1. Jun 26, 2004

### plover

I found this http://home1.gte.net/deleyd/random/probprdx.html#CHILD a day or two ago, but I don't think I believe the reasoning used in the solution that's given.

There are four situations involving a family with 2 children. In each case, a statement is made about one of the children, and the question is "What is the probability that the other child is a girl?"* (It is assumed that, in general, boys and girls are equally probable, and that the probability of each child is independent.)

The four cases are:
1. One child is a younger daughter named Mary.
2. One child is an older daughter named Mary.
3. One child is a daughter.
4. One child is a daughter named Mary.
The first 2 are straight-forward: the probability is 1/2.

For the third one, the answer given is 1/3. The reasoning is that there are three cases: boy-girl, girl-boy, girl-girl (the order indicating relative age).

Here is the answer the page author gives for the fourth one:
It depends on the probability of the mother naming both her children Mary. If she names all her children Mary then knowing one of them is named Mary doesn't help us and the answer as you know from question 3 is 1/3. If she names only one child Mary, then this uniquely identifies the child and the probability is 1/2. That is, there are two mutually exclusive possibilities: Mary is the older child or Mary is the younger child. In either case the probability of the other child being a girl is 1/2.​

Now I don't think I accept the reasoning on case 3 (I don't accept case 4 as it stands either, but that's because of the problem with 3).

It seems to me that, for case 3, any reference to the relative ages of the children is irrelevant and the probability would still be 1/2.

Suppose case 3 were stated thus:
You meet a mother and daughter and they tell you there is one other child in the family. What is the chance that the other child is a girl?​
Why would this be different from case three?

Or to ask the question differently, isn't this an essentially identical problem:
There are two identical decks of cards. Out of your sight, someone draws a card from each deck, places them face down and removes the decks. You turn over one card and it's the 3 of diamonds, what is the chance that the other is a 3 of diamonds?​

To translate the answer to case 4 into the card scenario, it seems that the author is saying that the answer would change depending on whether or not you could tell the decks apart, which would make no sense, yes?

In order to get a probability of 1/3, you would need a question like:
A family has 2 children, one of which is named Mary (and the parents are stodgy types who wouldn't name a boy Mary; they also confuse easily - so they don't give their children identical names). You meet a girl from the family, what is the chance her name is not Mary?​

Does what I'm saying sound correct, or did I miss something?

* This sentence was edited for clarity. It used to read:
'In each case, some information is given about one child, and the question is "What is the probability that the second child is a girl?"'
The phrasing of the four statements was changed also - each used to start "There is..." rather than "One child is..."

Last edited by a moderator: Apr 21, 2017
2. Jun 27, 2004

### loseyourname

Staff Emeritus
I'm not going to address problem four, because you seemed to indicate that if your issues with problem three are resolved, then you don't have any issues with problem four, so here goes.

The case you stated is different from the case given by the problem. You simply state that one child is a daughter, then want to know the probability that the other child will be a daughter. The question itself asked what the probability of the second child being a daughter is. Now you don't know if the daughter identified is the first or second child. There are three possible sequences of children given the information you have (that one is a daughter). They run exactly as you quoted from the solution:

boy-girl
girl-boy
girl-girl

There is, of course, a fourth possible sequence if no information is given, boy-boy, but this can be eliminated because you know one child is a daughter.

So given that you now have three possibilities, you need only check to see how many include a daughter as the second child. As you can clearly see, the answer is 2, so the probability of the second child being a daughter, given that you know at least one child is a daughter, is 2/3. I have no idea how the text came to the conclusion that the answer is 1/3. This can be shown a little more formally as well.

Let P(A) = the probability that either child will be a daughter, where event A is at least one child being born a daughter.
Let P(B) = the probability that the second child be a daughter, where event B is the second child being born a daughter.

Let us again list out all the possible sequences of children:

girl-boy
boy-girl
boy-boy
girl-girl

We can see that P(A) = 3/4 and P(B) = 1/2

The conditional probability that B will occur, given that A has already occured, can be calculate thus:

$$P(B\vert{A}) = \frac{P(A and B)}{P(A)}$$

We can see that P(A and B), the probability that either child will be a daughter and the second child will be a daughter, is 1/2, the same as P(A), either child being born a daughter, so that

$$\frac{P(A and B)}{P(A)}$$ => $$\frac{.5}{.75} = .\overline{6}$$

or 2/3.

Last edited: Jun 27, 2004
3. Jun 27, 2004

### Hurkyl

Staff Emeritus
Well, case 4, in general, is a very delicate question; the answer could range anywhere from 0 to 1 depending on the tendancies of the parents.

4. Jun 28, 2004

### plover

Doh!

First of all, thank you for taking the time to answer this.

Ooh ick, I seem to have misled you. The solution you give analyzes an entirely plausible reading of what I wrote, which, however, was not the reading I intended. I thought it was best to restate the problem in my own words - mostly to avoid snatching excessive quotage off the web site where I found the problem - but apparently I was not careful enough.

The problem seems to lie in my phrasing of the question:
"What is the probability that the second child is a girl?"​
which, if I'm reading correctly, you took to mean:
"What is the probability that the younger (i.e. second-born) child is a girl?"​
whereas the meaning I intended was:
"Given statement X about one of the children, what is the probability that the other child (i.e. the one not mentioned in statement X) is a girl?"​

This ambiguity does not seem to be present in the original wording. Here are cases 3 and 4 as given by Deley (the author of the page where I found this):

I'll edit my statement in the original post to remove this ambiguity. Hopefully, it will better reflect my intent - and my various analogies will make more sense.

For the problem that you solve, I follow and agree with the reasoning just fine.

I am, however, left with my original question - am I justified in disagreeing with Deley's solution? Reading your analysis is at least reassuring that I'm thinking about the problem in the right way.

Last edited: Jul 4, 2004
5. Jun 28, 2004

### NateTG

Try looking at it backwards

Think of it this way:

Let's say that you have a population, for example, the entire US, and you compile a list of all of the families that have two childrent, and for conveniece, let's assume that there are 1,000,000 such familes.

So, let's take a look at the possibilities:

older brother younger brother
older brother younger sister
older sister younger brother
older sister younger sister

Clearly, we would expect each of these four possibilities to have about 250,000 occurances.

Now, let's take a look at the situations:

Question 1:
The answer given for this one is incorrect:
If the probability of the second daughter's names is independant (which is not true) then the probability is one half.
Consider that of the four possibilities listed above, the requirement that the younger sibling is a daughter leaves only:
older sister younger sister
and
older brother younger sister

Since there is insufficient information to identify how the name restriction applies, it's hard to move on from here.

However, parents are unlikely to name both of their children the same name. Let's say that each time that parents name a female child the probaility is 1% that the name will be mary if none of the siblings have the name mary, and zero otherwise then we have the following numbers:
Older brother, younger sister 2,500 familiess
Older sister, younger sister 2,4250 families

This is a simplification, but it's clear that the naming tendencies can affect the probabilities, and I would be inclined to say that the odds that the other child is a girl are slightly less than 1/2.

Question 2:
Since the naming of the second child does not affect the naming of the second, this is probably 1/2, but it still depends on naming probabilities which are not specified.

Question 3:
This is the one that you have trouble whith, so consider the possibilities:
Older brother, younger sister 250,000
Older sister, younger brother 250,000
Older sister, younger sister 250,000
so the sample indicates that the chance that the other child is a girl is indeed 1/3.

Question 4:

Assuming a 1% chance of females being named mary we have:
Older brother, younger sister 2,500
Older sister, younger brother 2,500
Older sister, younger sister 4,975

(The result is the same if we assume that both siblings cannoth be named Mary).

Of the four questions, the only one that you have sufficient information to answer is the third one. The three other questions require assumptions about how the children are named.

6. Jun 28, 2004

### AKG

I don't understand 4. I don't see why knowing the name of one of the daughter's helps. The reasoning given says that we have a uniquely identified daughter, Mary, so either mary is the oldest or youngest, either way leaving a 1/2 probability that the other is a girl. But for #3, it says, "there is a daughter." Now, this daughter must have a name, let's call her G. We know that if there is a daughter, and let her be G, then G is either the oldest or youngest, and so we can use the same reasoning to conclude that the probability that the other is a girl is 1/2.

7. Jun 28, 2004

### NateTG

In the post above, I briefly discussed the issues associated with naming the girl. In practice, there is insufficient information for gauging the effect of giving names on the problem -- for example, it's possible that only the oldest daughter is ever named Mary.

The simplest assumption to make is that every girl that is born has some non-zero chance of being named Mary. This is unrealistic because two sisters rarely have the same name, but if we proceed with that assumption we can observe some skew:

Let's say that, independant of all other factors, the chance that a girl is named mary is $$0<p<1$$ (I used $$<$$ instead of $$\leq$$ there are at least some girls named Mary, and at least some girls not named Mary.), and that boys are never named Mary.

Then the chance that a family with a daughter and a son have a daughter named mary is also p. The chance that a family with two girls has a daughter named mary is $$2p-p^2$$ since each of the daughters might be named Mary.

So, of families with two children:
$$\frac{1}{4}$$ will have two daughters
$$\frac{1}{4}$$ will have two sons
and
$$\frac{2}{4}$$ will have a son and a daughter.

So there will be $$\frac{3}{4}$$ with at least one daughter:
$$\frac{2}{4}$$ with a daughter and a son
and
$$\frac{1}{4}$$ with two daughters

Now, using the $$p$$ above we get that:
there are
$$\frac{2}{4} \times p$$ families with a son and a daughter named mary
and
$$\frac{1}{4} \times (2p-p^2)=\frac{2}{4}p(1 - \frac{p}{2})$$ familes with two daughter, at least one of which is named mary.

Now, the probability of the other child being a girl is:
$$\frac{\frac{2}{4}p(1 - \frac{p}{2})}{\frac{2}{4}p(1 - \frac{p}{2})+\frac{2}{4}p}$$
The $$\frac{2}{4}p$$ cancels nicely:
$$\frac{1-\frac{p}{2}}{(1-\frac{p}{2})+1}$$
which simplifies to:
$$\frac{2-p}{4-p}$$
So it's relatively easy to see that as the probability of a family with a dauther named girl being named mary goes from $$0$$ to $$1$$, the probability of a family with a daughter named mary having a son goes from $$\frac{1}{2}$$ to $$\frac{1}{3}$$

Last edited: Jun 28, 2004
8. Jun 29, 2004

### Alkatran

Maybe I can help here:

You've already heard about how there is GB, BG, GG, BB. Well, consider teh statement:

One is a girl.

Does it disprove BG? No. GB? No. BB? yes. GG? no.

There is only one permutation that is no longer plausible.

Or look at it this way: if you say one is a girl you don't know if both are girls, but you know that both of them aren't boys.

If you know there are no boys, it makes more sense to assume 1/3 chance of having two girls, no? But both statements are really the same thing!

9. Jun 30, 2004

### WannabeG133

Couldn't the fourth one be 1/3 too? It never says whether she is an older or younger child, so you could have BG, GG, or GB. It's been a while since I've had one of these kinds of problems.

10. Jul 2, 2004

### plover

Thanks for the input everyone.

I want to restate a lot of stuff here in order to make clearer some of the questions that were only implicit in my original post.

First of all, I should say that I no longer see any reason to doubt that the answer to case #3 is 1/3, so that issue is not behind what follows.

Second, I thought this statement was an important point:
This highlights the importance of examining underlying assumptions when solving these problems for real world populations as opposed to hypothetical samples that can be constructed to order.

However, insomuch as it is possible to separate the two, the concepts I'm trying to clarify for myself are questions of abstract probability theory, not statistical methodology. I suspect that this is why my instinct in the original post was to restate the problem in terms of cards.

One thing I'm still curious about is whether any meaningful distinction can be made between the three different statements of the problem:

3a - [Deley's original form]
A mother has two children. One of them is a daughter. What is the probability that the other child is a girl?​

3b - [My first form]
There are four situations involving a family with 2 children. In each case, a statement is made about one of the children, and the question is "What is the probability that the other child is a girl?"
Statement #3: One child is a daughter.​

3c - [My alternate form]
You meet a mother and daughter and they tell you there is one other child in the family. What is the chance that the other child is a girl?​

My current opinion is that these three forms can all be analyzed the same way. I realize that this may seem obvious, but I was wrong earlier, so it's good to have a second view.

I also think that my original card example functions as an analogous situation (albeit one that would require there to be children of 52 different genders. :surprise: ):
There are two identical (but separate*) decks of cards. Out of your sight, someone draws a card from each deck, places them face down and removes the decks. You turn over one card and it's the 3 of diamonds, what is the chance that the other is a 3 of diamonds?​
If I'm following the reasoning on the original case #3 correctly, the answer to this one would seem to be 1/103 (as opposed to the 1/52 that my earlier interpretation would have suggested). I.e. there are 103 pairs of cards that would contain at least one 3 of diamonds, but only one of those pairs contains two 3s of diamonds.

Well, there are a still a few more points I want to get to, but I'll post what I have for the moment and get back to this later.

Last edited: Jul 2, 2004
11. Jul 2, 2004

### Alkatran

The card analogy won't work because taking a six of spades lowers the probability of getting another six of spades. Having a daughter doesn't necessarily lower the probability of having another daughter.

12. Jul 2, 2004

### plover

That's why there are two separate decks of cards. The decks are identical but not mixed together. Pulling a 6 of spades from one deck doesn't effect the chances of pulling a 6 of spades from the other.

13. Jul 4, 2004

### plover

Since I'm no longer troubled by the answers to the original problem, and can also no longer convince myself that the "meta" questions I wanted to ask have any real relevance, I think I'll quit while I'm ahead (so to speak ).

Thanks everyone.