Two Children

**Xei** · 09-07-2009 07:37 PM

This'll kill you.

1. Mr. Jones has two children. The older of the two is a girl. What is the probability that both children are girls?

2. Mr. Smith has two children. At least one of his children is a girl. What is the probability that both children are girls?

**PhilosopherStoned** · 09-07-2009 07:49 PM

1. .5

2. .33

I'm playing because I don't know probability or statistics at all but that's the naive answer. Is there a trick?

**Xei** · 09-07-2009 07:53 PM

The trick is that 2. is ambiguous. It could be 1/2 or 1/3.

Mull it over. It took me quite a while to get to grips with it. Apparently senior mathematicians had huge rows over this, but it turned out to be because your interpretation changes the answer.

**PhilosopherStoned** · 09-07-2009 08:33 PM

I get it now. Cool.

**Licity** · 09-07-2009 11:17 PM

Originally Posted by Xei

This'll kill you.

1. Mr. Jones has two children. The older of the two is a girl. What is the probability that both children are girls?

2. Mr. Smith has two children. At least one of his children is a girl. What is the probability that both children are girls?

How would you have to interpret #2 to get 1/3 as an answer? I can't see any other way of looking at it initially, because it doesn't matter in the least which child is the girl.

**Photolysis** · 09-08-2009 02:54 AM

1. 1/2

2. 1/2

I can see the interpretation for #2 but using it is wrong; it doesn't matter which child is the girl, you only need to work out the chances of the other one being a girl too; the order is irrelevant. 1/3 is incorrect because this interpretation is erroneous.

How would you have to interpret #2 to get 1/3 as an answer? I can't see any other way of looking at it initially, because it doesn't matter in the least which child is the girl.

There are 3 possibilities:

The youngest is a girl, the oldest is a boy
The youngest is a boy, the oldest is a girl
The youngest and oldest are both girls.

The error however is assuming these are all equally likely; there are not 1/3rd. They are actually 1/4, 1/4 and 1/2 respectively.

If we look at the probabilities we can see why.

P(other child is a girl) = 0.5 = 2 girls
P(other child is a boy) = 0.5 = 1 girl, 1 boy

if we have one boy and one girl
P(first is a girl) = 0.5
P(first is a boy) = 0.5

Multiplying the probabilities we get the 3 outcomes:

P(both girls) = 0.5
P(1st girl, 2nd boy) 0.5x0.5 = 0.25
P(1st boy, 2nd girl) 0.5x0.5 = 0.25

We have to multiply the probabilities whilst boy/girl and girl/boy are equally likely, they are a 50% chance of a 50% chance, so 25%

I seem to recall that prominent mathematicians had problems with the "Monty Hall Problem" as well. Apparently some are just really bad at probability.

**Invader** · 09-08-2009 03:31 AM

50/50, for both. Age has no relevance to what the gender of the second child would be.

You know that one child is a girl in both cases. In order for the statement "Both
children are girls" to be true, you only need the other child to be a girl. The
probability of the other child's gender is equal to the probability that both children are
girls when you already know that one of them is a girl. What is the probability
of the second child being a girl? .5

Is something wrong with my reasoning?

**Xei** · 09-08-2009 07:24 AM

I can see the interpretation for #2 but using it is wrong; it doesn't matter which child is the girl, you only need to work out the chances of the other one being a girl too; the order is irrelevant. 1/3 is incorrect because this interpretation is erroneous.

Aaaactually you're wrong.

It depends on how you decide that there is at one least girl in the family. There are two distinct ways; either you look at one particular child, and if it is a girl, deduce that there is at least one girl in the family. Or, you look at both children, and if one or both are girls, you deduce that there is at least one girl in the family.

For the former, you have the following possibilities:

C N
G B
G G

C is the chosen child, N the not-chosen child. The probability of both being girls is hence 1/2.

For the latter, we have the following:

1 2
G G
G B
B G

1 is one child, 2 is the other. The probability of both being girls is now 1/3.

**Photolysis** · 09-08-2009 07:51 AM

Or, you look at both children, and if one or both are girls, you deduce that there is at least one girl in the family

Why would we be estimating the probability of something we know?

The whole idea is to work out the probability of something we don't know due to the limited information.

It's like asking "work out the chance that X will have a better poker hand than you"

"well, if we look at his cards..."

For the latter, we have the following:

1 2
G G
G B
B G

1 is one child, 2 is the other. The probability of both being girls is now 1/3.

As I said, this is incorrect; the probabilities are not equally likely. I explained why already.

**Xei** · 09-08-2009 07:57 AM

Originally Posted by Photolysis

Why would we be estimating the probability of something we know?

The whole idea is to work out the probability of something we don't know due to the limited information.

It's like asking "work out the chance that X will have a better poker hand than you"

"well, if we look at his cards..."

This is a pure mathematical problem. I mean in real life why would you only look at one child in the first place? Just look at both. It's not a real problem, it's just an example of a mathematical paradox.

As I said, this is incorrect; the probabilities are not equally likely. I explained why already.

No, each has 1/3 a chance of being selected.

Think about it. You analyse a family; does this family have one or two girls? If yes, then there are three distinct situations, equally likely, listed above.

Also I checked it on Wikipedia. :B

**Replicon** · 09-08-2009 08:30 AM

The 1/3 interpretation works like this:

The solution space (with girl = 1 and boy = 0) for two kids looks like this:

0 0
0 1
1 0
1 1

By saying "at least one is a girl" you're eliminating the "0 0" case, leaving three options. Out of those three, only one is the "both are girls" case.

**Xei** · 09-08-2009 08:33 AM

Yes exactly.

And in the other interpretation, you only look at, say, the left column, and choose the rows with a 1 in.

**Photolysis** · 09-08-2009 10:07 AM

Having thought about it some more, it seems you're right Xei and co.

If you know that a specific child is a girl, the other one has a 50% chance of also being a girl.

If you merely know that there is at least one girl, it's a 33% chance, since the amount of people with 1 girl and 1 boy is double that with 2 girls.

So, my apologies for that. Looks like I learnt something today :p

**Alric** · 09-08-2009 11:48 AM

If you only look at the second child, then both are 50%. Even if you look at the three possibilities, if you remove a girl from each then its 50%.

Girl Boy - remove the girl and you only got Boy
Girl Girl - remove the girl and you only got Girl
Boy Girl - remove the girl and you only got Boy

So the possibilities becomes.

Boy
Girl
Boy

You remove the second boy because its a repeat and then it is 50%. Either boy or girl.

It only becomes 1/3, if you are factoring in additional things, like the second girl that you already know is a girl.

**PhilosopherStoned** · 09-08-2009 11:57 AM

but you can't remove the second boy because it represents a unique state of the system under consideration. You're just translating the one interpretation into the other.

**Alric** · 09-08-2009 12:18 PM

If you know for a fact, that one is a girl then you are only looking at the second child. The second child only has two choices, either being a girl or a boy. There is no third choice when you look at it that, ever.

**slash112** · 09-08-2009 12:52 PM

Originally Posted by Xei

This'll kill you.

1. Mr. Jones has two children. The older of the two is a girl. What is the probability that both children are girls?

2. Mr. Smith has two children. At least one of his children is a girl. What is the probability that both children are girls?

1) there is no chance that they are both girls

2) 50% chance they are both girls

**Xei** · 09-08-2009 01:17 PM

Originally Posted by Photolysis

Having thought about it some more, it seems you're right Xei and co.

If you know that a specific child is a girl, the other one has a 50% chance of also being a girl.

If you merely know that there is at least one girl, it's a 33% chance, since the amount of people with 1 girl and 1 boy is double that with 2 girls.

So, my apologies for that. Looks like I learnt something today :p

Maybe you can empathise a little more with those mathematicians now, eh?

If you know for a fact, that one is a girl then you are only looking at the second child. The second child only has two choices, either being a girl or a boy. There is no third choice when you look at it that, ever.

Yes, and the situation changes when you look at both children when deciding there is at least one girl.

**PhilosopherStoned** · 09-08-2009 01:45 PM

Originally Posted by Alric

If you know for a fact, that one is a girl then you are only looking at the second child. The second child only has two choices, either being a girl or a boy. There is no third choice when you look at it that, ever.

Think of it like this. If you have every pair of siblings in the world such that one of them is a girl gathered for you to look at all at once, you will see that 1/3 of them are m-f, 1/3 of them are f-m and 1/3 are f-f.

The observation that one of them is a girl does nothing more then to tell you that the object that you are looking at is a member of the above set so it has 1/3 probability of being any of the three.

**Licity** · 09-08-2009 06:33 PM

Originally Posted by Xei

This'll kill you.

1. Mr. Jones has two children. The older of the two is a girl. What is the probability that both children are girls?

2. Mr. Smith has two children. At least one of his children is a girl. What is the probability that both children are girls?

I love problems like this. You think you get it, and then you get it a second time. Makes all the math worthwhile.

**Invader** · 09-09-2009 01:29 AM

I went ahead and proved the 1/3rd probability for scenario 2 via the use of a random number generator, simple text editing, and a word counter. Anyone else can repeat what I've done exactly with similar results.

http://www.random.org/integers/

Generate 10,000 random integers (for 5,000 pairs)
Each integer should have a value between 1 and 2 (to represent both genders).
Format in 2 columns.

Values 1 and 2 were used to represent 'girl' and 'boy' respectively.
Your list of number pairs should look similar to what's posted in the following spoiler tag. 5,000 pairs makes a long list:

Spoiler for 5,000 pairs:

Now, my text editor wouldn't count the number pairs, but it would at least highlight them. I replaced:
1 1 with GIRLPAIR
2 2 with BOYPAIR
1 2 with GIRLBOY
2 1 with BOYGIRL

Spoiler for New list will appear thusly:

This will allow a word counter to give you an accurate number for each pair listed.
I used http://www.writewords.org.uk/word_count.asp for this.

For this test I got the following results:

1 1 = 1235 pairs
2 2 = 1264 pairs
1 2 = 1290 pairs
2 1 = 1211 pairs

Each pair then has a 25% chance to occur (the numbers will vary slightly with the RNG, but with enough trials the deviation will become smaller and smaller). That essentially means that the following quote...

Originally Posted by PhilosopherStoned

If you have every pair of siblings in the world such that one of them is a girl gathered for you to look at all at once, you will see that 1/3 of them are m-f, 1/3 of them are f-m and 1/3 are f-f.

...is absolutely true.

PS's explanation made sense the first time over, but I felt it necessary to generate some
solid evidence. I'm still a bit baffled, and have a new sudden interest in statistics.

I hope that helps anyone else having trouble understanding.

**Photolysis** · 09-09-2009 02:38 AM

Maybe you can empathise a little more with those mathematicians now, eh?

Heh, I'm more bemused by the large error I made, considering I believe I'm particularly adept at probability. I'd even go as far to say that it's probably my second best branch of mathematics, outside of mental arithmetic.

It's really also psychological as well in this case, I'd venture, because the first question makes it very easy to hastily jump to the wrong conclusion, because the two questions appear so similar. I think I would have been less likely to fall in to that trap if it had been a stand alone question, at any rate.

I do still seem to recall that mathematicians struggled with the Monty Hall problem though. Or at least some of them did.

What is truly amazing though is the sheer amount of people who are incredibly bad at probability. Whether that be "if there are X outcomes, the probability of one must be 1/X", people who display the gambler's fallacy, people who think that events are not independent ("if you get 10 heads in a row, is heads or tails more likely for the 11th toss?"), or even people who think that the probability changes in situations like the Monty Hall problem.

**PhilosopherStoned** · 09-09-2009 06:44 AM

Originally Posted by Invader

I'm still a bit baffled, and have a new sudden interest in statistics.

Me too. I've always dismissed it as not being "real" math. I also thought that stuff like this only happened in foundations when choosing axioms. Probability/statistics seems very interesting. My lack of interest in it has kept me from making a proper study of quantum physics to get up to calculational capacity with it but I'm going to pick a good book on probability and statistics and inhale it when I get it. It's always so much easier to learn math when one has an actual interest in it, no?

What I wonder is under what conditions one would choose to use one or the other interpretation? I'll have to think about that.

**PhilosopherStoned** · 09-09-2009 10:34 AM

I think that I've figured out the fundamental difference between the two interpretations of the second question and what makes the two questions so different.

Originally Posted by Xei

1. Mr. Jones has two children. The older of the two is a girl. What is the probability that both children are girls?

In this one, we have the definite information that the older sibling is a girl. This allows us to only consider the younger one so the probability is reduced to the probability of one person being a girl. It could just as well say that the heavier one is a girl and the same inference would result. The key feature is that we can say "this" sibling is a girl, what is the probability that "that" one is?

Originally Posted by Xei

2. Mr. Smith has two children. At least one of his children is a girl. What is the probability that both children are girls?

Here we can regard the known girl as being "this" sibling and ask the probability of "that" sibling being a girl, in which case the problem reduces as above or we can take it to only say that "a" sibling is a girl, in which case there are three equiprobable ways that that could happen.

So the key distinction is that when I divide the siblings as "this" sibling and "that" sibling, in any way we see fit, it eliminates one of the possible ways in which one of the siblings could be a girl. Specifically, it eliminates the possibility in which "this" sibling is a boy and "that" sibling is a girl.

Does that make any sense? I'm sure somebody else could put it more simply.

**Xei** · 09-09-2009 12:07 PM

Yeah that sounds about right.

In the problems with answer 1/2, you determine that at least one child is a girl by looking at a specific child and asking if she is a girl.

In the problems with answer 1/3, you determine that at least one child is a girl by looking at both children.

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