Contrast these three statements and questions:

1. I have two children. WE KNOW AT LEAST the older one is a boy. What’s the probability they are both boys?
2. I have two children. WE KNOW AT LEAST one is a boy. What’s the probability they are both boys?
3. I have two children. WE KNOW AT LEAST one is a boy. What is the probability my youngest child is a boy?

Answer to question 1 is ½
Answer to question 2 is 1/3
Answer to question 3 is 2/3

So far it all seems settled. But what if we rephrase question 3 like this:

We we are walking down the street and we meet Mr Jones and he introduces us to his son. Later we are told that Mr Jones has two children. What is the probability that Mr Jones's younger child is a boy?

I can't put my finger on why but for reason "walking along and meeting Mr Jones with his son" feels like it adds a different dimension than just simply knowing he has a son. What do you webdiplomacy maths geniuses think?

The reason I ask, is that this question was discussed on another forum and while most agreed with the standard responses one poster presented this argument:

Surely the answer to this particular question is 75%.
Say there are 200 families then:
a) 50 are GG
b) 50 are BB
c) 100 are GB/BG

The probability of meeting Mr Jones with a son are respectively:
a) 0%
b) 100%
c) 50%

that leaves 50 BB families and 50 GB/BG familes of which 75 have a younger boy, or 75% probability.

^ And this is equivalent to making the probability that he has two sons 50%.
Is his reasoning sound or have I just been bamboozled by the way he has presented his calculations.

I didn't think you were a troll before this, but now, I am convinced that you are one.

Well done, sir. Well done. :)

That isn't a rephrase of question 3. In question 3, the given is that he has at least one son. In your rephrase the given is that when you randomly met one of his children, it was a son. Do you see how that is fundamentally different?

Thanks uclabb - so it is fundamentally different and it is right the probability that he hast two sons is 75%? (Sorry I am not very good at maths).

It never occurred to me that the two questions were not actually the same. Weird.

Mhmm.

With only passing involvement in the other thread on probability, I seem to have missed one point there on the original questions 1, 2, 3.

How come we are involving the oldest son in the probability? We don't NEED probability, because we KNOW. The probability of the oldest being a son == 1. So the probability of the youngest being a son must be 50%.

If you compare it to dice, let's say you rolled one die, it comes up a six. You put it aside. Now you throw a second die, and on that roll, you question: what are the odds of them BOTH being a six? The answer is: 1/6.

Meeting one child makes ghe odds of two boys 1/2. The odds of the younger being a boy are still 2/3 because we don't know the relative age of the one we met. But because we have met one then we are only dealing with BB and BG where the order is the order we met them. That is why the odds of two boys moves to 1/2.

I also missed the other thread, but I agree with MoW: the answer to #1 and #2 should be 0.5 in both cases.

@MoW - We are introduced to a son. Nothing indicates it is his oldest. It could be his youngest.

Same mistake. We don't know if the son we met was the oldest or youngest child.

Nope. Not gonna do it. You can't suck me in for a third time.

@Draug I was referring to the original questions, 1, 2, 3.

Actually, correction, my point only applies to 1 and 2.

How is the probability of the oldest 1? We have not in any questions been given am age.

Correction on question one the older is a boy. But question two makes no such clarification.

Actually, the question in Q2 has no bearing on age. The question is simply: what is probability of them both being boys? For one of the boys we do not need probability. His gender is given. That means to calculate the probability of BOTH being boys, we only need the probability of the SECOND being a boy. That probability is simply 50%.

Again, see my parallel with dice. You don't have to throw two dice, just one.

Q3 is different since it introduces age as a factor.

TMOW, I would direct you to the other thread, but I'm too lazy to find it.

Think of it this way:
We have two children, let's call them X and Y. They have a distinct order because they are distinct entities, so let's call that order XY (the method of determining this order doesn't matter). Now, with any family of two children, we have the following options with equal probability:
X-Y:
G-G
G-B
B-G
B-B

Of these options, we can rule out G-G, as neither X nor Y is female. We cannot, however, rule out any of the other options, as we haven't given any indication as to which child (X or Y) might be male, only that at least one of them is. Thus we are left with:
B-G
G-B
and B-B

Of these three, only one meets our end condition of both being male. Hence, the probability is 1/3.

I disagree. Including the GG option and then excluding it in the probability makes no sense. It has zero probability of happening. You say that X and Y have a distinct order. This is true for question 3, but certainly not for 1 and 2 where order is not included at all. As such, for the purposes of these two Qs, option nr 1 never existed and G-B == B-G. There are only two options.

Taking your XY example, let's *fix* the gender of X as it has been in the phrasing of the scenario.

Given X=B, the options are:

Y=G
Y=B

This makes it 50%. Again, look at my dice example. One die has already been cast.

X hasn't been fixed though, we only know that either X or Y is fixed.

Ignore that GG is a possibility if you want, but you still have to accept that it is twice as likely to have one boy and one girl as it is to have two boys (assuming perfectly equal probability of B and G on each birth, which is how the question was posed in the original thread). Since GG is out of the question, the odds are one third.


As for your dice example, it is wrong. You are attributing order to the dice where none is given. You are saying "OK, there is one die with a 6 on it, now let me roll another and see if it is also a 6." The situation in question is more accurately analogized as "I've rolled two dice and hidden them from you, but I'm telling you that either the first die or the second die is a 6, what's the probability that they both are?" This situation leaves us with the options, 16, 26, 36, 46, 56, 66, 61, 62, 63, 64, and 65. As there are eleven such options and we are specifically looking for one of them, our odds are 1/11, not 1/6.

Aaah. Thanks for that, that's a good explanation.

uclabb said: "That isn't a rephrase of question 3. In question 3, the given is that he has at least one son. In your rephrase the given is that when you randomly met one of his children, it was a son. Do you see how that is fundamentally different? "

I have thought about it some more, and while I was briefly swayed by the idea that the answer might be 3/4 instead of 2/3, I think the "we meet Mr Jones walking with his son" is essentially the same as question number 3. That is, we know just two things about Mr Jones, he has two children, and one of them is a boy.

It is different. Your newest question has the same relationship with question 1 as question 3 has with question 2. Saying the older is a boy is the same as saying that the kid that you met was a boy.

"I can't put my finger on why but for reason "walking along and meeting Mr Jones with his son" feels like it adds a different dimension than just simply knowing he has a son. What do you webdiplomacy maths geniuses think?"
I think if you do not approach from a mathematical angle, the odds change.
If Mr. Jones had two sons, he would be likely to introduce as "This is my older son, Jonathan." In the absence of such a qualifier, "This is my son," implies that it is his only son. Not with 100% certainty of course, but more than the mathematical average.

"We we are walking down the street and we meet Mr Jones and he introduces us to his son. Later we are told that Mr Jones has two children. What is the probability that Mr Jones's younger child is a boy?"

So uclabb are you saying you think the answer to the above question is 1/2? Or have I misunderstood you?

The above statement doesn't tell you whether the boy you just met was Mr Jones older or younger son. -- The phrasing could be confusing as you might interpret it as "younger son than the boy you just met", but that is not what I mean.

No. It is 3/4. There is a half chance that the person you met was the younger son, and a half chance that he isn't: 1/2 + 1/4 = 3/4. This is assuming there is no reason to believe that he would choose to introduce you to a son over a daughter for some reason.

Sorry I meant 3/4 (1/2 for two boys).