Only if the train did not pass by.

[Steven G]

I asked this question when I first joined the forum but no one answered it. I will try again.

Suppose you meet me on the street corner with my son and I tell you that I have another child at home. P(that other child is a girl) = 2/3. No problem here.

Now suppose you meet me on the corner and I introduce you to my son who I inform you is my 1st born and I tell you that I have another child at home. p(that other child is a girl) = 1/2.

Similarly, suppose you meet me on the corner and I introduce you to my son who I inform you is my 2nd born and I tell you that I have another child at home. p(that other child is a girl) = 1/2.

Based on the above, it does not matter if I tell you the son you met is my 1st or 2nd born as the p(that other child is a girl) = 1/2 in either case.


Now here is where I get confused. Suppose you meet me on the corner and I introduce you to my son who i inform you is my xxx born and I tell you that I have another child at home. Maybe a loud train passed by and you could not tell whether I said 1st or 2nd.

I want to say that p(that other child is a girl) = 1/2 as it does not matter which order I tell you my son was born in. On the other hand I want to say that p(that other child is a girl) = 2/3 since if you did not hear me than this becomes the 1st case above.

What am I missing!!

 

[Dr.Peterson]

I asked this question when I first joined the forum but no one answered it. I will try again.

Suppose you meet me on the street corner with my son and I tell you that I have another child at home. P(that other child is a girl) = 2/3. No problem here.
...
What am I missing!!

This problem is notoriously easy to misstate so that the answer is not what you think it is.

Before I look at the rest, can you convince me that the probability in this first situation is really 2/3?

One thing that troubles me is that you said, "my son", which implies that the other is a girl (with probability 1!). But that's mostly a quibble.

Another issue is that you didn't clearly say that you have only two.

But the main issue with these problems is determining the sample space, and particularly how the boy I meet is chosen. Note that if you have two boys, then I might meet you with either of them. Do you see what effect that has?

 

[Steven G]

Yes, I am assuming that I have exactly two children.

Before I look at the rest, can you convince me that the probability in this first situation is really 2/3? You know (now) that I have two children and at least one is a boy. The sample space is bb bg gb. There are two cases out of three where I have a girl.

 

[pka]

To be honest I am puzzled by your title "Only if the train did not pass by"?
The logical construction If P then Q has the equivalencies:
not P or Q
not Q implies not P
P only if Q
Q is necessary for P
P is sufficient for Q
The above is stand fair for symbolic logic courses.
How do these apply if at all to this thread?

 

[Cubist]

I like this problem. I had to think for a while before I agreed with the first situation probability of 2/3

Probability can be very non-intuitive. I think this is an example of Bayes' theorem(click). But the way that Jomo states the second half of the question it becomes a bit like "Schrodinger's Cat". I'll have to think about this (and have more tea) before saying any more!

 

[Dr.Peterson]

As I said, the wording of this problem is very sensitive; the wording of your question makes the usual surprising answer of 2/3 wrong.

Suppose that there are eight families with two children each, whose children are respectively BB, BG, GB, GG. Each one meets me on the street; in the first four cases the father is with the older child, and in the last four cases the younger child. So the child I meet in each case is the bold one here: BB, BG, GB, GG, BB, BG, GB, GG. These are equally likely, right?

My sample space, given that I meet a boy, consists of families BB, BG, BB, GB, where again the bold one is the child I meet. That is, I meet four boys with their fathers, and out of those four, two have sisters.

So the probability that the other child is a girl is 1/2.

The trick is that with your specific scenario, it is the child (the one I meet) who is randomly selected, not the family. A family with two boys is twice as likely to be selected as a family with a boy and a girl. The 2/3 probability would be appropriate if I just meet the father and ask if he has a boy, and he says yes. Then the sample space is what you said, BB, BG, GB, and the probability that he also has a girl is 2/3.

I discussed this in my blog here; there I discuss a number of questions and answers from the site I formerly volunteered with, where different formulations of the problem came up. The issue is whether the choice is made "family first" (which leads to the 2/3 probability), or "child first" (which leads to 1/2); your version is child-first, as I read it.

I think this deals with all your questions.

 

[Cubist]

As I said, the wording of this problem is very sensitive; the wording of your question makes the usual surprising answer of 2/3 wrong.
...
I discussed this in my blog here; ...

This is a very thorough answer. I followed your logic of the 8 families. Thank you. I probably need to re-read it all myself (a few times!).

If you don't mind, can you please explain the difference between Jomo's first 2/3 scenario, which you say is wrong (and I am confused so I don't have a strong opinion), and the bold sections below...

From this page (linked from your blog)
"Remember: information that creates conditional probability can dramatically affect common sense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a mother of two who says she has a daughter, a basic knowledge of probability tells you there's a 2/3 probability that the daughter mentioned has a brother. If she says she's an older daughter, you know there's a 1/2 probability that the daughter has a younger brother."

In other words, is there a "trick" to knowing the correct "probability scenario". A particular word to look out for perhaps, or maybe it is just something that comes with experience?

 

[Cubist]

If you don't mind, can you please explain the difference between Jomo's first 2/3 scenario, which you say is wrong (and I am confused so I don't have a strong opinion), and the bold sections below.

I think I understand the difference now.

In the scenario from the blog that I highlighted, there are only two random events. The mother is looking at the outcome of the two random events (births) and if either one of them, or both of them, are girls then she says "I have a daughter". etc, etc, and the probability of the other child being a son GIVEN she has a daugter is 2/3

But in Jomo's scenario, there are three random events:- two births, and a meeting (did I meet with the first or second born). This explains the size 8 sample space {BB, BG, GB, GG, BB, BG, GB, GG} that Dr Peterson mentions (each random event has 2 possibilities and 2^3=8). And I agree that the probability of the "not met" child being a girl is 1/2.

I've learnt something (I think)

 

[Dr.Peterson]

From this page (linked from your blog)
"Remember: information that creates conditional probability can dramatically affect common sense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a mother of two who says she has a daughter, a basic knowledge of probability tells you there's a 2/3 probability that the daughter mentioned has a brother. If she says she's an older daughter, you know there's a 1/2 probability that the daughter has a younger brother."

In other words, is there a "trick" to knowing the correct "probability scenario". A particular word to look out for perhaps, or maybe it is just something that comes with experience?

I think the main "trick" is to be very cautious in probability (or combinatorics), knowing how subtle it can be! I've often said that I am not sure of my answer in either field until I get the same answer using two different methods.

The specific trick is to pay close attention to how things are selected, and what the resulting sample space is. This problem is just a good example of that. I don't think there are any key words to use; if anything, you need to look beyond words and focus on meaning.

I think I understand the difference now.

In the scenario from the blog that I highlighted, there are only two random events. The mother is looking at the outcome of the two random events (births) and if either one of them, or both of them, are girls then she says "I have a daughter". etc, etc, and the probability of the other child being a son GIVEN she has a daughter is 2/3

But in Jomo's scenario, there are three random events:- two births, and a meeting (did I meet with the first or second born). This explains the size 8 sample space {BB, BG, GB, GG, BB, BG, GB, GG} that Dr Peterson mentions (each random event has 2 possibilities and 2^3=8). And I agree that the probability of the "not met" child being a girl is 1/2.

I've learnt something (I think)

That is not a bad way to look at it.

Also, in the first case what is given is just that there is a son; in the second, it is that this child is a son (regardless of whether "this" means "the one you see" or "the firstborn" or whatever. A particular boy is part of the given (your third event).

 

[firemath]

I like this problem. I had to think for a while before I agreed with the first situation probability of 2/3

Probability can be very non-intuitive. I think this is an example of Bayes' theorem(click). But the way that Jomo states the second half of the question it becomes a bit like "Schrodinger's Cat". I'll have to think about this (and have more tea) before saying any more!

I love Schrodinger's cat.

But I was of the persuasion that since he said "my son," he either has a daughter or an unabashed favoritism toward another son, thus making the problem a bit like Schrodinger's.

 

[JeffM]

I must admit I find the argument about selecting families or children first hard to follow. Just following the axioms and definitions tends to keep me right rather than creating sampling scenarios. Here it is the definition of conditional probability that is the main lifeline.

X has two children, P(bb) = 1/4, P(bg) = 1/4, P(gb) = 1/4, P(gg) = 1/4, where sequence indicates order of birth.

P(at least one b) = (1/4) + (1/4) + (1/4) = 3/4

P(one b and one g) = (1/4) + (1/4) = 1/2.

All very intuitive up to here.

P(one b and one g given at least one b) = P( one b and one g) / P(at least one b) = (1/2) / (3/4) = 2/3.

It may well be unintuitive, especially to those who have never studied probability, but it follows mechanically from the definition of conditional probability.

P(b first) = P(bb) + P(bg) = (1/4) + (1/4) = 1/2. No adjustment needed because bb and bg are mutually exclusive.

P(g second given b first) = P(b first and girl second) / P(b first) = (1/4) / (1/2) = 1/2.

All that is needed is the definition of conditional probability.

All well and good but I think what Jomo is getting at is the weirdness that the degree of information available changes probabilities. And this does relate to Bayes Theorem as well as conditional probabilities. I think it was Poincare who said that a probability is a measure of ignorance. Therefore, information reduces ignorance and thereby changes probabilities. I know that the degree of belief theory of probability is not problem free, but it addresses this issue quite nicely.