were talking about times 2 child that showed up first.

Here is an example, I roll 2 dice. I look at the dice and tell you one of the die is even or odd. Is this a good one to one bet? If I only tell you odd when both are odd of course but if am equally likely to say odd or even when they are split then no.

If you assume that once I tell you odd that the odds of the other being odd are 2 to 3 and that once I tell you even that the odds are 2 to 3.

1 Die A is even Die B is odd you win
2 Die A is even Die B is even I win
3 Die A is odd Die B is odd you win
4 Die A is odd Die B is even I win

I didn’t fully explain cases 1 and 3 but reqardless of what I tell you. You are going to quess the opposite.

If oppose to me tell you if it is odd or even you ask me if it is odd or even then that changes the outcome. In that case you eliminate the chance that is both odd or both even without doubling the chance that it is the other.

The best way I can discrible what is happening is that the odds at the start are equally like for all 4 cases. When one case is eliminated the odds weren’t effected for cases 1 and 3 but the doubled for case 2.

The question is worded vaguely so that it can be resonably interpreted either way.

If your friend said: “I met a couple and one of their children, a girl; they have another child but I don’t know if it’s a girl or a boy,” then the 2 in 3 option disappears.

If your friend said: “I met a couple with two children; I asked if they had any girls and they said ‘yes’ but didn’t indicate how many,” then the 1 in 2 option disappears.

The question is only a puzzle because of its vague wording–if you state the question clearly the mystery disappears. In my opinion, the first example above is the common-sense interpretation of the question, which is the point: you ask it that way expecting to get a common-sense answer so you can say “WRONG! the correct answer is this non-intuitive one backed up by this proof.” In fact, a proof can be constructed for either interpretation.

Indeed it is, but the issue here is that the question you are addressing is also different from the original question, and different in the same way as the paragraph kk fusion alludes to. (Or, to be fair, is supposed to be different from the original question–the semantics can be misleading.)

There’s a subtle but real difference between the cases where the children are differentiated and when they’re not.

Whenever the children are differentiated in some way–in other words, when you are told that “the eldest child” is a girl, or “the first child to enter the door” is a girl, or “the child that I am telling you about” is a girl–then the odds that “the other” child is a boy is simply 1/2.

However, when the children are not differentiated (which, vagueness and semantics aside, is supposed to be the point of this question)–when all you know is that the couple has at least one girl, and that information is not tied to a specific child in any way-- then the odds of the couple also having a boy is 2/3.

(Note the subtle difference in wording here, which, to be fair, is not in Cecil’s original column. Since the information about “at least one girl” is not tied to a particular child, there is no “other child.”)

We are told a couple has two children, one of them a girl. Drachman then asks what the odds are the other child is a boy, assuming the biological odds of having a male or female child are equal?

We are told a couple has two children, one of them a boy. Drachman then asks what the odds are the other child is a girl, assuming the biological odds of having a male or female child are equal?

We are told a couple has two children, one of their sexes. Drachman then asks what the odds are the other child is the opposite sex, assuming the biological odds of having a male or female child are equal?

All 3 questions should have the same answer?

Think about a room with a 100 couples.

Yes.

I cannot prove the answer to the first 2 questions.

The last one is easy.

1 boy-girl
2 boy-boy
3 girl-boy
4 girl-girl

None of the four cases are removed. Two of the cases are same sex and two are opposite sex. So the answer is 1 to 2.

If all 3 questions have the same answer then the first question also is 1 to 2.

If you’re told the sex of one of the children, then you actually have to account for being told the sex of one of the children. Ignoring the semantical argument and assuming that the children are undifferentiated, either case 2 or case 4 will be removed, with equal probability.

Four couples are in a room. You ask all four to tell you the sex of one of there children and then quess the other is opposite each time. How many times are you going to be correct?

Presumably 50% of the time, as the question implies a differentiation of the children. As I said in my first post.

Can I ask where you’re going with this? The conversation is starting to go in circles, and you’re either trying to make some subtle point I’m missing, attempting to semantically parse the question in a way that forces the answer, or misunderstanding the statistics. If the first, please say it plainly. If the second, yes you’re right, as I already pointed out. If the third, I’m happy to explain, but you’ll need to specify why you think whatever you think differs from whatever Cecil thinks.

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)
Dear Cecil:

The answers to the logic questions submitted by Jordan Drachman were illogical. In the first problem he says there is an equal chance the card placed in a hat is either an ace of spades or a king of spades. An ace of spades is then added. Now a card is drawn from the hat–an ace of spades. Drachman asks what the odds are that the original card was an ace. Drawing a card does not affect the odds for the original card. They remain 1 in 2 that it was an ace, not 2 in 3 as stated.

In the second problem we are told a couple has two children, one of them a girl. Drachman then asks what the odds are the other child is a boy, assuming the biological odds of having a male or female child are equal. His answer: 2 in 3. How can the gender of one child affect the gender of another? It can’t. The answer is 1 in 2. --Adam Martin and Anna Davlantes, Evanston, Illinois

This is where we are going wrong I think

Cecil is correct and Adam Martin and Anna Davlantes of Evanston, Illinois are incorrect. This is, I think, easier to visualize with the card question, as it eliminates some of the semantical difficulties.

Yes it does, because you have eliminated the 1-in-4 case where the initial card was a king and you drew it.

Nope – 2 in 3.

Here’s a program in Ruby to illustrate it:

``````
def run_a_round
#Empty the hat
hat = []
#Randomly put an ace or a king into it
hat.push([:King, :Ace][rand(2)])
hat.push(:Ace)
#Draw one of the two cards in the hat
drawn = hat.delete_at(rand(2))
#If we drew a king, return nil. Otherwise, the card left in the hat
drawn == :King ? nil : hat[0]
end

aces, kings, drew_a_king = 0, 0, 0
100000.times do
result = run_a_round()
#We drew an ace, and the remaining card was an ace
if result == :Ace
aces += 1
#We drew an ace, and the remaining card was a king
elsif result == :King
kings += 1
#We drew a king -- discard this run
else
drew_a_king += 1
end
end
puts "Aces=#{aces}; Kings=#{kings}; Drew a king=#{drew_a_king}"

``````

If each couple is known to have exactly two children, and you ask each of the four couples, “Do you have at least one son?” – then (on average) three of the four will say, “Yes,” and (on average) two of those three will have a daughter, not two sons.

It depends on the population that you are sampling, and on the question you ask.

It’s always the implied level of simplicity, it ALWAYS has to do with the implications that are made.

Suzie had a baby. Suzie’s baby is a boy. Suzie is pregnant, there is a 50% chance that the next child is a boy.

Suzie and fred have 2 children. Their first child is a boy. Their second child has a 50% chance of being a boy.

Suzie and fred have 2 children. Today they are at school talking with the principle because Bobby (one of their children) got in a fight. Betty the babysitter is babysitting a child that is 2/3 chance a girl and 1/3 chance a boy.

Now start bringing in more variables.
makes up an arbitrary statistic to prove a point
98% of boys that get in fights at school have at least one brother

By nature of the fact that on a given day they have to go through this experience, the odds are drastically placed in the favor of it being a boy.

Double or Nothing.
you bet 5 dollars.
you have a 50% chance to have the oppurtunity to leave the table with 10 dollars.
25% to leave with 20
12.5 : 40
6.25 : 80
if you now have 80 dollars in the pot. You have a 50% chance to have the oppurtunity to walk away with 160 dollars.

My point is you will have a difficult time attributing human nature to statistics. There are always hundreds of assumptions. Who is there to prove there is anything but a goat behind all three doors? How do you know they aren’t moving them? Who’s to say all gameshows aren’t rigged? and Why would the host care if you win or lose, he is just supposed to make it entertaining, I dont think he gets to keep the ferrari if you lose. (I would hazard a guess that ratings go up when they are given away, and the shows dont pay full price for their gifts.)

The point of statistics was never to consider for human nature, you could endlessly throw curve balls at any scenario that involves another person making a choice based on knowledge of the result.

No, the odds are still 50-50. Any scenario which differentiates between the children gives you enough information to consider them individually. If you know the firstborn is a boy, or the child walking through the door first is a boy, or the child at school is a boy, or rhe child we’re discussing is a boy–all of those scenarios allow you to independently consider the sex of each child.

The point of statistics is to estimate what outcomes might be based on what you know. Clearly having additional information changes the answer to the situation–that’s sort of the point of the girl-boy problem. If the situation is not what you thought it was, then your statistical answer will likely be wildly off. That’s true, but not, I think, surprising to anyone.

This is not correct, because the act of drawing the card (and seeing what it is) has given you new information.

There is a 50/50 chance of the original card being a King or an Ace. After you add the new Ace and draw a card, there is a 50/50 chance you’ll draw the original card versus drawing the new card. These two events are independent, so there are potentially four outcomes, all having equal probability:
[ol]
[li]The original card was a King; you draw the original card.[/li][li]The original card was a King; you draw the new (Ace) card.[/li][li]The original card was an Ace; you draw the original card.[/li][li]The original card was an Ace; you draw the new card.[/li][/ol]
However, we are given that you in fact drew an Ace, which eliminates scenario #1. That leaves the other three scenarios, still equally probable, two of which have the original card being an Ace.

Here’s another way of looking at this. Imagine doing the following over and over again: put in an Ace, and draw a card at random. If it’s a King, you’re done. If it’s an Ace, put it in again and draw again. Repeat up to 100 times.

If you make it all the way through 100 trials and only ever see Aces, would you still conclude that the probability of the original card being an Ace is only 50% ?

If a couple has two children, there are four different possibilities:
1) The oldest is a girl, the youngest is a girl.
2) The oldest is a girl, the youngest is a boy.
3) The oldest is a boy, the youngest is a girl.
4) The oldest is a boy, the youngest is a boy.

If one of the children is a girl, possibility 4 gets taken away, leaving only three possibilities. Of those three, two options consist of having a boy, so the answer is two out of three.

If one of the children is a boy, possibility 1 gets taken away, leaving only three possibilities. Of those three, two options consist of having a girl, so the answer is two out of three.

Once you’ve been told the sex of one of the children, see the appropriate paragraph above.

Yes, you’re right. All three questions have the same answer, which is 2/3.

While this is correct, it doesn’t seem to make sense. Here’s why. Suppose you know the sex of the oldest child. So you meet a couple and they say their first-born was a girl, what are the chances of the youngest being a boy? In that case, you can get rid of options 3 and 4. Then there is a one in two chance of the younger being a boy. Similarly, suppose a different couple tells you that their youngest is a boy. What is the chance of the oldest being a girl? This time you can eliminate possibilities 1 and 3, leaving you with a one in two chance of the oldest being a boy. It may seem confusing, but it’s very important in the question whether you know the sex of a specific child, or just that of one of the kids, regardless of birth order.

Here’s a simpler way to think of it. You know that for two kids there’s a 25% chance of them both being boys, a 25% chance of them both being girls, and a 50% chance that there’s one of each. So there’s twice as many families with one of each. Now if somebody told you the sex of one of the kids and asked you to guess the other you could think “I know that there’s more families with one of each, so chances are the sex of the kid I don’t know is the opposite of the one I do know about.” And you’d be right!

This is the key - do you know the birth order of the one child, or just the gender?
Those are 2 separate pieces of information; each affects the odds, by eliminating one or more of the 4 possibilities. For example, if they tell you the oldest child is different gender from younger - you still don’t know anything (50-50 boy or girl) but you eliminate both scenarios boy-boy and girl-girl.

Similarly, if they say “one child is a girl, what are the odds the oldest is a boy?”. Now, the knowledge that one child is a girl is irrelevant to the scenarios. That may or may not be the oldest. So the odds are still 50-50? So if they say “one child is a boy, what are the odds the oldest is a boy?” - same odds?

Hmmm…