Probability question

[Just thinking]

If the mother pull the boy out from behind a curtain and said, "look I have a boy", then you would be back to 1/3 again.

And that is what my initial intent was. Hey, I admit I worded it poorly, and gave a totally different problem from what I had in mind -- so I apologize for all the confusion this started. And I also appreciate and wish to thank all the explanations everyone offered to set me straight. So, I'd like to offer the problem this way to set things straight ... (I hope ):

There is a store in town where if a mother wishes to shop there with her child she is only allowed to bring one -- and it must be a boy child. Everyone in the universe knows of this rule. When someone walks up to her and discovers she has exactly 2 children, what is the probability that she will say the other child is a boy also?

 

[patnray]

The probability that one specific child ("the other child") is a boy is 1/2.

As long as we don't know if the child with her is the oldest or youngest, then the probability that both her children are boys is 1/3.

Not the same thing. The first is a probability about one child. The second is a probability about two children.

 

[CurtC]

No, patnray, it doesn't matter whether we know if the child is the oldest or youngest. If she has a boy with her (and there is not some other condition about her having to bring a boy if she has one like in Just Thinking's hypothetical), then when you see a boy with her, the probability for the other kid is 1/2.

 

[69dodge]

I realize this is a quick-and-dirty implimentation, but if you were planning on running this over a large number of iterations, it's inefficient and vulnerable to (short-term) deadlock.

It's a fairly tight loop, and these are usually worth the effort of optimizing.

Instead of randomizing both children and checking to see if they fit the criteria for inclusion, I would do something like this:

[pseudocode]

int child[2]; // an array

// Pick a child to be a boy.
knownchild := random(2)
child [knownchild] := gender[BOYval]

// gender is an enum. BOYval and GIRLval map to either 0 or 1. Doesn't really matter which you choose, so long as you code the rest of the program accordingly.

// pick a gender for the other child.
child [1 - knownchild] := gender[random(2)]

[/pseudocode]

This is guaranteed to run in one pass per trial, does not compromise (pseudo)-randomness and requires no comparisons.

It also gives a different answer.

oops.

 

[patnray]

No, patnray, it doesn't matter whether we know if the child is the oldest or youngest. If she has a boy with her (and there is not some other condition about her having to bring a boy if she has one like in Just Thinking's hypothetical), then when you see a boy with her, the probability for the other kid is 1/2.

That's what I said : The probability of the other child being male is 1/2 (one specific child).

But the probabilty of BOTH children being male is 1/3 so long as we don't know if the child with her is the oldest or youngest. The fact that he is male gives us information about the SET OF BOTH CHILDREN, not one particular child. (OK, we know the sex of the particular child standing next to her, but if we don't know which of her children he is then he doesn't give any more information about the set of both children).

Both proababilities are correct, but they are probabilities about different things (one child vs both children).

 

[treble_head]

Whoa, you guys are still here?

 

[Moose]

This is guaranteed to run in one pass per trial, does not compromise (pseudo)-randomness and requires no comparisons.

It also gives a different answer.

oops. [/B]

Huh? What did I mess up?

 

[Earthborn]

That's what I said : The probability of the other child being male is 1/2 (one specific child).

But the probabilty of BOTH children being male is 1/3 so long as we don't know if the child with her is the oldest or youngest.

Please explain why it is even relevant which one is oldest or youngest. As far as I see, the OP only tells us the set of both children consists of a known child and the other child.

If the other child is also a boy, then both children are boys. Always. Therefore the chance that both children are boys is the same as the chance that the other child is a boy.

 

[Moose]

Earthborn, how you discriminate between the two children isn't important, only the fact that you do, in this case.

Here's the difference:

If I show you a mother and her two sons, then ask you what the odds are that the next child will be a boy, the odds, barring any unusual factors, will be 50-50, as you correctly seem to expect. This is because you're evaluating a single potential event, in isolation.

If I show you a mother and one of her children, a boy, then ask you the probability that her other child is also a boy, then you're not looking at a single independant event, but a pair of linked events, after the fact, and asking what are the odds that this could have occurred. Such a problem has to be examined as a whole.

As such, there are four (nearly) equal possibilities. B-B, B-G, G-B, and G-G. Knowing that both children are not girls, you eliminate that possibility, leaving one possibility in three that both children are boys.

 

[Earthborn]

If I show you a mother and one of her children, a boy, then ask you the probability that her other child is also a boy, then you're not looking at a single independant event, but a pair of linked events, after the fact, and asking what are the odds that this could have occurred. Such a problem has to be examined as a whole.

I fail to see why. I fail to see why we need to consider 'events' or 'linked events' at all. If the other child is a boy, than both are boys. So whether both are boys depends only on the sex of the other child.

The question is not what the odds are that a chain of events could have occured. The question is what the chance is that the other child is (also) a boy.

Please explain why people are constantly dragging age and 'events' into the issue.

If I show you a mother and her two sons, then ask you what the odds are that the next child will be a boy, the odds, barring any unusual factors, will be 50-50

Suppose she now has 3 sons. What is the chance that the next will (also) be a boy. 50-50. What is the chance that the fifth will (also) be a boy. 50-50. It's always the same chance.

In the problem as it is stated in the OP, time is irrelevant. It makes no difference whether it asks for the chance the next child is a boy, or whether the other child is a boy or whether the child of a neighbour is a boy.

 

[Just thinking]

I fail to see why. I fail to see why we need to consider 'events' or 'linked events' at all. If the other child is a boy, than both are boys. So whether both are boys depends only on the sex of the other child.

The question is not what the odds are that a chain of events could have occured. The question is what the chance is that the other child is (also) a boy.

Because the "other child" is part of a linked event. When you already have a pair of events established (in this case two separate births from the same mother which already happened) you must consider all the possible ways a given pair of offspring could have happened -- as well as how much information you know about each one (if given any).

When someone considers the probability of the 'other child' being male, it all depends on how that event is percieved. If you look at it as a completely separate event unto itself, then it's 50-50. But that's not what this thread is dealing with. If you know that the mother has 2 kids (not with her) and at least one is a male (one you know nothing about), then the 'other child' being male is now 33-67 (1/3). If you randomly see her with a male child (also with her at random from her litter of 2), the odds are again 50-50 that the 'other child' is male. If she was told to bring one male child with her (if she has at least one to bring), and then you see her with one male child, then it's back to 33-67 for the 'other' being male.

See, it all depends on how the problem is constructed and what parameters are known and met. Think about it -- if she was told to bring one male child (if she has one) to an event and you see her with no child, but she says she has 2 kids, guess what -- you now know she has 2 girls. The probability of either child being male is 0, even though the chances of a single male birth are still 50%.

 

[treble_head]

well, I took the time to physically do the test. I had a friend flip 2 coins 300x without telling him why (he'll do anything for a free meal). I had decided ahead of time to count heads as boys (since coins have boys on their heads) and tails were rated as girls.

I categorized the data as BG(GB), BB, and GG. There was no need to seperate the data for BG or GB as they were thrown at the same time.

The data came out to:

BG = 147
BB = 73
GG = 80 N/A

The two remaining applicable numbers were BG=147 and BB=73

What's that, you say? But 73 is 1/2 of 147! (.4965986)

BUT it is approximately 1/3 of the total (220). You have to take the total number of throws into account before figuring out the percentages of groups (minus the non-applicable data GG).

hope this helps some people out. After this, i am done.

 

[Earthborn]

Because the "other child" is part of a linked event.

But it isn't relevant. The same question as the OP can be asked without any 'linked event'.

Suppose the question was: two children live in houses next to eachother. One of them is a boy. What is the chance that the other is also a boy? It would be the exact same problem. You can argue until you see blue in the face that we must consider a totally irrelevant thing, such as which one is older or which one lives in the house with the highest number, but that doesn't make it suddenly relevant.

All that is relevant is:
KnownChild = Boy
UnknownChild = Boy or Girl

The OP doesn't ask anything else.

you must consider all the possible ways a given pair of offspring could have happened

No, I don't.

If you know that the mother has 2 kids (not with her) and at least one is a male (one you know nothing about), then the 'other child' being male is now 33-67 (1/3).

No, it isn't. The sex of the 'other child' is independent of the child I'm told is male. Therefore I must conclude that the chance of the other child being male is 50-50.

If you randomly see her with a male child (also with her at random from her litter of 2), the odds are again 50-50 that the 'other child' is male.

If we assume the mother tells the truth, it should make no difference whether we see the child or whether we are told about the child.

If she was told to bring one male child with her

But she wasn't.

See, it all depends on how the problem is constructed and what parameters are known and met.

Yes, I agree. I just can't understand how anyone could interpret the OP as anything else than asking what the chance is that the other child is male.

 

[CurtC]

But the probabilty of BOTH children being male is 1/3 so long as we don't know if the child with her is the oldest or youngest.

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.

 

[69dodge]

Originally posted by Earthborn
As far as I see, the OP only tells us the set of both children consists of a known child and the other child.

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. Let's talk about CurtC's census version instead. You ask someone, whom you don't know, if she has exactly two children. She says, "Yes." Then you ask her, "Are they both girls?". She says, "No." Now, what's the probability that they're a boy and a girl, and what's the probability that they're both boys?

Of parents with two children, twice as many have a boy and a girl as have two boys. So the probabilities are 2/3 and 1/3, respectively.

If the other child is also a boy, then both children are boys. Always. Therefore the chance that both children are boys is the same as the chance that the other child is a boy.

Yes, but that chance isn't 1/2 unless the determination of which child will be called "the other one" is made totally at random. In this case, it is not totally random: if the parent has one boy and one girl, by saying "one is a boy" she has guaranteed that it is the girl who is the "other one". So, the "other one" is more likely to be a girl than a boy, not because of any mysterious influence at the time of conception, but simply because if there is a girl we've chosen to call her the "other one".

 

[casebro]

Treble head, flipping 2 coins at a time is not the question. The question is " We flipped one coin. It is heads. What is the probability the the next coin will be heads?" answer: 1/2

 

[Earthborn]

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

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

would not have said "one is a boy" if the child in question had in fact been a girl

I'm assume the parent is honest and does not tell us that the randomly chosen child is a boy if it was a girl.

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.

That may be true, but I don't understand why they are supposing it. What in the OP tells us that the parent will only tell us anything about boys? Absolutely nothing, as far as I can see.

The OP is not totally clear.

That's why I try to presume what I consider the simplest interpretation: parent tells us the sex of one child, and asks for the sex of the other child. The Yersinia may have meant something different, but then she (I guess) should have mentioned it. Maybe the people who say the answer is 1/3 have once heard a similar puzzle in which the answer was correctly 1/3 because it mentioned a few other things, and confuse it with this one. I try to take the OP literally.

Of parents with two children, twice as many have a boy and a girl as have two boys. So the probabilities are 2/3 and 1/3, respectively.

Duh!

In this case, it is not totally random: if the parent has one boy and one girl, by saying "one is a boy" she has guaranteed that it is the girl who is the "other one". So, the "other one" is more likely to be a girl than a boy, not because of any mysterious influence at the time of conception, but simply because if there is a girl we've chosen to call her the "other one".

That assumes Yersinia could not have told us "one is a girl". But we don't know that, therefore we must not assume it. Unless we have evidence of such rules, we must assume that Yersinia was equally likely to say "one is a girl" as "one is a boy" if there is a boy and a girl.

Here are the possibilities:
"I have 2 children, one is a boy." KnownChild = Boy UnknownChild = Boy
"I have 2 children, one is a boy." KnownChild = Boy UnknownChild = Girl
"I have 2 children, one is a girl." KnownChild = Girl UnknownChild = Boy
"I have 2 children, one is a girl." KnownChild = Girl UnknownChild = Girl

The OP excludes the last two possibilites, so only two are left. Simply assuming that only boys are ever mentioned is... well... sexism.

And it also doesn't explain why people are constantly dragging to order of birth into the issue.

 

[treble_head]

Treble head, flipping 2 coins at a time is not the question. The question is " We flipped one coin. It is heads. What is the probability the the next coin will be heads?" answer: 1/2

That is not the question. the original question is "I have 2 children, one is a boy. what is the prob that the other is also a boy?" (from original quote)

ALSO A BOY. It does not imply that the other is or is not a boy, and it does NOT NOT NOT NOT NOT ask whether the other is a boy or a girl.

You DO NOT know which coin has been flipped first, only that one is heads. Both children are already born!

It asks whether the other child is also a boy. ALSO a boy.

BOTH HAVE TO BE BOYS for the equation to turn out in that favor.

According to the original question, one has to be a boy. That leaves out Tails + Tails, in my test or any other.

Could I just repeat this...
ALSO A BOY
ALSO A BOY
ALSO A BOY!!!!!!!
NOT WHAT ARE THE ODDS OF THE OTHER CHILD BEING A BOY.

According to the rules, if I flipped 2 coins at the same time, it does not matter what coin I flipped first. According to the original problem, both coins are already flipped. You know ONE is a boy.

If you already know that Girl Girl is not an option, the coin toss is my advantage. If I toss Tails Tails, it is not an option.

If I throw anything else, I can show you that one is Heads (male). It does not affect whether the other throw is male or female. It's a 50/50 deal as to what you throw, but it doesn't take into account ALL of the throws. Given all the throws, the likelyhood of Male Male is still 1/3. If you believe otherwise, we can test this over the phone and you can throw the coins. I will still win $66 out of every $100 we both bet.

 

[69dodge]

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

Isn't that the same as what I said? But I have no complaints about the way you phrased it here either. So, no problem.

I'm assume the parent is honest and does not tell us that the randomly chosen child is a boy if it was a girl.

Ok, I understand how you interpret the statement, and under that interpretation I agree that the answer is 1/2.

Of course we all assume the parent is honest. But saying "one of my children is a boy" is not a lie unless both children are girls. (Saying "I have chosen one of my children at random, and it's a boy" is a lie if the chosen child is a girl and the other child is a boy. But that's a different statement.) So, interpreting the original statement as ruling out only the two-girl situation is reasonable too. In fact, it might be considered to be the interpretation that makes the fewest assumptions: initially, all combinations of boys and girls are possible, and the only combinations we eliminate are those that we must eliminate because they would make the statement a lie.

 

[Earthborn]

You DO NOT know which coin has been flipped first

Why is it relevant which coin has been flipped first?

ALSO A BOY!!!!!!!
NOT WHAT ARE THE ODDS OF THE OTHER CHILD BEING A BOY.

Same thing. If the other child is a boy, then the other child is also a boy. I really don't see why you are making such a fuss over that word 'also'. It really doesn't make any difference.

BOTH HAVE TO BE BOYS for the equation to turn out in that favor.

That's right. And given that the known child is already a boy, whether they are both boys is solely dependent on the sex of the unknown child.

If you believe otherwise, we can test this over the phone and you can throw the coins. I will still win $66 out of every $100 we both bet.

Before you do, I think you'll first have to convince others that your interpretation of the question is the correct one. Or even a reasonable one.

Once you have done that, you'll find that everybody already agrees with you about the odds.