This is one of those questions that Intro to Probability teachers like to throw at their students:

Suppose you approach someone who you know has two children, but you don’t know their children’s genders. You ask the person “Is one of your children a boy?” and they answer “Yes.” What is probability that the person’s other child is also a boy? (In other words, once you know that one of their children is a boy, what are the odds that they have two boys?)

I think 1/3.

1/2
The trouble is is that people know the chance of both being a boy is 1/4, but once they hear 1 is a boy they incorrectly conclude the second is now more likely to be a girl.

Mathmaticaly:
Biologicaly it's a 1/2 chance for each gender. For two boys, we multiply 1/2 and 1/2 for each boy and get the intuitive answer of 1/4 chance of having 2 boys.
However, in this case, it is a 1/1 chance of the first one being a boy (we are absolutly certain he is) thus, when we multiply 1/1 by 1/2 and get the answer of 1/2 for the second one's probability of being male.

-Andrew

The answer is 1/3.

Here I am assuming the probability of any one child being a boy versus a girl is a fifty-fifty proposition. It's not; boys slightly outnumber girls at birth. But I'm ignoring that little fact.

With two children, there are four equiprobable ordered events: <G,G>, <G,B>, <B,G>, and <B,B>. The knowledge that one of them is a boy eliminates the two daughers case. What's left are two cases with one boy and one girl, and one case with two boys. These are equiprobable, mutually exclusive events than span the space. Two of the three cases have a girl. The odds are 2/3 that the other child is a girl, 1/3 that the other child is a boy.

This can also be solved with Baye's Law. It yields the same result.

This can also be solved with Monte-Carlo techniques. These yield the same result (to within statistical error). I got a girl in 6661 out of 10000 cases and a boy in the other 3339 cases.

Could the probability not be restated in the following way?

Assume a man is walking along and meets a high school friend. His friend is now married, and has his wife and son with him. The wife is also pregnant. The man knows one child is a boy, what is the probability that the yet-to-be born child will be a boy?

1/2
Mathmaticaly:
Biologicaly it's a 1/2 chance for each gender. For two boys, we multiply 1/2 and 1/2 for each boy and get the intuitive answer of 1/4 chance of having 2 boys.
However, in this case, it is a 1/1 chance of the first one being a boy (we are absolutly certain he is) thus, when we multiply 1/1 by 1/2 and get the answer of 1/2 for the second one's probability of being male.


No, you don't know whether the first one was a boy or not. You only know that ONE of them was a boy. Before you receive any information, there are four equally likely possibilities: BB, BG, GB and GG. Since you know that one of them is a boy, that rules out the GG possibility. That leaves BB, BG and GB as (equiprobable) possibilities, so the answer is 1/3.

Could the probability not be restated in the following way?

Assume a man is walking along and meets a high school friend. His friend is now married, and has his wife and son with him. The wife is also pregnant. The man knows one child is a boy, what is the probability that the yet-to-be born child will be a boy?

No, in this case you have extra information (that the first child was a boy). In the original problem, you only know that there is at least one boy, not whether he was the first-born or not. If you know the first-born is a boy, the odds that the second child is a boy are 1/2.

With two children, there are four equiprobable ordered events: <G,G>, <G,B>, <B,G>, and <B,B>. The knowledge that one of them is a boy eliminates the two daughers case. What's left are two cases with one boy and one girl, and one case with two boys. These are equiprobable, mutually exclusive events than span the space. Two of the three cases have a girl. The odds are 2/3 that the other child is a girl, 1/3 that the other child is a boy.
Perhaps you should read this article, it will explain why your logic is fallacious.
The gambler's fallacy is a formal fallacy. It is the incorrect belief that the likelihood of a random event can be affected by or predicted from other, independent events. -Wikipedia (http://en.wikipedia.org/wiki/Gambler's_fallacy)
While you can plot them as ordered events, you must realize that both <G,G> and <G,B> are eliminated, because the first event of both of them is a girl, which we confirmed is false.

Here I am assuming the probability of any one child being a boy versus a girl is a fifty-fifty proposition. It's not; boys slightly outnumber girls at birth. But I'm ignoring that little fact.
It is a 50/50 chance. The sperm carries either an x or a y chromosome, and there are always en equal number of each created.
Stitistically more males may be born, but this doesn't mean it isn't 50/50 (once again, Gamblers Fallacy.)

Could the probability not be restated in the following way?

Assume a man is walking along and meets a high school friend. His friend is now married, and has his wife and son with him. The wife is also pregnant. The man knows one child is a boy, what is the probability that the yet-to-be born child will be a boy?
Yep, exactly the same problem, 1 unknown with 2 possibilities, equal likelynes of each, and 1 known red hearring.

-Andrew

This is one of those questions that Intro to Probability teachers like to throw at their students:

Suppose you approach someone who you know has two children, but you don’t know their children’s genders. You ask the person “Is one of your children a boy?” and they answer “Yes.” What is probability that the person’s other child is also a boy? (In other words, once you know that one of their children is a boy, what are the odds that they have two boys?)
The probability teacher's answer is 1/3, of course.

Butl, in practice there's a considerable chance that the person would respond differently if they had two boys, eg:

"Is one of your children a boy?"
"No, they are both boys."

This is a legitimate response - it assumes an implied "exactly" rather than an implied "at least one".


This problem has a history of causing trouble in 'Net space, mainly because of the difficulty in contriving an unambiguous situation where you eliminate the possibility of two girls, without influencing the probability of two boys.

While you can plot them as ordered events, you must realize that both <G,G> and <G,B> are eliminated, because the first event of both of them is a girl, which we confirmed is false.
Not so.
In the case of <G,B>, "Yes" is clearly a legitimate response to the question as posed.

Not so.
In the case of <G,B>, "Yes" is clearly a legitimate response to the question as posed. Ah yes sorry. (quadraphonics posted while I was typing.)
However, since we are only concerned with the gender of the other boy, after the information, the events (G,B) and (B,G) become identical. If it were worded as "What is the probability that the second(or first) born child was a boy?" it would be 1/3.


-Andrew

Ah yes sorry. (quadraphonics posted while I was typing.)
However, since we are only concerned with the gender of the other boy, after the information, the events (G,B) and (B,G) become identical. If it were worded as "What is the probability that the second(or first) born child was a boy?" it would be 1/3.
The two events are separate events.

The mention of "the other child" is a linguistic trick which is (deliberately?) not mentioned in the OP.
As soon as you think or say "the other child", you're falsely implying that a particular child has already been identified, which is not the case.

We are only concerned with whether both children are boys.

No, in this case you have extra information (that the first child was a boy). In the original problem, you only know that there is at least one boy, not whether he was the first-born or not. If you know the first-born is a boy, the odds that the second child is a boy are 1/2.
'First-born' has nothing to do with it. The 'first-born' child could have been an unseen daughter at the movies. The only information I gave was that one child was a boy and the wife was pregnant. I gave less information than was in the original version.

Could the probability not be restated in the following way?

Assume a man is walking along and meets a high school friend. His friend is now married, and has his wife and son with him. The wife is also pregnant. The man knows one child is a boy, what is the probability that the yet-to-be born child will be a boy?

No, that's a different problem. The OP deals with dependent events, your restatement deals with independent events.
The events this child here is a boy and that unborn child is a boy are independent.
The events at least one of these two children are boys and both of these two children are boys are dependent.

'First-born' has nothing to do with it. The 'first-born' child could have been an unseen daughter at the movies. The only information I gave was that one child was a boy and the wife was pregnant. I gave less information than was in the original version.

My usage of the term "first-born" means, here, first out of the two children under consideration. Telling us that the wife has had a boy before the other child in question is more info than the original version, where we don't know the ordering. This is why the original version gives a probability of 1/3, while your version gives a probability of 1/2. It is MORE information, and it is a different problem with a different outome.

Is this going to turn into another ``can a fly stop a train'' thread?

The two events are separate events.

The mention of "the other child" is a linguistic trick which is (deliberately?) not mentioned in the OP.
As soon as you think or say "the other child", you're falsely implying that a particular child has already been identified, which is not the case.

We are only concerned with whether both children are boys
Alright allow me to put it this way:
The events (B,G) and (G,B) loose half their weight in this case; each is 25% likely to be true, while (B,B) is 50% likely.
Let's break the problem down furthur:
We incorporate the persons answer into the scenario, not just by eliminating the (G,G) possibility, but by taking which child she could be reffering to into account.
She is either reffering to child 1 or child 2. The new events appear as follows:
(B,G,1)
(G,B,2)
(B,B,1)
(B,B,2)
Events:
(B,G,2), (G,B,1), (G,G,1) (G,G,2) are contradictory and are therefore imppossible.
Now we see, there is a 1/2 chance of it being 2 boys and a 1/2 chance of there being a girl.

The fact is, the gender of the childs are independant.

-Andrew

My usage of the term "first-born" means, here, first out of the two children under consideration. Telling us that the wife has had a boy before the other child in question is more info than the original version, where we don't know the ordering. This is why the original version gives a probability of 1/3, while your version gives a probability of 1/2. It is MORE information, and it is a different problem with a different outome.

It's more than that - the OP doesn't identify a child in any way. In order to answer the question "Do you have [at least one] boy?", the responder has to consider her two children as a set. The response could refer to either or both.
In turn, the probability we want to calculate is a property of both children as a set, not just one of them.

Alright allow me to put it this way:
The events (B,G) and (G,B) loose half their weight in this case; each is 25% likely to be true, while (B,B) is 50% likely.
That doesn't make any sense at all.
We incorporate the persons answer into the scenario, not just by eliminating the (G,G) possibility, but by taking which child she could be reffering to into account.
Yes, that's the trap I referred to in my first response to the thread.
In the idealised probability teacher's scenario, the person isn't referring to one of her children, she is referring to them both as a set.

You can imagine that the person interprets the question in a different way (as I suggested), which would change the resulting answer.

She is either reffering to child 1 or child 2. The new events appear as follows:
(B,G,1)
(G,B,2)
(B,B,1)
(B,B,2)
Events:
(B,G,2), (G,B,1), (G,G,1) (G,G,2) are contradictory and are therefore imppossible.
Now we see, there is a 1/2 chance of it being 2 boys and a 1/2 chance of there being a girl.
Your analysis implies that the person has listened to the question, then chosen one of her children at random, and said whether that chosen child is a boy or not.
Ie you eliminate (B,G,2) and (G,B,1) because in those situations the person would have answered "No".

OK, let's restate the OP like this.

You approach somone and ask if they have any children. He states 'I have a son and my wife is pregnant'. What is the probability the unborn child will be a boy?

That's not restating the OP - it's a different problem. I thought I explained that pretty well in the previous post.
To answer the question, the probability the unborn child will be a boy is of course 50%.

That's not restating the OP - it's a different problem. I thought I explained that pretty well in the previous post.
To answer the question, the probability the unborn child will be a boy is of course 50%.
Of course, now what is the probability the man on the street will have two boys after the child is born, given that one is already known to be a boy?

50%, of course.
But that's still not the same problem as the OP.

Alright, I got my head around it now, thanks Pete... counterintuitive indeed.

-Andrew

Is this going to turn into another ``can a fly stop a train'' thread? OK, suppose you are in a dark train car with a mother and her two children of unknown sex.

1) You shine a flashlight on one of the children, and it is a boy. What is the probability that the other child is a boy?

2) You ask the mother to shine the light on a son of hers, if any be present. She does. What is the probability that the other child is a boy?

You haven't let me finish yet, Pete. :)

Now, assume a man approaches a casual friend on the street. He has heard the man has one child and his wife is pregnant and in the hospital. He asks the sex of his child and the friend replys 'A boy'. The friend is not sure if his friend is referring to his first-born or if the new child has already been born and is a boy. If his friend is referring to the first-born as a boy, there is a 50% probability that both children will be boys, correct? If his friend is referring to the new-born being a boy, the probability that both are boys is 1/3, correct?

If his friend is referring to the first-born as a boy, there is a 50% probability that both children will be boys, correct? If his friend is referring to the new-born being a boy, the probability that both are boys is 1/3, correct? 50% each time. As soon as you specify one of the children, the probability that the other one is a boy is 50%.

It's the lack of specificity in "one of the children is a boy" that allows two equiprobable ways of the other being a girl - compared with one equiprobable way of the other being a boy, netting 1/3.

Any specifying will do - if she says "the taller one is a boy", the odds become 50% on the other one - again, in math problem world, not reality.

Yes, and if the friend had said 'It's a boy' and handed him a cigar, the odds of both being boys is 50%. But remember, he did not specify which was a boy, so why is the probability still 50%?

Yes, and if the friend had said 'It's a boy' and handed him a cigar, the odds of both being boys is 50%. But remember, he did not specify which was a boy, so why is the probability still 50%? If the newborn - specified by the cigar ritual - is a boy, there are only two equiprobable events: the other child is a girl, the other child is a boy.

50%.
But remember, he did not specify which was a boy, so why is the probability still 50%? If he does not specify - and that is difficult, in real life rather than math problem world (the cigar gives it away) - then there are three equiprobable events: two that have girls in them, one that has two boys.

One of the girl possibilities would be eliminated by any specifying of which child is a boy for sure, (such as the newborn, the taller, the older, etc) leaving two equiprobable events.

iceaura,
If he does not specify - and that is difficult, in real life rather than math problem world (the cigar gives it away) - then there are three equiprobable events: two that have girls in them, one that has two boys.

Your 'three' equiprobable events are (1) boy/girl, (2) girl/boy, and (3) boy/boy. If one child is a boy, then there are only two possibilities, boy/girl or boy/boy. Naming a specific child makes no difference in a boy/girl or girl/boy possibility.
One of the girl possibilities would be eliminated by any specifying of which child is a boy for sure, (such as the newborn, the taller, the older, etc) leaving two equiprobable events.
Why would that be true? It is not logical to assume that specifying a particular boy would reduce the boy/girl and girl/boy possibilities to one probability, while not specifying a particular boy leaves both possibilities open.

Your 'three' equiprobable events are (1) boy/girl, (2) girl/boy, and (3) boy/boy. If one child is a boy, then there are only two possibilities, boy/girl or boy/boy. In all three equiprobable events, one child is a boy.
It is not logical to assume that specifying a particular boy would reduce the boy/girl and girl/boy possibilities to one probability, while not specifying a particular boy leaves both possibilities open. Specifying one of the children as a boy rules out the boy/girl combo that had that child as a girl. Only one boy/girl combo is left.

Not specifying either child leaves both children possibly female, and we have two possible boy/girl pairs.

You haven't let me finish yet, Pete. :)

Now, assume a man approaches a casual friend on the street. He has heard the man has one child and his wife is pregnant and in the hospital. He asks the sex of his child and the friend replys 'A boy'. The friend is not sure if his friend is referring to his first-born or if the new child has already been born and is a boy.

This is equivalent to a specific interpretation of the OP described earlier:
She is either reffering to child 1 or child 2. The new events appear as follows:
(B,G,1)
(G,B,2)
(B,B,1)
(B,B,2)
Events:
(B,G,2), (G,B,1), (G,G,1) (G,G,2) are contradictory and are therefore imppossible.
Now we see, there is a 1/2 chance of it being 2 boys and a 1/2 chance of there being a girl.
Your analysis implies that the person has listened to the question, then chosen one of her children at random, and said whether that chosen child is a boy or not.
Ie you eliminate (B,G,2) and (G,B,1) because in those situations the person would have answered "No".
It's a valid interpretation, but it's not the probability teacher's intended interpretation.

If his friend is referring to the first-born as a boy, there is a 50% probability that both children will be boys, correct? If his friend is referring to the new-born being a boy, the probability that both are boys is 1/3, correct?
It's 50% either way.

I can see this descending into an impenetrable discussion about interpretations, further clouded by misunderstandings. It's not going to be pretty.

If he does not specify - and that is difficult, in real life rather than math problem world (the cigar gives it away) - then there are three equiprobable events: two that have girls in them, one that has two boys.
Your 'three' equiprobable events are (1) boy/girl, (2) girl/boy, and (3) boy/boy. If one child is a boy, then there are only two possibilities, boy/girl or boy/boy. Naming a specific child makes no difference in a boy/girl or girl/boy possibility.
It most certainly does. The statements "I have two children", "At least one of my two children is a boy", and "The older of my two children is a boy" are different statements with different amounts of information. Information is very closely connected to probability. The latter two statements must yield different probabilities for the gender of the unspecified child because the two statements have different information content.

Without any knowledge of gender, there are four equiprobable gender events for the two children: <boy,boy>, <boy, girl>, <girl,boy>, and <girl,girl>. Here I am using angle brackets to denote an ordered pair. Now suppose you are given additional information that one of the children is a boy. That information lets you remove the <girl,girl> pair from the realm of possibilities and nothing more.

Information regarding order and gender is more information than gender information alone. If you are told the older is a boy you can now remove the <girl,boy> pair from the realm of possibilities.

DH,
Information regarding order and gender is more information than gender information alone. If you are told the older is a boy you can now remove the <girl,boy> pair from the realm of possibilities.
I have understood the algebra 'logic' behind this all along. My question is why is the child listed first in the set assumed to be older (or taller, or more stupid)? The full sets were <boy,boy>, <boy,girl>, <girl,boy>, and <girl,girl>. Would the results be the same if I stated 'my most stupid child is a boy'? Why would this eliminate the <girl,boy> pair from the realm of possibilities? Would that be different than stating 'I have a stupid boy' when it is known that I have two children?

If I give no information other than 'one of my two children is a boy', then the remaining child must be either a boy or a girl, regardless if I mentioned one boy was stupid. I must have two children of the same sex, or two children of opposite sexes. Focusing on eliminating one of the <boy,girl> and <girl,boy> sets by specifying the 'stupidest' is a boy seems a fallacious step to me. If I state I have two children, then there is a 1/4 probability they could be any of the original four sets. But once I specify that one of the children is a boy, then I have either two children of the same sex or two children of different sexes. The probability that both are boys is 1/3. The probability that they are of different sexes is 2/3. Stating that the 'stupidest' is a boy doesn't change those probabilities as far as I can tell.

And finally. :D Let's say I have a boy. My wife is pregnant. What is the probability the child will be a boy? 50%, correct? What is the probability that I will have two children of different sexes after the child is born? It is not 'additional information' that is important, I believe it is the specific sequence that the events take place, the order.

1/2 * 1/2 = 1/4

Weird, I found that I struggled with the original statement of the problem. But once I transformed it into a statement about coin flips the solution became obvious.

I suppose this thread was a success in that it generated a lot of discussion, but it’s clear that I should have been more explicit in my original question. I thought it was pretty clear that in my original formulation of the question the “questioner” doesn’t know if the parent is referring to their first of second child when they say “One of my children is a boy,” but obviously it wasn’t clear enough.

So, as other people have stated, if you only know that one child is a boy then the odds of the other child being a boy are only 1/3. But if you know that the first child is a boy, or that the tallest child is a boy, etc., then the odds of the other child also being a boy are 1/2.

I suppose this thread was a success in that it generated a lot of discussion, but it’s clear that I should have been more explicit in my original question. I thought it was pretty clear that in my original formulation of the question the “questioner” doesn’t know if the parent is referring to their first of second child when they say “One of my children is a boy,” but obviously it wasn’t clear enough.

Nah, don't blame yourself. Certain posters deliberately chose to be obtuse about it.

I suppose this thread was a success in that it generated a lot of discussion, but it’s clear that I should have been more explicit in my original question. I thought it was pretty clear that in my original formulation of the question the “questioner” doesn’t know if the parent is referring to their first of second child when they say “One of my children is a boy,” but obviously it wasn’t clear enough.

So, as other people have stated, if you only know that one child is a boy then the odds of the other child being a boy are only 1/3. But if you know that the first child is a boy, or that the tallest child is a boy, etc., then the odds of the other child also being a boy are 1/2.
I had thought about coins myself, as mapsdnasggeyerg mentioned.

You know I have flipped two coins. You ask me if one of the coins came up heads, to which I replied 'yes'. At that point, the probability both flips came up heads is 1/3. If you ask me if the first flip came up heads and I replied 'yes', then the probability that both coins are heads is 50%. But if you ask me if one of the coins was a dime instead, the probability that both coins came up heads remains at 1/3. That's what I object to when you state 'the tallest child is a boy, etc.'. That gives no more information than stating one of the coins was a dime.

I had thought about coins myself, as mapsdnasggeyerg mentioned.

You know I have flipped two coins. You ask me if one of the coins came up heads, to which I replied 'yes'. At that point, the probability both flips came up heads is 1/3. If you ask me if the first flip came up heads and I replied 'yes', then the probability that both coins are heads is 50%. But if you ask me if one of the coins was a dime instead, the probability that both coins came up heads remains at 1/3. That's what I object to when you state 'the tallest child is a boy, etc.'. That gives no more information than stating one of the coins was a dime.
"One of the coins is a dime" is equivalent to "One of the children is 4'2" tall." Tis adds no relevant info, as you say. But...
"The tallest child is a boy" is equivalent to "The dime is a head."

"One of the coins is a dime" is equivalent to "One of the children is 4'2" tall." Tis adds no relevant info, as you say. But...
"The tallest child is a boy" is equivalent to "The dime is a head."
Sure Pete, but it is not equivalent to "The dime is a head and the first flip". "The tallest child is a boy" does not eliminate a first-born girl.

Sure Pete, but it is not equivalent to "The dime is a head and the first flip". "The tallest child is a boy" does not eliminate a first-born girl.

It doesn't matter. ANY ordering of the children will change the problem, regardless of whether it's by height, weight, color, odor or whatever.

It doesn't matter. ANY ordering of the children will change the problem, regardless of whether it's by height, weight, color, odor or whatever.
But it doesn't change the probabilities unless you state whether the first-born was a girl or a boy.

But it doesn't change the probabilities unless you state whether the first-born was a girl or a boy.

Yes, it does. If you don't believe me, write a compute program to simulate it. Or persist in your ignorance; it doesn't matter to me.

Yes, it does. If you don't believe me, write a compute program to simulate it. Or persist in your ignorance; it doesn't matter to me.
When you write a program, you are basing it on algebra and sets. That is what I am arguing, that the algebra can lead to false probabilities because one of the <B,G> or <G,B> sets is eliminated when there is no logical reason to do so.

You do argree that tossing two separate coins is the same exercise as using boys and girls?
When we flip two coins, then specify one of the coins is heads, the probability that both are heads is 1/3. When we have two children, then specify one of them is a boy, the probability that both are boys is 1/3. If we specify the larger coin is a heads, that does not change the probability that both are heads. The probability that both are heads is still 1/3. Do you disagree? If we specify that the taller child is a boy, that does not change the probability that both are boys. The probability that both are boys is still 1/3. Do you disagree?

If we specify the larger coin is a heads, that does not change the probability that both are heads. The probability that both are heads is still 1/3. Do you disagree?

Yes. When you tell us "the larger coin is heads" you give us two pieces of information: one coin is larger than the other, and the large one is heads. The first piece of information lets us group all possible events as follows: <H,h>, <H,t>, <T,h> and <T,t> (where capital letters correspond to the large coin, and lower-case letters correspond to the small coin). When you tell us that the large coin is heads, we eliminate the last two possibilities (in which the large coin is tails), leaving two equiprobable possibilities. Thus, the answer is 1/2 for this problem. If you had only told us that one of the coins was heads (but not which one), we'd only have been able to eliminate the last possibility (<T,t>), which would leave three equiprobable events. That latter version is equivalent to the original problem; any time that you distinguish one coin from the other, you change the problem.


If we specify that the taller child is a boy, that does not change the probability that both are boys. The probability that both are boys is still 1/3. Do you disagree?

Yes, for the exact same reasons given above.

The thing is, I had already flipped the coins before I told you the larger coin was a heads. ;)

The thing is, I had already flipped the coins before I told you the larger coin was a heads. ;)

Yes, obviously. So what?

Careful - there could be a bit of the Monty Hall problem creeping in.

2inq, try this experiment.
Flip 2 coins. If the larger coin is a head, note whether both are heads.
Repeat 29 times, for 30 tosses altogether.

I'm doing this right now, and will post the results when I'm done.
I'm guessing that the larger coin will be a head about 15 times, and that both coins will be ahead about half that number.

Sorry... I got called away partway through.

Here's my rsults:
/---------------------------------------\
| | Larger coin | Both coins |
| Trial | is a head? | are heads? |
|---------------------------------------|
| 1 | Y | N |
| 2 | N | |
| 3 | Y | N |
| 4 | N | |
| 5 | Y | N |
| 6 | Y | N |
| 7 | Y | Y |
| 8 | N | |
| 9 | Y | N |
| 10 | Y | Y |
| 11 | Y | Y |
| 12 | Y | N |
| 13 | N | |
| 14 | N | |
| 15 | Y | Y |
| 16 | N | |
| 17 | N | |
| 18 | Y | N |
| 19 | Y | N |
| 20 | N | |
| 21 | N | |
| 22 | Y | N |
| 23 | N | |
| 24 | Y | Y |
| 25 | N | |
| 26 | N | |
| 27 | Y | Y |
| 28 | Y | N |
| 29 | N | |
| 30 | N | |
|---------------------------------------|
| Total | 16 | 6 |
\---------------------------------------/

You could do larger trials more easily with a computer simulation, as quadraphonics suggested.
This doesn't mean using the computer "basing it on algebra and sets." It means simulating coin flips with a random number generator, and counting the results.

...For example, it's trivial to get a computer to generate 1000 pairs of random numbers, then count those where the larger number is even and those where both are even.
I've done it just now, with the following results:
Larger is even: 482
Both are even: 238 (49.3%)

It's just as easy to do the same for as many trials as you like. Here's the result of a million trials:
Larger is even: 500216
Both are even: 250444 (50.06%)

Here's the javascript code to do it. If you copy this code into Notepad and save it as "Test.htm", you can run it yourself:
<html>
<head>
<title>Probability test</title>
<script type="text/javascript">
function runTrials () {
var iterations = document.getElementById("Iterations").value;
var largerIsEven = 0;
var bothAreEven = 0;

for (var loop=0; loop<iterations; loop++) {
var firstNumber = Math.round(100*Math.random());
var secondNumber = Math.round(100*Math.random());

if (firstNumber > secondNumber) {
if (firstNumber % 2 == 0) {
largerIsEven++;
if (secondNumber % 2 == 0) {
bothAreEven++;
}
}
} else {
if (secondNumber % 2 == 0) {
largerIsEven++;
if (firstNumber % 2 == 0) {
bothAreEven++;
}
}
}
}

document.getElementById("largerIsEven").value = largerIsEven;
document.getElementById("bothAreEven").value = bothAreEven;
document.getElementById("proportion").value = Math.round(bothAreEven*10000/largerIsEven)/100 + "%";
}

function clearResults() {
document.getElementById("largerIsEven").value = "";
document.getElementById("bothAreEven").value = "";
document.getElementById("proportion").value = "";
}
</script>
</head>
<body>
<form>
<p>How many trials? <input type="text" id="Iterations" size="10" value="10000" onchange="clearResults();" />
<input type="button" value="Go!" onclick="runTrials();" /></p>
<hr />
<p>Larger is even: <input type="text" id="largerIsEven" size="8" value="" readonly="true" /></p>
<p>Both are even: <input type="text" id="bothAreEven" size="8" value="" readonly="true" /></p>
<p>Proportion: <input type="text" id="proportion" size="8" value="" readonly="true" /></p>
</form>
</body>

My results were a little different due to chance. But here they are:

---- small coin------ large coin ----- both heads
(1)-- H -------------- H ------------Y
(2)-- H -------------- T ----------- N
(3)-- T -------------- H ----------- N
(4)-- H -------------- T ----------- N
(5)-- H -------------- H ----------- Y
(6)-- T -------------- H ----------- N
(7)-- H -------------- T ----------- N
(8)-- H -------------- H ----------- Y
(9)-- H -------------- H ----------- Y
(10)- T -------------- H ----------- N
(11)- H -------------- H ----------- Y
(12)- H -------------- T ----------- N
(13)- H -------------- H ----------- Y
(14)- T -------------- T ----------- N
(15)- T -------------- T ----------- N
(16)- T -------------- T ----------- N
(17)- H -------------- H ----------- Y
(18)- H -------------- T ----------- N
(19)- H -------------- T ----------- N
(20)- T -------------- H ----------- N
(21)- H -------------- T ----------- N
(22)- H -------------- H ----------- Y
(23)- H -------------- H ----------- Y
(24)- H -------------- T ----------- N
(25)- H -------------- T ----------- N
(26)- H -------------- H ----------- Y
(27)- H -------------- T ----------- N
(28)- T -------------- H ----------- N
(29)- T -------------- T ----------- N
(30)- T -------------- H ----------- N
=============================
---- 20 ------------- 16 ---------- 10

OK, what do we see here? After 30 flips of the coins, we see 'twin heads' came up 10 times, for a probabitity of 1/3. If we add the stipulation that 'the large coin was a heads', the large coin was a heads 10 times out of the 10 times 'twin heads' came up. The probability that both coins are heads is still 1/3, 10 times out of 30 flips. Nothing was gained in predicting if both coins are heads. What you are predicting is 'what probability is the smaller coin a heads when, and only when, the large coin is a heads'. That restriction was not made in the problem, it was only stated that the large coin was a heads.

The probability of getting both coins heads is 1/4, not 1/3, which is the same as the odds of getting two tails. There are two independent events here (small and large coin), each of which has two equiprobable outcomes (heads or tails). The odds of getting both coins heads (or both coins tails) is 1/2*1/2=1/4, not 1/3. 2inq, it seems are taking the three outcomes both heads, both tails, and one heads/one tails (or 2 boys, 2 girls, mixed gender) as equiprobable. This is not the case. Think of the sum on a pair of randomly rolled dice. It is much easier to get a sum of seven (1/6) than a boxcars (1/36). The one heads/one tails outcome is twice as likely as two heads.

So what gives here? The answer is simple: Thirty is way too small a number for a Monte-Carlo experiment. Here's a simple rule of thumb: You need about one hundred trials to yield answers with one signficant digit of accuracy, ten thousand trails for two significant digits, one million for three significant digits, and so one. With only thirty trials you get about 1/2 significant digit (essentially garbage).

D H,
The probability of getting both coins heads is 1/4, not 1/3, which is the same as the odds of getting two tails.
Yes, I knew that was obvious. I wrote down my actual flips, which were skewed mostly due to the small coin comming up heads 20 times out of 30 flips, when it should have been 15 on a large number of samples.

I still stand by my assertion that simply stating 'the large coin is a heads' adds no useful information to change the probability that both coins are heads. It is not 1/3 or 1/2, it is still 1/4.
2inq, it seems are taking the three outcomes both heads, both tails, and one heads/one tails (or 2 boys, 2 girls, mixed gender) as equiprobable.
On the contrary, that is what the people who responded to the OP as a 1/3 probability were doing. By stating 'one of the children is a boy', you and others were doing that very thing by eliminating one of the <b,g>, <g,b> combinations. The true probability for the OP is 1/4, the same as for the coin problem I posted. It is correct that the answer is not intuitive. :D If the OP had stated 'the oldest is a boy, what are the probabilities of the other child being a boy', then the answer is obviously 1/2.

I still stand by my assertion that simply stating 'the large coin is a heads' adds no useful information to change the probability that both coins are heads. It is not 1/3 or 1/2, it is still 1/4.

Stand by it then. You're still wrong. Don't you see that saying "the large coin is heads" is more specific (and hence has greater information content) than saying "at least one of the coins is heads"?

On the contrary, that is what the people who responded to the OP as a 1/3 probability were doing. By stating 'one of the children is a boy', you and others were doing that very thing by eliminating one of the <b,g>, <g,b> combinations.

Wrong again. The statement "I have two children" means the possible gender outcomes for these two children are <g,g>, <g,b>, <b,g>, and <b,b>. These four events are mutually exclusive and equiprobable assuming a 50-50 split between boys and girls. (This is not true, as the actual ratio at birth is about 105 boys for every 100 girls. As I said in my first post, I'm ignoring that little detail.)

The additional statement "At least one is a boy" eliminates the <g,g> outcome. Three equiprobable events are left. Two of the three events includes a girl, one does not. The odds are 1/3 that the other child is a boy.

The statement "the oldest child is a boy" (or any other specific information) eliminates the <g,g> and the <g.b> event. Two equiprobable events are left, one of which includes a girl and one does not. The odds are 1/3 that the other child is a boy.

Nah, don't blame yourself. Certain posters deliberately chose to be obtuse about it.
You sure had that right.

You are right, I was wrong.

Let us state the problem differently.Suppose you pick a random child from a data base containing information about familes with two children and discover that the chosen child is a boy. What is the probability that his sibling is also a boy?Oddly enough, this question has two answers because the following phrase is ambiguous: pick a random child

It seems that there are two ways to pick a random child. Assume that there are N families assigned numbers from one to N. Also assume that the children are assigned numbers from 1 to 2N.Choose a family by picking a random number in the range 1 to N. Then flip a coin to decide on the first or second born child as the random child. If the child is a boy, 1/3 of the time his sibling will also be a boy. As indicated in other posts, picking a boy this way excludes the familes with two girls. 1/3 of the remaining familes have two boys and 2/3 have a boy & a girl.


Choose a child by picking a random number in the range 1 to 2N. If the child is a boy, 1/2 of the time his sibling will be a boy. Note that half the boys have brothers: They come from the 1/4 of the families that have two boys. Half the boys have sisters: They come from the half of the families that have a boy & a girl.The term random can be tricky.

It seems that there are two ways to pick a random child. Assume that there are N families assigned numbers from one to N. Also assume that the children are assigned numbers from 1 to 2N.[list]Choose a family by picking a random number in the range 1 to N. Then flip a coin to decide on the first or second born child as the random child. If the child is a boy, 1/3 of the time his sibling will also be a boy. As indicated in other posts, picking a boy this way excludes the familes with two girls. 1/3 of the remaining families have two boys and 2/3 have a boy & a girl.
Choosing a child in this way is exactly the same as choosing a random number from 1 to 2N, and in each case the probability that the other child is a boy is 50%.

Your restatement describes a different problem to the one in the OP.
In the OP, you're not choosing a random child and testing if it is a boy, you're choosing a random family and testing both children to see if at least one of them is a boy.

It seems that there are two ways to pick a random child. Assume that there are N families assigned numbers from one to N. Also assume that the children are assigned numbers from 1 to 2N.[list]Choose a family by picking a random number in the range 1 to N. Then flip a coin to decide on the first or second born child as the random child. If the child is a boy, 1/3 of the time his sibling will also be a boy. As indicated in other posts, picking a boy this way excludes the familes with two girls. 1/3 of the remaining familes have two boys and 2/3 have a boy & a girl.

The key to the odds of the other child also being a boy equaling 1/3 in my original question is that by asking "Is one of your children a boy?" you are blindly testing both children for boyness at the same time, and if you get an affirmative response you don’t know which of the two children is a boy.

If you label the children Child A and Child B and then ask "Is child B a boy?" the answer doesn't have anything to do with the gender of child A, so the odds of A being a boy are 1/2 regardless of child B's gender. In order for the odds to be 1/3, you have to ask "Is either child A or child B a boy?" and then not know which child the parent is referring to when they say "Yes".

You are right, I was wrong. Anyone else favorably impressed? May we all be high class people.

It's the math forum, of course, so it's expected, but still - -

There are subtly erroneous posts being made here. One in particular gives the following possibilities when it is known that one coin is larger than the other, implying that all possibiliteies have been included.(H,h)
(H, t)
(T,h)
(T,t)The above assumes that the large coin was tossed first. What about the following complete list?(H,h) & (h,H)
(H,t) & (t,H)
(T,h) & (h,T)
(T,t) & (t,T)Probability is trickier than you might expect if you do not have much experience in this field. I have had a lot of experience and still mess up more often than I like to admit.

The above type of error is not surprising. An analysis is made using an unfounded assumption without realizing the assumption has been made.

BTW: Apparently nobody seems to understand my post about there being two differenerent random processes for picking a child.

Both processes are equally likely to pick any particular child, but the two processes result in different probabilities for the sex of the sibling of the child chosen. I am too lazy to repost and further explain the two methods.

Those who did not understand the first time are not likely to understand if I repeat the post.


Again I point out that choose a random child is an ambiguous phrase. One must specify a random process to eliminate the ambiguity.

That's what I did when I realized my mistake. I couldn't logically understand why one of the <b,g> & <g,b> sets was eliminated and not he other. I finally started to write down the sets when there was a large and a small coin, then realized there were 8 sets. I had been overlooking the additional two-head and two-tail sets. It became clear after I wrote all the sets down. And yes, this was tricky for me for I have never done any probability analysis before.

There are subtly erroneous posts being made here. One in particular gives the following possibilities when it is known that one coin is larger than the other, implying that all possibiliteies have been included.(H,h)
(H, t)
(T,h)
(T,t)The above assumes that the large coin was tossed first. What about the following complete list?(H,h) & (h,H)
(H,t) & (t,H)
(T,h) & (h,T)
(T,t) & (t,T)

No, those are different problems. In the first one, the coins were ordered by size only. In your second version they're ordered by both size and the order of flipping.

An example of the probilities of two coins of the same size would be unordered sets as follows:

<h,h> -------25%
<h,t & t,h> --50%
<t,t> --------25%

If one coin is larger than the other, the problem automatically changes to ordered sets:

<H,h> -------12 1/2%
<h,H> -------12 1/2%
<H,t> -------12 1/2%
<h,T> -------12 1/2%
<T,h> -------12 1/2%
<t,H> -------12 1/2%
<T,t> -------12 1/2%
<t,T> -------12 1/2%

From there, it is easy to see which sets are eliminated when specifies something like 'The large coin is a heads'. It only happens in 4 of the 8 sets, for a probability of 1/2. At least, this was the way I did it.

An example of the probilities of two coins of the same size would be unordered sets as follows:

<h,h> -------25%
<h,t & t,h> --50%
<t,t> --------25%

If one coin is larger than the other, the problem automatically changes to ordered sets:

<H,h> -------12 1/2%
<h,H> -------12 1/2%
<H,t> -------12 1/2%
<h,T> -------12 1/2%
<T,h> -------12 1/2%
<t,H> -------12 1/2%
<T,t> -------12 1/2%
<t,T> -------12 1/2%

From there, it is easy to see which sets are eliminated when specifies something like 'The large coin is a heads'. It only happens in 4 of the 8 sets, for a probability of 1/2. At least, this was the way I did it.

Yeah, if you're only given information about the size of the heads coin, and not the order of flippings, then the extra ordering based on which coin is first is spurious. So you end up with the same answer. You could keep going literally forever with more and more complicated orderings (by weight, age, smell, etc.), but if the problem doesn't contain any information that relates to them, there's no point. You only need to include orderings as they relate to the information in the problem; any further ordering will simply wash out in the analysis.

This is one of those questions that Intro to Probability teachers like to throw at their students:

Suppose you approach someone who you know has two children, but you don’t know their children’s genders. You ask the person “Is one of your children a boy?” and they answer “Yes.” What is probability that the person’s other child is also a boy? (In other words, once you know that one of their children is a boy, what are the odds that they have two boys?)

Hmm I misread the first time I guess.
Reading the question again I have to ask, how is it different from asking a person with one child if the child is a boy ?

OP problem:
1 * 1/2 = 1/2
50% chance the child you don't know about is a boy. :shrug:

Please read the thread. The answer is 1/3.