Probability Puzzle

[Gargamel]

OK, wow, I get sick for a couple days, and this thread explodes.

Excellent!    As long as we kept it civil.  I only glanced at it.

I have a hard time arguing against the 50/50 answer, as it makes total sense.  But it depends on your point of view.

There are two children, hence there are 4 possible combinations of children. (please excuse this horribly formatted chart)

		Child2
		Boy	Girl
Child 1	Boy	Boy/Boy	Boy/Girl
	Girl	Girl/Boy	Girl/Girl

Since we know that one of the children is a Boy, that excludes the Girl/Girl option.

Boy/Boy	Boy/Girl
Girl/Boy	Girl/Girl

Therefore, the answer is 1/3.

 

BUT!  An argument can therefore be made that since the first child is a boy, that also excludes the Girl/Boy option.

Boy/Boy	Boy/Girl
Girl/Boy	Girl/Girl

Bringing us back to the 50/50 answer.

 

BUT!!!  A counterargument to the counterargument can then be made that order doesn't matter, as we're not sure which one is the first child, they could be running around in circles!

 

And now I have typed/read the words Boy and Girl so much, they look weird and no longer have meaning. 

 

Annnd.... continue...

[YNM]

13 hours ago, K^2 said:

Because if both are boys, he's guaranteed to tell you that one's a boy, but if one is a girl, it's a 50/50 chance.

Let's say you happen to live in a universe where they like a boy (so if there's any boy they'd say that first).

With what you were saying that'd lead to 1/4 actually. It's just asking "what's the probability of having 2 boy siblings", not necessarily in the way the question was meant to say.

[Cheif Operations Director]

7 hours ago, Tullius said:

If we agree that "one of them is a boy" means "at least one of them is a boy", we have the following:

case	kid A	kid B
1	girl	girl
2	boy	girl
3	girl	boy
4	boy	boy

So there are three cases, where the parent will say "one of them is a boy", but only one case, where the second kid is also a boy.

As we assume for simplicity that all cases are equally likely, the answer is 1/3.

The reason is that the boy is not identified, i.e. you don't know in which of the two prams the boy is (nor do you know anything else identifying the boy)

If the parent would point at a pram and say "this one contains a boy" (or he would just say "Joey is a boy"), the kid is identified (i.e. in terms of the table above, we have "kid A is a boy") and so the correct answer is 1/2.

Your logic is wrong for the following reasons.

Case 1 is not even in the purview of the question. 

Case 2 & 3 can be group together. We no that one of the two kids is a boy not a "group A" or a "group B"

The problem simpily says "other" not Kid B or Kid A we do not know which even know which one is which. 

This means that:

1 is a boy

the other can either be a boy or a girl 

7 hours ago, Gargamel said:

OK, wow, I get sick for a couple days, and this thread explodes.

Excellent!    As long as we kept it civil.  I only glanced at it.

I have a hard time arguing against the 50/50 answer, as it makes total sense.  But it depends on your point of view.

There are two children, hence there are 4 possible combinations of children. (please excuse this horribly formatted chart)

		Child2
		Boy	Girl
Child 1	Boy	Boy/Boy	Boy/Girl
	Girl	Girl/Boy	Girl/Girl

Since we know that one of the children is a Boy, that excludes the Girl/Girl option.

Boy/Boy	Boy/Girl
Girl/Boy	Girl/Girl

Therefore, the answer is 1/3.

 

BUT!  An argument can therefore be made that since the first child is a boy, that also excludes the Girl/Boy option.

Boy/Boy	Boy/Girl
Girl/Boy	Girl/Girl

Bringing us back to the 50/50 answer.

 

BUT!!!  A counterargument to the counterargument can then be made that order doesn't matter, as we're not sure which one is the first child, they could be running around in circles!

 

And now I have typed/read the words Boy and Girl so much, they look weird and no longer have meaning. 

 

Annnd.... continue...

For example If I hold up two hands with two crayons in each and say one is red and other may be red or orange and I ask you to find the probability that it is a red or orange crayon What differance does it make as to wether the red is in my left or right hand. You already know I one is red. You are find out the second.

If I said the same thing the next day and then just switched the hand on you it does not make a differance.

Edited July 22, 2018 by Cheif Operations Director

[Cheif Operations Director]

The 1/3 probability is wrong because we do not have the girl girl option available.

Your arguement if you believe 1/3 is:

boy-boy

girl-boy

girl-girl

You arguement if you believe 1/4 is:

boy-boy

boy-girl

girl-boy

girl-girl

The girl option obviously not relevant and the boy-girl arguement is the same as the girl-boy arguement because non are assigned a number, hence they can be GROUPED as the same on. Leaving you with two options:

1. boy-boy

2. boy-girl/girl-boy

[Tullius]

18 minutes ago, Cheif Operations Director said:

Your logic is wrong for the following reasons.

Case 1 is not even in the purview of the question. 

Case 2 & 3 can be group together. We no that one of the two kids is a boy not a "group A" or a "group B"

The problem simpily says "other" not Kid B or Kid A we do not know which even know which one is which. 

This means that:

1 is a boy

the other can either be a boy or a girl 

Cases 2 and 3 are NOT equal. They are the two possibilities to get 1 boy and 1 girl.

For you example with the crayons: You have one crayon in each of your hands and they can be red or orange. The possible cases are of course:

case	right hand	left hand
1	orange	orange
2	red	orange
3	orange	red
4	red	red

There are three cases in which you will say that you have at least one red crayon (cases 2, 3 and 4), but only in one of them (case 4) the two crayons will be red, i.e. if the probability of each case is equal, we have

P (2 red crayons | at least 1 red crayon) = 1/3

[Cheif Operations Director]

Just now, Tullius said:

Cases 2 and 3 are NOT equal. They are the two possibilities to get 1 boy and 1 girl.

For you example with the crayons: You have one crayon in each of your hands and they can be red or orange. The possible cases are of course:

case	right hand	left hand
1	orange	orange
2	red	orange
3	orange	red
4	red	red

There are three cases in which you will say that you have at least one red crayon (cases 2, 3 and 4), but only in one of them (case 4) the two crayons will be red, i.e. if the probability of each case is equal, we have

P (2 red crayons | at least 1 red crayon) = 1/3

Your Missing the point.

Let me put it this way

I walk up to you right now

I ask you In one of my hands is a red crayon. The my other hand their is either and orange crayon or a red crayon what is the probability that it is orange or red? 

[Tullius]

1 minute ago, Cheif Operations Director said:

Your Missing the point.

Let me put it this way

I walk up to you right now

I ask you In one of my hands is a red crayon. The my other hand their is either and orange crayon or a red crayon what is the probability that it is orange or red? 

Well, if you have at least one red crayon in your hands, we have the following possible situations:

case	right hand	left hand
1	red	orange
2	orange	red
3	red	red

So the probability of two red crayons is 1/3.

[Cheif Operations Director]

Just now, Tullius said:

Well, if you have at least one red crayon in your hands, we have the following possible situations:

case	right hand	left hand
1	red	orange
2	orange	red
3	red	red

So the probability of two red crayons is 1/3.

How is Case 1 and Case 2 the same?

Let me elaborate:

I never said anything about hands what if I'm an amputee (I'm not serious about that)

All I'm saying is that I said in one of my hands nothing about which hand it frankly does not matter.

I asked you to find the probability that the other crayon is red or orange

[Tullius]

4 minutes ago, Cheif Operations Director said:

How is Case 1 and Case 2 the same?

Let me elaborate:

I never said anything about hands what if I'm an amputee (I'm not serious about that)

All I'm saying is that I said in one of my hands nothing about which hand it frankly does not matter.

I asked you to find the probability that the other crayon is red or orange

Cases 1 and 2 are not the same.

Let's imagine you flip a coin to decide if you put a red (heads) or orange (tails) crayon into a box (a process which I can't see). And you do this twice, as we want two crayons in the box. Then you shake the box, so it is impossible to know which crayon was put in first. You look into it and say to me: "There is at least 1 red crayon in the box".

Now, I know that either on the first flip, on the second flip or on both flips, you got a heads, which means that there are 3 different possible cases and only in 1 we end up with two red crayons.

If instead you told me that the first crayon you put in the box was red (or the crayon in your right hand is red), then of course the probability of two red crayons would be 1/2, as there only two possibilities:

• the first is red and the second is red
• the first is red and the second is orange

[Cheif Operations Director]

@Tullius

Ahh!

What I'm saying is that you inky have two options for hand two wether hand two is the left or right hand

Graphs only confuse this

The left or right hand arguement is not valid!

We know 1 crayon is red. 

We know that in the other hand there is either a Blue crayon or a Orange crayon. 

What is the probability that IN HAND TWO there is a Blue crayon or an Orange crayon.

You ONLY HAVE TWO OPTIONS!

Blue or Orange! 

50%-50%

 

 

[Tullius]

1 hour ago, Cheif Operations Director said:

@Tullius

Ahh!

What I'm saying is that you inky have two options for hand two wether hand two is the left or right hand

Graphs only confuse this

The left or right hand arguement is not valid!

We know 1 crayon is red. 

We know that in the other hand there is either a Blue crayon or a Orange crayon. 

What is the probability that IN HAND TWO there is a Blue crayon or an Orange crayon.

You ONLY HAVE TWO OPTIONS!

Blue or Orange! 

50%-50%

 

 

But which one is hand one or hand two?

If you keep both hands behind your back at all times, I don't know in which one the first red crayon is. It can either be the right hand or the left hand. And then the question is, if there is a red crayon in the second hand. In that case, it will be 1/3.

If you show me your first hand with the red crayon, there are only two possibilities left, as I know which one is the first hand.

And, if graphs confuse you, maybe some mathematical computations don't:

On 7/19/2018 at 9:54 AM, Tullius said:

[...] We are computing the probability that there are two boys, given that there is at least one boy, or in mathematical notation

P(two boys | at least one boy)

So we can use the definition of conditional probability to get

P(two boys | at least one boy) = P(two boys AND at least one boy) / P(at least one boy) = P(two boys) / P(at least one boy) = (1/4) / (3/4) = 1/3

Edited July 22, 2018 by Tullius

[Cheif Operations Director]

10 minutes ago, Tullius said:

But which one is hand one or hand two?

If you keep both hands behind your back at all times, I don't know in which one the first red crayon is. It can either be the right hand or the left hand. And then the question is, if there is a red crayon in the second hand. In that case, it will be 1/3.

If you show me your first hand with the red crayon, there are only two possibilities left, as I know which one is the first hand.

And, if graphs confuse you. Maybe some mathematical computations don't confuse you:

Tullius...

the graphs do not confuse me.

 let me put it to you this way...

In a box I have a 1 blue crayon or a 1 Orange  crayon. What are the odds that it is a blue crayon.

50% right? 

Back to the child problem. 

If you have 1 child that is a boy

and a mystery child (the box) you have who options boy (blue) girl (oranage). What is the probability that the mystery child is a boy(Blue) 

50%!

Edited July 22, 2018 by Cheif Operations Director

[Tullius]

2 minutes ago, Cheif Operations Director said:

Tullius...

the graphs do not confuse me.

 let me put it to you this way...

In a box I have a 1 blue crayon or a 1 Orange  crayon. What are the odds that it is a blue crayon.

50% right? 

Back to the child problem. 

If you have 1 child that is a boy

and a mystery child (the box) you have who options boy (blue) girl (oranage). What is the probability that the mystery child is a boy(Blue) 

50%!

It is NOT independent!

You are not telling me that the oldest kid is a boy, but that at least one of the kids is a boy. It could be either the oldest or the youngest, and I DON'T know.

And that is the important bit, as you say that at least one of the kids is a boy, if either the oldest or the youngest or both is a boy.

And so there are THREE cases of equal probability, in which you say that there is at least one boy. And there is only ONE case, in which we have two boys (i.e. the second one is a boy too). Therefore the probability is 1/3.

And, if you still don't believe me, just look at the calculation I posted in my previous post.

[Cheif Operations Director]

1 minute ago, Tullius said:

It is NOT independent!

You are not telling me that the oldest kid is a boy, but that at least one of the kids is a boy. It could be either the oldest or the youngest, and I DON'T know.

And that is the important bit, as you say that at least one of the kids is a boy, if either the oldest or the youngest or both is a boy.

And so there are THREE cases of equal probability, in which you say that there is at least one boy. And there is only ONE case, in which we have two boys (i.e. the second one is a boy too). Therefore the probability is 1/3.

And, if you still don't believe me, just look at the calculation I posted in my previous post.

Ok let me understand this

1 is a Boy- 1 is a girl

1 is a boy- 1 is a boy

1 is a girl -1 is a boy

1/3 what is what you are saying

 

[Tullius]

1 minute ago, Cheif Operations Director said:

Ok let me understand this

1 is a Boy- 1 is a girl

1 is a boy- 1 is a boy

1 is a girl -1 is a boy

1/3 what is what you are saying

 

I would rather say, if we use A and B as the identifiers of the kids (and you need some form of identifiers):

• A is boy, B is girl
• A is girl, B is boy
• A is boy, B is boy

These are the three possible cases in which you will say that there is at least one boy.

[Cheif Operations Director]

Just now, Tullius said:

I would rather say, if we use A and B as the identifiers of the kids (and you need some form of identifiers):

• A is boy, B is girl
• A is girl, B is boy
• A is boy, B is boy

These are the three possible cases in which you will say that there is at least one boy.

Why is 1 and 2 not the same thing seeing as the problem does not PRSENT ANY IDENTIFIERS. They could be twins

[Tullius]

Just now, Cheif Operations Director said:

Why is 1 and 2 not the same thing seeing as the problem does not PRSENT ANY IDENTIFIERS. They could be twins

First, even for twins, one is usually older than the other by a few minutes. However, you could just replace oldest and youngest by any other choice of identifiers (e.g. genderneutral names),

But, let's just start with a much simpler problem and just look at the different possibilities parents can have two children:

case	oldest kid	youngest kid
1	girl	girl
2	girl	boy
3	boy	girl
4	boy	boy

And we can hopefully agree that each of these cases is distinct and in our mathematical context equally likely, i.e. each one has probability 1/4.

So what is the probability of two girls? 1 case among 4, and so the probability is 1/4.

What is the probability of one boy and one girl? 2 cases among 4, and so the probability is 2/4 = 1/2.

And now the million dollar question: In how many cases of these four cases will you say that there is at least one boy?

[Cheif Operations Director]

3 minutes ago, Tullius said:

First, even for twins, one is usually older than the other by a few minutes. However, you could just replace oldest and youngest by any other choice of identifiers (e.g. genderneutral names),

But, let's just start with a much simpler problem and just look at the different possibilities parents can have two children:

case	oldest kid	youngest kid
1	girl	girl
2	girl	boy
3	boy	girl
4	boy	boy

And we can hopefully agree that each of these cases is distinct and in our mathematical context equally likely, i.e. each one has probability 1/4.

So what is the probability of two girls? 1 case among 4, and so the probability is 1/4.

What is the probability of one boy and one girl? 2 cases among 4, and so the probability is 2/4 = 1/2.

And now the million dollar question: In how many cases of these four cases will you say that there is at least one boy?

Ok so your saying is that 

Case 2 and Case 3 are differant If you forget the whole age part 

 

It's just the way you say it, it's the order 

[Tullius]

13 minutes ago, Cheif Operations Director said:

Ok so your saying is that 

Case 2 and Case 3 are differant If you forget the whole age part 

 

It's just the way you say it, it's the order 

You could call them kid A and kid B based on the order they came in through the door yesterday or based on the grades of their last math tests. It doesn't really matter.

You could even take one dollar coin and one Euro coin, if you want to think of coin flipping. If there is exactly one heads after flipping both, either the dollar is heads and the Euro is tails, or the dollar is tails and the Euro is heads. And, of course these two possibilities are very distinct, although both amount to one heads and one tails. And coins or kids, doesn't make a difference.

Another way you may look at it: You could just compute that

P (two girls) = P (kid A girl) * P (kid B girl) = (1/2) * (1/2) = 1/4

P (two boys) = P (kid A boy) * P (kid B boy) = (1/2) * (1/2) = 1/4

using that the gender of both kids is independent. So you have

P (same gender) = P (two girls) + P (two boys) = 1/4 + 1/4 = 1/2

Since, probabilities always have to add up to 1, we therefore must have

P (different gender) = 1 - P (same gender) = 1 - 1/2 = 1/2

So, it is twice as likely that the kids have a different gender (i.e. one boy and one girl) than the kids being both boys.

Edited July 22, 2018 by Tullius

[Cheif Operations Director]

I just can not understand your arguement 

1 is a boy one is a mystery it's 50-50

you saying boy-girl {not equal} to girl-boy is obsured! It's the same thing. 

[Tullius]

Let us use coins instead of kids.

For two coin tosses, we have the following possible outcomes:

Each of the possible outcomes is of course equally likely, i.e. any single one of the cases HH, HT, TH and TT has probability 1/4. So the likelihood of obtaining exactly 1 heads is 1/2.

And you cannot change the probabilities by mixing the coins after the toss. (The coins are flipped, so you won't change the result by simply moving the coins around)

So according to the diagram above, I flip two coins. Now, I tell you that at least one of the coins is showing heads, i.e. number of heads >= 1. And nothing more, I neither show you the coin with heads, nor do I tell you if it was the first toss or the second toss.

What is the probability that the two coins are showing heads?

[K^2]

5 hours ago, YNM said:

Let's say you happen to live in a universe where they like a boy (so if there's any boy they'd say that first).

With what you were saying that'd lead to 1/4 actually. It's just asking "what's the probability of having 2 boy siblings", not necessarily in the way the question was meant to say.

Absolutely agreed. Again, I'm not saying this isn't a valid interpretation of question in OP. Just that I don't see why it should be the default interpretation. There are two different setups that can lead to exactly the same question being asked, and the answer is 1/3 for one setup and 1/2 for other.

[YNM]

8 hours ago, Tullius said:

What is the probability that the two coins are showing heads?

1/4.

But this is more like sifting through a database of two coin flips and gone :

"Get me all the results that have heads" -> gets rid of 1/4 of the database

"Get me all the results that have two heads from the reduced database"

> gets rid of 2/3 of the remaining database

Hence the answer meant was 1/3.

8 hours ago, K^2 said:

Just that I don't see why it should be the default interpretation.

TBH it's kind of is in a fraternal twin situation - it's more likely to be a pair rather than doubles of male or doubles of female (pairs is as likely as any doubkes though). So if someone says "oh, one is of a certain gender" then it's more likely the other will be the opposite pair.

Then you can actualy extend this to every case of siblings. Just that it doesn't apply for a sequential question.

[Xeorm]

7 hours ago, Tullius said:

Let us use coins instead of kids.

For two coin tosses, we have the following possible outcomes:

Each of the possible outcomes is of course equally likely, i.e. any single one of the cases HH, HT, TH and TT has probability 1/4. So the likelihood of obtaining exactly 1 heads is 1/2.

And you cannot change the probabilities by mixing the coins after the toss. (The coins are flipped, so you won't change the result by simply moving the coins around)

So according to the diagram above, I flip two coins. Now, I tell you that at least one of the coins is showing heads, i.e. number of heads >= 1. And nothing more, I neither show you the coin with heads, nor do I tell you if it was the first toss or the second toss.

What is the probability that the two coins are showing heads?

1/3. Though I take the question to be saying "You see one child is a boy,  what is the probability that the other is a boy?" Mostly as other implies some sort of identification. 

The identification is key. If I see one of the options, then the only options for the other are boy or girl, so 1/2. If I know only that there is a boy, then that means less information, leading to a less accurate guess.