Probability question

[ReFLeX]

There is no difference -- the outcome of one child's sex is not dependent on any other child's sex. Where do you think this is happening?

Aha. Not in real life, but it is dependent in the actual problem (older child/younger child).

 

[Just thinking]

Aha. Not in real life, but it is dependent in the actual problem (older child/younger child).

How is it dependent in this problem? Each child is given an ideal probability of 0.5 for being male or female upon birth.

 

[Drooper]

Aha. Not in real life, but it is dependent in the actual problem (older child/younger child).

It is independent in the problem as well!!!


Aaaaaahhhhh......

.......must ......

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.......to.......

........stay........

..........away......

.........from......

............this............

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............driving.............

.................me..............

.......bananas..................

 

[Dr Adequate]

Not this again.

I'll go with ambiguous.

Consider the following two games:

(1) I toss two coins. I choose one of them at random and show you which way it came down, and invite you to bet on what the other is. What are your odds of success?

(2) I toss two coins. If at least one of them comes down heads, I'll tell you so and invite you to bet on the what the other is. What are your odds of success?

The question is, which of these two games are we playing here? And the answer seems to be unclear. If it is a condition of the puzzle that the proud parent must be able to say "at least one is a boy", then we're playing game (2). On the other hand, if the parent might have had the option of revealing that at least one of the children is a girl, but just happened randomly to reveal that at least one is a boy, then we're playing game (1).

We need to distinguish between: (1) "A person is drawn randomly from the set of all parents of two children. Pick a child to meet, either child. It turns out to be a boy. What is the probability..." and (2) "A person is drawn randomly from the set of all parents of two children one of whom is a boy. What is the probability..."

But the question as usually stated does not distinguish between the two cases. Indeed, neither applies. The parent in question was not drawn randomly from any set of people, but invented for the purposes of the puzzle.

 

[ReFLeX]

Oops. The probability of the outcome of the other's sex is dependent.

 

[Just thinking]

The question is, which of these two games are we playing here?

We are (I believe) playing game 2. And I am stipulating that it does not matter whether you are told one is a boy or shown one is a boy. (And when told or shown, you don't know if he is the older or younger.)

 

[Just thinking]

Oops. The probability of the outcome of the other's sex is dependent.

Dependent on just how much you know of the other male child. Just knowing (or seeing) he is male is not the same as knowing (for example) that he is the older of the two.

PS: Oooops accepted and understood. This can be taxing.

 

[Just thinking]

Aaaaaahhhhh......

.......must ......

........try......

.......to.......

........stay........

..........away......

.........from......

............this............

........thread..........

............driving.............

.................me..............

.......bananas..................

It's no use ... I'm stuck here too. Give in to the force.

 

[Moose]

I wrote some code in visual basic to test this out.

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.

 

[ReFLeX]

It is independent in the problem as well!!!


Aaaaaahhhhh......

.......must ......

........try......

.......to.......

........stay........

..........away......

... You know what drove me nuts was reading the infamous .999 =1 thread far after it happened. The original proof was absolutely ridiculous yet it spawned tens of thousands of words of pointless discussion.
As I remember it, the third step was 9x = 9x.
Laughable! That's a tautology... you don't need two steps to prove that. And next, even worse. From that they got x = 1. Problem is, x could also equal 666, for all that tells you. Oy.

 

[Drooper]

Not this again.

I'll go with ambiguous.

Consider the following two games:

(1) I toss two coins. I choose one of them at random and show you which way it came down, and invite you to bet on what the other is. What are your odds of success?

(2) I toss two coins. If at least one of them comes down heads, I'll tell you so and invite you to bet on the what the other is. What are your odds of success?

The question is, which of these two games are we playing here? And the answer seems to be unclear. If it is a condition of the puzzle that the proud parent must be able to say "at least one is a boy", then we're playing game (2). On the other hand, if the parent might have had the option of revealing that at least one of the children is a girl, but just happened randomly to reveal that at least one is a boy, then we're playing game (1).

We need to distinguish between: (1) "A person is drawn randomly from the set of all parents of two children. Pick a child to meet, either child. It turns out to be a boy. What is the probability..." and (2) "A person is drawn randomly from the set of all parents of two children one of whom is a boy. What is the probability..."

But the question as usually stated does not distinguish between the two cases. Indeed, neither applies. The parent in question was not drawn randomly from any set of people, but invented for the purposes of the puzzle.

Who needs politics when you can argue for hours about "the money or the box".


The ambiguity is always a problem with this teaser, but even under stricter descriptions of the process people still get stumped by the apparent counter intuitive result, as a reading of this thread confirms.

 

[Just thinking]

... You know what drove me nuts was reading the infamous .999 =1 thread far after it happened.

... but 0.99999 --> is 1.

 

[Dr Adequate]

We are (I believe) playing game 2.

In that case, the answer would be 1/3. I think everyone would agree that if we are playing game 2, the answer is 1/3. But some people think that we are in fact playing game 1. Hence all the arguments. No-one disagrees about how to calculate probabilities, and so this is an entirely different kind of argument from the one that attends the Monty Hall Lemon Problem --- you're disagreeing about what the question is. I say it's ambiguous, and unless you can all arrive at a consensus on what it means --- I'm right.

 

[Earthborn]

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

The age of a child (meaning younger or older) is relevant in that it establishes the order of events.

I don't see why we need to worry about an order of events. All the information that is given is that one is a boy, and the other is unknown.

You would be right if the OP told us one is a boy and the firstborn is unknown. But it doesn't do that, so the order of events is completely irrelevant. They might as well be born at exactly the same time.

If you know the younger to be a male, it's 50-50 that the older is a male.

If you know one is male, it's 50-50 that other is male. That's all you need to know.

Not true -- we must determine all the possible outcomes for that unknown individual and compare how many result in it being a male against how many result in it being female. Then we calculate the probability based on that.

There are only 2 possible outcomes:

The child we know of is a boy and the unknown child is a girl
The child we know of is a boy and the unknown child is a boy

The following possibilities are excluded:

The child we know of is a girl and the unknown child is a girl
The child we know of is a girl and the unknown child is a boy

 

[Drooper]

In that case, the answer would be 1/3. I think everyone would agree that if we are playing game 2, the answer is 1/3. But some people think that we are in fact playing game 1. Hence all the arguments. No-one disagrees about how to calculate probabilities, and so this is an entirely different kind of argument from the one that attends the Monty Hall Lemon Problem --- you're disagreeing about what the question is. I say it's ambiguous, and unless you can all arrive at a consensus on what it means --- I'm right.

I think you have misread the thread.

Some people can't calculate probabilities (it's a well studied phenomenon). Just look at the post above mine.

 

[Orangutan]

CurtC, I'm a convert!

Your census taker example convinced me.

There could be an additional qestion like "What is the probablility that she could not present a boy", which would be 1/4.

So anyway I switched, I now go for 1/3.

If she had shown you a girl the chances are that the other child is a boy would be 2/3 right?

Wow this is a good problem!

O.

 

[Just thinking]

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

I don't see why we need to worry about an order of events. There are only 2 possible outcomes:

The child we know of is a boy and the unknown child is a girl
The child we know of is a boy and the unknown child is a boy

The following possibilities are excluded:

The child we know of is a girl and the unknown child is a girl
The child we know of is a girl and the unknown child is a boy

You are assuming (from the first 2 outcomes) that each has an equal probability of happening. They do not. Also, when you say "The child we know of ... " just what exactly do you mean? I think you mean to say "We know that at least one child ... ".

Given that, there are two ways to have one boy and one girl from separate births but only one way to have two boys. You are correct to exclude all combinations where you know of one to be a girl, because we see a boy. Now, since there are 3 outcomes (not just 2) from only knowing one was a boy, and only one results with 2 boys, the chances for that are 1/3.

 

[egslim]

I have 2 children, one is a boy. what is the prob that the other is also a boy?

If you know one is male, it's 50-50 that other is male. That's all you need to know.

No, that's only true if you know a specific one (oldest, youngest) is male.
All the opening post tells us is that at least a random one of the two is male.

There is 50/50 chance that child1 (as determined by birth sequence) is either a boy or a girl, likewise for child2.
But thats not the question: We know we saw a boy, but we don't know wether is was child1 or child2. Therefore, the only thing we dó know is they can't both be girls.
Which leaves 3 options of equal probability and therefor any chance of 1/2 is precluded.

(1) I toss two coins. I choose one of them at random and show you which way it came down, and invite you to bet on what the other is. What are your odds of success?

(2) I toss two coins. If at least one of them comes down heads, I'll tell you so and invite you to bet on the what the other is. What are your odds of success?

What additonal information would you need to determine which game we're playing?
It seems to me the only difference between the two questions is slightly less information in (2). Since we lack that information in the opening post as well, the question is only answerable if we assume it's (2).

 

[Just thinking]

If she had shown you a girl the chances are that the other child is a boy would be 2/3 right?

Yes.

 

[Jellby]

Hold on ... why do you assume that? Where was that ever revealed?

It was an implicit premise: if you see a woman giving birth to a son and you know she already has another child, it's obvious the baby you're seeing is younger than his sibling.

You would need to know the frequency of Identical vs. Fraternal given a twin birth. I do not know this, and it may differ for two boys vs. two girls as well.

Well, yes, ignoring all that (as we ignore differences in boy/girl frequencies and hermaphrodites).

In this case. it is. We can have (as outcomes for two separate individual births) BB, BG (older boy), GB (older girl) or GG. Note, that 3 of the 4 result in her having at least 1 boy. Therefore, if you're told a woman has at least 1 boy, the chance she has two is 1/3. But if you're told the older is a boy, then only BB and BG are possible outcomes, and her chances of having 2 boys is 1/2. (The same holds true if you're told the younger is a boy -- BB and GB.)

As I understand it, if you can label unambiguously both children and you know which one is a boy, then the probability is 1/2. If you know one is a boy but can't assign that piece of information to either of the labeled children, then it's 1/3.

So, if we know that the oldest is a boy, we have 1/2. But would it be different if we knew the tallest is a boy? I don't think so (assuming their height is independent of sex). What if we know somehow that one of the children is out of the city when we see mother and child? We can call the one in this city A and the one away B, we know A is a boy... the probability of B being a boy too is 1/2 (I'd say). What are the restrictions we have when giving a label the the children? (Surely, we can't say "let's call the one we know is a boy A", can we? This would be what Earthborn suggested).

Probability is a man-made concept, it's not intrinsic to the phenomenon but dependent on the knowledge we have (or can have) about it. The problem is: what do we exactly know in this case? (or which game are we playing?).