Probability question

[Jellby]

The problem is you don't know what information the parent is giving, unless you asked for it. If you ask a mother:

- Do you have two children?
- Yes
- Is the oldest a boy?
- Yes

Then the probability of the youngest being a boy is 1/2, we all agree here.

But if the mother tells you without being asked: "I have two children and the oldest is a boy". We can't say whether the probability is 1/2 or 1/3, it depends on why she said that. Maybe she thought "I'll reveal the sex of my oldest kid", then it's 1/2. But maybe she thought "I'll tell who's one of my boys, if I have any", then it's 1/3.

This is best seen with coins. If someone tosses two coins without letting you see them, and then always shows you the coin under the left hand, the probabilities for the other coin are, of course, 50:50. But if he choses to show you whatever coin gave heads (if any), then the probabilities are 33:67 (unless he shows no coin).

Or someone tosses 100 coins and says "there are at least 50 heads". Why should we assume he could have said "there are at least 50 tails"? Maybe his strategy is always saying "there are at least X heads", with X being the highest possible ten (still being true). Or maybe his strategy is always saying "there are at least Y heads" with Y the exact number of heads. On the other hand, if we are allowed to ask "are there at least 50 heads?", that's different.

This raises the question, what would be the best strategy for both players in this kind of game?

So, if someone says something without being asked, don't trust him/her.

 

[Dr Adequate]

You are supposing that the parent in the OP who said "one [of my two children] is a boy" had a particular child in mind, chosen independently of its sex, and would not have said "one is a boy" if the child in question had in fact been a girl, even if the other child were a boy. Those who give the answer as 1/3 are supposing that the parent would have said "one is a boy" in that case also, so that only if both her children were girls would she not have said it.

The OP is not totally clear.

At last, I have a disciple.
_____________________________

Everyone else --- listen up --- this is not a maths problem, and it is not a question about probability theory. It's a question about what you think the question is about. And since the question is not exactly stated, it is meaningless.

Now STFU and resume your normal daily lives, before I get angry.

 

[Alkatran]

No, I am assuming the parent chose to tell us the sex of a randomly chosen child.

That's wrong though. She didn't tell us the sex of a randomly chosen child. Think of it like this.
Boy = 1, Girl = 0

Your questio: Is child(rnd[1 or 2]) a boy? And the other?

ACTUAL Question: Is child1 + child2 > 0? Is child1 + child2 = 2?

The trick here is they said "other" child and throw you off.

 

[CurtC]

Everyone else --- listen up --- this is not a maths problem, and it is not a question about probability theory. It's a question about what you think the question is about. And since the question is not exactly stated, it is meaningless.

There was one aspect of it that was a math/logic problem: when some here were maintaining that if you see a lady with her son in the grocery store, and learn that she has two children, that the probability of the other being a boy would be 1/3. I spent some time trying to correct that. I think I finally succeeded, but I'm not sure.

Now STFU and resume your normal daily lives, before I get angry.

But this is my normal daily life!

 

[patnray]

We still are disagreeing on this one point. It doesn't matter when you see a son with her whether we know he's the oldest or youngest. If we see a son with her, the probability for the other kid is 1/2. We don't need to know which kid it is.

You still don't see the distinction. I agree that the probability that "the other kid" is male is 1/2. That is the probability that any one child is male.

I think everyone posting in this thread agrees that the probability of any one child being male is 1/2.

Where people seem to get confused is that we are also considering a set of two children and probabilities about the SET, not the individual children. When discussing the set, we can have information about the set that does not give us information about either member ("one of them is a boy"). Or we can have information about the members of the set ("The oldest is a boy").

If you know that the first member of the set is a boy, then there are only two possibilities for the set: BB or BG. Thus, in this case, the probability that the set is BB is 1/2. Similarly, if you know that the second member of the set is a boy then there are only 2 possibilities: BB or GB. Again, the probability that the set is BB is 1/2.

But if we know only that one of the members of the set is a boy, without knowing which one, then there are 3 equally likely possibilities: BB, BG, and GB. Thus the probability THAT THE SET CONTAINS 2 BOYS is 1/3.

We know that one child is male, but we do not know which member of the set he is (even if he is standing next to his mom).

The statment that "the probability of the other child being male is 1/2" is true.

The statement that "the probability that both children are male is 1/3" is ALSO true.

There is no contradiction here, because they are probabilities about DIFFERENT things. The first is a statement about a single child. The second is a statement about both children.

In addition to the ambiguities in the original question, there has been some ambiguity in the way people have attempted to answer the question. The answer depends on whether you are considering a single child or the set of both children.

 

[patnray]

Here's another way to look at it;

In the set of all women who have 2 children, 1/4 will have 2 boys.

In the set of all women who have 2 children and at least 1 boy, 1/3 will have 2 boys.

In the set of all women who have 2 children and whose oldest is a boy, 1/2 will have 2 boys.

In the set of all women who have 2 children and whose youngest is a boy, 1/2 will have 2 boys.

(Excludes adoptions, identical twins, and artificial means of selecting the sex of the children....)

 

[Jellby]

In the set of all women who have 2 children, 1/4 will have 2 boys.

In the set of all women who have 2 children and at least 1 boy, 1/3 will have 2 boys.

In the set of all women who have 2 children and whose oldest is a boy, 1/2 will have 2 boys.

In the set of all women who have 2 children and whose youngest is a boy, 1/2 will have 2 boys.

So, if a woman tells you:

"I have 2 children, at least one of them if a boy"

We are in 1/3 for both being boys, right?

What if she then adds: "he's my oldest, he's blond and his name is Fred"? Does this bring the probability up to 1/2, just because she said he's the oldest? I say no, because she chose to give you the information, she told you about the boy (or one of the boys) she has, if the boy were the youngest, she would have told you about the youngest.

To have a 1/2 probability, you have to make sure you are getting information about the oldest kid first, before getting the actual information, otherwise you are not getting information about the oldest kid but about the boy.

And if she just says "I have two kids and the oldest is a boy" you don't know what is the situation. Is she telling about the boy or about the oldest kid? Of course, of all the women with two kids whose oldest is a boy, 1/2 will have two boys, but we have only 1 woman and we don't know why she's saying what she's saying.

 

[patnray]

It comes down to if we have information about the set alone or information about specific members of the set. And, again, you must specify if the probability applies to individual members or to the set as a whole.

If you have information that specifies the sex of an individual member of the set, then the probability for both the set and the other member is 1/2.

If she says "I have two children, one of them is a boy named Fred", she hasn't specified which member of the set Fred is, so the probability that both children are boys is 1/3 (there are still three possibilities).

If she says "I have two children, the oldest is a boy named Fred" then she has identified which member of the set Fred is, and the probability that she has 2 boys is 1/2 (there are now only two possibilities).

If she says "I have two children, and this one is Fred", we can see Fred is a boy. But we do not know which member of the set of 2 children Fred specifies, so the probability that she has 2 boys is again 1/3.

I'm not sure what you are getting at, aside from the ambiguity of the statement "one of them is a boy". We are assuming here that she means at least one is a boy...

 

[Moose]

What if she then adds: "he's my oldest, he's blond and his name is Fred"? Does this bring the probability up to 1/2, just because she said he's the oldest? I say no, because she chose to give you the information, she told you about the boy (or one of the boys) she has, if the boy were the youngest, she would have told you about the youngest.

No, not quite.

See, patnray's right. For any set of two children, the probability of B-B is 1/4. 1/4 of B-G, 1/4 of G-B, and 1/4 of G-G.

This is the base probability.

Each piece of information we add alters the odds somewhat.

To say that there is at least one boy among the children is the same as saying that there is not two girls. So we can eliminate the G-G probability, which adjusts the three remaining probabilities to 1/3 each.

To then add that the eldest is a boy also would eliminate the G-G possibility, but also the G-B possibility.

This leaves one way out of the remaining two for both children to be boys. Hence, the probability in this case increases to 1/2.

Note that since we have information that the eldest child is definitely a boy (named Fred), so we know that the odds of the eldest child being a boy is 1/1. This, in itself, is independent of the gender of the second child which remains its base 1/2.

No matter what, however, we have to consider the set of two children as a whole at all times. We're simply not allowed to ignore a child in the set. It's just that when we know the specific gender of a specific child, the probability relating to that child becomes 1/1.

This pretty much gets to the core of the Schrodinger's Cat thought experiment. While the gender of either of the children is unknown, for the purposes of probability, that child has both genders at once. (This is the same as knowing a gender, but not who it belongs to.) *

When we "open the box" and look, the probability assigned to that ambiguity is immediately reduced to 1/1.

(I don't think I really understood Shrodinger's Cat until just now. I'm glad I jumped into this thread.)


[Edit to add:] * Note that this is not the same thing as the child literally having two genders. The child presumably knows what gender it is. In its frame of reference, the odds are 1/1. In the frame of reference of imperfect knowledge, we cannot assume a gender and must consider both genders as simultaneously valid.

 

[CurtC]

We know that one child is male, but we do not know which member of the set he is (even if he is standing next to his mom).

Yes you do know which - he's the one with his mom. Therefore the probability of two boys is 1/2. There is no need to know his birth order.

The statment that "the probability of the other child being male is 1/2" is true.

The statement that "the probability that both children are male is 1/3" is ALSO true.

Given that you see a boy with her at the store, there's absolutely no difference between "the other child being male" and "both children are male." If you're getting different probabilities for the same situation, it's a sure sign that you're doing something wrong.

Just to make sure we're not getting hung up on semantics here, let me restate it. If you see a woman at the store with a boy, and you learn that she has two kids, the probability that both are boys is 1/2. Do you agree or disagree?

 

[Soapy Sam]

I'm sitting here with tears of laughter running down my face, having just ploughed through the whole thread again.


I can think of no better argument in favour of contraception.

We really do need to get out more.

 

[Jeff Corey]

... If you see a woman at the store with a boy, and you learn that she has two kids, the probability that both are boys is 1/2. Do you agree or disagree?

Not if she's at the store with someone else's child. Then the probability would be 1/4.

 

[Jellby]

I've found some words about this in Martin Gardner's "Aha! Gotcha. Paradoxes to puzzle and delight".

First there's a story about two parrots. Then there's a game with two coins, where both players must agree first on what kind of information will one of them give. Then he says (re-translated from Spanish):

Sometimes, when introducing the two parrots paradox, it is presented in such an ambiguous form than it's impossible to give an answer. For example, you meet a stranger and he says: "I have two children. At least one of them is a boy". What is the probability of both being boys?

This problem is not well defined, since you don't know the circumstances that made the stranger tell you that sentence. The other could just as well have told us "at least one of them is a girl", picking the sex at random if the two children are different, and giving the right one if they are the same. If that is his behaviour, the probability of both being boys is 1/2.

I see the problem comes when counting the "possible cases". Someone says "at least one is a boy", what is the probability of both being boy? We count how many couples of boys are among... which population? Do we consider all the parents who could have said that, or those who would have said that?. Those who could are, of course, all those who have at least one boy. Those who would (assuming they would say either "one is a boy" or "one is a girl") are all those who have two boys and half of those who have one boy and one girl (the other half would have said one is a girl). If there is nothing forcing the parent to say something about boys, the probability is 1/2.

Similarly, when the information is "the oldest is a boy", those who could have said that are all the BB and all the BG (so 1/2 probability). Those who would are half the BB (the other half would have talked about the youngest) and half the BG (the other half would have talked about the girl) (that's 1/2 again). But if something forces the parent to talk about boys, and not about the oldest or youngest, those who would are half the BB and all the BG, and we now have 1/3 probability.

So, to have 1/3 there must be something to restrict the information only to boys, just a direct question is enough. Then, additional unrestricted information about the age, hair colour or whatever doesn't change anything.

Does this make any sense?

 

[Just thinking]

There was one aspect of it that was a math/logic problem: when some here were maintaining that if you see a lady with her son in the grocery store, and learn that she has two children, that the probability of the other being a boy would be 1/3. I spent some time trying to correct that. I think I finally succeeded, but I'm not sure.

Well ... in my case you succeeded in making me clarify that my condition was that if the mom has at least one boy she must bring him with her. This was my intent all along -- I just didn't word it well enough. I do not believe that 'patnray' is convinced of that yet.

Here is a comment from 'patnray' ... "We know that one child is male, but we do not know which member of the set he is (even if he is standing next to his mom)."

Of course, this all depends on whether or not the mother was required to only display a male child. If she was, it's 1/3 -- if not, then it's 1/2.

 

[Rockon]

According to John Paulos

Although he uses a girl instead of a boy, and computes the odds of the sibling being a boy, in his book "Innumeracy," (pg 86) he claims the odds should be 1/3 (based on the way the original problem was described.)

But he states the problem a little differently:

Assumption: A family only has two children. You knock on the door of their house and a female child answers the door.

"In any case, given that a family has at least one daughter, what is the conditional probablility that it also has a son? The perhaps surprising answer is 2/3, since there are three equally likely possibilities - older boy, younger girl; older girl, younger boy; older girl, younger girl-and in two of them the family has a son. The fourth possibilitiy-older boy, younger boy-is ruled out by the fact that a girl answered the door."

Strangely enough, I was just about to create a post about this question into this forum because I disagree with that answer and I wanted to see what people in here thought....and lo and behold, there's already a huge thread raging about it!

Anyway, according to Paulos, and Marilyn vos Savant, who also posed this in her weekly column some time ago, the answer should be 1/3.

I happen to believe the answer should be 1/2, for the same reason "bmillsap" gave way back on the first page of this thread: I think they oversimplify the case of the "younger girl, older girl." I think birth order matters in this case, Paulos and vos Savant (and probably a bunch of other mathemeticians) don't.

Tim

 

[Dr Adequate]

* bangs head on desk repeatedly until it hurts *

 

[Rockon]

Hey Doctor,

I didn't mean to give you a headache....

The possible combinations seem to be:

BG
GB
BB

Since two out of the three possibilities contain girls, the answer to the original question is 1/3.

I believe the actual number of possibilities are:

BG
GB
B(1)B(2)
B(2)B(1)

Making it four possibilities and therefore 1/2. In other words, since birth order matters when you have a boy/girl combo, why shouldn't it matter when you have a boy/boy combo?

Or, look at it this way:

A family with two children has one boy. That boy could have the following siblings:

Older sister
Younger sister
Older brother
Younger brother

I count 4 possibilities.

Tim

 

[patnray]

Just to make sure we're not getting hung up on semantics here, let me restate it. If you see a woman at the store with a boy, and you learn that she has two kids, the probability that both are boys is 1/2. Do you agree or disagree?

Disagree.

Consider the following game: I flip two coins, then cover them up. Without any additional information, the possibilities are HH, HT, TH, and TT

If I tell you "At least 1 coin is H", what is the probability that both are H? I think we agree it is 1/3. Note that here the question "What is the probability that the other one is H?" is meaningless, since we haven't identified either one (there is no "other" one).

Suppose, instead I tell you "The first coin flipped is H." What is the probability that both are heads? 1/2. In this case "What is the probability that the other one is heads?" is meaningful, since it refers to the second coin. 1/2 it is...

Suppose I slide one coin out from under the cover without letting you know whether it is the first or second coin. You see that it is heads. How is this any different from the first case? You know only that at least one coin is H. What is the probability that both coins are H? We can eliminate TT, but HT, TH, and HH are still possibilities. Of those 3, HH is the only one in which the unseen coin is H. Thus 1/3. Here, "the other coin" has some meaning: it is the one we can't see. But we still don't know if it was the first or the second coin. Thus we do not have enough information to eliminate either HT or TH from the 3 possible outcomes.

Suppose I flip 2 coins and cover them, then I slide one coin out from under the cover at random without telling you if it was the first or second. If the covered coin is the same as the uncovered coin I pay you a dollar. If they are different, you pay me a dollar. We do this 1000 times. I think, on average, you will pay me $166 ((1/2 - 1/3) * 1000)...

 

[CurtC]

Disagree.

I'm glad to see there is discussion about the logic and not just semantics!

Suppose I slide one coin out from under the cover without letting you know whether it is the first or second coin. You see that it is heads. How is this any different from the first case? You know only that at least one coin is H. What is the probability that both coins are H? We can eliminate TT, but HT, TH, and HH are still possibilities.

No, you know specifically which coin was heads, so TT and TH are eliminated (the first letter of the pair in my usage indicates which one we've seen).

Suppose I flip 2 coins and cover them, then I slide one coin out from under the cover at random without telling you if it was the first or second. If the covered coin is the same as the uncovered coin I pay you a dollar. If they are different, you pay me a dollar. We do this 1000 times. I think, on average, you will pay me $166 ((1/2 - 1/3) * 1000)...

It depends - when you decide which to slide out, are you first looking at both coins? Because that's not the same scenario. If you instead can't see them, but slide one out, then we play the game only when you've slid out a heads, I would definitely play, because that's the situation analogous to the lady in the store with her son, and the probability is 1/2. (Actually, if we played by your rules, I would expect only to break even)

 

[Jellby]

Suppose I flip 2 coins and cover them, then I slide one coin out from under the cover at random without telling you if it was the first or second. If the covered coin is the same as the uncovered coin I pay you a dollar. If they are different, you pay me a dollar. We do this 1000 times. I think, on average, you will pay me $166 ((1/2 - 1/3) * 1000)...

No, that seems a fair game.

Half the times both coins will be the same, half the times they won't.

You mean we play only when the uncovered coin is heads? It's still fair. 1/4 of the times it'll be HH and you'll uncover H. 1/4 of the times it'll be HT, but only 1/8 of the times you'll uncover H. Similarly for TH. In total, you'll uncover H 1/4+1/8+1/8=1/2 of the times, the rest we discard them, of these, 1/4 (i.e., 1/2 of 1/2) are HH. [edit: I mean, 1/4 in total are HH, that's one half of the remaining 1/2 cases.]

Or maybe you mean you don't select a coin at random, but knowingly select the one that's H in the case both are different (and still, we play only when you can, and do, uncover H)? Then I agree, 1/3 of the times they'll be HH.