Probability question

[Camillus]

Let ME restate the question: There is a particular child who is not here.

1) What are the chances of this child being male ?

Because you haven't specified the sex of the child who is here the question, as you pose it, can have two possible answers (assuming that we limit the number of sexes to two). It is either 1/3 (if the child who is here is a boy) or 2/3 (if the child who is here is a girl). Since the problem cannot be solved unless we know the sex of at least one child your question cannot be answered more precisely.

2) How could any sibling's sex have any possible effect on the sex of the child in question?

In reality it does not, since the sex of the child is not determined mathematically. However, although I thought when I first read it that the OP was asking what the actual biological chance of a second child being male was, it would appear that most posters are treating it as a mathematical question and reality, therefore, does not enter into it.

 

[casebro]

How do you get 3 outcomes?

There are only 2 possible outcomes: the other child is a boy, or the other child is a girl.

How you folks come up with 3 or 4 outcomes is some kind of faulty logic. Entertaining, but faulty.

 

[Just thinking]

Re: Re: Re: Re: Probability question

1/2, for both of those.

Well, after giving these two questions I posed more thought, I do believe that the man on the street has a 50% chance of having a brother -- but the grocery store pair being BB is 1/3.

1) The man on the street : From a perfectly random population sample of 400 pairs of offspring in a perfect world, 100 pairs will be MM, 200 will be one of each and 100 will be WW. We can immediately discard all 100 of the WW pairs since the person was a man. So that leaves us with 300 pairs. In that population we have 400 men and 200 women. If one man was to come up to you at random, would it not be the same as you picking a man at random? (We need not concern ourselves of the probability of a woman coming up to us because this is a conditional probability problem. We only determine the chances of event B occuring under the conditions set forth in event A -- not the chance of event A happeneing.) If so, then 200 will be available from a MM pair and 200 from a mixed pair. This brings the probability to 1/2 for him to be from the MM group.

2) The grocery store child : In keeping with similar numbers, assume that there is a total population of 400 mothers with 2 offspring each. 100 mothers will have BB children, 200 will have one of each and 100 will have GG. Immediately we dismiss the GG mothers because we see that one is male (this is the stated condition -- event A). We now have a population of 300 mothers, one of which is our mother in the store. What is the chance that she is a BB mother (event B)? Well, only 100 out of 300 fit that description, therfore the probability is 1/3. We need not determine the chances of her walking in with a girl -- it's not part of the problem. (If you're still not sure, consider if of the 400 mothers, 1 had BB, 2 had one of each and 397 had GG. If one of those 400 mothers showed up with a boy, you'd give her a 1/3 probability of other one being a boy, no?)

Now, who's with me on this?

 

[Drooper]

How do you get 3 outcomes?

There are only 2 possible outcomes: the other child is a boy, or the other child is a girl.

Ask yourself this:

which "other" child? There are two "other" children in this example.


If I am making a flaw in logic, how come it works exactly as predicted in real life?

DO my suggested experiment and prove it to yourself.

 

[John Jackson]

How do you get 3 outcomes?

There are only 2 possible outcomes: the other child is a boy, or the other child is a girl.

How you folks come up with 3 or 4 outcomes is some kind of faulty logic. Entertaining, but faulty.

If a couple have 2 children they could have them in one of the following combinations:

Boy then boy

Boy then girl

Girl then boy

Girl then girl


If you are given the information that one of the children is a boy, without specifying the order of birth, then they could have had them in one of the following combinations:

Boy then boy

Boy then girl

Girl then boy

That’s where the 3 combinations come from. As can be seen, where one has to be a boy, there is one chance of another boy, but 2 chances of a girl.

Obviously if you did this test on one couple, it would be found that the sibling would be either a boy or a girl – which is why it looks like it must be a 50-50 chance. Try it on 100 couples, however, and you would find that (around) 67% had a boy and a girl and 33% would have 2 boys.


Interestingly, if you meet a woman and you know that she has one sibling, ask her how her brother is doing. You have a 67% chance of being right. Handy to know if you like to fake psychic ability!

 

[Orangutan]

I'm still not convinced it's anything other than 1/2.

Let me re-word the question again.

I have two children.

I present to you one of them, you now know it's sex.

What is the probability that my other child is the same sex?

Is this not the same question?

 

[Just thinking]

I just realised wat I wrote there might not be clear enough. When I said "reveal the other outcome" I meant reveal both the coins (to the observer). It is important that the observer is only given the information that one is heads - nothing can be revealed about which of the coins is heads (if it is the probability collapses to a 1/2 chance).

This can actually be done by even showing the observer one of the coins is heads. Strange as it may sound (at first) it's OK. All the observer sees (or knows -- there is no difference here) is that one coin is heads, the other has a 1/3 chance being heads also and 2/3 chance being tales.

 

[Just thinking]

I'm still not convinced it's anything other than 1/2.

Let me re-word the question again.

I have two children.

I present to you one of them, you now know it's sex.

What is the probability that my other child is the same sex?

Is this not the same question?

No -- you must specify a sex.

BTW, the answer to the above is 50%.

 

[Soapy Sam]

Drooper- I don't say it doesn't work.

I say it shouldn't work.

Wouldn't work in any universe I designed, anyway.

It implies that if God exists, he's even stupider than I am.

 

[CurtC]

Re: Re: Re: Re: Re: Probability question

Now, who's with me on this?

Definitely not me. When you see one kid in the grocery store, and you ask what sex the other one is, it's a prob of 1/2 that it's a boy, not 1/3.

At the risk of confusing things by making analogies, consider this:

In a town, you have 100 families with two kids, and they're distributed evenly:

Group A, 25 families, had a boy then a boy.
Group B, 25 families, had a boy then a girl.
Group C, 25 families, had a girl then a boy.
Group D, 25 families, had a girl then a girl.

A census worker approaches one of these mothers, and asks whether she has a boy. She answers "yes." The census worker now knows that she's dealing with a mom in Group A, B, or C (not D), a total of 75 families. Only in Group A is the other child a boy as well. The chance that the census worker has picked a mother in Group A is 25/75, or 1/3.

Can this be any clearer?

And Just Thinking, if you're looking at one of the kids and you see it's a boy, this is a different situation, because the question is then "is that other specific kid a boy," and not what it was before, when you're asking whether out of two kids, is either one of them a boy.

 

[Just thinking]

Re: Re: Re: Re: Re: Re: Probability question

Definitely not me.

Can this be any clearer?

And Just Thinking, if you're looking at one of the kids and you see it's a boy, this is a different situation, because the question is then "is that other specific kid a boy," and not what it was before, when you're asking whether out of two kids, is either one of them a boy.

I agree with you on the census worker problem -- but seeing one boy is no different than being told one is a boy. It infers no extra information either way. That "other specific kid" is not so specific as you may think. You do not know if it is the first or second child. There still exists equally likely chances of BB, BG or GB.

Try this thought experiment: I am going to flip 2 coins 400 times together as a pair and on occasion I will show you one coin being heads. On average, how many times will the other coin be heads? I think it will be 1/3 of the time.

 

[CurtC]

Re: Re: Re: Re: Re: Re: Re: Probability question

...but seeing one boy is no different than being told one is a boy. It infers no extra information either way.

Yes it is different. When you are told that one is a boy, you don't know specifically which one it might be. When you are looking at a kid, it doesn't matter whether he's the older or younger, but he is one or the other specifically, and the probability of the other kid being a boy is 1/2.

Try this thought experiment: I am going to flip 2 coins 400 times together as a pair and on occasion I will show you one coin being heads. On average, how many times will the other coin be heads? I think it will be 1/3 of the time.

Let's do it this way - we flip two coins, one of which I can see, one I can't. If I see that one is heads, the probability that the other is also heads is 1/2.

 

[drkitten]

Re: Re: Re: Re: Re: Re: Re: Re: Probability question

Let's do it this way - we flip two coins, one of which I can see, one I can't. If I see that one is heads, the probability that the other is also heads is 1/2.

That's doing it the wrong way; it's a confusion of a priori vs
a posteriori probabilities.

In short, if you know beforehand which coin you are going to see,
it's a different situation than if you don't.

 

[John Jackson]

Re: Re: Re: Re: Re: Re: Re: Re: Probability question

Let's do it this way - we flip two coins, one of which I can see, one I can't. If I see that one is heads, the probability that the other is also heads is 1/2.

If you know which coin is a head then yes it leaves it as 50:50.

It's the same if you know which order the children were born in. If you know the boy was the eldest (or the youngest) then the sex prediction for the sibling is a 50:50 one.

With the coins example though. If you know that one of the coins is a head but you don't know which one then you're left with the following possibilities:

H - H

H - T

T - H

Double the chance of getting a tale. It's the same with the children example. If you know the order of events the next event is 50:50. If you do not know the order of the events then the resulting probability is 33/67.



Edited for clarity.

 

[Just thinking]

Re: Re: Re: Re: Re: Re: Re: Re: Probability question

Yes it is different. When you are told that one is a boy, you don't know specifically which one it might be. When you are looking at a kid, it doesn't matter whether he's the older or younger, but he is one or the other specifically, and the probability of the other kid being a boy is 1/2.

Really? Let's say that you're in the store with your brother, but he is in a different isle. After you're done speaking with the woman you tell him you met a woman with two children, one of which is male. The only difference is that he didn't see the child, he only heard of him, through you. Do you somehow know more of this child than he does? Why is seeing any better than some other way of knowing his existance? How does seeing him affect the other child's sex? Would your brother somehow calculate a different probability for the same woman's other child just because you saw one of them? We know the woman has one male child -- your seeing her with a male child does not give you any more information.

Let's say that there are exactly 4 women. Woman 1 (W1) has 2 children, BB. W2 has GB (older girl). W3 has BG (older boy) and W4 has GG. You see one of them in the store with a male child -- right off you know it's not W4. I hope we can agree on that. Now, what is the chance that she's W1? Since it was established that she has a male child with her we see that she can be either W1, W2 or W3. Each is equally likely because the initial condition was predetermined -- a woman with one male child with her. Therefore the probability is 1 out of 3.

Let's do it this way - we flip two coins, one of which I can see, one I can't. If I see that one is heads, the probability that the other is also heads is 1/2.

Not fair -- you now know the order of events. This is not the same as what I suggested. Look at what Jackson says ... "If you do not know the order of the events then the resulting probability is 33/67." I agree with that 100%.

I think I'm beginning to see the flaw in CurtC's thinking. He seems to feel that if I flip 2 coins and show him one of them as heads (after I flipped both) that because the second coin has two possible outcomes it's 50/50 for it being heads or tails. Not true. Although there are only two outcomes there were three ways to get there. I could have flipped HH, HT or TH -- each allows me to show him 1 coin as having heads. And each was equally likely. So to get HH, even seeing one coin as being heads, is 1/3.

 

[patnray]

Case 1: I flip a coin. It comes up heads. I flip a second coin. What's the probability that it comes up heads? 1/2

Case 2: I flip two coins without looking at the result. What's the probability they are both heads? 1/4

Case 3: I flip two coins without looking at the result. You look at one of them (I don't know which) and tell me that it is heads. What is the probability that the other is heads? 1/3, because 2 tails is no longer possible.

The confusion stems from the fact that the probability of heads on any individual coin flip is always 1/2. But when we flip two coins there are 4 possible outcomes. AFTER the coins have been flipped, and without any additional information (Case 2) we can say the probability of either coin being heads is still 1/2. But once we know that one of them is heads (but not which one), we do have additional information: they are not 2 tails (Case 3). We still do not know the status of either coin, but, in this case, the possible outcomes are HT, TH, or HH, with only one in three being both heads. If, on the other hand, we know that coin A is heads the problem is different (Case 1). There is only one coin we are unsure about and only 2 possibilities: H or T.

But I still maintain that the original problem does not contain enough information. Her children could be identical twins or adopted. Or she could have meant that exactly one of her children is a boy...

I vote E) Poorly worded question that requires subjective determination.

 

[gnome]

But I still maintain that the original problem does not contain enough information. Her children could be identical twins or adopted. Or she could have meant that exactly one of her children is a boy...

I vote E) Poorly worded question that requires subjective determination.

It is sometimes vaguely worded... but there is guidance in the fact that any other possibility makes the question trivial. By the fact that it's being asked and considered a puzzle, you can generally resolve the ambiguity.

 

[patnray]

It is sometimes vaguely worded... but there is guidance in the fact that any other possibility makes the question trivial. By the fact that it's being asked and considered a puzzle, you can generally resolve the ambiguity.

Generally. But sometimes recognizing the ambiguity is the point of the puzzle...

 

[69dodge]

Re: Re: Re: Re: Re: Probability question

Originally posted by Just thinking
consider if of the 400 mothers, 1 had BB, 2 had one of each and 397 had GG. If one of those 400 mothers showed up with a boy, you'd give her a 1/3 probability of other one being a boy, no?

No, 1/2.

There are four boys in all, and the one I see might be any of them with equal probability. Two of them have a brother, and two of them have a sister.

Consider two mothers, each with ten children. One has ten boys, the other has one boy and nine girls. You meet a woman with a son of hers. What's the probability that she's the one with ten boys? It's a lot more than 1/2, because if she were the one with nine girls, she probably would have been out with one of them instead of with her only boy.

Google for Bayes's theorem. The probability of A given B does depend on the probability of B given A.

 

[Just thinking]

Re: Re: Re: Re: Re: Re: Probability question

No, 1/2.

There are four boys in all, and the one I see might be any of them with equal probability.

I do not believe this to be the case. Here's why.

Each woman represents a group -- one group with 2 boys and 2 groups with one of each. You may see any of the three women with equal probability, but not each boy. Since we know that each woman has one boy with her, the probability for seeing a specific boy from either of the mixed groups is 1/3 each, the same for seeing his mother. The probability for seeing a specific boy from the first woman is 1/6 (1/3 x 1/2 -- a 1/3 chance of it being the two-boy mother times the 1/2 probability for a specific son). And since she has two boys, it's 1/6 for the other one as well. Now let's add up all 4 boys to see what we get ...

1/6 + 1/6 + 1/3 + 1/3 = 1

This shows that there is a 100% chance of seeing one of the four boys.

Are we on agreement with this?

I think what you are implying is that if I see a mother with a son, it's more likely she is the one with two boys rather than one with a boy and girl -- all things being equal. But my problem was in response to the initial post's problem where we know one of them to be a son. I simply allowed you to see the son instead of just hear about him -- each mother must therefore have a son with her (if she has at least one) if and when she's in the store. (This is how you know one of her children is male.) Remember, my problem starts with "A woman is in a store with her son ... " It does not start by saying "What is the chance a woman with 2 children is in a store with her son ..." In this way I am forcing her to bring (and show) her son, if she has one. This is really no different than the 2 coin toss where I show you one coin is heads and ask what is the chance both are heads -- 1/3.