Probability question re: boy/girl riddle

Factual Questions
1

If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

Now, I understand why the answer is 1/3.

What I can’t wrap my head around is this:

If I am going to visit a couple that has two children, and a son answers the door, the probability that his sibling is a boy is 1/2, right?

Known child: boy

Unknown child has an equal chance of being: boy-girl

Why can’t I use the same logic in the riddle?

Known child: boy

Unknown child has an equal chance of being: boy-girl

2

I think I got it:

In the riddle, we’re assuming we were randomly given the information about one of the children, in the case where a son answers the door, we know the sex of a specific child?

3

[QUOTE=TazMan]
If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

Now, I understand why the answer is 1/3.

What I can’t wrap my head around is this:

If I am going to visit a couple that has two children, and a son answers the door, the probability that his sibling is a boy is 1/2, right?

Known child: boy

Unknown child has an equal chance of being: boy-girl

Why can’t I use the same logic in the riddle?

Known child: boy

Unknown child has an equal chance of being: boy-girl
[/QUOTE]

Okay, try this. The above riddle is analogous to the following situation, along the vein of the ‘visiting’ example.

You didn’t visit the family yourself. A friend of yours did, and reported: “The couple has two children. One answered the door. The other was out in the backyard playing at the time. At least one of them is a boy.”

So, you have the following possibilities:

  • Boy at the door, girl in the backyard
  • Girl at the door, boy in the backyard
  • Boy at the door, boy in the backyard

Whereas if you knew that the boy answered the door, you have only possibilities 1 and 3 still in line with the evidence.

That’s the general rule. If you know that one specific child is a boy, it is different from knowing that ‘the kids are not both girls’, (which is the same thing as saying that at least one of the children is a boy but not specifying which in any way.)

Like a lot of cases with probabilities, it’s easy to trip up. If your friend, in the previous example, WANTED to be able to report to you that a boy had been in one place, (at the front door or in the backyard,) and would have chosen the place accordingly if there were one child of each gender, then you’re not at 1/2 probability anymore.

Hope that this helps.

4

[QUOTE=chrisk]
Okay, try this. The above riddle is analogous to the following situation, along the vein of the ‘visiting’ example.

You didn’t visit the family yourself. A friend of yours did, and reported: “The couple has two children. One answered the door. The other was out in the backyard playing at the time. At least one of them is a boy.”

So, you have the following possibilities:

  • Boy at the door, girl in the backyard
  • Girl at the door, boy in the backyard
  • Boy at the door, boy in the backyard

Whereas if you knew that the boy answered the door, you have only possibilities 1 and 3 still in line with the evidence.

That’s the general rule. If you know that one specific child is a boy, it is different from knowing that ‘the kids are not both girls’, (which is the same thing as saying that at least one of the children is a boy but not specifying which in any way.)

Like a lot of cases with probabilities, it’s easy to trip up. If your friend, in the previous example, WANTED to be able to report to you that a boy had been in one place, (at the front door or in the backyard,) and would have chosen the place accordingly if there were one child of each gender, then you’re not at 1/2 probability anymore.

Hope that this helps.
[/QUOTE]

It does help, but I’m still not so sure why we can’t do this:

The child we know the sex of: girl
The child we don’t know the sex of: girl/boy

5

[QUOTE=TazMan]
It does help, but I’m still not so sure why we can’t do this:

The child we know the sex of: girl
The child we don’t know the sex of: girl/boy
[/QUOTE]

Because it needs to be an INDEPENDENT criterion from the information that we have. Specifying the criterion in terms of our information is the opposite, it’s the most dependent it could possibly be. The example of the friend picking whether to tell me about the kid answering the door or the kid in the backyard is another example of dependence - he knows both and is choosing what to tell us based on a preference in his own head and the full information.

For full independence, we should ideally find a way of distinguishing the kids and then arbitrarily choose which one we want to test before any information is gathered.

6

[QUOTE=TazMan]
If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

Now, I understand why the answer is 1/3.
[/QUOTE]

Wait, I haven’t even gotten this far.

If at least one of them is a boy, then you could have

GB
BB

What other option is there?

7

[QUOTE=TazMan]
If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

Now, I understand why the answer is 1/3.

What I can’t wrap my head around is this:

If I am going to visit a couple that has two children, and a son answers the door, the probability that his sibling is a boy is 1/2, right?

Known child: boy

Unknown child has an equal chance of being: boy-girl

Why can’t I use the same logic in the riddle?

Known child: boy

Unknown child has an equal chance of being: boy-girl
[/QUOTE]

because the sex of 1 child can’t influence the sex of another, if the sex of any child is unknown, the probability of being a boy or girl will always be 1/2 (barring known family history of only having girls, etc.). So if 1 is known to be a boy, the probability that both are boys is…1/3 (possibilities are older boy-younger boy, older boy-younger girl, older girl- younger boy). The reason it’s confusing is a fine logical distinction between the 2 riddles.
The 1st is picking the physical possibilites from the expanded set of all possible permutations, while the 2nd is zeroing in on a specific individual. To put it another way, the 1st is looking at % of total population (which you can easily imagine like that of a census), while the 2nd, down to the granularity of an individual person, will always be a binary state - excepting cases like Michael Jackson & Amy Winehouse, of course. :smiley:
Basically, taking % of a sample population <> one binary state value for an individual.

8

[QUOTE=Freudian Slit]
Wait, I haven’t even gotten this far.

If at least one of them is a boy, then you could have

GB
BB

What other option is there?
[/QUOTE]

BG

9

[QUOTE=TazMan]
I think I got it:

In the riddle, we’re assuming we were randomly given the information about one of the children, in the case where a son answers the door, we know the sex of a specific child?
[/QUOTE]

Well, not quite, but close. The key is, in the riddle with the answer “1/3”, the probability distribution you are looking at is over the genders of the kids in two-children families and the only information you are conditionalizing on is that the family has at least one son. That is, you start with four equiprobable possibilities SS, SD, DS, DD and you filter out precisely the last one with the information you have, leaving three equiprobable situations, two of which are all boys.

In the son answering the door situation with the answer “1/2”, the probability distribution “widens” a bit, and you also have further information to conditionalize on: now the probability distribution isn’t just over the genders of the kids in two-children families, but also over the genders of the kids who answer the door; furthermore, the information you have isn’t just that the family has at least one son, but also that a son answered the door. That is, you start with eight equiprobable possibilities SS[older S answers], SS[younger S answers], SD[S], SD[D], DS[D], DS[S], DD[oldest D], and DD[youngest D], and the information you have filters out all the ones where a daughter answers the door, leaving four equiprobable situations, two of which are all boys.

The key is that you’re looking at more in the second case, and also, in doing so, take advantage of the opportunity to have more information; specifically, the information of who answers the door. And further information can always cause probabilities to shift wildly.

10

[QUOTE=Freudian Slit]
Wait, I haven’t even gotten this far.

If at least one of them is a boy, then you could have

GB
BB

What other option is there?
[/QUOTE]

In a sense, there are no other options (either a boy and a girl, or two boys, yeah). But just because you’ve categorized a situation into certain options, it doesn’t follow that all those options are equiprobable. I could win the lottery tomorrow or I could lose, but the probabilities aren’t 50-50.

The usual implicit model of birth genders, however, is that every birth has independent equiprobable likelihood of being a boy or a girl. Using that fact, we can categorize two-children families into four situations, based on the gender of the oldest child and the gender of the youngest child, and be justified in concluding them all equiprobable (being able to make this conclusion is key for “probability as counting”). Then, we conditionalize on the information we know, which is that at least one child is male. That tosses out one of the four equiprobable situations, leaving three equiprobable situations, in only one of which both children are male.

11

[QUOTE=Indistinguishable]
Well, not quite, but close. The key is, in the riddle with the answer “1/3”, the probability distribution you are looking at is over the genders of the kids in two-children families and the only information you are conditionalizing on is that the family has at least one son. That is, you start with four equiprobable possibilities SS, SD, DS, DD and you filter out precisely the last one with the information you have, leaving three equiprobable situations, two of which are all boys.
[/QUOTE]

Sorry; I meant, of course, to say “only one of which is” in place of the bolded text.

12

Is this essentially the “Let’s Make A Deal” thing? Because I never, ever, no matter how it was explained to me, understood that one.

13

[QUOTE=Indistinguishable]
In the son answering the door situation with the answer “1/2”, the probability distribution “widens” a bit
[/QUOTE]

It might have been better to word this as the sample space “widening”; that is, regardless of the probability distribution in play, the possibilities you are considering are ones which draw finer distinctions. But this isn’t really essential; what’s essential is that you have the further information, in the second case, of who answers the door, whereas in the first case, you did not (whether analyzed as you not caring or caring but just not knowing or whatever else).

14

[QUOTE=Zsofia]
Is this essentially the “Let’s Make A Deal” thing? Because I never, ever, no matter how it was explained to me, understood that one.
[/QUOTE]

No, that’s the Monty Hall problem. It’s a little different, but I’m sure that if you search through past threads, you should be able to find a suitable explanation. The key is that the host always opens a door without a prize behind it; therefore, by opening a door, he is giving you new information. And new information can always cause probabilities to shift widely…

15

If a boy answers the door, the probability that the other child is boy or girl is the same in both cases, 1/2. Because in the first case, “at least one of them is a boy”, you would have eliminated GB, leaving just BG and BB.

You are comparing the situation before somebody opens the door to the situation after somebody opens the door. They are different, because you know more after you learn the gender of the first child.

16

[QUOTE=Indistinguishable]
No, that’s the Monty Hall problem. It’s a little different, but I’m sure that if you search through past threads, you should be able to find a suitable explanation. The key is that the host always opens a door without a prize behind it; therefore, by opening a door, he is giving you new information. And new information can always cause probabilities to shift widely…
[/QUOTE]

Sorry, I should have worded this better. Obviously, when Monty opens a door, you gain (irrelevant) new information of the sort “The door he opened has no prize behind it.” What’s more important is the information you gain from the fact that Monty choose to open that door in the first place. That is, what you get when reflecting on “But, wait, why’d he pick that door instead of the other one? Why, he appears to have been avoiding the other door. Maybe it’s because he’s not allowed to open the door with the prize behind it.”. Formalizing this, one gains the desired counterintuitive result.

17

In a way, the usual boy-girl problem is the opposite of the Monty Hall problem. The typical naive answer to the boy-girl problem is “Two boys and mixed gender are equiprobable” instead of the correct “Mixed gender is twice as likely” because people are mistakenly thinking of the situation where they have slightly more information; that where they not only know that the family has at least one boy, but also that a boy was chosen in some random selection from the two children. (The OP’s second case is the situation where they actually do have such extra information)

The typical naive answer to the Monty Hall problem is “Switching and staying are equiprobable to win” instead of the correct “Switching is twice as likely to win” because people are mistakenly thinking of the situation where they have slightly less information; that where they know that no prize is behind Door #X, but they don’t also know that Monty Hall (when required to choose a losing door other than the contestant’s) chose Door #X. (If, after picking a door on the Monty Hall show, you use magic X-ray glasses to reveal that the second door has no prize, then staying or switching to the third door are equally good; this is the situation where you actually do lack such extra information)

18

[QUOTE=ultrafilter]
BG
[/QUOTE]

But isn’t that only assuming that order matters? The original problem never said anything about it mattering if the boy was older or younger.

19

[QUOTE=Freudian Slit]
But isn’t that only assuming that order matters? The original problem never said anything about it mattering if the boy was older or younger.
[/QUOTE]

Expanding on post #10:
Roll a die twice. What’s the probability that it will add up to 5?

Well, there’s 36 possibilities: 1-1, 1-2, 1-3, …, 6-4, 6-5, 6-6. Four of them add up to 5. That gives a ratio 4/36.

Though, order doesn’t matter… so that gives 21 possibilities. Two of them add up to 5. That gives a ratio 2/21.

Though, actually, nothing matters but the sum. So that gives 11 possibilities (from 2 through 12). One of them is 5. That gives a ratio 1/11.

Though, actually, nothing matters but whether the sum equals 5. So that gives 2 possibilities (it does or it doesn’t). One of them is “it does”. That gives a ratio 1/2.

Well, all that is true. But not all those ways of classifying outcomes use equiprobable possibilities. So not all of those “probability as counting” arguments give the actual probability. In this case, only in the first categorization are all the possibilities equiprobable (a conclusion supported by the implicit assumption that die rolls are independent, with each face equiprobable). Similarly, when dealing with children’s genders, it’s not the case that “two boys” is equiprobable with “a boy and a girl”; however, it is the case that “two boys” is equiprobable with “an older boy and a younger girl” and with “an older girl and a younger boy” (a conclusion supported by the implicit assumption that birth genders are independent, with each gender equiprobable).

20

[QUOTE=TazMan]
It does help, but I’m still not so sure why we can’t do this:

The child we know the sex of: girl
The child we don’t know the sex of: girl/boy
[/QUOTE]

There are two children who could be the child we know the sex of.