Hi all,

I thought I might post up this question for what is obviously a very smart place regarding probability. It seems to result in spirited discussion on most forums I've posted it on / seen it posted on - so I'd like to see the result here with a lot more education.

Assuming 50/50 birth rate.

You are in a park and strike up a conversation with a mother and her son. You learn in her family there are two children, what is the probability that the other child is a daughter?

I have searched here to see if it was posted before, however sections of the question and "park" resulted in no matches, although it may have been reworded differently, sorry if it has been.

Look forward to seeing your answers.

2/3.

There are 4 possible combination of children's gender. MM, MF, FM, FF. You eliminate the last one, and 2 of the 3 have the female as the other child.

The question posed is not answerable unless you know what rationale was used to decide which child to take to the park. Assuming that a fair coin was flipped to decide which child to take the answer is 0.5.

A family can be MM, MF, FM, or FF, each with probability 0.25. The following situations result in a son being taken to the park:

Family was MM and any child was chosen. Probability 0.25
Family was FM and the son was chosen. Probability 0.125
Family was MF and the son was chosen. Probability 0.125

Of these situations, half come from a MF family and half from a MM family, so the probability is 0.5.

Two thirds is the answer to some daughter/son questions, but not this one. Perhaps the OP meant it to be or he is deliberately being cunning

We have no information about the other child, so it's a 50/50 chance as per the birth rate.

The same essential problem setup comes up regularly on discussion forums, although it's usually a variation where information about both children is given, and the correct answer is 2/3 , which is supposed to be unintuitive. The first responder on this thread must be harking back to one of those problems!

(For example, "the mother is by herself, but tells you she has at least one son, and that one of her children is at the shop and one is at home. what is the chance that the child at the shop is a boy?" answer 2/3).

this problem seems exactly the same as saying there is a woman at the park with a boy. The woman is pregnant. What is the probability that her unborn child will turn out to be male, given that there is a 50/50 chance of a child being a boy. Not counting some genetic predisposition to crank out boys, the existing of the first boy should have no impact on the probability of the second child.

I have to say MtnHawk got it in the first post. The answer should be 2/3.

Pyro - I know where your logic is coming from, however since we have no information as to how the children as chosen, we can make no assumptions doing so. Hence, we must ignore that area of the problem and base the answer without including it.

Sorry but you are completely wrong, and me, Kittens, and VBAces are correct.

Here is a question for which the answer is 2/3rds. You ask someone with two children if they have a daughter. They tell you yes. The chance that they have a son is 2/3. Totally different question to what you asked.

Let me give you one more explanation as to why the answer is 1/2. We write the possible ways of having two children and taking one to the park as follows:

{ Boy at park, Boy at home || Boy at Park, Girl at home || Girl at Park, Boy at home || Girl at Park, Girl at home }. These possibilities are equally likely.

since we have seen a boy at the park this sample space reduces to

{ Boy at park, Boy at Home || Boy at Park, Girl at Home }. These possibilities are equally likely.

The chance the child at home is a boy is 50-50

If you still don't believe us you can easily google the answer. Your question is identical to '1. You know that a family has 2 children. You learn that the older child is a boy. Now what is the probability that the other child is a girl?' as VBaces pointed out. Google then leads you to http://en.wikipedia.org/wiki/Boy_or_Girl_paradox, where this question is answered inside the top 5 lines.

Actually, no, I have to disagree with you respectfully there.

The only way you can ever shrink the probabilty options avaliable, is when there is a unique indentifier on the subject you are viewing. Older / Younger etc descrimiate against the two subjects in the data range, isolating specific examples so you can rule them out in the probability tree.

For instance, all older son's are named Tim, younger son's named James, older daughers named Mary, younger daughters named Sarah.

If I told you that the older daughter was in the park, you can specifically eliminate mary, changing the probability tree to the 50/50 resultant for the second child. However if I told you I see a daugher, you do not know if it's Mary or Sarah, hence you cannot eliminate them on the probablity tree as you would get two different answers. Obviously with all problems, you must be able to represent them on a single probabilty tree.

I agree that if I gave specific information as to which subject you were viewing (oldest/youngest, etc) then the answer becomes 50/50. Seen / Unseen however is invalid, because it does not specifically nominate a single data set.

The same thing can be said for two coin flips.

(1) If I flip a coin, then flip a second coin, and tell you at least one of the result's is heads, but you don't know which coin toss (1st or 2nd) that had the H result, the chance of the second coin being a T is 2/3, yeah?

(2) If I flip a coin, then flip a second coin, and tell you the first flip was heads, the chance of the second coin being a T is 1/2, yeah?

(3) If I flip a coin, then flip a second coin, and tell you at least one of the result's is heads (same as problem 1), but then go one step further and physically show you one of the results, without revealing what toss it was from, the answer is still 2/3's, agreed? Visual information is only useful when we can decriminate between the options.

http://mathforum.org/dr.math/faq/faq.boy.girl.html

This has to be a level because a question that is absolutely equivalent to the one you originally asked is on this page and the answer is given as 0.5.


Suppose we're told that the oldest child is a boy? Let's look at this new information using the same two methods:

Combinations in the Sample Space

When the oldest child is a boy, the original sample space:

{BB, BG, GB, GG}

shrinks to:

{BB, BG}

because the two other possibilities, {GB, GG}, show girls as the oldest child and thus are no longer possible.

Since only one of the possibilities in the new sample space, {BG}, includes a girl, the probability that the second child in the family is a girl is 1/2.

This is EXACTLY the same thing as what you asked with 'oldest' being substituted for 'in park'. There is nothing special at all about 'oldest', it can be 'oldest', 'tallest', 'worst at understanding simple probability questions', 'in park' or anything else.

How on earth can you say that oldest is a unique identifier but 'being in park' isn't? I mean I get that the average guy in the street isn't very good at Maths but jesus christ.

Do me a favour, and tell me if you agree with (1), (2) and (3) in the coin example. I think that will demonstrate what the difference is.

This should be pretty straightforward.

Label the child as B1.

If he has another brother, then he is either the eldest or the youngest. The sample space is,

{B1B2} if he is the eldest or {B2B1} if he is the youngest.

If he has a sister, then he is either the oldest or the youngest. The sample space is,

{B1G} if he is the oldest or {GB1} if he is the youngest.

Two of the events are favorable to him having a sister out of 4 possible events. Thus, the probability is 1/2.

Okay sure:

1. I agree, 2/3

2. I agree 1/2

3. I likely agree with 2/3 though your wording is very unclear, you don't mention that you reveal a head for instance. It's certainly possibly you are trying to explain a situation where the answer would be 2/3.

Now do me a favor and try this simple thought experiment.

1)Flip two coins.

2) They land on the same side 50% of the time.

3)Pretend you take one of the coins to the park and show someone what side it landed on. It makes no difference whatsoever how you choose which coin to take to the park.

4) Ask the person how likely it is that the coin at home landed on the same side as the coin at the park.

5) Since we established in 2 that the coins land on the same side 50% of the time, their answer is 50%. They are correct !

6) Pretend that instead of flipping coins you were giving birth to children. No difference in the question at all - heads is as likely as boy, tails is as likely as girl.

7) This is a joke step not intended as an insult. You showed a tail at the park to me... I answered 50% chance that your coin at home is also tails.... you try and argue that it's 33% as the possiblities are HT, TH, and TT... but you are wrong.

Really you should get it now I think. If you disagree, tell me with which step.

edit: If you actually still don't get it, actually try the experiment. Flip two coins, choose one to take to the park however you like, see if the one you didn't take to the park has the same side showing, repeat until the probability converges to close enough to 0.5 to convince you. (hopefully you don't need to actually go the park...)

Pyromantha, firstly I wish to say thank you for talking civily, and making an educated argument for your view's, I'm happy to try and do the same which makes a lot nicer discussion. I'll try and answer any questions you put forward to me, if you could do the same

In regards to my example.

I appologise for not being clear, but yes the result I show you in (3) is a head, and always a head. We do not know what would have happened if the result was TT, possibly not shown you anything, we are simply dealing with a situation where a H result has occured. You agree that in this case it's 2/3? (hopefully with the ambiguity cleared up).

In this coin flip case, where the probability is 2/3. Why can I not simply say

{coin that was shown}{coin that was not shown} and give a result of 1/2? What stops me in this case breaking it up into seen / unseen like the park question?

In regards to your example, the step I have a small disagreement over is your step 3). If we randomly selected a coin, and no matter what took it into the park then yep, absolutely, the probability would be 1/2 - (I think I understand where our differences in opinion are now). However in this case we would also take the girl to the park every so often also (or a T), which we know didn't happen in the example.

Lets say however (and we don't know this is the case) that it was take a son to the park day (or no matter what I always take a H, or I don't go), and mothers only go to the park with their son, or they aren't there. In this case, we understand that all the mothers in the park have one son, otherwise they wouldn't be here. This effectively rules out the girl/girl possibility, leaving the result as 2/3. Just the same as in the coin flip example, one of the results shown is a H. We don't know what would have happened if the flip result was TT, but we simply must ignore it because 1st it wasn't the case, 2nd we don't know what would have happened if it was the case, 3rd we must make assumptions to do so.

Which is right? We'll I feel that to get 1/2, you must make an assumption that the children are selected on an equal basis, and daughters are going to the park as well. To get 2/3, you simply ignore the section of the problem we do not have information about, or have the probabilities to factor it in. Which for a mathematical problem I feel is more appropriate than making assumptions.

You don't need to randomly select a coin. You can choose whichever one you want to take every time and it makes no difference. That is the whole point

I think you are saying that families with two girls never take one to the park... this would indeed make the answer to your OP 2/3rds. Note that this is extremely different to merely noting that you didn't see such a family on this occasion. Your 'take a son to the park day' question has an answer 2/3rds.

But if we start making bizarre assumptions like this we can make the answer whatever we like, for example we could assume that families with any girls never take a child to the park, in which case the answer would be 100%, or we could assume that families with two boys never take a child to the park the answer would be 0%.

If you make no assumptions whatsoever about how a child goes to the park then the answer is 50%. If it was 'take a son to the park day' we would have answered 2/3rd immediately, but do you not see that this is you making assumptions about which child goes? At any rate it isn't the same as your original question.

We're making not necessarily making it "son day", we're simply not calculating a part of the problem that is incalculable. For instance -

(1) If I flip a coin, then flip a second coin, and tell you at least one of the result's is heads, but you don't know which coin toss (1st or 2nd) that had the H result, the chance of the second coin being a T is 2/3, yeah?

You answered that the result for this is 2/3. Did you take into account how the coins were selected? Of course not, you simply took the information from the problem and drew a conclusion.

In the park problem, you simply take the information from the problem (there is a boy) and draw a conclusion, not factor in things such as how the people were selected, which are unknown.

P.s. Do you also see why you can't break it up into seen / unseen?

No. The answer is 1/2 as long as you would have revealed a Tail if you had flipped two Tails.

If you would have flipped both coins again until at least one of them came up heads then the answer is 2/3rds. In which case the question is exactly the same as 'take a son to the park day', i.e. it involves totally bizarre assumptions and you can make equally bizarre assumptions and get any answer you like.

Call Heads 'boy'. How big a proportion of the time you take a boy to the park is the other coin also a boy? It has *ABSOLUTELY* nothing to do with how you select a coin/child to go to the park, you can use *ANY* methodology you like.

Ok, I flip two coins, and take one to the park. When I have a H I ask someone what the other coin is. 2/3 of the time the answer is a T. When I take a T to the park I just walk around and don't ask anyone anything.

I repeat this 100 times, out of the people I asked, 50 had the answer of T as the opposite coin, 25 had the answer of H for the opposite coin, and 25 times I just walked around and didn't ask anyone anything, so the times aren't counted.

You're degenerating your argument into insisting your right, rather than reasoning your right.

- You still aren't answering if you still think <boy I see><boy I don't see> is valid.

- You still haven't pointed out how my question in the OP is different to coin example #3. Which is mearly an extention of example #1.

"How on earth can you say that oldest is a unique identifier but 'being in park' isn't? I mean I get that the average guy in the street isn't very good at Maths but jesus christ."

Don't say stuff like this, it just makes you look dumb when your wrong.

You are making that assumption. You don't know. Hence you can't include it.

It's conditional probablity.

Just as in the example where a H was shown to you in coin flip example #1, you can't say the chance is 1/2, because if I had flipped a T I would have shown you.

Let me step this out in an easier way for you to understand.

(1) If I flip a coin, then flip a second coin, and tell you at least one of the result's is heads, but you don't know which coin toss (1st or 2nd) that had the H result, the chance of the second coin being a T is 2/3, yeah?

You are in a park and strike up a conversation with a mother and her son. You learn in her family there are two children, what is the probability that the other child is a daughter?

What is the difference between these two questions? We have two occurances of a 50/50 chance event occuring. We know the results of one. We do not know what result "first" or "second" it came from. The only difference is the method of discovery. In one you are told it's a result (heads), in another you're visually seeing it's a result (a boy). In both cases it does not descriminiate between if it's the first or second result. An example of which would be showing you the coin flip, as proof of the result actually being a head. As long as I still didn't tell you if it was the first or second flip the probability would not change.

Feel free to highlight the difference or where I am wrong.

No. The answer depends on what methodology you used to decide to which result to inform me of. If you picked at random and happened to inform me of a head then the answer is 1/2. This is because the sample space of {Hr H : Hr T : Tr H: Tr T} reduces to {Hr H : Hr T} where r is the revealed coin and both are equally likely.

If it was 'inform a person that you flipped a Head at the park day and flip and try again if you flip two Tails' then the answer is 2/3rds because the sample space is {Hr H : Hr T : Hr T}. As I mentioned in my post above you can make any set of bizarre assumptions about how you reveal the results to get any answer between 0 and 1.

This just seems to be degenerating into a pointless argument. I am totally certain that I'm correct and you just keep posting more and more wrong things leading me to believe you don't understand probability at all so I doubt I will be able to change your viewpoint. Let's just leave it - you are welcome to have the last word if you like.

You're right, I really hate appealing to authority - but I fully understand the question, it seems that you don't understand it at all, and this *exact* question can be found online with the answer given as 1/2, correctly.

I say *exact*, because in the version I found online you knocked on the door of a house with two children and the door was opened by a girl. This is in a Maths textbook, google probability boy girl door and you will find it.

http://books.google.co.uk/books?id=_...esult&resnum=4

Pages 57-58.

Edit found another link with a similar question:

http://www.jimworthey.com/puzzle3.html

Here we again have the walking through the door rather than taking to the park. If you think walking through a door differs in some substantive way from taking to the park then so be it, I can't find an exact replica of your problem online.

Anyway, that's absolutely the last thing I have to say in this thread, go ahead and post whatever.

So you're telling me that if I flip two coins, and I tell you what one of the results is, you're willing to bet me even money the second coin will not be the opposite value?

What if he flips two coins, Tells you what one of the results is, Do you then think its better than even chance that the other coin will be the opposite value?

It's not clear to me why older/younger discriminates between two subjects, but at park/not at park doesn't.